RMT-MCQ

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MCQ on Breathing and Exchange of Gases — Biology Notes

Microwave Engineering Questions and Answers – Parallel Plate Waveguide

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Parallel Plate Waveguide”.

1. The modes of wave propagation that a parallel plate waveguide can support are:

a) TEM, TE, TM modes

b) TM, TE modes

c) TEM, TM modes

d) TEM, TE modes

View Answer

Answer: a

Explanation: Parallel plate waveguide is the simplest type of waveguide that can support TE and TM modes. It can also support a TEM mode since it is formed from two flat conducting plates.

2. The fringe effect can be neglected in a parallel plate waveguide is because of:

a) The dielectric material used

b) Width of the plates is greater than the distance between the plates

c) Material of the parallel plate waveguide used

d) None of the mentioned

View Answer

Answer: b

Explanation: The strip width W of the parallel plate waveguide is assumed to be much greater than the separation d, hence the fringe effect or the fringing fields can be neglected.

3. The characteristic impedance of a parallel plate waveguide is given by:

a) η*D/W

b) η*W/D

c) D/ η*W

d) η*√(D/W)

View Answer

Answer: a

Explanation: Characteristic impedance of a waveguide is the ratio of voltage and current. Defining voltage and current in the integral form of electric field and magnetic field respectively and solving the characteristic impedance is η*D/W. here η is the intrinsic impedance of the medium in the waveguide, D is the distance between the plates and W is the width of the rectangular plate.

4. If the width of a parallel plate waveguide is 30 mm and the distance between the parallel pates is 5 mm, with an intrinsic impedance of 377Ω, then the characteristic impedance of the wave is:

a) 50 Ω

b) 62.833 Ω

c) 100 Ω

d) None of the mentioned

View Answer

Answer: b

Explanation: The expression for intrinsic impedance of a parallel plate waveguide is η*D/W. substituting the given values of intrinsic impedance and distance between plates and width of the plates, intrinsic impedance is 62.833Ω.

5. In TM mode, if the direction of wave propagation is in ‘z’ direction, then:

a) HZ=0

b) EZ=0

c) EY=0

d) HY=0

View Answer

Answer: a

Explanation: In TM mode (transverse magnetic), when the wave propagation is along Z direction, magnetic field is absent in Z direction since the fields are transverse in nature. Hence HZ=0.

6. The wave impedance of a TM mode in a parallel plate waveguide is a:

a) Function of frequency

b) Independent of frequency

c) Proportional to square of frequency

d) Inversely proportional to square of frequency

View Answer

Answer: a

Explanation: The wave impedance of a parallel plate waveguide in TM mode is β/k which is a function of frequency. The wave impedance is real for f>fC and purely imaginary for f<fC.

7. In a parallel plate waveguide, for a propagating mode, the value of β is:

a) Real

b) Complex

c) Imaginary

d) Cannot be defined

View Answer

Answer: a

Explanation: The phase velocity and guide wavelength for a parallel plate waveguide are defined only for propagating modes. Propagating modes are those modes for which β are always positive. Hence β is always real for a parallel plate waveguide.

8. For TM2 mode, if the distance between two parallel plates of a waveguide are 40 mm, then the cut off wavelength for TM2 mode is:

a) 20 mm

b) 80 mm

c) 40 mm

d) 60 mm

View Answer

Answer: c

Explanation: The cutoff wavelength of a TMn mode in a parallel plate waveguide is 2d/n, where d is the distance between the plates and n signifies the mode of operation. For the given condition, substituting the given values, cut off wavelength is 40 mm.

9. For a parallel waveguide, the dominant mode for TM propagation is:

a) TM0 mode

b) TM1 mode

c) TM2 mode

d) Dominant mode does not exist

View Answer

Answer: b

Explanation: The mode of propagation for which the cutoff wavelength for wave propagation is maximum is called dominant mode. In TM mode of propagation, TM0 mode is similar to TEM mode of propagation. Hence, TM1 mode is the dominant mode.

10. Phase velocity of the plane waves in the two direction in a parallel plate waveguide is:

a) 1/√(μ∈)secant θ

b) 1/√(μ∈)cosecant θ

c) 1/√(μ∈)tangent θ

d) 1/√(μ∈)cosine θ

View Answer

Answer: a

Explanation: The phase velocity of each plane wave along its direction of propagation (θ direction) is 1/√(μ∈), Which is the speed of light in the material filling the guide. But, the phase velocity of the plane waves in the z direction is 1/√(μ∈)secant θ.

11. For a parallel plate waveguide, which of the following is true?

a) No real power flow occurs in the ‘z’ direction

b) Power flow occurs in ‘z’ direction

c) No power flow occurs in any direction

d) Wave propagation in z direction is not possible in any mode

View Answer

Answer: a

Explanation: The superposition of the two plane waves in Z direction is such that complete cancellation occurs at y=0 and y=d, to satisfy the boundary conditions that EZ=0 at these planes. As f decrease to fc, β approaches 0 so that θ approaches 90⁰. The two plane waves are then bouncing up and down, with no motion in +z direction, and no power flow occurs in the z direction.

12. TE mode is characterized by:

a) EZ=0

b) HZ=0

c) Ex=0

d) Ey=0

View Answer

Answer: a

Explanation: In TE mode of wave propagation, the electric field is in transverse direction and hence electric field component in the direction of wave propagation is 0. Hence, EZ=0.

13. If in a parallel plate waveguide, PL=4 mW/m and Pₒ=10 mW/m, then what is the conduction loss?

a) 0.5

b) 0.4

c) 0.1

d) 0.2

View Answer

Answer: d

Explanation: Conductor loss of a parallel plate waveguide is given by PL/2Pₒ. Substituting the given values in the above equation, conductor loss is 0.2.

14. If the distance between the two plates of a parallel plate waveguide is 20 mm and is operating in TE₂ mode, then the cut off frequency of TE₂ mode is:

a) 20 MHz

b) 15 GHz

c) 5 GHz

d) None of the mentioned

View Answer

Answer: b

Explanation: The cutoff frequency for TEn mode is n/2d√(∈μ) for a parallel plate waveguide. Substituting the given values, the cutoff frequency is 15 GHz.

15. The wave impedance for a non-propagating mode in TE mode is:

a) K/β

b) Imaginary

c) Zero

d) Non-existing

View Answer

Answer: b

Explanation: Wave impedance of a parallel plate waveguide for TEN modes is k/β. This expression is valid and real only for propagating modes. For non propagating modes, impedance becomes imaginary.

Microwave Engineering Questions and Answers – Rectangular Waveguide

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Rectangular Waveguide”.

1. The modes of propagation supported by a rectangular wave guide is:

a) TM, TEM, TE modes

b) TM, TE

c) TM, TEM

d) TE, TEM

View Answer

Answer: b

Explanation: A hollow rectangular waveguide can propagate TE and TM modes. Since only a single conductor is present, it does not support TEM mode of propagation.

2. A hollow rectangular waveguide cannot propagate TEM waves because:

a) Of the existence of only one conductor

b) Of the losses caused

c) It is dependent on the type of the material used

d) None of the mentioned

View Answer

Answer: a

Explanation: A rectangular hollow waveguide can propagate both TE and TM modes of propagation. But due the presence of only one conductor, rectangular waveguide does not support the propagation of TEM mode.

3. For any mode of propagation in a rectangular waveguide, propagation occurs:

a) Above the cut off frequency

b) Below the cut off frequency

c) Only at the cut-off frequency

d) Depends on the dimension of the waveguide

View Answer

Answer: a

Explanation: Both TE and TM modes of propagation in rectangular waveguide have certain separate and specific cut off frequencies below which propagation is not possible. Hence propagation of signal occurs above the cut off frequency.

4. In TE mode of wave propagation in a rectangular waveguide, what is the equation that has to be satisfied?

a) (∂2/ ∂x2 + ∂2/ ∂y2+ kC2).HZ(x, y) =0

b) (∂2/ ∂x2 + ∂2/ ∂y2– kC2).HZ(x, y) =0

c) (∂2/ ∂x2 – ∂2/ ∂y2+ kC2).HZ(x, y) =0

d) None of the mentioned

View Answer

Answer: a

Explanation: For TE mode of propagation in a rectangular waveguide, electric field along the direction of propagation is 0. Hence for propagation, the above partial differential equation in terms of magnetic field along Z direction has to be satisfied.

5. Dominant mode is defined as:

a) Mode with the lowest cut off frequency

b) Mode with the highest cut off frequency

c) Any TEM mode is called a dominant mode

d) None of the mentioned

View Answer

Answer: a

Explanation: Among the various modes of propagation in a rectangular waveguide, the mode of propagation having the lowest cutoff frequency or the highest wavelength of propagation among the other propagating modes is called dominant mode.

6. For TE1ₒ mode, if the waveguide is filled with air and the broader dimension of the waveguide is 2 cm, then the cutoff frequency is:

a) 5 MHz

b) 7.5 MHz

c) 7.5 GHz

d) 5 GHz

View Answer

Answer: c

Explanation: The cutoff frequency for TE 10 mode of propagation in a rectangular waveguide is 1/2a√(∈μ) where ‘a’ is the broader dimension of the waveguide. Substituting for the given value and 1/√(∈μ)=3*108. The cutoff frequency is 7.5 GHz.

7. TEₒₒ mode for a rectangular waveguide:

a) Exists

b) Exists but defined only under special cases

c) Does not exist

d) Cannot be determined

View Answer

Answer: c

Explanation: The field expressions for TEₒₒ mode disappears or becomes zero theoretically. Hence, TEₒₒ mode does not exist.

8. For dominant mode propagation in TE mode, if the rectangular waveguide has a broader dimension of 31.14 mm , then the cutoff wave number:

a) 100

b) 500

c) 50

d) 1000

View Answer

Answer: a

Explanation: The cutoff wave number for the dominant mode of a rectangular waveguide is given by π/a where ‘a’ is the broader dimension of the waveguide, substituting the given values, the wave number 100.

9. The lowest mode of TM wave propagation is:

a) TM10 mode

b) TM01 mode

c) TM11 mode

d) TM12 mode

View Answer

Answer: c

Explanation: The field components for other lower modes of propagation in TM mode disappear for other lower modes of propagation. Hence, the lowest mode of propagation is TM11 mode.

10. The cutoff frequency for the dominant mode in TM mode propagation for a rectangular waveguide of dimension of 30mm*40mm is:

a) 2 GHz

b) 1 GHz

c) 2 MHz

d) 4 MHz

View Answer

Answer: a

Explanation: The cutoff frequency of dominant mode in TM mode is √((π/a)2 + (π/b)2). Here, ‘a’ and ‘b’ are the dimensions of the waveguide. Substituting the corresponding values, the cutoff frequency is 2 GHz.

11. In TE10 mode of wave propagation in a rectangular waveguide, if the broader dimension of the waveguide is 40 cm, then the cutoff wavelength for that mode is:

a) 8 cm

b) 6 cm

c) 4 cm

d) 2 cm

View Answer

Answer: a

Explanation: In TE10 mode of propagation in a rectangular waveguide, the cutoff wavelength of the waveguide is given by 2a where ‘a’ is the broader dimension of the waveguide. Substituting, the cutoff wavelength is 8 cm.

12. In TE01 mode of wave propagation in a rectangular waveguide, if the smaller dimension of the waveguide is 2 cm, then the cutoff wavelength for that mode is:

a) 4 cm

b) 8 cm

c) 1 cm

d) 2 cm

View Answer

Answer: a

Explanation: For TE01 mode of wave propagation in a rectangular wave guide, if the smaller dimension of the wave guide is 2 cm, then the cut off wavelength is 2b where b is the smaller dimension of the waveguide. substituting, the cutoff wavelength is 4 cm.

Microwave Engineering Questions and Answers – Circular Waveguide

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Circular Waveguide”.

1. In TE mode of a circular waveguide, EZ=0. The wave equation is:

a) ∇2HZ+k2HZ=0

b) ∇2HZ-k2HZ=0

c) ∇2HZ-HZ=0

d) ∇2HZ+HZ=0

View Answer

Answer: a

Explanation: In TE mode, EZ=0. Hence, when we substitute it in the wave equation, we get ∇2HZ+k2HZ=0.

2. Bessel’s differential equation for a circular waveguide is:

a) ρ2(d2R/ dρ2) + ρ(dR/dρ) + (ρ2kC2– n2) R=0

b) n2(d2R/ dρ2) + n(dR/dρ) + (ρ2kC2– n2) R=0

c) d2R/ dρ2 + dR/dρ + (ρ2kC2– n2) R=0

d) None of the mentioned

View Answer

Answer: a

Explanation: After solving the wave equation ∇2HZ+k2HZ=0 in TE mode by making suitable assumptions and making appropriate substitutions, the final equation obtained is ρ2(d2R/ dρ2) + ρ(dR/dρ) + (ρ2kC2– n2) R=0.

3. The lowest mode of TE mode propagation in a circular waveguide is:

a) TE10 mode

b) TE00 mode

c) TE01 mode

d) TE11 mode

View Answer

Answer: c

Explanation: A circular waveguide can support various modes of propagation. Among these, the lowest mode of propagation supported by the waveguide is TE10 mode of propagation.

4. What is the cutoff frequency for TE₁₁ mode in a circular waveguide of radius 2 cm with P’₁₁= 1.841?

a) 5.5 GHz

b) 4.3 GHz

c) 7.7 GHz

d) 8.1 GHz

View Answer

Answer: b

Explanation: The cutoff frequency for TE11 mode of propagation in a circular waveguide is given by Pnm/ 2πa√μϵ. Substituting the given values in the above expression, the cutoff frequency is 4.3 GHz.

5. In a circular waveguide, if the propagation is in TE21 mode with P21=3.054, with a diameter of 60 mm, then the cutoff frequency for the mode is:

a) 5.6 GHz

b) 6.4 GHz

c) 3.5 GHz

d) 4.8 GHz

View Answer

Answer: d

Explanation: The cutoff frequency for TE21 mode of propagation in a circular waveguide is given by Pnm/ 2πa√μϵ. Substituting the given values in the above expression, the cutoff frequency is 4.8 GHz.

6. For a circular waveguide in TM11 mode of propagation with inner radius of 30mm, and the phase constant being equal to 0.3, then the wave impedance is equal to:

a) 0.16 Ω

b) 0.15 Ω

c) 0.5 Ω

d) 0.4 Ω

View Answer

Answer: a

Explanation: For a given mode of propagation in a circular waveguide, wave impedance is given by the expression k/β, where is the intrinsic impedance of the air k is the wave number, β is the phase constant. Determining the wave number and substituting in the given equation, the wave impedance is 0.16 Ω.

7. For TM mode. The wave equation in cylindrical co ordinates is:

a) (∂2/∂ρ2+1/ρ ∂/∂ρ + 1/ρ2 (∂2/∂∅2 + kc2) =0

b) ∂2E2/∂ρ2 + 1/ρ ( ∂E/∂ρ)=0

c) ∂2E2/∂ρ2 + 1/ρ2 (∂2E2/∂∅2 ) = 0

d) None of the mentioned

View Answer

Answer: a

Explanation: The wave propagation in a cylindrical waveguide in TM mode of propagation is governed by the equation (∂2/∂ρ2+1/ρ ∂/∂ρ + 1/ρ2 (∂2/∂∅2 + kc2) =0. This is a second order differential equation.

8. In TM mode, what is the first propagating mode?

a) TM01 mode

b) TM11 mode

c) TM12 mode

d) TM10 mode

View Answer

Answer: a

Explanation: TM mode in a circular waveguide supports various modes of propagation. Among these modes of propagation, the first or the lowest mode of propagation is TM01 mode.

9. For TM01 mode of propagation in a circular waveguide with P01=2.405, with the inner diameter of the circular waveguide being equal to 25 mm. What is the cut off frequency for this mode of propagation?

a) 2.8 GHz

b) 6 GHz

c) 3.06 GHz

d) 4 GHz.

View Answer

Answer: c

Explanation: The cutoff frequency for TM01 mode of propagation in a circular waveguide is given by Pnm/ 2πa√μϵ. Substituting the given values in the above expression, the cutoff frequency is 3.06 GHz.

10. If β is 0.3 for a circular wave guide operating in TM12 mode with P21=5.315, with the radius of the circular waveguide being equal to 25 mm, then the intrinsic impedance of the wave is:

a) 0.55 Ω

b) 0.4 Ω

c) 0.3 Ω

d) 1.2 Ω

View Answer

Answer: a

11. The cutoff frequencies of the first two propagating modes of a Teflon on a filled circular waveguide with a=0.5 with ∈r=2.08 is:

a) 12.19 GHz, 15.92 GHz

b) 10 GHz, 12 GHz

c) 12 GHz, 15 GHz

d) 15 GHz, 12 GHz

View Answer

Answer: a

Explanation: The cutoff frequencies are given by the expression p*C/2πa√∈. Substituting the given values in the above expression, the cutoff frequencies are 12.19 GHz, 15.92 GHz.

Microwave Engineering Questions and Answers – Co-axial Lines

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Co-axial Lines”.

1. What are the modes of propagation that a co axial line supports?

a) TM, TE mode

b) TM, TE, TEM mode

c) TM, TEM mode

d) TE, TEM mode

View Answer

Answer: b

Explanation: Certain propagating media support specific modes of propagation. Coaxial cables support all the three modes of propagation. They are TM, TE, and TEM modes.

2. The dominant waveguide mode of a co axial line is:

a) TE11 mode

b) TE01 mode

c) TM01 mode

d) TEM mode

View Answer

Answer: a

Explanation: Co-axial cable many modes of propagation. Among those supported modes of propagation, the dominant mode, the mode with lowest propagating wavelength is TE11 mode.

3. In a co axial line with inner and outer diameters of 0.0645 and 0.0215 inches and a Teflon di electric with ∈r=2.2. The highest usable frequency before the TE11 waveguide mode starts to propagate is:

a) 16.8 GHz

b) 117.7 GHz

c) 15.3 GHz

d) 8.4 GHz

View Answer

Answer: a

Explanation: Before the wave propagation starts, the unstable frequency is given by the expression Ck/2π√∈. Here, C is the velocity of light. Substituting the given values in the above expression, the frequency is 16.8 GHz.

4. If a wave guide has he inner and outer conductor diameters of 0.0645 and 0.0215 inches respectively for a co axial lines then the cut off wave number is:

a) 298

b) 300

c) 285

d) 123

View Answer

Answer: a

Explanation: The cutoff wave number for wave propagation is given by 2/ (a + b). a, b are the inner and outer diameter respectively. Substituting in the above expression, cut off wave number is 298.

5. The commercially used co axial cable and connectors used has a characteristic impedance is:

a) 50Ω

b) 100Ω

c) 33.34Ω

d) 66.6Ω

View Answer

Answer: a

Explanation: All commercial manufacturer s of coaxial cables and connectors have set a standard for all the manufactured products. The standard value is 50 Ω.

6. In television systems the characteristic impedance of the cables used is:

a) 75Ω

b) 150Ω

c) 100Ω

d) 50Ω

View Answer

Answer: a

Explanation: All cable manufacturers of the television system follow a set of standards. As per these set standards, the characteristic impedance of the line is 75 Ω.

7. SWR standing wave ratio has to be ________for co axial connector.

a) Low

b) High

c) Infinite

d) Cannot be calculated

View Answer

Answer: a

Explanation: Higher the value of the standing wave ratio more is the reflection which implies mismatch. Hence, standing wave ratio has to be low.

8. What are the connectors used in pairs called?

a) Jack and plug

b) Male and female connectors

c) Both the mentioned

d) None of the mentioned

View Answer

Answer: c

Explanation: There are certain special connectors that can be used to connect two devices operating devices. These special connectors are to be used in pairs and are with both the names.

9. The frequency range for N type co axial connector is:

a) 8-12 GHz

b) 11-18 GHz

c) 14-20 GHz

d) 2-8 GHz

View Answer

Answer: b

Explanation: N type connectors are coaxial connectors used at microwave frequency range. These type of connectors can be used in the frequency range of 11-18 Hz.

10. The frequency range of SMA co-axial connector used most commonly is:

a) 18-25 GHz

b) 25-50 GHz

c) 11-18 GHz

d) 8-12 GHz

View Answer

Answer: a

Explanation: SMA coaxial cable connectors are designed to operate at high frequencies. The frequency ranges from 18-25 GHz.

Microwave Engineering Questions and Answers – Surface Waves on Grounded Di-electric Sheet

This set of Microwave Engineering Questions and Answers for Freshers focuses on “Surface Waves on Grounded Di-electric Sheet”.

1. Surface waves are typical by a field that decays ______away from the dielectric surface, with most of the field contained in or near the dielectric.

a) Linearly

b) Exponentially

c) Cubical

d) Field remains a constant

View Answer

Answer: b

Explanation: Surface waves are typified by a field that decays exponentially away from the dielectric surface, with most of the field contained in or near the dielectric. At higher frequencies, the field generally becomes more tightly bound to the dielectric, making such waveguides practical.

2. Because of the presence of the dielectric, the phase velocity of a surface wave is:

a) Greater than that in vacuum

b) Lesser than that in vacuum

c) Independent of the presence of dielectric

d) Insufficient data

View Answer

Answer: b

Explanation: The fields are stronger and concentrated near the dielectric, and hence because of the presence of the dielectric, the phase velocity of a surface wave is lesser than that in vacuum.

3. For wave propagation on grounded dielectric sheet, the equation to be satisfied by Ez , in the region of presence of dielectric 0≤x≤d for the propagation to be in Z direction

a) (∂2/∂x2 + ∈rk02– β2) eZ(x,y)=0

b) (∂2/∂x2 + k02– β2) eZ(x,y)=0

c) (∂2/∂x2 – k02+β2) eZ(x,y)=0

d) (∂2/∂x2 + ∈rk02) eZ(x,y)=0

View Answer

Answer: a

Explanation: The equation describes the variation of the electric field along the direction of propagation that is the Z direction. In the equation, it is clear that the relative permittivity of the dielectric is also a part of the second term of the equation.

4. The cut off wavenumber for the region of dielectric in a grounded dielectric sheet is:

a) kC2= ∈rk02-β2

b) kC2= ∈rk02+β2

c) h2= -k02+β2

d) kC2= k2 + β2

View Answer

Answer: a

Explanation: Cutoff wave number signifies the minimum threshold wave number required for propagation. Here the expression kC2= ∈rk02-β2 gives the cutoff wave number for the propagation of waves on a grounded dielectric sheet.

5. For surface waves on a dielectric sheet the cutoff frequency of the TM mode can be given as:

a) fC = nC/2d√(ϵr-1)

b) fC = C/2nd√(ϵr-1)

c) fC = C/2nd√ϵr

d) fC = 2C/nd√ϵr

View Answer

Answer: a

Explanation: Grounded dielectric sheets allow TM mode of propagation on them. The cut off frequency for the propagation of TM mode is given by the expression fC = nC/2d√(ϵr-1).

6. The cutoff frequency in TM1 mode for the propagation of EM waves on dielectric slab of relative permittivity 2.6 and thickness 20 mm is:

a) 6.5 GHz

b) 5.92 GHz

c) 4 GHz

d) 2 GHz

View Answer

Answer: b

Explanation: The expression for cutoff frequency for wave propagation in TMn mode is fC = nC/2d√(ϵr-1).here n represents the mode. Substituting the given values, cutoff frequency is 5.92 GHz.

7. In TE of propagation, HZ must obey the below equation for wave propagation in the region of presence of dielectric:

a) (∂2/∂x2 + kc2) hZ(x,y)=0

b) (∂2/∂x2– h2)hZ(x,y)=0

c) (∂2/∂x2 – Kc2)hZ(x,y)=0

d) (∂2/∂x2 +h2)hZ(x,y)=0

View Answer

Answer: a

Explanation: In TE mode of propagation,, electric field does not exist in the direction of wave propagation. Hence only magnetic field exists in the direction of wave propagation. This magnetic field must obey the equation (∂2/∂x2 + kc2) hZ(x,y)=0

8. Cutoff frequency fC for TEM mode of propagation is:

a) Fc= (2n-1)c/4d√εr -1

b) Fc= (2n-c)/2d(√εr-1)

c) Fc= (2n-1)/4d(√εr)

d) Fc= (2n-1)/8d√εr – 1

View Answer

Answer: a

Explanation: Grounded dielectric sheets allow TE mode of propagation on them. The cut off frequency for the propagation of TM mode is given by the expression (2n-1)c/4d√εr -1 .

9. What is the cutoff frequency of TE₁ mode of propagation if the relative permittivity of the slab is 3.2 and the thickness of the slab is 45 mm?

a) 2.24 GHz

b) 4 GHz

c) 1.12 GHz

d) 8 GHz

View Answer

Answer: c

Explanation: The expression for cutoff frequency for wave propagation in TEN mode is (2n-1)c/4d√(εr -1). substituting the given values in the above expression, the cutoff frequency for TE₁ mode of propagation is 1.12 GHz.

10. The first propagating mode on a grounded dielectric is:

a) TMO mode

b) TM1 mode

c) TM2 mode

d) TM3 mode

View Answer

Answer: b

Explanation: Since for TMO mode of propagation on a dielectric sheet the cutoff frequency is 0, it is not practically possible for propagation. Hence, TM1 mode is the first propagating mode.

Microwave Engineering Questions and Answers – Striplines

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Striplines”.

1. Which mode of propagation is supported by a strip line?

a) TEM mode

b) TM mode

c) TE mode

d) None of the mentioned

View Answer

Answer: a

Explanation: Since a stripline has 2 conductors and a homogeneous dielectric, it supports a TEM wave, and this is the usual mode of operation.

2. The higher order wave guide modes of propagation can be avoided in a strip line by:

a) Restricting both the ground plate spacing and the sidewall width to less than λd/2

b) Restricting both the ground facing plate spacing and the sidewall width to less than λd

c) Filling the region between 2 plates with di electric

d) Restricting both the ground plate spacing and the sidewall width between λg and λg/2

View Answer

Answer: a

Explanation: When stripline is used as a media for propagation, it is always preferred that only certain modes of propagation are allowed. Hence, in order to avoid the higher order modes, it is achieved by restricting both the ground plate spacing and the sidewall width to less than λd/2.

3. Stripline can be compared to a:

a) Flattened rectangular waveguide

b) Flattened circular waveguide

c) Flattened co axial cable

d) None of the mentioned

View Answer

Answer: c

Explanation: A stripline has an enter conductor enclosed by an outer conductor and are uniformly filled with a dielectric medium, these are similar to a coaxial cable. Hence it can be compared to a flattened coaxial cable.

4. If the dielectric material filled between the round plates of a microstrip line has a relative permittivity of 2.4, then the phase velocity is:

a) 1.3*108 m/s

b) 1.9*108 m/s

c) 3*108 m/s

d) 2*108 m/s

View Answer

Answer: b

Explanation: Phase velocity is given by the expression C/√∈ for a stripline. Substituting the given values, the phase velocity for the above case is 1.9*108 m/s.

5. Expression for propagation constant β of a strip line is:

a) ω(√µ∈∈r)

b) ω(√µₒ/√∈r)

c) ω/(√µₒ∈ₒ∈r)

d)c/(√µₒ∈ₒ∈r)

View Answer

Answer: a

Explanation: Propagation constant is associated with the propagating wave in the strip line. This propagation constant for a wave is defined by the expression ω(√µ∈∈r).

6. If the phase velocity in a stripline is 2.4*108m/s, and the capacitance per unit length of a micro stripline is 10pF/m, then the characteristic impedance of the line:

a) 50 Ω

b) 41.6 Ω

c) 100 Ω

d) None of the mentioned

View Answer

Answer: b

Explanation: Characteristic impedance of a stripline is given by 1/ (vPc). Substituting the given values of phase velocity and capacitance, the characteristic impedance of the line is 41.6 Ω.

7. The expression for characteristic impedance Zₒ of a stripline is:

a) (30πb/√∈r)(1/We+0.441b)

b) (30πb) (1/We+0.441b)

c) 30π/√∈r

d) (1/We+0.441b)

View Answer

Answer: a

Explanation: Characteristic impedance of a stripline is a function of the various parameters of the stripline. They are effective width, thickness and relative permittivity of the dielectric material. Changing any one of these parameters results in changing the characteristic impedance of the line The derived expression is hence (30πb/√∈r)(1/We+0.441b).

8. If the effective width of the center conductor is 3 mm and the distance between the two ground plates is 0.32 cm with the material of the dielectric used having a relative permittivity of 2.5, then what is the characteristic impedance of the strip line?

a) 50Ω

b) 71.071Ω

c) 43.24Ω

d) 121Ω

View Answer

Answer: c

Explanation: The characteristic impedance of a stripline is given by the expression (30πb/√∈r)(1/We+0.441b). Substituting the given values in the given expression and hence solving, the characteristic impedance of the line is 43.24 Ω.

9. The wave number of a stripline operating at a frequency of 10 GHz is:

a) 401

b) 155

c) 206

d) 310

View Answer

Answer: d

Explanation: The wave number of a microstrip line is given by the expression 2πf√∈r/c, c is the speed of light in space, ∈r is the relative permittivity of the dielectric medium. Substituting the given values in the equation, the wave number is 310.

10. If the loss tangent is 0.001 for a stripline operating at 12 GHz with the relative permittivity of the dielectric material being used equal to 2.6, then the conductor loss is:

a) 0.102

b) 0.202

c) 0.001

d) 0.002

View Answer

Answer: b

Explanation: Conductor loss in a stripline is given by the expression k*tanδ/2. K is given by the expression 2πf√∈r/C which is the wave number. Substituting the values in the above two equations, conductor loss is 0.202.

11. If the dielectric material used between the grounded plates of a stripline is 2.2, when the strip line operating at 8 GHz, the wavelength on stripline is:

a) 1.2 cm

b) 2.52 cm

c) 0.15 cm

d) 3.2 cm

View Answer

Answer: b

Explanation: The propagating wavelength on the stripline is defined by the relation C/f√∈r. substituting in the above relation, the propagating wavelength on the microstrip line is 2.52 cm.

12. Fields of TEM mode on strip line must satisfy:

a) Laplace’s equation

b) Ampere’s circuital law

c) Gaussian law

d) None of the mentioned

View Answer

Answer: a

Explanation: If φ(x,y) is the function of potential in the stripline varying along the width and thickness, this potential function must satisfy the Laplace’s equation.

Microwave Engineering Questions and Answers – Micro-Strip Lines

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Micro-Strip Lines”.

1. Micro strip can be fabricated using:

a) Photo lithographic process

b) Electrochemical process

c) Mechanical methods

d) None of the mentioned

View Answer

Answer: a

Explanation: Microstrip lines are planar transmission lines primarily because it can be fabricated by photolithographic processes and is easily miniaturized and integrated with both passive and active microwave devices.

2. The mode of propagation in a microstrip line is:

a) Quasi TEM mode

b) TEM mode

c) TM mode

d) TE mode

View Answer

Answer: a

Explanation: The exact fields of a microstrip line constitute a hybrid TM-TE wave. In most practical applications, the dielectric substrate is very thin and so the fields are generally quasi-TEM in nature.

3. Microstrip line can support a pure TEM wave.

a) True

b) False

c) Microstrip supports only TM mode

d) Microstrip supports only TE mode

View Answer

Answer: b

Explanation: The modeling of electric and magnetic fields of a microstrip line constitute a hybrid TM-TE model. Because of the presence of the very thin dielectric substrate, fields are quasi-TEM in nature. They do not support a pure TEM wave.

4. The effective di electric constant of a microstrip line is:

a) Equal to one

b) Equal to the permittivity of the material

c) Cannot be predicted

d) Lies between 1 and the relative permittivity of the micro strip line

View Answer

Answer: d

Explanation: The effective dielectric constant of a microstrip line is given by (∈r + 1)/2 + (∈r-1)/2 * 1/ (√1+12d/w). Along with the relative permittivity, the effective permittivity also depends on the effective width and thickness of the microstrip line.

5. Effective dielectric constant of a microstrip is given by:

a) (∈r + 1)/2 + (∈r-1)/2 * 1/ (√1+12d/w)

b) (∈r+1)/2 + (∈r-1)/2

c) (∈r+1)/2 (1/√1+12d/w)

d) (∈r + 1)/2-(∈r-1)/2

View Answer

Answer: a

Explanation: The effective dielectric constant of a microstrip line is (∈r + 1)/2 + (∈r-1)/2 * 1/ (√1+12d/w). This relation clearly shows that the effective permittivity is a function of various parameters of a microstrip line, the relative permittivity, effective width and the thickness of the substrate.

6. The effective dielectric constant of a micro strip line is 2.4, then the phase velocity in the micro strip line is given by:

a) 1.5*108 m/s

b) 1.936*108 m/s

c) 3*108 m/s

d) None of the mentioned

View Answer

Answer: b

Explanation: The phase velocity in a microstrip line is given by C/√∈r. substituting the value of relative permittivity and the speed of light in vacuum, the phase velocity is 1.936*108 m/s.

7. The effective di electric constant of a micro strip line with relative permittivity being equal to 2.6, with a width of 5mm and thickness equal to 8mm is given by:

a) 2.6

b) 1.97

c) 1

d) 2.43

View Answer

Answer: b

Explanation: The effective dielectric constant of a microstrip line is given by (∈r + 1)/2 + (∈r-1)/2 * 1/ (√1+12d/w). Substituting the given values of relative permittivity, effective width, and thickness, the effective dielectric constant is 1.97.

8. If the wave number of an EM wave is 301/m in air , then the propagation constant β on a micro strip line with effective di electric constant 2.8 is:

a) 602

b) 503.669

c) 150

d) 200

View Answer

Answer: b

Explanation: The propagation constant β of a microstrip line is given by k0√∈e. ∈e is the effective dielectric constant. Substituting the relevant values, the effective dielectric constant is 503.669.

9. For most of the micro strip substrates:

a) Conductor loss is more significant than di electric loss

b) Di electric loss is more significant than conductor loss

c) Conductor loss is not significant

d) Di-electric loss is less significant

View Answer

Answer: a

Explanation: Surface resistivity of the conductor (microstrip line) contributes to the conductor loss of a microstrip line. Hence, conductor loss is more significant in a microstrip line than dielectric loss.

10. The wave number in air for EM wave propagating on a micro strip line operating at 10GHz is given by:

a) 200

b) 211

c) 312

d) 209

View Answer

Answer: d

Explanation: The wave number in air is given by the relation 2πf/C. Substituting the given value of frequency and ‘C’, the wave number obtained is 209.

11. The effective dielectric constant ∈r for a microstrip line:

a) Varies with frequency

b) Independent of frequency

c) It is a constant for a certain material

d) Depends on the material used to make microstrip

View Answer

Answer: b

Explanation: The effective dielectric constant of a microstrip line is given by (∈r + 1)/2 + (∈r-1)/2 * 1/ (√1+12d/w). The equation clearly indicates that the effective dielectric constant is independent of the frequency of operation, but depends only on the design parameters of a microstrip line.

12. With an increase in the operating frequency of a micro strip line, the effective di electric constant of a micro strip line:

a) Increases

b) Decreases

c) Independent of frequency

d) Depends on the material of the substrate used as the microstrip line

View Answer

Answer: c

Explanation: As the relation between effective permittivity and the other parameters of a microstrip line indicate, effective dielectric constant is not a frequency dependent parameter and hence remains constant irrespective of the operation of frequency.

Microwave Engineering Questions and Answers – Impedance and Admittance Matrices

This set of Microwave Engineering Assessment Questions and Answers focuses on “Impedance and Admittance Matrices”.

1. The one below among others is not a type TEM line used in microwave networks:

a) Co-axial wire

b) Micro strip line

c) Strip lines

d) Surface guide

View Answer

Answer: d

Explanation: Coaxial micro strip and strip lines all support TEM mode of propagation through them. But surface guides do not support TEM mode of propagation in them. Hence it cannot be called a TEM line.

2. The one below is the only micro wave network element that is a TEM line:

a) Co-axial cable

b) Rectangular wave guide

c) Circular wave guide

d) Surface wave guide

View Answer

Answer: a

Explanation: Coaxial cables support TEM mode of propagation in them and rectangular waveguide, circular wave guide, surface waveguides do not support TEM mode of propagation in them.

3. The relation between voltage, current and impedance matrices of a microwave network is:

a) [V] = [Z][I].

b) [Z] = [V][I].

c) [I] = [Z][V].

d) [V] = [Z]-[I].

View Answer

Answer: a

Explanation: In microwave networks, at any point in a network, the voltage at a point is the product of the impedance at that point and current measured. This can be represented in the form of a matrix.

4. The relation between voltage, current and admittance matrices of a microwave network is:

a) [I] = [Y] [V].

b) [Y] = [V] [I].

c) [I] = [Z] [V].

d) [V] = [Z]-1[I].

View Answer

Answer: a

Explanation: The relation between voltage current and admittance matrices is [I] = [Y] [V]. here I represents the current matrix, Y is the admittance matrix and V is the voltage matrix.

5. Admittance and impedance matrices of a micro waves network are related as:

a) [Y] = [Z]-1.

b) [Y] = [Z].

c) [V] = [Z] [Z]-1.

d) [Z] = [V] [V]-1.

View Answer

Answer: a

Explanation: Both admittance and impedance matrix can be defined for a microwave network. The relation between these admittance and impedance matrix is [Y] = [Z]-1. Admittance matrix is the inverse of the impedance matrix.

6. The element of a Z matrix, Zij can be given in terms of voltage and current of a microwave network as:

a) ZIJ = VI/IJ

b) ZIJ = VIIJ

c) 1//ZIJ = 1/JIVI

d) VIJ = IJ/JI

View Answer

Answer: a

Explanation: The element Zij of a Z matrix is defined as the ratio of voltage at the ith port to the current at the jth port given that all other currents are set to zero.

7. In a two port network, if current at port 2 is 2A and voltage at port 1 is 4V, then the impedance Z₁₂ is:

a) 2 Ω

b) 8 Ω

c) 0.5 Ω

d) Insufficient data

View Answer

Answer: a

Explanation: Z12 is defined as the ratio of the voltage at port 1 to the current at port 2. Substituting the given values in the above equation, Z12 parameter of the network is 2 Ω.

8. In a 2 port network, if current at port 2 is 2A and voltage at port 1 is 4 V, then the admittance Y₂₁ is:

a) 0.5 Ʊ

b) 8 Ʊ

c) 2 Ʊ

d) 4 Ʊ

View Answer

Answer: a

Explanation: Admittance parameter Y12 is defined as the ratio of current at port 1 to the voltage at port 2. Taking the ratio, the admittance Y12 is 0.5 Ʊ.

9. For a reciprocal network, Z matrix is:

a) A unit matrix

b) Null matrix

c) Skew symmetric matrix

d) Symmetric matrix

View Answer

Answer: d

Explanation: For a reciprocal matrix, the impedance measured at port Zij is equal to the impedance measured at port Zji. Since these parameters occupy symmetric positions in the Z matrix, the matrix becomes symmetric.

10. For a lossless network, the impedance and admittance matrices are:

a) Real

b) Purely imaginary

c) Complex

d) Rational

View Answer

Answer: b

Explanation: For a network to be lossless, the network should be purely imaginary. Presence of any real component implies the presence of resistance in the network from which the network becomes lossy. So the matrices must be purely imaginary.

11. The matrix with impedance parameters Z₁₁=1+j, Z₁₂=4+j, Z₂₂=1, Z21=4+j is said to be

a) Reciprocal network

b) Lossless network

c) Lossy network

d) None of the mentioned

View Answer

Answer: a

Explanation: In the given case, Z12=Z21. This condition can be satisfied only by reciprocal networks. Hence the given network is a reciprocal network.

Sanfoundry Global Education & Learning Series – Microwave Engineering.

Microwave Engineering Questions and Answers – Scattering Matrices

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Scattering Matrices”.

1. S parameters are expressed as a ratio of:

a) Voltage and current

b) Impedance at different ports

c) Indecent and the reflected voltage waves

d) None of the mentioned

View Answer

Answer: c

Explanation: S matrix can be used to represent any n port network. S parameters are defined for microwave networks. Hence instead of voltage and current measurement, the amplitude of the incident and reflected voltage waves is measured.

2. The relation between incident voltage matrix , reflected voltage matrix and S matrix for a microwave network:

a) [v-] = [s] [v+].

b) [v+] = [s] [v-].

c) [v-] [v] = [s].

d) [s] = [v] [v-].

View Answer

Answer: a

Explanation: S parameter for a microwave network is defined as the ratio of reflected voltage wave to the incident voltage wave. When represented in the form of a matrix, reflected voltage matrix is the product of S parameter and the incident voltage wave at that port.

3. The specific element Sij of the scattering matrix can be determined as:

a) SIJ= Vi-/Vj+

b) SIJ= Vi+/Vj-

c) S= Vj+/Vi-

d) None of the mentioned

View Answer

Answer: a

Explanation: The parameter Sij is found by driving port j with an incident wave of voltage Vj+ coming out of ports i. The incident waves on all ports except the jth port are set to zero.

4. The device used to get the measurement of S parameters of n- port micro wave network is:

a) CRO

b) Network analyzer

c) Circulator

d) Attenuator

View Answer

Answer: b

Explanation: Network analyzer is a device to which any microwave network can be externally connected with the help of probes and the s parameters of the network can be obtained.

5. For a one port network , the scattering parameter S₁₁ in terms of impedance parameter Z₁₁ is:

a) (Z11-1)/ (Z11+1)

b) (Z11+1)/ (Z11-1)

c) (Z11+1) (Z11-1)

d) Z11

View Answer

Answer: a

Explanation: If Z matrix of a one port network is computed, then the s matrix of the same can be computed using the Z11 coefficient. To compute the S11 parameter of the network, the relation used is (Z11-1)/ (Z11+1).

6. Scattering matrix for a reciprocal network is:

a) Symmetric

b) Unitary

c) Skew symmetric

d) Identity matrix

View Answer

Answer: a

Explanation: For a reciprocal network, the input to port I and output at port j is the same as the input at port j and output measured at port i. Hence, the ports are interchangeable. As the ports are interchangeable, this is reflected in the matrix and the matrix becomes symmetric.

7. S₁₂=0.85-45⁰ and S₁₂=0.85 +45⁰ for a two port network. Then the two port network is:

a) Non-reciprocal

b) Lossless

c) Reciprocal

d) Lossy

View Answer

Answer: a

Explanation: For a reciprocal network, the S matrix is symmetric. For the matrix to be symmetric, Sij=Sji. Since this condition is not satisfied in the above case, the matrix is non reciprocal.

8. Scattering matrix for a lossless matrix is:

a) Unitary

b) Symmetric

c) Identity matrix

d) Null matrix

View Answer

Answer: a

Explanation: For a lossless network, the scattering matrix has to be unitary. That is, the law of conservation of energy is to be verified for this case. Using appropriate formula, this condition can be verified.

9. If the reflection co efficient of a 2 port network is 0.5 then the return network loss in the network is:

a) 6.5 dB

b) 0.15 dB

c) 6.020 dB

d) 10 dB

View Answer

Answer: c

Explanation: Given the reflection coefficient of the network, return loss of the network is calculated using the formula -20 log │Г│. Substituting for reflection coefficient, the return loss of the network is 6.02 dB.

10. If the reflection co efficient of a 2 port network is 0.25 then the return network loss in the network is:

a) 12.05 dB

b) 0.15 dB

c) 20 dB

d) 10 dB

View Answer

Answer: a

Microwave Engineering Questions and Answers – Transmission Matrix(ABCD)

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Transmission Matrix(ABCD)”.

1. ABCD matrix is used:

a) When there is two or more port networks in the cascade

b) To represent a 2 port network

c) To represent a 2 port network

d) To represent the impedance of a microwave network

View Answer

Answer: a

Explanation: The Z, Y, and S parameter representation can be used to characterize a microwave network with an arbitrary number of ports. But most microwave networks consist of cascade of two or more two port networks. In this case it is convenient to use ABCD matrix for network representations.

2. The voltage equation for a 2 port network that can be represented as a matrix is:

a) V1=AV2 + BI2

b) V1=CV2 + DI2

c) V1=BV2 +AI2

d) V1=DV2+CI2

View Answer

Answer: a

Explanation: In the equation, V1 is the voltage measured at port 1 and V2 is the voltage measured at port 2 and I2 is the current measured at the second port. A and B are the network constants.

3. ABCD matrix of the cascade connection of 2 networks is equal to:

a) Product of ABCD matrices representing the individual two ports

b) Sum of the ABCD matrices representing the individual two ports

c) Difference of the ABCD matrices representing the individual two ports

d) Sum of transpose of ABCD matrices representing the individual two ports

View Answer

Answer: a

Explanation: When two networks are connected in cascade, each of the two networks are represented as a 2×2 square matrix. Then to obtain the equivalent matrix of the cascade, the product of the ABCD matrices of each stage is taken.

4. For simple impedance Z, the ABCD parameters are:

a) A=1, Z=B, C=0, D=1

b) A=0, B=1, C=1, D=0

c) A=Z, B=1, C=1, D=0

d) A=1, B=0, C=Z, D=1

View Answer

Answer: a

Explanation: If simple impedance or an equivalent impedance of a network is represented as a ABCD matrix, writing the equations in terms of voltage and current and setting each variable to zero, the four constants are obtained. For an impedance Z, the constants are A=1, Z=B, C=0, D=1.

5. For a simple admittance Y, the ABCD parameters are:

a) A=1, B=0, C=Y, D=1

b) A=Z, B=1, C=1, D=0

c) A=1, B=0, C=Z, D=1

d) A=1, Y=B, C=0, D=1

View Answer

Answer: a

Explanation: If simple admittance or an equivalent admittance of a network is represented as a ABCD matrix, writing the equations in terms of voltage and current and setting each variable to zero, the four constants are obtained. For an admittance Y, the constants are A=1, Z=B, C=0, D=1.

6. C parameter for a transmission line of characteristic impedance Zₒ, phase constant β and length ‘l’ is:

a) j Yₒ Sin βl

b) j Zₒ Sin βl

c) j Zₒ tan βl

d) j Yₒ tan βl

View Answer

Answer: a

Explanation: If a transmission line is represented as two port network, constants can be derived in terms of the A, B, C, D constants for the network. But setting each electrical parameter to zero, this constant is found. By doing so, the C parameter of transmission line is j Yₒ Sin βl.

7. For a 2 port network if Z₁₁=1.5 and Z₁₂=1.2, A parameter for the same 2 port network is:

a) 1.5

b) 1.25

c) 0.75

d) 1.75

View Answer

Answer: b

Explanation: A parameter for the two port network is the ratio of the impedance Z11 and the impedance Z12. Substituting in this equation,’ A’ parameter of the network is 1.25.

8. For a 2 port network, if the admittance parameter Y₁₂=0.4, then B among the ABCD, parameters for the 2 port network is:

a) 2.5

b) 4.5

c) 5

d) 6

View Answer

Answer: a

Explanation: For a two port network, B parameter is defined as the reciprocal of the admittance Y12. Taking the reciprocal of the given value, the B parameter of the network is 2.5.

9. If D=1.6 and B=2.8 for a 2 port network, then Y₁₁=?

a) 0.5714

b) 0.987

c) 0.786

d) 1.75

View Answer

Answer: a

Explanation: The admittance Y11 of the network is defined as the ratio of B parameter to the D parameter of the network. Taking the ratio of the given values, admittance Y11 is 0.5714.

10. If A=2.8 and B=1.4 for a 2 port network then Z₁₁=?

a) 0.5

b) 2

c) 4.2

d) 2.7

View Answer

Answer: b

Explanation: Z11 parameter of a two port network is the ratio of the A parameter of the network to the B parameter of the network. Taking the ratio of the given values, Z11 is 2.

Microwave Engineering Questions and Answers – Aperture Coupling

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Aperture Coupling”.

1. The matched network is placed between:

a) load and transmission line

b) source and the transmission line

c) source and the load

d) none of the mentioned

View Answer

Answer: A

Explanation: At microwave frequencies, for maximum power transmission, the characteristic impedance of the transmission line must be matched to the load impedance with which the line is terminated. Hence to match these impedances, the matched network is laced between load and transmission line.

2. When a transmission line is matched to a load using a matched network, reflected waves are present:

a) between the load and the matched network

b) between the matched network and the transmission line

c) between the source and the transmission line

d) between the matched network and source

View Answer

Answer: A

Explanation: The matching circuit is used to match the transmission line and the load. This circuit prevents the reflection of the waves reaching the source. Hence, reflected waves are present between the load and the matched network.

3. Impedance matching sensitive receiver components may improve the _____ of the system.

a) noise

b) SNR

c) amplification factor

d) thermal noise

View Answer

Answer: B

Explanation: SNR (signal to noise ratio) of the system defines the ratio of signal power to noise power. An increase in this value results in increase of the signal strength. Impedance matching certain sensitive receiver components helps in delivering maximum power to the load and increased signal strength.

4. One of the most important factors to be considered in the selection of a particular matching network is:

a) noise component

b) amplification factor

c) bandwidth

d) none of the mentioned

View Answer

Answer: C

Explanation: Any type of matching network can ideally give a perfect match at a single frequency. But it is desirable to match a load over a band of frequencies. Hence, bandwidth plays an important role in the selection of the matching network.

5. The simplest type of matching network, L section consists of _______ reactive elements.

a) one

b) two

c) four

d) six

View Answer

Answer: B

Explanation: As the name of the matching circuit indicates, ‘L’ section consists of 2 reactive elements, one element vertical and another horizontal. 2 types of ‘L’ sections exist. The best one is chose based on the normalized value of the load impedance.

6. The major limitation of a lumped elements matching ‘L’ network is:

a) they are not equally efficient at higher frequencies as they are at lower frequencies.

b) size of the network

c) they restrict flow of current

d) none of the mentioned

View Answer

Answer: A

Explanation: Since we use lumped elements like inductors and capacitors as the components of the matching network, they behave differently at frequencies higher than 1GHz, because of the frequency dependent factor of inductive and capacitive reactance. This is one of the major limitations.

7. An ‘L’ network is required to match a load impedance of 40Ω to a transmission line of characteristic impedance 60Ω. The components of the L network are:

a) 28.28+j0 Ω

b) 28.28+j1 Ω

c) 50Ω

d) 48.9Ω

View Answer

Answer: A

Explanation: Since both load impedance and characteristic impedance are resistive (real), the imaginary part of the matching network is 0. Real part of the matching network is given by the expression ±√(RL(Z0– RL))-XL. Substituting the values given, the matching network impedance is 28.28Ω.

8. The imaginary part of the matching network is given by the relation:

a) ±(√(Z0– RL)/RL)Z00

b) ±(√(Z0– RL)/RL)

c) ±(√(Z0– RL)/ Z0

d) None of the mentioned

View Answer

Answer: A

Explanation: By theoretical analysis, the expressions for real and imaginary parts of the impedance of the matching network are derived in terms of the load impedance and the characteristic impedance of the transmission line. This expression derived is ±(√(Z0– RL)/RL)Z0 .

9. Which of the following material is not used in the fabrication of resistors of thin films?

a) nichrome

b) tantalum nitride

c) doped semiconductor

d) pure silicon

View Answer

Answer: D

Explanation: Certain physical properties are to be met in order to use a material to make thin film resistors. These properties are not found in pure silicon which is an intrinsic semiconductor.

10. Large values of inductance can be realized by:

a) loop of transmission line

b) spiral inductor

c) coils of wires

d) none of the mentioned

View Answer

Answer: B

Explanation: Loop of transmission lines are used to make inductors to realize lower values of inductance. Coils of wires cannot be used to realize inductors at higher frequencies. Spiral conductors can be used to realize inductors of higher values at higher frequencies.

11. A short transmission line stub can be used to provide a shunt capacitance of:

a) 0-0.1µF

b) 0-0.1pF

c) 0-0.1nF

d) 1-10pF

View Answer

Answer: B

Explanation: Since a transmission line consists of two wires, which can act plates of a capacitor, they can be used as a capacitor of very low values of the range 0-0.1pF.

Microwave Engineering Questions and Answers – Impedance Matching Using Slotted Lines

This set of Microwave Engineering Interview Questions and Answers for freshers focuses on “Impedance Matching Using Slotted Lines”.

1. Slotted line is a transmission line configuration that allows the sampling of:

a) electric field amplitude of a standing wave on a terminated line

b) magnetic field amplitude of a standing wave on a terminated line

c) voltage used for excitation

d) current that is generated by the source

View Answer

Answer: a

Explanation: Slotted line allows the sampling of the electric field amplitude of a standing wave on a terminated line. With this device, SWR and the distance of the first voltage minimum from the load can be measured, from this data, load impedance can be found.

2. A slotted line can be used to measure _____ and the distance of _____________ from the load.

a) SWR, first voltage minimum

b) SWR, first voltage maximum

c) characteristic impedance, first voltage minimum

d) characteristic impedance, first voltage maximum

View Answer

Answer: a

Explanation: With a slotted line, SWR and the distance of the first voltage minimum from the load can be measured, from this data, load impedance can be found.

3. A modern device that replaces a slotted line is:

a) Digital CRO

b) generators

c) network analyzers

d) computers

View Answer

Answer: c

Explanation: Although slotted lines used to be the principal way of measuring unknown impedance at microwave frequencies, they have largely been superseded by the modern network analyzer in terms of accuracy, versatility and convenience.

4. If the standing wave ratio for a transmission line is 1.4, then the reflection coefficient for the line is:

a) 0.16667

b) 1.6667

c) 0.01667

d) 0.96

View Answer

Answer: a

Explanation: ┌= (SWR-1)/ (SWR+1). Substituting for SWR in the above equation for reflection co-efficient, given SWR is 1.4, reflection co-efficient is 0.16667.

5. If the reflection coefficient of a transmission line is 0.4, then the standing wave ratio is:

a) 1.3333

b) 2.3333

c) 0.4

d) 0.6

View Answer

Answer: b

Explanation: SWR= (1+┌)/ (1-┌). Where ┌ is the reflection co-efficient. Substituting for the reflection co-efficient in the equation, SWR is 2.3333.

6. Expression for ϴ means phase angle of the reflection co efficient r=|r|-e^jθ, the phase of the reflection co-efficient is:

a) θ=2π+2βLmin

b) θ=π+2βLmin

c) θ=π/2+2βLmin

d) θ=π+βLmin

View Answer

Answer: b

Explanation: here, θ is the phase of the reflection co-efficient. Lmin is the distance from the load to the first minimum. Since voltage minima repeat every λ/2, any multiple of λ/2 can be added to Lmin .

7. In the expression for phase of the reflection coefficient, Lmin stands for :

a) distance between load and first voltage minimum

b) distance between load and first voltage maximum

c) distance between consecutive minimas

d) distance between a minima and immediate maxima

View Answer

Answer: a

Explanation: Lmin is defined as the distance between the terminating load of a transmission line and the first voltage minimum that occurs in the transmission line due to reflection of waves from the load end due to mismatched termination.

8. If SWR=1.5 with a wavelength of 4 cm and the distance between load and first minima is 1.48cm, then the reflection coefficient is:

a) 0.0126+j0.1996

b) 0.0128

c) 0.26+j0.16

d) none of the mentioned

View Answer

Answer: a

Explanation: ┌= (SWR-1)/ (SWR+1). Substituting for SWR in the above equation for reflection co-efficient, magnitude of the reflection co-efficient is 0.2. To find θ, θ=π+2βLmin, substituting Lmin as 1.48cm, θ=86.4⁰. Hence converting the polar form of the reflection co-efficient into rectangular co-ordinates, reflection co-efficient is 0.0126+j0.1996.

9. If the characteristic impedance of a transmission line 50 Ω and reflection coefficient is 0.0126+j0.1996, then load impedance is:

a) 47.3+j19.7Ω

b) 4.7+j1.97Ω

c) 0.26+j0.16

d) data insufficient

View Answer

Answer: a

Explanation: ZL=Z0 (1+┌)/ (1-┌). Substituting the given values of reflection co-efficient and characteristic impedance, ZL is 47.3+j19.7Ω .

10. If the normalized load impedance of a transmission line is 2, then the reflection co-efficient is:

a) 0.33334

b) 1.33334

c) 0

d) 1

View Answer

Answer: a

Explanation: ZL=Z0 (1+┌)/ (1-┌), this is the expression for load impedance. Normalized load impedance is the ratio of load impedance to the characteristic impedance, taking ZLL/Z0 as 2, the reflection co-efficient is equal to 0.33334.

Sanfoundry Global Education & Learning Series – Microwave Engineering.

Microwave Engineering Questions and Answers – Single Stub Matching

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Single Stub Matching”.

1. The major advantage of single stub tuning over other impedance matching techniques is:

a) Lumped elements are avoided

b) It can be fabricated as a part of transmission line media

c) It involves two adjustable parameters

d) All of the mentioned

View Answer

Answer: d

Explanation: Single stub matching does not involve any lumped elements, it can be fabricated as a part of transmission media and it also involves to adjustable parameters namely length and distance from load giving more flexibility.

2. Shunt stubs are preferred for:

a) Strip and microstrip lines

b) Coplanar waveguides

c) Circular waveguide

d) Circulators

View Answer

Answer: a

Explanation: Since microstrip and strip lines are simple structures, impedance matching using shunt stubs do not increase the complexity and structure of the transmission line. Hence, shunt stubs are preferred for strip and microstrip lines.

3. The two adjustable parameters in single stub matching are distance‘d’ from the load to the stub position, and _________

a) Susceptance or reactance provided by the stub

b) Length of the stub

c) Distance of the stub from the generator

d) None of the mentioned

View Answer

Answer: a

Explanation: Reactance or susceptance of the matching stub must be known before it used for matching, since it is the most important parameter for impedance matching between the load and the source.

4. In shunt stub matching, the key parameter used for matching is:

a) Admittance of the line at a point

b) Admittance of the load

c) Impedance of the stub

d) Impedance of the load

View Answer

Answer: a

Explanation: In shunt stub tuning, the idea is to select d so that the admittance Y, seen looking into the line at distance d from the load is of the form Yₒ+jb) Then the stub susceptance is chosen as –jB, resulting in a matched condition.

5. For series stub matching, the parameter used for matching is:

a) Impedance of the transmission line at a point

b) Voltage at a point on the transmission line

c) Admittance at a point on the transmission line

d) Admittance of the load

View Answer

Answer: a

Explanation: In series sub matching, the distance‘d’ is selected so that the impedance, Z seen looking into the line at a distance‘d’ from the load is of the form Zₒ+jX. Then the stub reactance is chosen as –jX resulting in a matched condition.

6. For co-axial lines and waveguides, ________ is more preferred.

a) Open circuited stub

b) Short circuited stub

c) Slotted section

d) Co-axial lines cannot be impedance matched

View Answer

Answer: b

Explanation: For co-axial cables and waveguides, short-circuited stub is usually preferred because the cross-sectional area of such an open-circuited line may be large enough to radiate, in which case the stub is no longer purely reactive.

7. For a load impedance of ZL=60-j80. Design of 2 single-stub shunt tuning networks to match this load to a 50Ω line is to be done. What is the normalized admittance obtained so as to plot it on smith chart?

a) 1+j

b) 0.3+j0.4

c) 0.4+j0.3

d) 0.3-j0.4

View Answer

Answer: b

Explanation: To impedance match a load to a characteristic impedance of the transmission line, first the load has to be normalized. That is, zL=ZL/Z0. For impedance matching using shunt stubs, admittance is used. Taking the reciprocal of impedance, normalized load admittance is 0.3+j0.4.

8. If the normalized admittance at a point on a transmission line to be matched is 1+j1.47. Then the normalized susceptance of the stub used for shunt stub matching is:

a) 1Ω

b) 1.47 Ω

c) -1.47 Ω

d) -1 Ω

View Answer

Answer: c

Explanation: When shunt stubs are used for impedance matching between a load and transmission line, the susceptance of the shunt stub must be negative of the line’s susceptance at that point for impedance matching.

9. After impedance matching, if a graph is plot with frequency v/s reflection co-efficient of the transmission line is done, then at the frequency point for which the design is done, which of the following is true?

a) There is a peak at this point of the curve

b) There is a dip at this point of the curve

c) The curve is a straight line

d) Such a plot cannot be obtained

View Answer

Answer: b

Explanation: Since the plot is frequency v/s reflection co-efficient, after impedance matching the reflection co-efficient will be zero or minimum. Hence, there is a dip at that point of the curve.

10. In series stub matching, if the normalized impedance at a point on the transmission line to be matched is 1+j1.33. Then the reactance of the series stub used for matching is:

a) 1 Ω

b) -1.33 Ω

c) -1 Ω

d) 1.33 Ω

View Answer

Answer: b

Explanation: The reactance of the series stub is negative of the reactance of the line at the point at which it has to be matched. That is, if the line reactance is inductive, the series stub’s reactance is capacitive.

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Microwave Engineering Questions and Answers – Two Port Power Gains

Microwave Engineering Questions and Answers – Reflections

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Reflections”.

1. Discontinuities in the matching quarter wave transformer are not of considerable amount and are negligible.

a) True

b) False

View Answer

Answer: b

Explanation: Discontinuities in the matching network cause reflections which result in considerable attenuation of the transmitted signal. Hence, discontinuities in transformers are not negligible.

2. The overall reflection coefficient of a matching quarter wave transformer cannot be calculated because of physical constraints.

a) True

b) False

View Answer

Answer: b

Explanation: Though the computation of total reflection is complex, the total reflection can be computed in two ways. They are the impedance method and the multiple reflection method.

3. In the multiple reflections analysis method, the total reflection is:

a) An infinite sum of partial reflections

b) An infinite sum of partial reflection and transmissions

c) Constant value

d) Finite sum of partial reflections

View Answer

Answer: b

Explanation: The number of discontinuities in the matching circuit (quarter wave transmission line) is theoretically infinite since the exact number cannot be practically determined. Hence, the total reflection is an infinite sum of partial reflections and transmission.

4. The expression for total reflection in the simplified form is given by:

a) Г=Г1+ Г3e-2jθ

b) Г=Г11+Г3

c) Г=Г12+ Г3e-2jθ

d) Г= Г1+ Г2e-2jθ

View Answer

Answer: a

Explanation: This expression dictates that the total reflection is dominated by the reflection from the initial discontinuity between Z1 and Z2 (Г1), and the first reflection from the discontinuity between Z2 and ZL (Г3e-2jθ).

5. The e-2jθ term in the expression for total reflection in a single section quarter wave transformer impedance matching network Г=Г1+ Г3e-2jθ signifies:

a) Phase delay

b) Frequency change

c) Narrowing bandwidth

d) None of the mentioned

View Answer

Answer: a

Explanation: The term e-2jθ in Г=Г1+ Г3e-2jθ accounts for phase delay when the incident wave travels up and down the line. This factor is a result of multiple reflections.

6. If the first and the third reflection coefficients of a matched line is 0.2 and 0.01, then the total reflection coefficient if quarter wave transformer is used for impedance matching is:

a) 0.2

b) 0.01

c) 0.21

d) 0.19

View Answer

Answer: d

Explanation: The total reflection co-efficient of a matched line due to discontinuities is given by Г=Г1+ Г3e-2jθ. Given that Г1=0.2 and Г3=0.01, β=2π/λ, l=λ/4. θ=βl, Substituting the given values in the above 2 given equations, the total reflection coefficient is 0.19.

7. If a λ/4 transmission line is used for impedance matching, then always Г1> Г3.

a) True

b) False

View Answer

Answer: a

Explanation: Since the load is matched to the transmission line the reflection from the load towards the source will be very less (Г3). Г1 is the reflection from the junction of the transmission line and the λ/4 matching section. Since this end will have some improper matching and discontinuities, Г1 is always greater than Г3.

8. To compute the total reflection of a multi-section transmission line, the lengths of the transmission lines considered are all unequal.

a) True

b) False

View Answer

Answer: b

Explanation: The computation of total reflection of a matched line due to discontinuities is theoretically complex. In order to obtain an approximated simple expression, the lengths of the multi section matching transformers is a constant or all of them are equal.

9. If ZL< Z0, then the reflection coefficient at that junction is:

a) ГN<0

b) ГN>0

c) ГN>1

d) None of the mentioned

View Answer

Answer: a

Explanation: When there is no proper matching between load impedance and the characteristic impedance of a transmission line and given the condition that ZL< Z0, then the reflection coefficient at that junction is always negative. That is, ГN<0.

10. The total approximate reflection coefficient is a finite sum of reflection co-efficient of individual matching section of the matching network.

a) True

b) False

View Answer

Answer: a

Explanation: In a multi section transformer there are N sections, if the reflection from each section is ГN, then the total reflection is the sum of reflections that occur due to individual sections. There is an exponential component associated with each reflection coefficient that decays exponentially.

11. Using the relation for total reflection co-efficient certain designs of matching networks can be made as per practical requirements.

a) True

b) False

View Answer

Answer: a

Explanation: We can synthesize any desired reflection coefficient response as a function of frequency by properly choosing the ГN and using enough sections (N).

Microwave Engineering Questions and Answers – Binomial Multi-section Matching Transformers

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Binomial Multi-section Matching Transformers”.

1. The passband response of a binomial matching transformer can be called optimum:

a) if the roll off in the response curve is high

b) if the response is flat in the impedance matched region

c) if the matching network is frequency sensitive

d) none of the mentioned

View Answer

Answer: B

Explanation: The response curve of a binomial matching transformer ( θ v/s │Г (θ) │) must be flat at the frequency for which impedance matching is performed and for those frequencies that lie in the required bandwidth. This is one of the most important characteristic of a good matching circuit.

2. If a quality binomial matching transformer gives a good flat response curve, it is called “maximally flat”.

a) true

b) false

View Answer

Answer: A

Explanation: A binomial matching section can be termed efficient when it is less frequency sensitive and gives a constant gain over a wide range of frequencies. This constant gain implies a flat curve over a wide range of frequencies. This is termed as “maximally flat”.

3. The response curve of a binomial matching transformer is plotted for each section of the matching network individually and then analyzed for optimum solution.

a) true

b) false

View Answer

Answer: B

Explanation: The response curve of a binomial multisection transformer is determined for an N-section transformer by setting the first N-1 derivatives of │Г (θ) │ to zero at the center frequency, fₒ.

4. To obtain a flat curve in the response of a binomial multisection transformer, N-1 derivatives of │Г (θ) │are set to zero. This implies:

a) the frequency sensitivity of the matching section is increased linearly

b) the frequency sensitivity of the matching section is increased exponentially

c) roll off in the curve is increased

d) none of the mentioned

View Answer

Answer: D

Explanation: The derivatives of │Г (θ) │ show the rate of change of reflection co-efficient with distance. If this derivative is not zero, the matching circuit becomes more sensitive and a higher bandwidth cannot be obtained. Hence to make the matching network frequency independent, the derivatives are set to zero.

5. The condition │Г (θ) │=0 for θ=π/2 of a binomial multi section transformer corresponds to the:

a) upper cutoff frequency

b) lower cutoff frequency

c) center frequency

d) none of the mentioned

View Answer

Answer: C

Explanation: θ=π/2 corresponds to the center frequency at which │Г (θ) │=0. θ=βl. β=2 π/λ and l=λ/4. Substituting for β and λ in the equation for θ, θ=π/2. This is the center frequency at which impedance matching is done at which the reflection coefficient is zero and perfect match is achieved.

6. The reflection co-efficient magnitude of a binomial multisection transformer is:

a) 2N│A││cos (θ)│N

b) 2N│A│

c) 2N│cos (θ) │N

d) none of the mentioned

View Answer

Answer: A

Explanation: The reflection co-efficient of a binomial multisection transformer is dependent on the length of the matching section, operating frequency and load impedance and characteristic impedance. A is a constant defined as A=2-N (ZL– Z0)/ (ZL+ Z0).

7. The reflection coefficient ГN in terms of successive impedances Zn and Zn+1 when multisection transformers are used in a binomial matching transformer is given by:

a) 0.5ln (Zn+1/Zn)

b) ln (Zn+1/Zn)

c) 0.5ln (Zn/Zn+1)

d) (Zn/Zn+1)

View Answer

Answer: A

Explanation: After binomial expansion of the equation for Г(θ), the maximum power is N, where N is the number of the sections in the transformer. After making suitable approximations so that the approximated values are in well agreement with actual values, the expression for reflection coefficient is 0.5ln (Zn+1/Zn).

8. In the plot of normalized frequency v/s reflection co-efficient for a binomial multisection filter, the curve has a dip at:

a) center frequency

b) upper cutoff frequency

c) lower cutoff frequency

d) none of the mentioned

View Answer

Answer: A

Explanation: Since the impedance matching circuit is used to match the load to the transmission line, there will be perfect match in the circuit resulting in zero or low reflection. Hence, there is a dip at the center frequency.

9. As the number of sections in the binomial multisection transformer increases the plot of normalized frequency v/s reflection co-efficient has a wider open curve.

a) true

b) false

View Answer

Answer: A

Explanation: When more number of sections are used for matching, the reflection co-efficient is low for neighboring frequencies as well. Hence, the network can be used for a wide range of operating frequencies. Hence, this increases the bandwidth.

10. A three section binomial transformer is used to match a 100Ω transmission line to a 50Ω transmission line. Then the value of the constant ‘A’ for this design is:

a) -0.0433

b) 0.0433

c) 0.01

d) -0.01

View Answer

Answer: A

Explanation: ‘A’ is given by the expression 2-(n+1)ln (ZL/Z0), Where N is the number of sections in the matching network. Substituting the given values in the equation for ‘A’, the value of A is -0.0433.

Microwave Engineering Questions and Answers – Chebyshev Multi-section Matching Transformers

This set of Microwave Engineering Questions and Answers for Experienced people focuses on “Chebyshev Multi-Section Matching Transformers”.

1. The major disadvantage of binomial multi section transformer is higher bandwidth cannot be achieved.

a) true

b) false

View Answer

Answer: a

Explanation: In some applications, a flat curve in the operating frequency is a major requirement. This requirement can be satisfied using a binomial transformer. But the disadvantage is that a higher bandwidth can be achieved.

2. Advantage of chebyshev matching transformers over binomial transformers is:

a) higher gain

b) low power losses

c) higher roll-off in the characteristic curve

d) higher bandwidth

View Answer

Answer: d

Explanation: Chebyshev transformers when designed to operate at a certain frequency called center frequency, the reflection co-efficient is low for a large frequency range implying that they have a higher operating range. This is the major advantage of chebyshev filters.

3. There are passband ripples present in the chebyshev characteristic curve.

a) true

b) false

View Answer

Answer: a

Explanation: This is a major difference between chebyshev and binomial transformer. Binomial transformers have a flat curve in the passband while chebyshev transformers have ripples in the transformer passband.

4. Chebyshev matching transformers can be universally used for impedance matching in any of the microwave networks.

a) true

b) false

View Answer

Answer: b

Explanation: Chebyshev transformers have passband ripples in the characteristic curve. In some critical applications, these ripples are not tolerable in the operating bandwidth. Hence, chebyshev transformers cannot be used for all the microwave networks for impedance matching.

5. The 4th order chebyshev polynomial is:

a) 8x4-8x2+1

b) 4x3-4x2+1

c) 4x3-3x

d) none of the mentioned

View Answer

Answer: a

Explanation: nth order polynomial for a chebyshev polynomial is generated using lower polynomials by the expression Tn (x) = 2xTn-1(x) – Tn-2(x). T2(x) = 2x2-1, T3(x)= 4x3-3x. Substituting the lower level polynomials in the given expression, T4(x) = 8x4-8x2+1.

6. Chebyshev polynomials do not obey the equal-ripple property.

a) true

b) false

View Answer

Answer: b

Explanation: For -1≤x≤1,│T(x)│≤ 1. In this range, the chebyshev polynomials oscillate between±1. This is the equal ripple property. Chebyshev polynomials obey the equal-ripple property.

7. Chebyshev polynomial can be expressed in trigonometric functions as:

a) Tn(cos θ)=cos nθ

b) Tn(sin θ)= sin nθ

c) Tn(cos θ)=cos nθ.sin nθ

d) none of the mentioned

View Answer

Answer: a

Explanation: If the chebyshev polynomial variable x is equated to a trigonometric variable cos θ, then the higher order chebyshev polynomials can be defined in terms of the same function with multiples of θ. This can be theoretically proved and function generation becomes simpler.

8. For values of x greater than 1, the chebyshev polynomial in its trigonometric form cannot be determined.

a) true

b) false

View Answer

Answer: b

Explanation: Since cosine function is defined for values of x between -1 and +1, for x values greater than 1, hyperbolic function is used to define the chebyshev polynomial. Tn(x)=cosh (n cosh-1x).

9. Reflection co-efficient Гn in terms of Zn and Zn+1, successive impedances of successive sections in the matching network are:

a) 0.5 ln (Zn+1/Zn)

b) 0.5 ln (Zn/Zn+1)

c) ln (Zn+1/Zn)

d) ln (Zn/Zn+1)

View Answer

Answer: a

Explanation: When multiple sections are used in the chebyshev matching network, the reflection co-efficient of the nth matching section, given the impedances at the ends of the section, reflection co-efficient can be obtained using the expression 0.5 ln (Zn+1/Zn).

10. In a 3 section multisection chebyshev matching network, if Z3 = 100Ω, and Z2=50Ω, then the reflection co-efficient Г2 is:

a) 0.154

b) 0.3465

c) 0.564

d) none of the mentioned

View Answer

Answer: b

Explanation: Гn for ‘n’ section matching chebyshev network is given by Гn=0.5 ln (Zn+1/Zn). substituting the given values in the expression, Г2 is 0.3465.

11. If Г3=0.2 and Z3=50Ω, then the impedance of the next stage in the multi-section transformer is:

a) 100Ω

b) 50Ω

c) 74.6Ω

d) 22.3Ω

View Answer

Answer: c

Explanation: Гn for ‘n’ section matching chebyshev network is given by Гn=0.5 ln (Zn+1/Zn). Substituting the given values in the expression, the impedance of the next stage is Z4=74.6Ω.

Microwave Engineering Questions and Answers – Terminated Lossless Transmission Lines – 2

Microwave Engineering Questions and Answers – Tapered Lines

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Tapered Lines”.

1. A single section tapered line is more efficient in impedance matching than a multisection tapered line for impedance matching.

a) True

b) False

View Answer

Answer: b

Explanation: As the number N of discrete transformer sections increases, the step changes in the characteristic impedance between the sections become smaller, and the transformer geometry approaches a continuous tapered line. Thus multisection are preferred over single section for impedance matching.

2. Passband characteristics of tapered lines differ from one type of taper to another.

a) True

b) False

View Answer

Answer: a

Explanation: The impedance of the tapered line varies along the line depending on the type of the tapering done. Thus impedance is a function of the type of taper. Hence passband characteristics depend on the type of taper.

3. For a continually tapered line, the incremental reflection co-efficient is:

a) ∆Z/2Z

b) 2Z/∆Z

c) ∆Z0/2Z0

d) None of the mentioned

View Answer

Answer: a

Explanation: The incremental reflection co-efficient ∆Г is a function of distance. If a step change in impedance occurs for z and z+∆z, then the incremental reflection co-efficient is given by ∆Z/2Z.

4. The variation of impedance of an exponentially tapered line along the length of the line is given by:

a) Z(z)=Z0eaz

b) Z(z)=Z0e-az

c) Z(z)=Z0e2az

d) Z(z)=Z0e-2az

View Answer

Answer: a

Explanation: The variation of impedance along the transmission line is a positive growing curve and is given by Z(z)=Z0eaz. The constant ‘a’ is defined as L-1 ln(ZL/Z0).

5. The value of constant ‘a’ for an exponentially tapered line of length 5 cm with load impedance being 100Ω and characteristic impedance of the line is 50Ω is:

a) 0.1386

b) 0.265

c) 0.5

d) 0.2

View Answer

Answer: a

Explanation: The constant ‘a’ for a tapered transmission line is given by L-1 ln(ZL/Z0). ‘a’ is a function of the tapered length, load and characteristic impedance. Substituting the given values in the above expression, ‘a’ has the value 0.1386.

6. Reflection co-efficient magnitude response is an exponential curve for tapered line.

a) True

b) False

View Answer

Answer: b

Explanation: The reflection co-efficient magnitude response of a exponential tapered line resembles only positive valued sinc function or can be called as a function with multiple peaks.

7. Triangular taper is the best solution for any impedance matching requirement.

a) True

b) False

View Answer

Answer: b

Explanation: Klopfenstein taper is the best and most optimized solution for impedance matching because reflection co-efficient has minimum value in the passband.

8. The maximum passband ripple in a Klopfenstein taper matching section is:

a) Г0/cos h A

b) Г0/sin h A

c) Г0/ tan h A

d) None of the mentioned

View Answer

Answer: a

Explanation: The maximum passband ripple in a Klopfenstein taper matching section is Г0/cos h A. Here, Г0 is the reflection co-efficient at zero frequency. A is a trigonometric function relating reflection co-efficient at zero frequency and maximum ripple in the passband.

9. For any load impedance, perfect match can be obtained and the minimum reflection co-efficient achieved can be zero.

a) True

b) False

View Answer

Answer: b

Explanation: From Bode-Fano criterion, there is a theoretical limit on the minimum achievable reflection co-efficient for a given load impedance. Hence, perfect match cannot be obtained.

10. For a given load (a fixed RC product), a broader bandwidth can be achieved with a low reflection co-efficient in the passband.

a) True

b) False

View Answer

Answer: b

Explanation: Based on the theoretical results of Bode-Fano criterion, a broader bandwidth can be achieved only at the expense of a higher reflection coefficient in the passband.

11. A perfect match can be obtained in the passband for any impedance matching circuit around the center frequency for which it is defined.

a) True

b) False

View Answer

Answer: b

Explanation: The passband reflection co-efficient cannot be zero unless the bandwidth is zero. Thus a perfect match can be obtained only at a finite number of discrete frequencies.

Microwave Engineering Questions and Answers – Series and Parallel Resonant Circuits

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Series and Parallel Resonant Circuits”.

1. In a series LCR circuit, at resonance point the energy stored in the inductor and capacitor in the form of magnetic and electric energies are equal.

a) True

b) False

View Answer

Answer: b

Explanation: At resonant frequency of a series LCR circuit, reactance of the capacitor is equal to the reactance of the inductor. The energy stored in the capacitor in the form of electric energy and the energy stored in the inductor in the form of magnetic energy is both equal.

2. Quality factor-Q of a resonant circuit signifies:

a) Loss in the resonant circuit

b) Gain in the resonant circuit

c) Magnetic energy stored in the circuit

d) Electric energy stored in the circuit

View Answer

Answer: a

Explanation: Quality factor of a resonant network is defined as the ratio of average energy stored to the energy loss/ second. Hence, lower loss implies a higher quality factor.

3. The total energy stored in a series RLC circuit is equal to the energy stored in the inductor.

a) True

b) False

View Answer

Answer: b

Explanation: The total energy in an RLC circuit is the sum of the energy stored in the magnetic field of the inductor and the electric energy stored in the capacitor. Loss in the circuit occurs due to the resistive component.

4. Higher the quality factor of a series LCR circuit, greater is the operating bandwidth of the resonant circuit.

a) True

b) False

View Answer

Answer: b

Explanation: Band width and quality factor of a series RLC circuit are both inversely related. Higher the quality factor, lower the operating bandwidth.

5. The plot of input impedance magnitude v/s frequency has a peak at the resonant frequency.

a) True

b) False

View Answer

Answer: b

Explanation: At resonant frequency, the capacitive reactance is equal to the inductive reactance cancelling each other’s effect. Hence, there is a dip at the center frequency in the plot of input impedance magnitude v/s frequency.

6. For a parallel resonance circuit, the plot of input impedance magnitude v/s frequency has a dip at the resonant frequency.

a) True

b) False

View Answer

Answer: b

Explanation: In parallel RLC circuit, the input impedance is highest at resonant frequency since the reactive components are in parallel. Hence, there is a peak at the resonant frequency in the plot of input impedance magnitude v/s frequency.

7. The relation between unloaded Q and external Q is:

a) External Q> unloaded Q

b) Unloaded Q> external Q

c) External Q = unloaded Q

d) None of the mentioned

View Answer

Answer: b

Explanation: To compute unloaded Q only the resistance in the resonant circuit is considered. But to calculate external Q, the resistance and other load in the external load is also considered. Sine Q and R are inversely proportional, as R increases Q decreases. Since R is greater for external Q computation, unloaded Q> external Q.

8. Loaded Q and External Q are 2 different parameters.

a) True

b) False

View Answer

Answer: a

Explanation: loaded Q and external Q are 2 different parameters. They are related by the expression

QL-1=Qe-1+ Q0-1, where QL is the loaded Q, Qe is external Q and Q0 is the unloaded Q.

9. The bandwidth of a series RLC circuit is 200 Hz. Then the quality factor of the circuit is:

a) 200

b) 100

c) 0.005

d) 0.5

View Answer

Answer: c

Explanation: The relation between quality factor and bandwidth is given as bandwidth=Q-1. Substituting for bandwidth in this expression, the quality factor of the resonant circuit is 0.005.

10. If a parallel RLC circuit is excited with a source of 8v, 50 Hz and the circuit has an inductor of 1mH, capacitor of 1µF and a resistor of 50Ω, then the power loss that occurs in the circuit is:

a) 6.4mW

b) 3.2mW

c) 12.8mV

d) None of the mentioned

View Answer

Answer: a

Explanation: The power loss in a parallel RLC circuit is 0.5│V│2/R. given the values of source voltage and resistance in the circuit, the power loss in the parallel RLC circuit is 6.4mW.

Microwave Engineering Questions and Answers – Transmission Line Resonators

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Transmission Line Resonators”.

1. Lumped elements can be used to make resonators that rare to be operated at microwave frequencies.

a) True

b) False

View Answer

Answer: b

Explanation: Lumped elements cannot be used at microwave frequencies since their behavior is not deterministic at these frequencies and the required response cannot be achieved.

2. Short circuited λ/2 transmission line has a quality factor of:

a) β/2α

b) 2β/α

c) β/α

d) Z0/ZL

View Answer

Answer: a

Explanation: Quality factor of a short circuited transmission line is a function of attenuation constant and phase constant of the transmission line. Higher is the attenuation in the transmission line, lower is the quality factor of the transmission line.

3. Quality factor of a coaxial cable transmission line is independent of the medium between the wires of the transmission line.

a) True

b) False

View Answer

Answer: b

Explanation: Quality factor is dependent on the permeability of the medium between the inner and outer conductor of the co-axial cable. For example, air has twice the quality factor as that of Teflon filled co-axial fiber.

4. A coaxial cable is air filled with air as dielectric with inner and outer radius equal to 1 mm and 4 mm. If the surface resistivity is 1.84*10-2Ω,then the attenuation due to conductor loss is:

a) 0.011

b) 0.022

c) 0.11

d) 0.22

View Answer

Answer: a

Explanation: Conductor loss in a coaxial cable is given by Rs(a-1+b-1)/2ln (b/a). Here ‘a’ and ‘b’ are the inner and outer radii of the coaxial cable. is the intrinsic impedance of the medium, for air is 377Ω. Substituting the given values in the equation, conductor loss is 0.022 Np/m.

5. An air coaxial cable has attenuation of 0.022 and phase constant of 104.7, then the quality factor of a λ/2 short circuited resonator made out of this material is:

a) 2380

b) 1218

c) 1416

d) Insufficient data

View Answer

Answer: a

Explanation: Quality factor of a λ/2 short circuited transmission line is β/2α. β is the phase constant and α is the attenuation constant of the line, substituting the given values, the quality factor of the transmission line is 2380.

6. The equivalent resistance of a short circuited λ/4 transmission line is independent of the characteristic impedance of the transmission line.

a) True

b) False

View Answer

Answer: b

Explanation: The equivalent resistance of a short circuited λ/4 transmission line is dependent of the characteristic impedance of the transmission line. The expression for equivalent resistance is Z0/αl. Resistance of a short circuited line is directly proportional to the characteristic impedance of the transmission line.

7. A microstrip patch antenna has a width of 5.08mm and surface resistivity of 1.84*10-2. Then the attenuation due to conductor loss is:

a) 0.0724

b) 0.034

c) 0.054

d) None of the mentioned

View Answer

Answer: a

Explanation: Attenuation due to conductor loss of a microstrip line is given by Rs/Z0W. Substituting the given values, attenuation due to conductor loss is 0.0724 Np/m.

8. If the attenuation due to dielectric loss and attenuation due to conductor loss in a microstrip transmission line is 0.024Np/m and 0.0724 Np/m, then the unloaded quality factor if the propagation constant is 151 is:

a) 150

b) 783

c) 587

d) 234

View Answer

Answer: b

Explanation: Unloaded Q for a microstrip line is given by β/2α. Α is the sum of attenuation due to conductor loss and dielectric loss. Substituting the given values the equation, unloaded Q is 783.

9. The equivalent capacitance of a short circuited λ/4 transmission line is dependent on the characteristic impedance of the transmission line.

a) True

b) False

View Answer

Answer: a

Explanation: Equivalent capacitance of a short circuited λ/4 transmission line is dependent on the characteristic impedance of the transmission line. It is inversely proportional to the characteristic impedance of the transmission line. Equivalent capacitance is π/4ω0Z0.

10. Inductance of an open circuited λ/2 transmission line is dependent on the characteristic impedance of the transmission line.

a) True

b) False

View Answer

Answer: a

Explanation: Inductance of an open circuited λ/2 transmission line is dependent on the characteristic impedance of the transmission line. Expression for inductance is 1/ω02c, C is the equivalent capacitance of the open circuited line. C has the expression π/4ω0Z0.

Microwave Engineering Questions and Answers – Rectangular Waveguide Cavity Resonators

This set of Microwave Engineering Questions and Answers for Aptitude test focuses on “Rectangular Waveguide Cavity Resonators”.

1. Microwave resonators can be constructed from open sections of waveguide.

a) True

b) False

View Answer

Answer: b

Explanation: For resonance to occur in waveguides, a closed structure is required. They resonate between the walls of the rectangular waveguide. Also radiation loss from an open ended waveguide can be significant.

2. There is no energy stored inside a rectangular waveguide cavity resonator.

a) True

b) False

View Answer

Answer: b

Explanation: Energy is stored in a waveguide resonator in the form of electric field and magnetic field. Power is dissipated in the metallic walls of the cavity as well as in the dielectric material that may fill the cavity.

3. A rectangular cavity supports:

a) TEM mode of resonance

b) TM mode of resonance

c) TE mode of resonance

d) TE, TM modes of resonance

View Answer

Answer: d

Explanation: A rectangular wave guide supports both TE and TEM mode of propagation. Likewise, when a rectangular waveguide is used as resonator, it supports both TE and TM modes of resonance.

4. A waveguide is open circuited at both the ends to use it as a waveguide resonator.

a) True

b) False

View Answer

Answer: b

Explanation: A closed cavity structure is required in order to bring resonance in the rectangular cavity. Also open ended waveguides result in radiation losses. Hence the waveguide is short circuited to form a resonator.

5. In order to obtain the resonant frequency of a rectangular waveguide, the closed cavity has to satisfy:

a) Gaussian equation

b) Helmholtz equation

c) Ampere’s law

d) None of the mentioned

View Answer

Answer: b

Explanation: Helmholtz wave equation is considered and solved using variable separable form. Then the boundary conditions are applied to the wave equation considering the walls of the cavity. Solving this gives the expression for resonant frequency.

6. Given the dimension of the waveguide as b<a<d, no resonant mode exists for this specification of dimensions.

a) True

b) False

View Answer

Answer: b

Explanation: For the given dimensional specification b<a<d, the dominant resonant mode (lowest resonant frequency) will be the TE101 mode, corresponding to the TE10 dominant waveguide mode in a shorted guide of length λg/2.

7. Unloaded Q of a rectangular waveguide cavity resonator:

a) Does not exist

b) Defined as the ratio of length of the waveguide to breadth of the waveguide

c) Defined as the ratio of stored energy to the power dissipated in the walls

d) None of the mentioned

View Answer

Answer: c

Explanation: Quality factor signifies the power loss in the circuit. It is defined as the ratio of stored energy to the power dissipated in the walls. Higher the power dissipation in the walls, lower is the quality factor of the waveguide resonator.

8. Find the wave number of a rectangular cavity resonator filled with a dielectric of 2.25 and designed to operate at a frequency of 5 GHz.

a) 157.08

b) 145.2

c) 345.1

d) 415.08

View Answer

Answer: a

Explanation: The wave number of rectangular wave resonator is 2πf√∈r/C, substituting the given values in the above equation, the wave number of the rectangular cavity resonator is 157.08.

9. The required length of the cavity resonator for l=1 mode (m=1, n=0) given that the wave number of the cavity resonator is 157.01 and the broader dimension of the waveguide is 4.755 cm:

a) 1.10 cm

b) 2.20 cm

c) 2.8 cm

d) 1.8 cm

View Answer

Answer: b

Explanation: The required length of the cavity resonator for the given mode is given by the expression d=lπ/√(k>sup>2-(π/a)2. Substituting the given values in the equation, the required length of the waveguide is 2.20 cm.

10. If the loss tangent of a rectangular waveguide is 0.0004, then Q due to dielectric loss is:

a) 1250

b) 2450

c) 2500

d) 1800

View Answer

Answer: c

Explanation: Q of a rectangular waveguide due to dielectric loss is given by 1/tanδ. Substituting for tanδ in the above equation, Q due to dielectric loss is 2500.

Microwave Engineering Questions and Answers – Ferrite Circulators

Microwave Engineering Questions and Answers – Circular Waveguide Cavity Resonators

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Circular Waveguide Cavity Resonators”.

1. A cylindrical cavity resonator can be constructed using a circular waveguide.

a) shorted at both the ends

b) open at both the ends

c) matched at both the ends

d) none of the mentioned

View Answer

Answer: a

Explanation: A cylindrical cavity resonator is formed by shorting both the ends of the cylindrical cavity because open ends may result in radiation losses in the cavity.

2. The dominant mode in the cylindrical cavity resonator is TE101 mode.

a) true

b) false

View Answer

Answer: b

Explanation: The dominant mode of propagation in a circular waveguide is TE111 mode. Hence, the dominant mode of resonance in a cylindrical cavity made of a circular waveguide is TE111 mode. In a cylindrical resonator, the mode of propagation depends on the length of the cavity.

3. Circular cavities are used for microwave frequency meters.

a) true

b) false

View Answer

Answer: a

Explanation: Circular cavities are used for microwave frequency meters. The cavity is constructed with a movable top wall to allow the mechanical tuning of the resonant frequency.

4. The mode of the circular cavity resonator used in frequency meters is:

a) TE011 mode

b) TE101 mode

c) TE111 mode

d) TM111 mode

View Answer

Answer: a

Explanation: Frequency resolution of a frequency meter is determined from its quality factor. Q factor of TE011 mode is much greater than the quality factor of the dominant mode of propagation.

5. The propagation constant of TEmn mode of propagation for a cylindrical cavity resonator is:

a) √ (k2-(pnm/a)2)

b) √ pnm/a

c) √ (k2+(pnm/a)2)

d) none of the mentioned

View Answer

Answer: a

Explanation: The propagation constant for a circular cavity depends on the radius of the cavity, and the wave number. If the mode of propagation is known and the dimension of the cavity is known then the propagation constant can be found out.

6. A circular cavity resonator is filled with a dielectric of 2.08 and is operating at 5GHz of frequency. Then the wave number is:

a) 181

b) 151

c) 161

d) 216

View Answer

Answer: b

Explanation: Wave number for a circular cavity resonator is given by the expression 2πf011√∈r/C. substituting the given values in the above expression; the wave number of the cavity resonator is 151.

7. Given that the wave number of a circular cavity resonator is 151 (TE011 mode), and the length of the cavity is twice the radius of the cavity, the radius of the circular cavity operating at 5GHz frequency is:

a) 2.1 cm

b) 1.7 cm

c) 2.84 cm

d) insufficient data

View Answer

Answer: d

Explanation: For a circular cavity resonator, wave number is given by √( (p01/a)2 +(π/d)2). P01 for the given mode of resonance is 3.832. Substituting the given values the radius of the cavity is 2.74 cm.

8. The loss tangent for a circular cavity resonator is 0.0004.Then the unloaded Q due to dielectric loss is:

a) 1350

b) 1560

c) 560

d) 2500

View Answer

Answer: d

Answer: Unloaded Q due to the dielectric loss in a circular cavity resonator is the reciprocal of the loss tangent. Hence, taking the reciprocal of the loss tangent, unloaded Q due to dielectric loss is 2500.

9. A circular cavity resonator has a wave number of 151, radius of 2.74 cm, and surface resistance of 0.0184Ω. If the cavity is filled with a dielectric of 2.01, then unloaded Q due to conductor loss is:

a) 25490

b) 21460

c) 29390

d) none of the mentioned

View Answer

Answer: c

Explanation: Unloaded Q of a circular resonator due to conductor loss is given by ka/2Rs. is the intrinsic impedance of the medium. Substituting the given values in the equation for loaded Q, value is 29390.

10. If unloaded Q due to conductor loss and unloaded Q due to dielectric loss is 29390 and 2500 respectively, then the total unloaded Q of the circular cavity is:

a) 2500

b) 29390

c) 2300

d) 31890

View Answer

Answer: c

Explanation: The total unloaded Q of a circular cavity resonator is given by the expression (Qc-1+ Qd-1)-1. Substituting the given values in the above expression, the total unloaded Q for the resonator is 2300.

Microwave Engineering Questions and Answers – Ferrite Isolators

Microwave Engineering Questions and Answers – Dielectric Resonators

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Dielectric Resonators”.

1. A dielectric material in the form of a small cube or disc can be used as a resonator.

a) true

b) false

View Answer

Answer: a

Explanation: A dielectric material in the form of a small cube or disc can be used as a resonator. It has the same operating principles as that of a rectangular waveguide resonator and a circular waveguide resonator.

2. Dielectric resonators use materials that are less lossy.

a) true

b) false

View Answer

Answer: a

Explanation: Dielectric resonators use materials that are less lossy and have high dielectric constant, ensuring that most of the fields will be contained in the dielectric.

3. The major disadvantage of dielectric resonators is:

a) complex construction

b) field fringing

c) requirement of high dielectric constant

d) none of the mentioned

View Answer

Answer: b

Explanation: The major disadvantage of dielectric resonators is field fringing or leakage from sides and ends of a dielectric resonator. This leakage of field energy results in high loss.

4. One of the most commonly used dielectric materials is:

a) barium tetratetanate

b) titanium

c) teflon

d) none of the mentioned

View Answer

Answer: a

Explanation: Materials having dielectric constant in the range of 10-100 are used in dielectric resonators and one more required characteristic property is low loss. Barium tetratetanate has all these properties. Hence it is most commonly used.

5. The resonant frequency of a dielectric resonator cannot be mechanically tuned.

a) true

b) false

View Answer

Answer: a

Explanation: By using an adjustable metal plate above the resonator, the resonant frequency can be mechanically tuned. As it has these desirable features, it is mostly used in integrated microwave filters and oscillators.

6. If a dielectric resonator has a dielectric constant of 49, then the reflection coefficient of the dielectric resonator is:

a) 0.5

b) 0.75

c) 0.1

d) 0.7

View Answer

Answer: b

Explanation: Reflection co-efficient of a dielectric resonator is given by (√∈r-1)/ (√∈r-1). Given that dielectric constant is 49, the reflection coefficient is 0.75.

7. Q factor does not exist for dielectric resonator.

a) true

b) false

View Answer

Answer: b

Explanation: Q factor exists for a dielectric resonator .it is defined as the ratio of the energy stored in the dielectric to the energy dissipated and other losses that may occur.

8. The approximate loaded Q due to dielectric loss for a dielectric resonator given the loss tangent is 0.0001 is:

a) 1000

b) 500

c) 2000

d) 10000

View Answer

Answer: a

Explanation: Loaded Q due to dielectric loss for a dielectric resonator is given by the reciprocal of the loss tangent. Taking the reciprocal of loss tangent, loaded Q due to dielectric loss is 1000.

9. The direction of propagation is in z direction outside the dielectric in the resonator.

a) true

b) false

View Answer

Answer: b

Explanation: In a dielectric resonator, the direction of propagation can occur along the Z direction in the dielectric at resonant frequency but the fields are cutoff in the air region around the dielectric.

10. A dielectric resonator is considered to be closed at both the ends.

a) true

b) false

View Answer

Answer: b

Explanation: For all analysis purpose, a dielectric resonator is considered to be of a short length L and termed as dielectric waveguide open at both the ends.

Microwave Engineering Questions and Answers – Excitation of Resonators

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Excitation of Resonators”.

1. The level of coupling required between a resonator and its attached circuitry is a standard and independent of the application where coupling is required.

a) true

b) false

View Answer

Answer: b

Explanation: The level of coupling required between a resonator and its attached circuitry depends on the application. A waveguide cavity to a frequency meter is loosely coupled to maintain high Q and good accuracy.

2. A measure of the level of coupling between a resonator and a feed is given by:

a) coupling coefficient

b) power transfer coefficient

c) voltage coefficient

d) reflection coefficient

View Answer

Answer: a

Explanation: Coupling coefficient tells how the resonator is coupled to the external circuitry. A resonator can be coupled in three ways. They can be under coupled, critically coupled or over coupled.

3. To obtain maximum power transfer between a resonator and feed line, the resonator should be matched to the load at:

a) resonant frequency

b) cutoff frequency

c) zero frequency

d) none of the mentioned

View Answer

Answer: a

Explanation: In order to obtain a maximum power transfer between a resonator and a feed line, the resonator should be matched to the feed line at the resonant frequency of the resonator which is coupled.

4. When impedance matching is done between a resonator and a feed line, the condition for impedance matching is:

a) R=Z0

b) R=Z0/2

c) R=2Z0

d) R=√Z0

View Answer

Answer: a

Explanation: If R is the resistance of the series RLC circuit and Z0 is the characteristic impedance of the feed line, for proper coupling their impedance has to be matched. This is the condition for impedance matching between a feed line and a resonator.

5. Coupling coefficient Q can be defined as the ratio of unloaded Q to external Q.

a) true

b) false

View Answer

Answer: a

Explanation: Coupling coefficient Q defined as the ratio of unloaded Q to external Q can be applied to both series resonance and parallel resonance circuits. For series resonant circuits, coupling coefficient is defined as the ratio of characteristic impedance of feed line to the resistance in the resonant circuit.

6. When the coupling coefficient is lesser than one, the resonator is over coupled to the feed line.

a) true

b) false

View Answer

Answer: b

Explanation: When the coupling coefficient is less than one, the resonator is under coupled to the feed line. For a series resonant circuit, resonator is under coupled implies that the resistance in the resonator is greater than the characteristic impedance of the transmission line.

7. Direct measurement of the unloaded Q of a resonator is not possible.

a) true

b) false

View Answer

Answer: a

Explanation: Direct measurement of the unloaded Q of a resonator is not possible because of the loading effect of the measurement system, but it is possible to determine unloaded Q from measurements of the frequency response of the loaded resonator when it is connected to a transmission line.

8. For practical applications cavity resonators can be modified as per the requirement of the application where it is used.

a) true

b) false

View Answer

Answer: a

Explanation: Small changes in the cavity resonator can be made by changing their shape, or by introducing small pieces of dielectric or metallic materials. The resonant frequency of a cavity resonator can be easily tuned with a small screw that enters the cavity volume or by changing the size of the cavity by a movable wall.

9. When coupling coefficient is 1, the resonator is ________ to the feed line.

a) under coupled

b) over coupled

c) critically coupled

d) none of the mentioned

View Answer

Answer: c

Explanation: The resonator is critically coupled to the feed line when the coupling coefficient is 1. Maximum power is transferred between the resonator and the feed line since the resistance of the resonator is equal to the characteristic impedance of the transmission line.

10. In aperture coupling, a small aperture in the transverse wall of the waveguide acts as:

a) shunt inductance

b) shunt capacitance

c) series inductance

d) series capacitance

View Answer

Answer: a

Explanation: In aperture coupling, a small aperture is made in the transverse wall of the cavity which is to be coupled to an external microwave circuit. This aperture made in the transverse wall of the cavity acts as a shunt inductance.

Microwave Engineering Questions and Answers – Properties of Dividers and Couplers

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Properties of Dividers and Couplers”.

1. Power dividers and couplers are ______ microwave components used for power division or power combining.

a) Passive

b) Active

c) Linear

d) Non linear

View Answer

Answer: a

Explanation: When a given input power is to be divided equally at microwave frequencies, power dividers are used. Since couplers and dividers do not store any energy they are called passive microwave components.

2. T- junction is an example for:

a) 2 port network

b) 3 port network

c) 4 port network

d) None of the mentioned

View Answer

Answer: b

Explanation: A T junction is a 3 port network where the junction is excited at one of the port, output is measured at another port, with the third port terminated with a known impedance.

3. A T junction has a 3×3 ‘s’ matrix.

a) True

b) False

View Answer

Answer: a

Explanation: ‘n’ port microwave network is represented as n×n s matrix. Characteristics of a microwave network can be known by analyzing his s matrix. A T junction has 3 ports. Hence, they are represented as a 3×3 matrix.

4. If a device is passive and contains no anisotropic elements, then the device is_______ network.

a) Reciprocal

b) Non reciprocal

c) Lossless

d) Lossy

View Answer

Answer: a

Explanation: If a device is passive, it does not store any energy and does not contain any sources. It then acts as a reciprocal network. That is, when the input and output ports are interchanged, the power delivered remains the same.

5. Scattering matrix of a reciprocal network is:

a) Symmetric

b) Asymmetric

c) Identity matrix

d) Null matrix

View Answer

Answer: a

Explanation: For a reciprocal network, when the output and input ports are interchanged/ the power delivered remains the same. Hence in the S matrix Sij=Sji. This condition is satisfied by a matrix that is symmetric.

6. If all the ports of a microwave network are matched, then the diagonal elements of the S matrix of the network is zero.

a) True

b) False

View Answer

Answer: a

Explanation: When all the ports of a microwave network are matched, no power is reflected back to the port that is excited by the source. Since no power is reflected back, the elements Sii of the S matrix become zero.

7. If a microwave network is lossless, then S matrix of the microwave network is:

a) Unitary

b) Symmetric

c) Identity matrix

d) Zero matrix

View Answer

Answer: a

Explanation: If a microwave network is lossless, law of energy conservation requires that the condition for example, │S12│2+│S13│2= 1. This condition has to be satisfied for a 3 port network. This can be generalized for any n port network.

8. A lossless reciprocal 3 port network can be matched at all the three ports.

a) True

b) False

View Answer

Answer: b

Explanation: A lossless three port network can be only matched at 2 ports. It not possible for the S matrix of a 3 port network to satisfy all the above conditions .This would violate the law of conservation of energy.

9. A circulator is a 3 port network that allows energy flow in clockwise direction only.

a) True

b) False

View Answer

Answer: b

Explanation: A circulator is a 3 port device that allows energy flow in only one direction. The direction of flow is either clockwise or counter clockwise. Bothe the directions are not simultaneously allowed in a circulator.

10. The diagonal elements of the S matrix of an improperly matched circulator is zero.

a) True

b) False

View Answer

Answer: b

Explanation: If the three ports of a circulator are not properly matched, it would result in the backward flow of power to the previous ports. Hence, the diagonal elements will not be zero due to improper matching as they signify the reflection of energy back to the same port.

11. Coupling factor of a directional coupler must be maximum and is a key factor that determines the performance of the coupler.

a) True

b) False

View Answer

Answer: a

Explanation: Coupling factor indicates the fraction of input power that is coupled to the output port. If a directional coupler has higher coupling factor, maximum power is obtained at the output port with low loss.

12. Directivity of a directional coupler signifies the direction of power flow in the coupler.

a) True

b) False

View Answer

Answer: b

Explanation: Directivity is the measure of the directional coupler’s ability to isolate forward and backward waves. Directivity is also defined as the ratio of power at the output port to the power at the isolated port.

13. Isolation of a directional coupler is a measure of the:

a) Power delivered to the uncoupled port

b) Power delivered to the coupled port

c) Power delivered to the second port

d) None of the mentioned

View Answer

Answer: a

Explanation: In a directional coupler, there exists a port called the isolated port through which no power is received and is isolated from the remaining ports of the coupler. Isolation of a directional coupler is a measure of the power delivered to the uncoupled port.

14. Insertion loss is the power delivered to the through port.

a) True

b) False

View Answer

Answer: a

Explanation: In a directional coupler, the second port is called is called the through port and the third port is called the output port. When power flows from port 1 to output port, the power measured at port 2 can be termed as insertion loss.

15. In a symmetric coupler, the power delivered to the through port and output port are equal.

a) True

b) False

View Answer

Answer: a

Explanation: A symmetric coupler is the one which delivers equal amount of power to both port 2 and port 3 of the directional coupler. The signals t these ports are out of phase by 900.

Microwave Engineering Questions and Answers – Power Dividers( T Junction)

This set of Microwave Engineering Interview Questions and Answers for Experienced people focuses on “Power Dividers( T Junction)”.

1. A T junction power divider can be used only for division of power.

a) True

b) False

View Answer

Answer: b

Explanation: A T junction power divider is a 3 port network that can be used either for power dividing or power combining. For power division, one of the ports is excited with the source and the other two ports are used to receive power. For power combining, 2 ports are excited with the source and output is taken at the third port.

2. The lossless T junction dividers can be can all be modeled as a junction of three transmission lines.

a) True

b) False

View Answer

Answer: a

Explanation: A T junction consists of three ports; hence they can be modeled with 3 transmission lines. However, they cannot be both lossless and matched simultaneously.

3. For the realization of lossless T-junction power divider using transmission lines, the characteristic impedance of the transmission line has to be real.

a) True

b) False

View Answer

Answer: a

Explanation: From various analyses, it is found that for a transmission line must have real characteristic impedance. If they are not real, they capacitive and inductive passive elements result in retaining some energy in the junction making them lossy.

4. The output power measured at the 2 ports of the T junction:

a) Is a constant

b) Variable

c) Is not real power

d) None of the mentioned

View Answer

Answer: b

Explanation: The output line impedance determines the power delivered to the 2 output ports in a T junction coupler. Depending on the value of the power desired, the impedance can be changed and the corresponding power is obtained.

5. Hybrid couplers are also a type of directional couplers.

a) True

b) False

View Answer

Answer: a

Explanation: Hybrid couplers are also a type of directional couplers that give a coupling factor of 3db) A coupling factor of 3dB means that 70.7% of the total input power is received at the output port.

6. If the input power is divided in the ratio of 2:1 in a T- junction coupler and the characteristic impedance of the 2 output lines is 150Ω and 75Ω, then the impedance of the input line is:

a) 100Ω

b) 50Ω

c) 150Ω

d) None of the mentioned

View Answer

Answer: b

Explanation: The input impedance of the T junction is the equivalent of the 2 output impedances in parallel. That is 150││75. Solving this, (150*75/(150+75)), the input impedance is 50 Ω.

7. A lossy T junction can be matched at all the three ports.

a) True

b) False

View Answer

Answer: a

Explanation: If a T junction is constructed using resistors, the T junction becomes lossy, but it can be simultaneously matched at all the three ports.

8. The diagonal elements of the s matrix of a resistive T junction are:

a) 0

b) 1

c) 0.5

d) 1.5

View Answer

Answer: a

Explanation: A resistive junction can be matched at all the three ports of the junction. Hence no power is reflected back. As a result, the diagonal elements are all 0 for a resistive T junction.

9. The power delivered to the input port of a resistive power divider is equal to the source voltage applied.

a) True

b) False

View Answer

Answer: b

Explanation: Since power is applied to a resistive power divider, there is loss and hence not all the supply power is delivered to the input port of the power divider.

10. The power input at the port 1 of resistive T junction is equally divided among the 2 output ports of the T junction.

a) True

b) False

View Answer

Answer: b

Explanation: The power division ratio of a resistive T junction depends on the resistance of the resistors used in forming those junctions. Depending on the resistors used, the power gets divided accordingly.

Microwave Engineering Questions and Answers – Wilkinson Power Dividers

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Wilkinson Power Dividers”.

1. A major disadvantage of the lossless T-junction power divider is:

a) Not matched at all the ports

b) Low power output

c) Complex construction

d) None of the mentioned

View Answer

Answer: a

Explanation: A T-junction hybrid cannot be matched at all the ports if the power divider is lossless. It can be matched only at 2 ports. This is one of the major disadvantages when they are to be used along with other microwave devices.

2. The Wilkinson power divider is a:

a) 2 port network

b) 3 port network

c) 4 port network

d) None of the mentioned

View Answer

Answer: b

Explanation: Wilkinson power divider is a 3 port network; if it is used as a divider it has one input port and 2 output ports. If it is used as coupler, it has two input port and one output port.

3. Wilkinson power divider is an equal split power divider.

a) True

b) False

View Answer

Answer: b

Explanation: Wilkinson power divider can be used to divide power in any ratio, but the most commonly used configuration is the equal split power divider.

4. If 10 watt is applied to the input port of a standard Wilkinson divider, then the sum of the power measured at the two output ports of the Wilkinson coupler is

a) 5 watt

b) 10 watt

c) 7.07 watt

d) 8 watt

View Answer

Answer: c

Explanation: For a standard Wilkinson coupler, the output power is 3 dB less than the total input power in decibels. That is, 70.7% of the total input power is delivered to the output port.

5. The analysis of Wilkinson coupler is done using:

a) Even-odd mode analysis

b) Symmetry

c) S matrix approach

d) None of the mentioned

View Answer

Answer: a

Explanation: Even-odd mode analysis is one of the simplest methods of analysis for Wilkinson coupler. This involves normalizing all impedances with the characteristic impedance of the transmission line used and carrying out some analysis.

6. A Wilkinson coupler designed can be operated at any frequency.

a) True

b) False

View Answer

Answer: b

Explanation: The length of the branches of a Wilkinson coupler is all wavelengths dependent and hence Wilkinson coupler designed to operate at one frequency cannot be used to operate at another frequency.

7. For an equal-split Wilkinson power divider of 50Ω system impedance, the characteristic impedance of quarter wave transmission line used is:

a) 70.7 Ω

b) 50 Ω

c) 100 Ω

d) None of the mentioned

View Answer

Answer: a

Explanation: The characteristic impedance of a Z Ω system is given by √2*Z. hence, the characteristic impedance of 50Ω system is 70.7 Ω.

8. The plot of frequency v/s S11 parameter of a Wilkinson coupler has a dip at the frequency at which it is designed to operate.

a) True

b) False

View Answer

Answer: a

Explanation: S11 parameter signifies the fraction of the power reflected back to port 1 when power is applied to port 1 of the coupler. Since the ports are matched at the frequency of design S11 is minimum and the curve has a dip.

9. The plot of S23 v/s frequency has the same curve as that of S11 v/s frequency.

a) True

b) False

View Answer

Answer: a

Explanation: When input is applied to port 1, output is measured at port 2 and port 3. S23 signifies the output at port 3 due to port 2 when input is applied at port 1. This parameter is minimum at the designed frequency.

10. S12 curve of a Wilkinson coupler when plotted versus frequency is a line passing through origin.

a) True

b) False

View Answer

Answer: b

Explanation: S12 gives the ratio of power at input port P1 to the power measured at port 2. Since the output power remains constant over a wide range of frequencies for a given input applied. Hence S11 is a curve parallel to X axis.

Microwave Engineering Questions and Answers – Quadrature Hybrid

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Quadrature Hybrid”.

1. Quadrature hybrids are those couplers which are:

a) 3 dB couplers

b) Directional couplers

c) They have a 900 phase difference between signals in through and coupled arms.

d) All of the mentioned

View Answer

Answer: d

Explanation: Quadrature hybrids are directional couplers that have a phase difference of 900 between the signals obtained at through and coupled ports.

2. Branch-line couplers are also popular as Quadrature hybrids.

a) True

b) False

View Answer

Answer: a

Explanation: Quadrature hybrids are also called as branch-line couplers. The other 3 dB couplers are coupled line couplers or Lange couplers. These couplers also can be Quadrature hybrids.

3. The S matrix of a Quadrature hybrid is of size 4×4 and the diagonal elements of a matched coupler are all:

a) 1

b) 0

c) Cannot be determined

d) None of the mentioned

View Answer

Answer: b

Explanation: In a matched coupler, since all ports are matched, no power goes back to the port from which the flow of energy in the coupler occurred. Since the backward power flow is zero for matched network, the diagonal elements are zero.

4. A branch-line coupler is an asymmetric coupler.

a) True

b) False

View Answer

Answer: b

Explanation: A branch-line coupler is a symmetric coupler that has all four ports placed symmetrically. Since the construction is symmetric, any port can be used as input and any port can be used as output.

5. Branch-line couplers are preferably made using waveguides so as to obtain high gain and simple construction.

a) True

b) False

View Answer

Answer: b

Explanation: Branch-line couplers are mostly done using microstrip lines. They also reduce the complexity of the circuit and can be easily integrated with other microwave devices in all large scale applications.

6. A 50 Ω branch-line Quadrature hybrid has to be designed to operate over a range of frequencies. The branch-line impedance of this coupler so designed is:

a) 70.7 Ω

b) 35.4 Ω

c) 50 Ω

d) 100 Ω

View Answer

Answer: b

Explanation: Each arm of a Quadrature hybrid is λ/4 long; λ is the wavelength at which the coupler is designed to operate. The branch-line impedance for a λ/4 line is Z0/√2. Substituting for Z0 in the equation, the impedance is 35.4 Ω.

7. The plot S11 v/s frequency for a branch-line coupler has a straight line characteristic for a wide range of frequency around the designed frequency range.

a) True

b) False

View Answer

Answer: b

Explanation: S11 parameter signifies the power measured at port 1 when port 1 is used as an input port. When the ports of the coupler are matched, no power is reflected back to the port 1. Hence S11 curve has a dip at the frequency for which the coupler was operated to design. A fall and rise in the curve is seen at this point.

8. The curve of S14 for a branch-line coupler is similar to that of the S11 curve of the branch-line coupler.

a) True

b) False

View Answer

Answer: a

Explanation: S14 parameter gives the power measured at the port 4 or the isolated port of the branch-line coupler. Since the signals that reach port 4 from 2 different arms are 900 out of phase with each other, theoretically power at port 4 is zero. Practically, it is zero for the designed frequency but some power is received at other frequencies.

9. S12 and S13 curves for branch-line couplers are almost a straight line parallel to X –axis. Both the curves are similar and follow same path.

a) True

b) False

View Answer

Answer: a

Explanation: S13 and S12 parameters give the power measured at port 2 and port 3 of the branch-line coupler when port 1 is used as the input port. Since these are the through and coupled ports, power measured across these ports is almost constant and resemble a straight line parallel to X axis.

10. If the branch-line impedance of a coupler designed to operate at 1 GHz is 70.70 Ω, then the characteristic impedance of the material of the arms of the branch-line coupler is:

a) 70.7 Ω

b) 50 Ω

c) 100 Ω

d) None of the mentioned

View Answer

Answer: c

Explanation: Given the branch impedance is 70.70 Ω; the characteristic impedance of the line is Z√2. This relation is used since all the arms of a branch-line coupler are λ/4 long. Substituting for Z, the characteristic impedance of the line is 100 Ω.

Microwave Engineering Questions and Answers – Coupled Line Directional Couplers

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Coupled Line Directional Couplers “.

1. In coupled line directional couplers, power from one line to another is transmitted through a microstrip line running between them.

a) true

b) false

View Answer

Answer: b

Explanation: In coupled line couplers, the power is transmitted between the 2 lines by coupling from one line to another due to the interaction of the electromagnetic fields. Hence, wireless power transmission occurs here.

2. The number of conductors used in the construction of coupled line couplers is fixed.

a) true

b) false

View Answer

Answer: b

Explanation: Since the method of power transmission in coupled line couplers is wireless power transmission by the interaction of electromagnetic fields, any number of wires can be used. But as a standard, 3 lines are used in the construction of these couplers.

3. The mode of propagation of propagation supported by coupled line couplers is:

a) TM mode

b) TE mode

c) TEM mode

d) quasi TEM mode

View Answer

Answer: c

Explanation: Coupled transmission lines are assumed to operate in TEM mode of propagation. TEM mode of propagation is mostly valid for coaxial and stripline structures while microstrip lines support quasi TEM mode of propagation.

4. Coupled line couplers are:

a) symmetric couplers

b) asymmetric couplers

c) in phase couplers

d) type of hybrid coupler

View Answer

Answer: a

Explanation: Coupled line couplers are symmetric three line couplers. Symmetric here means that the lines are of equal width and thickness. Their position with respect to ground is identical.

5. For coupled line coupler, if the voltage coupling factor is 0.1 and the characteristic impedance of the microstrip line is 50 Ω, even mode characteristic impedance is:

a) 50.23 Ω

b) 55.28 Ω

c) 100 Ω

d) 80.8 Ω

View Answer

Answer: b

Explanation: Even mode characteristic impedance of coupled line coupler is Z0√ (1+C) /√ (1-C).here C is the voltage coupling coefficient. Substituting the given values, even mode characteristic impedance is 55.28 Ω.

6. If the coupling coefficient of a coupled line coupler is 0.1 and the characteristic impedance of the material is 50 Ω, then the odd mode characteristic impedance is:

a) 45.23 Ω

b) 50 Ω

c) 38 Ω

d) none of the mentioned

View Answer

Answer: a

Explanation: Odd mode characteristic impedance of a coupled line coupler is Z0√ (1-C)/ √ (1+C). C is the voltage coupling co-efficient. Substituting the given values, odd mode characteristic impedance is 45.23.

7. Dielectric and conductor loss have no effect on the directivity of the coupled line coupler.

a) true

b) false

View Answer

Answer: b

Explanation: Both dielectric loss and conductor loss reduce the directivity of the coupled line coupler. In the absence of loss under matched conditions, the directivity of a coupler could be up to 70 dB.

8. Multisection couplers have a very narrow operational bandwidth which is a major disadvantage.

a) true

b) false

View Answer

Answer: b

Explanation: Multisection couplers have very high operational bandwidth. This high bandwidth can be achieved only when the coupling levels are low. In order to achieve these low coupling levels, stripline are used in their construction.

9. Three section binomial couplers have very low directivity as compared to other coupler designs.

a) true

b) false

View Answer

Answer: b

Explanation: Three section binomial couplers have very low conductor and dielectric losses. This low loss can be achieved by efficient design. Since the losses are low for a binomial coupler, they have directivity greater than 100 dB.

10. The capacitance per unit length of broadside parallel lines with width W and separation d is:

a) ∈W/d

b) ∈d/W

c) dW/∈

d) none of the mentioned

View Answer

Answer: a

Explanation: The capacitance of the line used in the construction of a coupled line coupler is directly proportional to the width of the line. As the width increase, capacitance increases. Capacitance varies inversely with distance d. as the separation increases, capacitance decreases.

Microwave Engineering Questions and Answers – Lange Coupler

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Lange Coupler”.

1. The major disadvantage of coupled line coupler is:

a) complex construction

b) low power gain

c) higher loss

d) none of the mentioned

View Answer

Answer: d

Explanation: Coupling in a coupled line coupler is too loose to achieve coupling factors of 3 or 6 dB. Since higher efficiency is not achievable, this a major disadvantage. One method of improving coupling is designing in such a way that the fringing field contributes to coupling.

2. Lange couplers consist of four parallel lines that are ______ coupled.

a) tightly

b) loosely

c) partially

d) none of the mentioned

View Answer

Answer: a

Explanation: Lange couplers consist of four parallel lines. These parallel lines are tightly coupled to achieve 3 dB coupling ratio. Achieving 3dB coupling implies that 70% of the applied power is coupled.

3. Lange coupler is a type of Quadrature hybrid.

a) true

b) false

View Answer

Answer: a

Explanation: Lange coupler ha two output ports. The output voltage measured at these ports is 900 out of phase with one another. Hence, they are called a type of Quadrature hybrid.

4. A major disadvantage of Lange coupler is:

a) high power loss

b) complex construction

c) low gain

d) none of the mentioned

View Answer

Answer: b

Explanation: The main disadvantage of Lange coupler is probably practical, as the lines are very narrow and close together, and the required bonding wires across the lines increases complexity.

5. For even mode excitation of a Lange coupler, all the four lines of the Lange coupler are at:

a) equal potential

b) zero potential

c) unequal potential

d) negative potential

View Answer

Answer: a

Explanation: during the analysis of Lange coupler, when they are excited in the even mode, all the four conductors of the coupler are at equal potential. Hence the capacitance Cm has no effect and need not be considered for further analysis.

6. Even mode characteristic impedance of a Lange coupler:

a) (Vp.Cep)-1

b) Vp.Cep

c) (Vp.Lep)-1

d) Vp.Lep

View Answer

Answer: a

Explanation: The even mode characteristic impedance of a Lange coupler is given by (Vp.Cep)-1. Here Vpp is the phase velocity and Cep is the capacitance of any of the four conductors of the Lange coupler measured during even mode of analysis.

7. In a Lange coupler having 4 lines, the adjacent lines are connected so as to act as two conductors in parallel.

a) true

b) false

View Answer

Answer: b

Explanation: In a Lange coupler consisting of 4 lines, the alternating lines are connected together. This simplifies the computation effort for finding equivalent capacitance and the construction is simpler. Also ports are easily distinguished and helps in flow of power in required direction.

8. For a 4 wire Lange coupler, if the odd mode and even mode capacitances are 12pF and 9pF respectively, then the equivalent capacitance in terms of two wired coupled line in even mode is:

a) 15.1 pF

b) 16.7 pF

c) 12 pF

d) 9 pF

View Answer

Answer: b

Explanation: Equivalent capacitance in terms of two wired coupled model is Ce (3Ce+C0)/(Ce+C0), Where Ce and C0 are the equivalent capacitances in even mode and odd mode for 4 wire model. Substituting the given values in the above expression, the equivalent capacitance is 16.7 pF.

9. For a 4 wire Lange coupler, if the odd mode and even mode capacitances are 12pF and 9pF respectively, then the equivalent capacitance in terms of two wired coupled line in odd mode is:

a) 18.56 pF

b) 25.71 pF

c) 42 pF

d) 31.6 pF

View Answer

Answer: b

Explanation: Equivalent capacitance in terms of two wired coupled model is C0 (3C0+Ce)/ (Ce+C0), Where Ce and C0 are the equivalent capacitances in even mode and odd mode for 4 wire model. Substituting the given values in the above expression, the equivalent capacitance is 25.71 pF.

10. The phase velocity of the waves in both even and odd mode of a Lange coupler is equal.

a) true

b) false

View Answer

Answer: b

Explanation: The phase velocity of the waves in the even mode and odd mode are not practically equal and there are lot of variations. But only for theoretical analysis, they are assumed to be equal.

Microwave Engineering Questions and Answers – 180 Degree Hybrids

This set of Microwave Engineering test focuses on “180 Degree Hybrids”.

1. 1800 hybrid is a network in which there is a phase shift of 1800 between the input signal applied and the output taken.

a) true

b) false

View Answer

Answer: b

Explanation: 1800 hybrid is a four port network that has one input port and two output ports. The phase difference between the 2 output ports is 1800.

2. Port 1 and port 4 of 1800 hybrid are called sum and difference ports respectively because of their behavior and action mechanism.

a) true

b) false

View Answer

Answer: a

Explanation: When a 1800 hybrid is used as a combiner, with input signals applied at port 2 and port 3, the sum of the inputs will be formed at port 1, while the difference will be formed at port 4. Hence they are referred to as sum and difference ports.

3. S matrix of 1800 hybrid consists of all diagonal elements zero.

a) true

b) false

View Answer

Answer: a

Explanation: If all the ports of 1800 hybrid are properly matched, no power is reflected back to the same port. Hence all the diagonal elements of the S matrix, Sii=0.

4. In 1800 hybrid, different power levels can be received at the two output ports of the hybrid.

a) true

b) false

View Answer

Answer: b

Explanation: 1800 hybrid is a symmetrical coupler. Hence, the input power applied at the input port can be divided equally and obtained at the 2 output ports. Unequal division of power is not possible in 1800 hybrid.

5. 18000 ring hybrid with system impedance of 50 Ω has to be designed. Then the characteristic impedance of the arms of the 1800 hybrid is:

a) 50 Ω

b) 70.70 Ω

c) 100 Ω

d) none of the mentioned

View Answer

Answer: b

Explanation: Given that the system impedance is 50 Ω, the characteristic impedance of the arms is 50√2. Hence the characteristic impedance is 70.70 Ω.

6. In a waveguide magic-T there is no coupling of power between port 1 and port 4.

a) true

b) false

View Answer

Answer: a

Explanation: Consider TE10 mode incident at port 1. There is odd symmetry about guide 4. Because the field lines of a TE10 mode in guide 4 would have even symmetry, hence there is no coupling between port 1 and port 4.

7. When a TE10 wave is incident on port 4 of a magic-T, all the power is coupled to port 1.

a) true

b) false

View Answer

Answer: b

Explanation: When port 4 of a Magic-T is excited, port 1 and port 4 are decoupled, due to symmetry. Port 2 and port 3 are excited equally by the incident wave with a phase difference of 1800.

8. The tapered coupled line 1800 hybrid can provide an arbitrary power division at the 2 output ports of the coupler.

a) true

b) false

View Answer

Answer: a

Explanation: When the arms of the coupled line is tapered, they result in division of power at the two output ports unequally and they offer a bandwidth of one decade or more.

9. The plot of frequency V/s S11 parameter for a tapered line coupler has a dip at the frequency at which it is designed.

a) true

b) false

View Answer

Answer: a

Explanation: When port 1 of a tapered coupler is excited, no power flows back to port 1 since the ports are matched. Hence, S11 value is almost zero or negligibly small and hence has a dip at the designed frequency.

10. For a tapered line coupler, the curves of S12 and S13 are identical and have the same magnitude at all frequencies.

a) true

b) false

View Answer

Answer: b

Explanation: Tapered line couplers result in unequal power division at the output ports. These are useful for various applications. Hence the S12 and S13 curves are not identical since the power outputs are not equal.

Microwave Engineering Questions and Answers – Other Couplers

This set of Microwave Engineering Quiz focuses on “Other couplers”.

1. Moreno crossed-guide coupler is a waveguide directional coupler consists of four waveguides at right angle.

a) True

b) False

View Answer

Answer: b

Explanation: Moreno crossed-guide coupler consists of two waveguides at right angles, with coupling provided by two apertures in the common broad wall of the guides.

2. Schwinger reversed phase coupler is a waveguide coupler designed so that the path lengths for the two coupling apertures are the same for_________

a) Coupled port

b) Uncoupled port

c) Back port

d) Isolated port

View Answer

Answer: b

Explanation: Schwinger reversed phase coupler is a waveguide coupler designed so that the path lengths for the two coupling apertures are the same for uncoupled port so that the directivity is essentially independent of frequency.

3. The in phase combining of power at the coupled port is achieved by means of a _______

a) Matching network

b) A small slot

c) Quarter Wave transformer

d) None of the mentioned

View Answer

Answer: b

Explanation: The λg/4 slot spacing leads to in-phase combining at the coupled port, but this coupling is very frequency sensitive. This is the opposite situation from that of the multi hole waveguide coupler.

4. Ribblet short-slot coupler consists of two waveguides that are separated by a distance “d”.

a) True

b) False

View Answer

Answer: b

Explanation: This coupler consists two waveguides with a common side wall. Coupling takes place in the region where part of the common wall has been removed.

5. Riblet-short slot coupler allows only:

a) TE10 mode of propagation

b) TE20 mode of propagation

c) Both a and b

d) None of the mentioned

View Answer

Answer: c

Explanation: In the part where common wall has been removed in this coupler, both the TE10 and TE20 mode are excited, and by proper design can be made to cause cancellation at the isolated port and addition at the coupled port.

6. Symmetric tapered coupled line couplers offer higher bandwidth when compared to other forms of couplers.

a) True

b) False

View Answer

Answer: a

Explanation: Multisection coupled line coupler can be extended to a continuous taper, yielding couple line couplers with good bandwidth characteristics.

7. The coupling and directivity of couplers with apertures in planar lines can be adjusted as per the requirement of the application.

a) True

b) False

View Answer

Answer: a

Explanation: In symmetric tapered coupled line coupler both the conductor width and separation between them can be adjusted to provide a synthesized coupling or directivity response. This can be tested with computer optimization of a stepped-section approximation to a continuous taper.

8. _________ is a key component in the scalar or vector network analyzer.

a) Reflectometer

b) Radiometer

c) Frequency meter

d) None of the mentioned

View Answer

Answer: a

Explanation: Reflectometer is a circuit that uses a directional coupler to isolate and sample the incident and reflected powers from the load. This is one of the key components in a scalar or vector network analyzer.

9. Reflectometer can also be used as a frequency meter.

a) True

b) False

View Answer

Answer: a

Explanation: Reflectometers can be used to measure the scattering parameters of a two port network. It is also used as SWR meter and also as power monitor in system applications.

10. A basic Reflectometer circuit can be used to measure the _____________ magnitude of the unknown load.

a) Reflection coefficient

d) Standing wave ratio

c) Transmission coefficient

d) None of the mentioned

View Answer

Answer: a

Explanation: A basic Reflectometer circuit can be used to measure the reflection coefficient magnitude of the unknown load. This is one of the major applications of Reflectometer. If the reflection coefficient is measured, the unknown load can be easily computed.

Microwave Engineering Questions and Answers – Properties of Ferrimagnetic Materials

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Properties of Ferrimagnetic Materials”.

1. Example of a non reciprocal device:

a) Branch line coupler

b) Wilkinson coupler

c) Magic-T hybrid

d) Circulator

View Answer

Answer: d

Explanation: Non reciprocal device is the one in which the response between any two ports I and j of a component depends on the direction of signal flow. Circulator is a device that allows power flow either in clockwise direction or counter clockwise direction.

2. A microwave network can be called non reciprocal only if it contains anisotropic materials like ferrite materials.

a) True

b) False

View Answer

Answer: b

Explanation: A microwave network consisting of active non linear devices like transistor amplifiers, ferrite phase shifters and more. Presence of active devices or anisotropic materials can make a microwave network non reciprocal.

3. This is not an example of anisotropic material:

a) Yttrium aluminum garnet

b) Aluminum

c) Cobalt

d) Silicon

View Answer

Answer: d

Explanation: Yttrium aluminum garnet is a ferromagnetic compound. Aluminum and cobalt are iron oxides that are anisotropic. Silicon is a non metal that is isotropic in nature.

4. The magnetic properties of a material are due to the existence of ___________

a) Electrons in atoms

b) Electric dipole moment

c) Magnetic dipole moment

d) None of the mentioned

View Answer

Answer: c

Explanation: The magnetic properties of ferromagnetic materials are due to the existence of magnetic dipole moments, which arise primarily from electron spin. The magnetic dipole moment of an electron is 9.27×10-24 A-m2.

5. ___________ is a measure of the relative contributions of the orbital moment and the spin moment to the total magnetic moment.

a) Lande’s factor

b) Gibbs factor

c) Newton’s ratio

d) None of the mentioned

View Answer

Answer: a

Explanation: An electron in orbit around a nucleus gives rise to an effective current loop and thus an additional magnetic moment, but this effect is negligible compared to the magnetic moment due to spin. Lande’s factor is a relative measure of these orbital moments.

6. Lande’s factor for all ferromagnetic materials is in the range of 0 to 1.

a) True

b) False

View Answer

Answer: b

Explanation: Lande’s factor (g) is one when the moment is only due to orbital motion and 2 when the moment is only due to spin. For most microwave ferrite materials, g lies between 1.98 and 2.01.

7. The variation of magnetic moment of a ferromagnetic material with applied bias field is linear.

a) True

b) False

View Answer

Answer: b

Explanation: With the increase in the applied bias field to a ferromagnetic material, the magnetic moment increases exponentially initially, after a certain applied bias field magnetic moment remains a constant.

8. A permanent magnet is made by placing the magnetic material in a strong magnetic field.

a) True

b) False

View Answer

Answer: a

Explanation: A permanent magnet is made by placing the magnetic material in a strong magnetic field and then removing the field to leave the material magnetized in a remanent state.

9. The operating point of a permanent magnet is in the:

a) First quadrant

b) Second quadrant

c) Third quadrant

d) Fourth quadrant

View Answer

Answer: b

Explanation: Unless the magnet shape forms a closed path, the demagnetization factors at the magnet ends will cause a slightly negative H field to be induced in the magnet. Thus the “operating point “of a permanent magnet will be in the second quadrant. This portion of the curve is called demagnetization curve.

10. After demagnetization of a magnetic material, the residual magnetization retained in the magnetic material is called:

a) Remanence

b) Residue

c) Retardation

d) None of the mentioned

View Answer

Answer: a

Explanation: The residual magnetization called remanence characterizes the strength of the magnet, so magnetic material with large remanence is chosen.

Microwave Engineering Questions and Answers – Ferrite Isolators

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Ferrite Isolators”.

1. Ferrite isolators are ____ port microwave devices.

a) Two

b) Three

c) Four

d) None of the mentioned

View Answer

Answer: a

Explanation: Ferrite isolators are two port devices having unidirectional transmission characteristics. Isolator provides isolation between the two ports and the power flow occurs only in one direction.

2. The matrix of an ideal isolator is not ______

a) Unitary

b) Symmetric

c) Lossless

d) None of the mentioned

View Answer

Answer: a

Explanation: Ferrite isolator allows wave propagation in only one direction and attenuates propagation in the other direction. So the isolator is lossy. Since the isolator is lossy, the scattering matrix of isolator is not unity.

3. A very common application of isolator is to provide isolation between a low power source and the load.

a) True

b) False

View Answer

Answer: b

Explanation: Isolators have a wide variety of applications. The most common among them is the use of an isolator between a high-power source and a load to prevent the possible reflections from damaging the source. An isolator can be used in place of a matching network, but it should be realized that any power reflected from the source is absorbed by the isolator.

4. The attenuation of a ________ is very large near the gyro magnetic resonance of the ferrite.

a) Linearly polarized wave

b) Circularly polarized wave

c) Left polarized wave

d) Right polarized wave

View Answer

Answer: b

Explanation: The attenuation of the circularly polarized wave is very large near the gyro magnetic resonance of the ferrite, while the attenuation of the wave propagating in the opposite direction is very small.

5. The isolators constructed using ferrite materials must operate at:

a) Gyro magnetic resonance

b) Magnetic resonance

c) Isolator resonance

d) None of the mentioned

View Answer

Answer: a

6. Forward attenuation provided by a resonance ferrite isolator is:

a) Zero

b) Low

c) High

d) None of the mentioned

View Answer

Answer: b

Explanation: Zero forward attenuation cannot be obtained in resonance isolators because the internal magnetic field is not truly circularly polarized. Because of this, there is some amount of forward attenuation in the isolator.

7. An isolator has a very large operating bandwidth and independent of any isolator parameter.

a) True

b) False

View Answer

Answer: b

Explanation: The bandwidth of an isolator is relatively narrow, dictated essentially by the line width ∆H of the ferrite material.

8. The length of a ferrite slab required operating with a minimum forward insertion loss and 30 dB reverse attenuation and the reverse attenuation at this point is:

a) 3 cm

b) 2.4 cm

c) 4 cm

d) 3.6 cm

View Answer

Answer: b

Explanation: Length of the ferrite slab required is equal to the ratio of the minimum forward insertion loss to the reverse attenuation at the point. Substituting the given values in the above equation, length of the ferrite slab is 2.4 cm.

9. The electric field distribution of the forward and reverse waves in a ferrite slab-loaded waveguide is quite different. This property is used in:

a) Field displacement resonator

b) Resonance isolator

c) Waveguide isolator

d) None of the mentioned

View Answer

Answer: a

Explanation: In a field displacement isolator, electric field distribution of the forward and reverse waves in a ferrite slab-loaded waveguide is different. The electric field for the forward wave can be made to vanish at the side of the ferrite slab.

10. Field displacement isolators require higher bias field than resonance isolators.

a) True

b) False

View Answer

Answer: b

Explanation: Field displacement isolators have much smaller bias field requirement since it operates well below gyro magnetic resonance. This property of field displacement isolator make it more preferred than resonance isolators.

Microwave Engineering Questions and Answers – Ferrite Phase Shifters

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Ferrite Phase Shifters”.

1. ______ is a device that produces a phase shift of a required amount of the input wave.

a) Phase shifter

b) Attenuator

c) Resonator

d) None of the mentioned

View Answer

Answer: a

Explanation: Ferrite phase shifter is a two port component that provides a variable phase shift by changing the bias field of the ferrite. Microwave diodes and FETs can also be used to implement phase shifters.

2. Phase shifters are used in _______ where the antenna beam can be steered in space by electronically controlled phase shifters.

a) Phased array antennas

b) Dipole array antennas

c) Slot antennas

d) Patch antennas

View Answer

Answer: a

Explanation: Phase shifters are used in phased array antennas where the antenna beam can be steered in space by electronically controlled phase shifters. Phase shifters can be of two types, Reciprocal phase shifters and non reciprocal phase shifters. These are chosen as per the application requirement.

3. Reciprocal phase shifters give different phase shift in different direction.

a) True

b) False

View Answer

Answer: b

Explanation: Reciprocal phase shifters are those devices which give the same phase shift in either direction. That is, either if port 1 or port 2 of the phase shifter is used as input port, the phase shifts at the output remains the same.

4. If a ferrite slab provides a phase shift of 48⁰/ cm, then the length of the ferrite slab required to produce a phase shift of 180⁰ is:

a) 4 cm

b) 3.75 cm

c) 4.5 cm

d) 3.5 cm

View Answer

Answer: b

Explanation: The given ferrite slab provides a phase shift of the 48⁰/ cm. hence the length of the required ferrite slab is 180/45, the required length is 3.75 cm.

5. If a ferrite slab provides a phase shift of 48⁰/ cm, then the length of the ferrite slab required to produce a phase shift of 90⁰ is:

a) 2.44 cm

b) 1.88 cm

c) 4.5 cm

d) 3.5 cm

View Answer

Answer: b

Explanation: The given ferrite slab provides a phase shift of the 48⁰/ cm. hence the length of the required ferrite slab is 90/45, the required length is 1.88 cm.

6. Gyrator is a device that produces a phase shift of ____ between the input and output.

a) 90⁰

b) 180⁰

c) 45⁰

d) None of the mentioned

View Answer

Answer: b

Explanation: Gyrator is a device that produces a phase shift of 180⁰ between the input and output of the gyrator. This is a special case of the ferrite phase shifter which gives a constant phase shift and cannot be changed.

7. The scattering matrix of a gyrator is:

a) Symmetric

b) Skew symmetric

c) Identity matrix

d) Null matrix

View Answer

Answer: b

Explanation: The scattering matrix of a gyrator is Skew symmetric. This is because of the 180⁰ phase shift that occurs in the device.

8. Ferrite phase shifters have more advantages over FETs and diodes in using them in microwave integrated circuits.

a) True

b) False

View Answer

Answer: Even though PIN diode and FET circuits offer a less bulky and more integratable alternative to ferrite components, ferrite phase shifters are cost effective; have high power handling capacity and power requirements.

9. A gyrator can be made a passive device by certain design methods so that they do not affect the power levels of the circuit in which they are used.

a) True

b) False

View Answer

Answer: a

Explanation: The gyrator can be implemented as a phase shifter with a 180⁰ phase shift; bias can be provided with a permanent magnet, making the gyrator a passive device.

10. If a ferrite slab produces a phase shift of 0.836 rad/ cm, then the length of the slab required to produce a phase shift of 135⁰ is:

a) 2.81 cm

b) 3 cm

c) 2 cm

d) 3.4 cm

View Answer

Answer: a

Explanation: Converting the given phase shift from radian scale to degree scale, the produced phase shift is 48⁰/ cm. To produce a phase shift of 135⁰, the required length is 135/45 this is equal to 2.81 cm.

Microwave Engineering Questions and Answers – Ferrite Circulators

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Ferrite Circulators”.

1. ________ is a three-port microwave device that can be lossless and matched at all spots.

a) Hybrid junction

b) Magic Tee

c) Circulator

d) Isolator

View Answer

Answer: c

Explanation: A circulator is a three-port microwave device that can be lossless and matched at all ports; by using the unitary properties of scattering matrix it is proved that such a device must be non-reciprocal.

2. The total number of ones in the scattering matrix of an ideal circulator is:

a) 4

b) 3

c) 2

d) 5

View Answer

Answer: b

Explanation: Since the circulator is matched at all the ports, the diagonal elements are zero. As the circulator allows power flow in only one direction, only one of the elements in each row has a 1 in the matrix. With three rows, there are three ones.

3. There is no method in which the scattering matrix of the opposite circularity can be obtained from the matrix we have.

a) True

b) False

View Answer

Answer: b

Explanation: By transposing the port indices of the existing matrix, the opposite circulatory can be obtained. For example, if S13 is a 1 in the given circulator, then S31 is automatically one in the opposite circulator.

4. Practically, opposite circulatory in a ferrite circulator can be obtained by:

a) Changing the order of port operation

b) Impedance matching the input ports

c) Changing the polarity of the magnetic bias field

d) None of the mentioned

View Answer

Answer: c

Explanation: For a ferrite circulator, opposite circulatory can be produced by changing the polarity of the magnetic bias field. This change in polarity causes power to flow in opposite direction but only in one direction.

5. A circulator device can also used as an isolator with a few modifications.

a) True

b) False

View Answer

Answer: a

Explanation: A circulator can be used as an isolator by terminating one of the circulator ports with known impedance so that the remaining two ports are used for operation. As power flow occurs only in one direction in these two ports, they can be used as isolators.

6. In the scattering matrix representation of a non-ideal circulator, the diagonal elements of the matrix are:

a) Zero

b) One

c) Reflection coefficient Г

d) None of the mentioned

View Answer

Answer: c

Explanation: In a non-ideal circulator, the three ports of the circulator are not properly matched and hence there will be some reflection back to the same ports. This impedance mismatch can be represented by the reflection co-efficient Г.

7. In a stripline junction circulator, the ferrite material is present in the form of a:

a) Slab

b) Ferrite disk

c) Ferrite material is not used in a microstrip circulator

d) Ferrite cubes

View Answer

Answer: b

Explanation: In a stripline junction circulator, two ferrite disks fill the space between the center metallic disk and the ground planes of the stripline. Three striplines are attached to the periphery of the center disk and the ground plane on the stripline.

8. The dielectric resonator in the circulator has a single highest order resonant mode.

a) True

b) False

View Answer

Answer: b

Explanation: In operation of a microstrip circulator, the ferrite disks form a dielectric resonator; in the absence of the bias field this resonator has a single lowest order resonant mode with a cos φ dependence.

9. In the plot of the magnitude of electric field around the periphery of the junction circulator, the curve has:

a) Three peaks

b) Two peaks

c) Four peaks

d) None of the mentioned

View Answer

Answer: a

Explanation: In the plot of the magnitude of electric field around the periphery of the junction circulator, the curve has three peaks. These three peaks are due to the three ports of the circulator where the field measured is maximum.

10. In the non-ideal scattering matrix representation of the circulator, the attenuation constant and phase constant α, β respectively are approximated as 1.

a) True

b) False

View Answer

Answer: b

Explanation: In the non-ideal scattering matrix representation of the circulator, the attenuation constant and phase constant α, β respectively are approximated in terms of the reflection coefficient which represents the impedance mismatch in the network. Α is approximated as 1-Г2 and β is approximated as Г.

Microwave Engineering Questions and Answers – Noise in Micro – Wave Circuits

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Noise in Micro – Wave Circuits”.

1. The type of noise caused by vibration of bound charges is called:

a) Thermal noise

b) Shot noise

c) Flicker noise

d) None of the mentioned

View Answer

Answer: a

Explanation: Thermal noise is the most basic type of noise, being caused by thermal vibrations of bound charges in a material. When an electron bound to atom gains energy and vibrates, thermal noise is produced. It is also called as Johnson noise.

2. ________ noise occurs due to the random fluctuation of charge in an electron tube.

a) Flicker noise

b) Shot noise

c) Thermal noise

d) White noise

View Answer

Answer: b

Explanation: Shot noise or Poisson noise is a type of electronic noise that can be modeled by Poisson process. This type of noise occurs due to the discrete nature of electric charge.

3. Flicker noise occurs in solid-state components and vacuum tubes.

a) True

b) False

View Answer

Answer: a

Explanation: Flicker noise is a form of noise that exhibits an inverse frequency power density curve. It has a pink noise power density spectrum. Since this noise is inversely proportional to the operating frequency, it is called 1/f noise.

4.________ noise is caused by random motion of charges in ionized gas.

a) Plasma noise

b) Quantum noise

c) Thermal noise

d) Flicker noise

View Answer

Answer: a

Explanation: Plasma noise is caused by random motion of charges in an ionized gas such as a plasma, ionosphere or sparking electrical contacts. A material medium is required to produce this type of noise.

5. The most insignificant form of noise is:

a) Plasma noise

b) Quantum noise

c) Shot noise

d) Flicker noise

View Answer

Answer: b

Explanation: Quantum originates due to the quantized nature of charge carriers and photons. This noise does not pose any problem in microwave circuits and also does not affect the signal strength. Hence they are often significant.

6. An X- band amplifier has a gain of 20 dB and a gain of 1GHz bandwidth. Noise figure of the amplifier is -62 dBm at 290 K and -64.7 dBm at 77 K. then the Y factor of the amplifier is :

a) 3 dB

b) 6.4 dB

c) 2.7 dB

d) 5.6 dB

View Answer

Answer: c

Explanation: The Y factor of an amplifier is given by the difference in noise figure of the amplifier measured at two different temperatures. Taking the difference of the two values, the Y factor is 2.7 dB.

7. The Y factor of an amplifier obtained by measuring the noise figure at the temperatures 77 K and 290 K is 2.7 db. Then the equivalent noise temperature of the amplifier is:

a) 100 K

b) 150 K

c) 170 K

d) None of the mentioned

View Answer

Answer: c

Explanation: The equivalent noise temperature of the amplifier is given by the relation (T1-YT2)/(Y-1). Substituting the given values in the above equation, equivalent noise temperature is 170 K.

8. An amplifier has a noise equivalent temperature of 170 K. If the amplifier is used with a source having an equivalent noise temperature of TS=450 K, the output noise power of the amplifier is:

a) -50 dBm

b) -60 dBm

c) -60.7 dBm

d) -55 dBm

View Answer

Answer: c

Explanation: Output noise power of an amplifier is given by the expression GkTSB + GkTeB Substituting the given values in the above expression, the output noise power of the amplifier is -60.7 dBm.

9. The noise power associated with a two port network modeled as thevinin equivalent is:

a) kTB

b) k/TB

c) TB/ K

d) None of the mentioned

View Answer

Answer: a

Explanation: Noise associated with a 2 port network modeled as thevinin equivalent is kTB) K is the Boltzmann constant, T is the temperature and B is the operating bandwidth of the circuit.

10. Excess noise ratio is defined as the ratio of generator noise to the noise associated with room temperature

a) True

b) False

View Answer

Answer: b

Explanation: Excess noise ratio is defined as the ratio of the difference in noise power of the generator and noise power associated with the room temperature to the noise power associated with room temperature.

11. Equivalent noise temperature associated with an arbitrary white noise source is:

a) N0/GKB

b) N0

c) N0/ kB

d) None of the mentioned

View Answer

Answer: a

Explanation: Equivalent noise temperature associated with an arbitrary white noise source is N0/GKb) Here N0 is the noise power delivered to the load resistor by the source, B is the operating bandwidth G is the gain of the circuit.

complete set of 1000+ Multiple Choice Questions and Answers.

Microwave Engineering Questions and Answers – Noise Figure

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Noise Figure”.

1. ___________ is defined as the ratio of desired signal power to undesired noise power.

a) Signal to noise ratio

b) Noise to signal ratio

c) Noise figure

d) Noise temperature

View Answer

Answer: a

Explanation: SNR is defined as the ratio of desired signal power to undesired noise power, and so is dependent on the signal power. When noise and a desired signal are applied to the input of a noise less network, both noise and signal will be attenuated or amplified by the same factor, so that the signal to noise ratio will be unchanged.

2. __________ is defined as the ratio of input signal to noise ratio to the output signal to noise ratio.

a) Noise figure

b) Noise temperature

c) SNRo

d) None of the mentioned

View Answer

Answer: a

Explanation: Noise figure is defined as the ratio of input signal to noise ratio to the output signal to noise ratio of a system or a receiver. SNRi is the signal to noise ratio measured at the input terminals of the device. SNR0 is the output signal to noise ratio measured at the output terminals of the device.

3. The equivalent noise temperature of a network given the noise figure of the network or system is:

a) T0(F-1)

b) T0(F+1)

c) T0(F)

d) T0/F

View Answer

Answer: a

Explanation: The equivalent noise temperature of a network given the noise figure of the network or system is given by T0(F-1). In this expression, F is the noise figure of the system. T0 has the value 290 K. T0 is the standard temperature considered.

4. Noise figure can be defined for any microwave network irrespective of any other constraints.

a) True

b) False

View Answer

Answer: b

Explanation: Noise figure is defined only for a matched input source and for a noise source equivalent to a matched load at a temperature T0= 290 K. noise figure and noise temperature are interchangeable noise properties.

5. Expression for noise of a two port network considering the noise due to transmission line and other lossy components is:

a) GkTB + GNadded

b) GkTB

c) GNadded

d) None of the mentioned

View Answer

Answer: a

Explanation: Expression for noise of a two port network considering the noise due to transmission line and other lossy components is GkTB + GNadded. Here, G is the gain of the system. Nadded is the noise generated by the transmission line, as if it appeared at the input terminals of the line.

6. Noise equivalent temperature of a transmission line that adds noise to the noise of a device is:

a) T (L-1)

b) T (L+1)

c) T (L)

d) T/L

View Answer

Answer: a

Explanation: Noise equivalent temperature of a transmission line that adds noise to the noise of a device is given by T (L-1). Here L is the loss factor of the line and T is the temperature at which the system is thermal equilibrium.

7. If the noise figures of the first stage of a two stage cascade network is 8 dB and the noise figure of the second stage is 7 dB and the gain of the first stage is 10, then the noise figure of the cascade is:

a) 8. 6 dB

b) 7.6 dB

c) 5.6 dB

d) 8.9 dB

View Answer

Answer: a

Explanation: Noise figure of a two stage cascade network is given by F1+ (F2-1)/G1. Here F1, F2 are the noise figure of the first and the second stage respectively. G1 is the gain of the first stage. Substituting the given values in the above equation, noise figure of the cascade is 8.6 dB.

8. Noise equivalent temperature of a 2 stage cascade network is given by:

a) Te1 + Te2/ G1

b) Te1 + Te1

c) Te1 / Te1

d) None of the mentioned

View Answer

Answer: a

Explanation: Noise equivalent temperature of a 2 stage cascade network is given by Te1 + Te1/ G1. Here, Te1 is the noise equivalent temperature of stage 1 and Te1 is the noise equivalent temperature of stage 2. G1 is the gain of the first stage of the amplifier.

9. When a network is matched to its external circuitry, the gain of the two port network is given by:

a) │S21│2

b) │S22│2

c) │S12│2

d) │S11│2

View Answer

Answer: a

Explanation: The gain of a two port network is given by the product of SS21 of the network and reflection co-efficient at the source end. But when the two port network is matched to the external circuitry, reflection coefficient becomes zero and gain reduces to │S21│2.

10. For a Wilkinson power divider of insertion loss L and the coupler is matched to the external circuitry, and then the gain of the coupler in terms of insertion loss is:

a) 2L

b) 1/2L

c) L

d) 1/L

View Answer

Answer: b

Explanation: To evaluate the noise figure of the coupler, third port is terminated with known impedance. Then the coupler becomes a two port device. Since the coupler is matched, ГS=0 and Гout=S22=0. So the available gain is │S21│2. This is equal to 1/2L from the available data.

11. Noise equivalent temperature of Wilkinson coupler having a gain of 1/2L is given as:

a) T (2L-1)

b) T (2L+1)

c) T (2L*1)

d) T / (2L-1)

View Answer

Answer: a

Explanation: Noise equivalent temperature of the Wilkinson coupler is found using the relation

T (1-G21)/G21. Substituting for G21 in the above expression, equivalent noise temperature is T (2L-1).

12. Expression for over all noise figure of a mismatched amplifier is:

a) 1+ (F-1)/ (1 -│Г│2)

b) 1

c) 1+ (F-1)

d) (F-1)/ (1 -│Г│2)

View Answer

Answer: a

Explanation: The overall noise figure of a mismatched amplifier is given by 1+ (F-1)/ (1 -│Г│2). Here F is the noise figure of the amplifier, when there is an impedance mismatch at the input of the amplifier; this impedance mismatch is given by Г.

complete set of 1000+ Multiple Choice Questions and Answers.

Microwave Engineering Questions and Answers – Non – Linear Distortion – 1

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Non – Linear Distortion – 1”.

1. Active devices like diodes and transistors become non-linear at high power levels due to:

a) Instability of transistor

b) Thermal noise

c) Gain compression

d) None of the mentioned

View Answer

Answer: c

Explanation: All practical components become non-linear at high power level operations. Active devices like diode and transistor become non-linear at high power levels due to effects such as gain compression or the generation of spurious frequency components due to device non-linearities, but all devices ultimately fail at very high power levels.

2. The ________property of devices like diodes and transistors is responsible for their usage in amplifiers and frequency convertors.

a) Symmetry

b) Non linearity

c) Linearity

d) None of the mentioned

View Answer

Answer: b

Explanation: Non linearity is of great utility for desirable functions such as amplification, detection and frequency and frequency conversion. But non-linearities in these devices lead to undesirable effects like gain compression and generation of spurious frequency components.

3. The output response of a non-linear circuit is given by the expression:

a) V0=a0+a1Vi1+a2Vi2+a3Vi3…

b) 0=a0+a1Vi1

c) V0=a2Vi2+a3Vi3

d) V0=a0

View Answer

Answer: a

Explanation: The output of non-linear devices in terms of the applied output voltage is given as the Taylor series in terms of the input signal voltage. This expansion is given by V0=a0+a1Vi1+a2Vi2+a3Vi3….

And higher order terms. The constant term leads to rectification converting an AC input signal to DC.

4. As the degree of non-linearity of a transistor or a diode increases, the power gain of the device:

a) Increases

b) Decreases

c) Remains a constant

d) None of the mentioned

View Answer

Answer: b

Explanation: Let V0cosωt be the input signal applies to a non linear device. The output of the device is given by V0=a0+a1Vi+a2Vi2+a3Vi3. Considering only up to power of 3 and expanding, we get terms, whose constants of third degree and higher order are negative. This results in fall in gain when the input signal levels are high.

5. The fall in gain of a device due to non-linearities is called:

a) Gain compression

b) Gain saturation

c) Impedance mismatch

d) None of the mentioned

View Answer

Answer: a

Explanation: The fall in gain of a device due to non-linearities is called gain compression. Physically, this is usually due to fact that the instantaneous output voltage of an amplifier is limited by the power supply voltage used to bias the active device.

6. 1 dB compression point is defined as the power level for which the output power of the non-linear device is 1 dB

a) True

b) False

View Answer

Answer: b

Explanation: 1 dB compression point is defined as the power level for which the output power is decreased by 1dB from the ideal linear characteristic. This power level is represented as P1dB and can be stated in terms of either input power or output power.

7. In a non-linear device, for a single input frequency the output will consist of:

a) Single frequency component

b) Harmonics of the input frequency

c) Constant gain

d) None of the mentioned

View Answer

Answer: b

Explanation: Let V0cosωt be the input signal applies to a non linear device. The output of the device is given by V0=a0+a1Vi1+a2Vi2+a3Vi3. Substituting in this equation gives higher frequency components which is associated with higher order terms in the expansion. When the input consists of only one frequency component the higher frequencies are filtered out using band pass filter.

8. When the input frequency consists of more than one closely spaced frequency, it results in:

a) Intermodal distortion

b) Gain compression

c) Signal fading

d) Signal attenuation

View Answer

Answer: a

Explanation: When the input frequency consists of more than one frequency, due to the non-linear effect of the device, higher frequency components are generated. When the two different frequencies are close enough to each other, due to the interference of multiple frequency components, intermodal distortion occurs.

9. Third-order intercept point is defined as the point at which the gain of a non linear device is 3 dB less than the maximum gain.

a) True

b) False

View Answer

Answer: b

Explanation: Third-order intercept point is defined as hypothetical intersection point at which the first order and third order powers would be equal is called the third-order intercept point.

10. The relation between an intercept points referenced at the input versus the output is given by the relation:

a) OIP3=G (IIP3)

b) (IIP3) = G (OIP3)

c) OIP3=G (IIP3)2

d) None of the mentioned

View Answer

Answer: a

Explanation: The relation between an intercept points referenced at the input versus the output is given by the relation OIP3=G (IIP3). Here OPI3 is the output third order intercept point; OIP3 is the input third order intercept point. G is the conversion loss in the device.

Microwave Engineering Questions and Answers – Non – Linear Distortion – 2

This set of Microwave Engineering Interview Questions and Answers for freshers focuses on “Non-Linear Distortion-2”.

1. If the third order input intercept point of a mixer is 13 dBm and if the mixer has a conversion loss of 6 dBm, then the third order intercept point at the output is:

a) 3 dBm

b) 7 dBm

c) 4 dBm

d) 2 dBm

View Answer

Answer: b

Explanation: Third order output intercept point is given by the relation OIP3=IIP3 (G). G is the conversion loss in the device. In dBm scale they have to be subtracted. Subtracting the given values, the output third order intercept point is 7 dBm.

2. In a coherent cascade network, if the third order intercept point of an amplifier is 158 mW and the third order intercept point of a mixer is 5 mW, with an insertion loss of 6 dB, then the third order intercept point of the coherent cascade is:

a) 2 mW

b) 4 mW

c) 4.4 mW

d) 5 mW

View Answer

Answer: c

Explanation: The third order intercept point of the cascade is given by the expression ((G2 (OIP3’)-1+ (OIP3’’)-1)-1. Substituting the values in the above expression for respective terms, the third order intercept point of the cascade is 4.4 mW.

3. In a non coherent cascade network, if the third order intercept point of an amplifier is 158 mW and the third order intercept point of a mixer is 5 mW, with an insertion loss of 6 dB, then the third order intercept point of the coherent cascade is:

a) 3 mW

b) 4.9 mW

c) 5.2 mW

d) 2 mW

View Answer

Answer: b

Explanation: The third order intercept point for a non coherent cascade network is given by ((G22 (OIP3’)-2+ (OIP3’’)-2)-2. Substituting the values in the above expression for respective terms, the third order intercept point of the cascade is 4.9 mW.

4. Inter modulation distortion occurs only in active non linear devices and they are the only source of inter modulation distortion.

a) True

b) False

View Answer

Answer: b

Explanation: It is also possible for inter modulation products to be generated by passive non linear effects in connectors, cables and antennas or almost every element where there is metal to metal contact. This effect is called passive inter modulation.

5. ________ is defined as the operating range for which a component or system has desirable characteristics.

a) Static range

b) Dynamic range

c) Characteristic gain

d) None of the mentioned

View Answer

Answer: b

Explanation: Dynamic range is defined as the operating range for which a component or system has desirable characteristics. For a power amplifier, this may be the power range that is limited at the low end by noise and at the high end by the compression point.

6. The expression for linear dynamic range is given by:

a) OP1 dB- No

b) OP1 dB/ No

c) OP1 dB+No

d) None of the mentioned

View Answer

Answer: a

Explanation: Linear dynamic range is defined as the ratio of 1 dB compression point to the noise level of the component. In dB scale, it is the difference of 1 dB compression point and the noise level of the component.

7. Passive intermodulation is significant only when:

a) The input signal levels are low

b) Input signal levels are high

c) Input signal is sinusoidal

d) Input signal is exponential

View Answer

Answer: b

Explanation: Because of the dependence of the third order intermodulation products with input power, passive intermodulation is usually only significant when input signal powers are relatively large.

8. A receiver has a noise figure of 7 dB, gain of 40 dB, noise temperature is 290 K. then the noise power at the receiver is:

a) 47. 4 dBm

b) -47.4 dBm

c) 23 dBm

d) -23 dBm

View Answer

Answer: b

Explanation: Noise power at the receiver is given by the expression GkB [TA + (F-1) T0]. G is the gain of the amplifier, TA is the noise temperature. F is the noise figure. Substituting the given values in the above equation, the power output at the receiver in dBm scale is -47.4 dBm.

9. For a receiver, 1 dB compression point is 25 dBm and the noise power at the receiver output is -47.4 dBm, the linear dynamic range is:

a) 70 dB

b) 72.4 dB

c) 68 dB

d) 85 dB

View Answer

Answer: b

Explanation: Linear dynamic range of the system is the difference of 1 dB compression point and the noise in the system. Substituting in the given equation, the linear dynamic range is 72.4 dB.

10. If the third order intercept point of a receiver is 35 dBm and the total noise in the receiver is -47.4 dBm signals to noise ratio is 10, then the spurious free dynamic range is:

a) 50.9 dB

b) 44.9 dB

c) 34.9 dB

d) 67 dB

View Answer

Answer: b

Explanation: The spurious free dynamic range of a receiver is given by the relation 0.6667 (OIP3_ No)-SNR. Substituting in the above equation, the spurious free dynamic range is 44.9 dB.

Microwave Engineering Questions and Answers – PIN Diodes

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “PIN Diodes”.

1. A PIN diode consists of ______number of semiconductor layers.

a) Three

b) Two

c) Four

d) One

View Answer

Answer: a

Explanation: PIN diode is a p-type, intrinsic, n-type diode consisting of narrow layer of p-type semiconductor and a narrow layer of n-type semiconductor material, with a thicker region of intrinsic or very lightly n doped semiconductor sandwiched between them.

2. The material out of which PIN diode is made is:

a) Silicon

b) Germanium

c) GaAs

d) None of the mentioned

View Answer

Answer: a

Explanation: Silicon is the semiconductor normally used because of its power handling capability and it offers high resistivity for the intrinsic region. But depending on the application, these days GaAs is also used in fabricating PIN diodes.

3. The behavior of a PIN diode is entirely different from normal diodes at all frequency of operation.

a) True

b) False

View Answer

Answer: b

Explanation: PIN diode acts as a ordinary diode at frequencies up to about 100MHz. at high frequencies it stops to rectify and then acts as a variable resistance.

4. The junction resistance and capacitance of the intrinsic region in a PIN diode are connected______ in the equivalent circuit of PIN diode.

a) Series

b) Parallel

c) Connected across package capacitance

d) None of the mentioned

View Answer

Answer: b

Explanation: The junction capacitance Cj and junction resistance Rj of a PIN diode are connected in parallel in the equivalent circuit of a PIN diode. The package resistance and package capacitance are connected in series to these junction parameters.

5. The resistance of the PIN diode with positive bias voltage:

a) Increases

b) Decreases

c) Remains constant

d) Insufficient data

View Answer

Answer: b

Explanation: When the bias is varied on the PIN diode, its microwave resistance RJ changes from a typical value of 6 KΩ under negative bias to perhaps 5 Ω under forward bias. Thus if the diode is mounted on a 50Ω coaxial line, it will not significantly load this line.

6. A PIN diode can be used in either a series or a shunt configuration to form a __________

a) Single pole single throw switch

b) Single pole double throw switch

c) Amplifier

d) Oscillator

View Answer

Answer: a

Explanation: A PIN diode can be used in either a series or a shunt configuration to form a single pole single throw switch. In the series configuration, the switch is on when the diode is forward biased and off when the diode is reverse biased.

7. The working principle of series and shunt configuration single pole single throw switch is the same.

a) True

b) False

View Answer

Answer: b

Explanation: In the series configuration, the switch is on when the diode is forward biased and off when the diode is reverse biased. In the shunt configuration, forward biasing the diode cuts-off the transmission while reverse biasing the diode ensures transmission from input to output.

8. Under ideal conditions, when a PIN diode is used as a switch, the switch must have _______ insertion loss in the ON state.

a) Maximum

b) Zero

c) Average

d) Insertion loss cannot be defined for a switch

View Answer

Answer: b

Explanation: Ideally, when PIN diode is used as switch, the switch should have zero insertion loss in the ON state and infinite attenuation in the OFF state. These are ideal conditions. But practically a good operating switch must have low insertion loss.

9. When PIN diode is used as a switch, the expression for insertion loss of the switch is given by:

a) 10 log (Po/PL)

b) 10 log (PL/P0)

c) 10 log (PL. Pₒ)

d) None of the mentioned

View Answer

Answer: a

Explanation: Insertion loss of a switch is defined as the ratio of incident power applied to the load when switch is absent to the actual power delivered to the load.

10. For a shunt configuration switch, the diode impedance is 40 Ω and the terminated line characteristic impedance is 50 Ω. Then the insertion loss of the switch is:

a) 2.2 dB

b) 4.2 dB

c) 8.4 dB

d) 3.6 dB

View Answer

Answer: b

Explanation: Insertion loss of a shunt configuration switch is given by 20 log (2ZD+Z0/2ZD). Substituting the given values in the above expression, the insertion loss of the shunt configuration switch is 4.2 dB.

11. In the series configuration of a PIN diode switch, the terminated load impedance was found to be 50 Ω and the diode impedance was 60 Ω. Then the insertion loss of the switch is:

a) 4 dB

b) 2 dB

c) 3.6 dB

d) 4.8 dB

View Answer

Answer: a

Explanation: Insertion loss of a shunt configuration switch is given by 20 log (2ZD+Z0/2ZD). Substituting the given values in the above equation, the insertion loss is 4 dB.

12. The number of PIN diodes used in SPST switch and SPDT switch are the same.

a) True

b) False

View Answer

Answer: b

Explanation: The number of PIN diodes in SPST switch is one, while the number of PIN diodes used in single pole double throw switch is two.

Microwave Engineering Questions and Answers – Varactor Diodes

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Varactor Diodes”.

1. Varactor diode is a semiconductor diode in which the _________ can be varied as a function of reverse voltage of the diode.

a) Junction resistance

b) Junction capacitance

c) Junction impedance

d) None of the mentioned

View Answer

Answer: b

Explanation: Varactors (variable-capacitor) have non-linearity of capacitance which is fast enough to follow microwaves. Varactor diode is a semiconductor diode in which the junction capacitance can be varied as a function of reverse voltage of the diode.

2. Any semiconductor diode has a junction capacitance varying with reverse bias. If such a diode has microwave characteristics, it is called:

a) IMPATT diode

b) TRAPITT diode

c) SKOTTKY diode

d) None of the mentioned

View Answer

Answer: d

Explanation: Any semiconductor diode has a junction capacitance varying with reverse bias. If such a diode has microwave characteristics, it is called varactor diode. With the reverse bias, the junction is depleted of mobile carriers resulting in a capacitance that is the diode behaves as a capacitance with the junction acting as dielectric between two conducting plates.

3. The width of depletion region of a varactor diode ________with increase in reverse bias voltage.

a) Increases

b) Decreases

c) Remains constant

d) None of the mentioned

View Answer

Answer: a

Explanation: The width of the depletion region goes on increasing with increase in reverse bias voltage of the varactor diode. As the width of depletion region is an inverse function for capacitance, as the width increases, capacitance decreases.

4. Diffused junction mesa silicon diodes are widely used at microwave frequencies.

a) True

b) False

View Answer

Answer: a

Explanation: Diffused junction mesa silicon diodes are widely used at microwave frequencies. They are capable handling large powers and large break down voltages. They have relative independence of ambient temperature and low noise.

5. Varactors made of ______ have higher frequency range of operation compared to silicon fabricated varactor diodes.

a) Germanium

b) GaAs

c) GaN

d) None of the mentioned

View Answer

Answer: b

Explanation: Varactor diodes made of silicon have frequency range of operation of 25 GHz. Varactor diodes made of gallium arsenide operate in the frequency range of 90 GHz. Varactors of gallium arsenide also have better at low temperature.

6. Varactor diodes are operated in _________ region to achieve maximum efficiency possible.

a) Cutoff region

b) Saturation region

c) Reverse saturation region

d) Active region

View Answer

Answer: c

Explanation: Varactors are used between the reverse saturation point and a point just above the avalanche region. The capacitance variation and the reverse voltage swing are limited to between the operating regions mentioned above.

7. The cutoff frequency for operation of a varactor diode at a specific bias is given by:

a) 1/2πRSCjv

b) 1/2πCSRjv

c) 1/2π√LC

d) None of the mentioned

View Answer

Answer: a

Explanation: Cutoff frequency for a certain bias voltage applied is given by 1/2πRSCjv. Here, Rs is the wafer resistance and Cjv is the junction capacitance measured in the varactor diode for a given specific bias voltage V.

8. ___________ is an amplifier constructed using a device whose reactance is varied to produce amplification.

a) Travelling wave tube

b) Parametric amplifier

c) Common emitter

d) Klystron amplifier

View Answer

Answer: b

Explanation: Parametric amplifier is an amplifier constructed using a device whose reactance is varied to produce amplification. Varactor diode is the most widely used element in a parametric amplifier.

9. Parametric amplifier is a ________ amplifier.

a) Low noise

b) High gain

c) Low gain

d) High noise

View Answer

Answer: a

Explanation: Parametric amplifiers are constructed using varactor diodes. Since they do not involve any resistance, they result in low noise levels. There will be no thermal noise, as the active element used involved is reactive and not resistive.

10. Parametric amplifiers find their application in long range RADAR and satellite ground stations.

a) True

b) False

View Answer

Answer: a

Explanation: Due to the advantage of low noise amplification, parametric amplifiers are used in applications when noise levels are high at the receiving end but the amplification of noise must not occur. Such applications include long range RADARS satellite ground stations, radio telescopes to name a few.

11. Gain of a parametric amplifier in terms of the frequencies involved in their operation is:

a) (fP – fS)/fS

b) fS/ (fP – fS)

c) fP/fS

d) None of the mentioned

View Answer

Answer: a

Explanation: Gain of a parametric amplifier in terms of the frequencies involved in their operation is (fP – fS)/fS. Here fP is the pump frequency, fS is the signal frequency and fi is the idler frequency.

Microwave Engineering Questions and Answers – GUNN Diodes

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “GUNN Diodes”.

1. Silicon and germanium are called ___________ semiconductors.

a) direct gap

b) indirect gap

c) band gap

d) indirect band gap

View Answer

Answer: b

Explanation: The forbidden energy gap for silicon and germanium are respectively 1.21 eV in Si and 0.79 eV in germanium. Silicon and germanium are called indirect gap semiconductors because the bottom of the conduction band does not lie directly above the top of the valence band.

2. GaAs is used in the fabrication of GUNN diodes because:

a) GaAs is cost effective

b) It less temperature sensitive

c) it has low conduction band electrons

d) less forbidden energy gap

View Answer

Answer: d

Explanation: In GaAs, the conduction band lies directly above the top of the valence band. The lowest energy conduction band in GaAs is called as primary valley. GaAs consists of six secondary valleys. The bottom of one of the secondary valley is at an energy difference of 0.35 eV with the bottom of the primary valley in conduction band.

3. In a GaAs n-type specimen, the current generated is constant irrespective of the electric filed applied to the specimen.

a) true

b) false

View Answer

Answer: b

Explanation: In a GaAs n-type specimen, when the electric field applied reaches a threshold value of Eth, the current in the specimen becomes suddenly oscillatory and with respect to time and these oscillations are in the microwave frequency range. This effect is called Gunn Effect.

4. When the electric field applied to GaAs specimen is less than the threshold electric field, the current in the material:

a) increases linearly

b) decreases linearly

c) increases exponentially

d) decreases exponentially

View Answer

Answer: a

Explanation: When the electric field applied is less than the threshold value of electric field, the electrons jump from the valence band to the primary valley of the conduction band and current increases linearly with electric field.

5. When the applied electric field exceeds the threshold value, electrons absorb more energy from the field and become:

a) hot electrons

b) cold electrons

c) emission electrons

d) none of the mentioned

View Answer

Answer: a

Explanation: When the applied electric field exceeds the threshold value, electrons absorb more energy from the field and become hot electrons. These electrons jump into the lowest secondary valley in the conduction band. When the electrons become hot, their mobility reduces.

6. GaAs is used in fabricating Gunn diode. Gunn diode is:

a) bulk device

b) sliced device

c) made of different type of semiconductor layers

d) none of the mentioned

View Answer

Answer: a

Explanation: A GUNN diode is a bulk device, that is, it does not contain any junction but it is a slice of n-type GaAs. P-type GaAs does not exhibit Gunn Effect. Hence it is a reversible and can be operated in both directions.

7. The electrodes of a Gunn diode are made of:

a) molybdenum

b) GaAs

c) gold

d) copper

View Answer

Answer: a

Explanation: Gunn diode is grown epitaxially onto a gold or copper plated molybdenum electrode, out of gallium arsenide doped with silicon, tellurium or selenium to make it n-type.

8. When either a voltage or current is applied to the terminals of bulk solid state compound GaAs, a differential ______ is developed in that bulk device.

a) negative resistance

b) positive resistance

c) negative voltage

d) none of the mentioned

View Answer

Answer: a

Explanation: When either a voltage or current is applied to the terminals of a sample of bulk solid state compound formed by group 5 and 3 elements of periodic table, a differential resistance is developed in the bulk device. This fundamental concept is called RWH theory.

9. The number of modes of operation for n type GaAs is:

a) two

b) three

c) four

d) five

View Answer

Answer: c

Explanation: n-type GaAs used for fabricating Gunn diode has four modes of operation. They are Gunn oscillation mode, limited space charge accumulation mode, and stable amplification mode bias circuit oscillation mode.

10. The free electron concentration in N-type GaAs is controlled by:

a) effective doping

b) bias voltage

c) drive current

d) none of the mentioned

View Answer

Answer: a

Explanation: The free electron concentration in n-type GaAs is controlled through effective doping so that they range from 1014 to 1017 per cc at room temperature. The typical specimen of n-type GaAs has the dimensions 150 µm by 150 µm.

11. The modes of operation of a Gunn diode are illustrated in a plot of voltage applied to the Gunn diode v/s frequency of operation of Gunn diode.

a) true

b) false

View Answer

Answer: b

Explanation: A graph of plot of product of frequency and the length of the device plotted along y-axis versus the product of doping concentration and length along X- axis. These are the parameters on which the four modes of operation of Gunn diode are explained.

12. The mode of operation in which the Gunn diode is not stable is:

a) Gunn oscillation mode

b) limited space charge accumulation mode

c) stable amplification mode

d) bias circuit oscillation mode

View Answer

Answer: a

Explanation: In Gunn oscillation mode, the device is unstable due to the formation of accumulation layer and field domain. This high field domain moves from cathode to anode.

13. The frequency of oscillation in Gunn diode is given by:

a) vdom/ Leff

b) Leff/ Vdom

c) Leff/ WVdom

d) none of the mentioned

View Answer

Answer: a

Explanation: In Gunn oscillation mode, the frequency of oscillation is given by vdom/ Leff, where vdom is the domain velocity, Leff is effective length that the domain moves from the time it is formed until the time a new domain is formed.

14. In Gunn diode oscillator, the Gunn diode is inserted into a waveguide cavity formed by a short circuit termination at one end

a) true

b) false

View Answer

Answer: a

Explanation: The Gunn diode is mounted at the centre of the broad wall of a shorted waveguide since for the dominant TE10 mode; the electric field is maximum at the centre.

15. In a Gunn diode oscillator, the electron drift velocity was found to be 107 cm/second and the effective length is 20 microns, then the intrinsic frequency is:

a) 5 GHz

b) 6 GHz

c) 4 GHz

d) 2 GHz

View Answer

Answer: a

Explanation: The intrinsic frequency for a Gunn oscillator is given by Vd/L. Here VD is the drift velocity and L is the effective length. Substituting the given values in the above equation, intrinsic frequency is 5 GHz.

Microwave Engineering Questions and Answers – IMPATT and BARITT Diodes

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “IMPATT and BARITT Diodes”.

1. The material used to fabricate IMPATT diodes is GaAs since they have the highest efficiency in all aspects.

a) true

b) false

View Answer

Answer: b

Explanation: IMPATT diodes can be fabricated using silicon, germanium, GaAs or indium phosphide. Out of these materials, GaAs have highest efficiency, low noise and high operating frequencies. But GaAs has a major disadvantage of complex fabrication process and higher cost. So, GaAs are not preferred over silicon and germanium.

2. When a reverse bias voltage exceeding the breakdown voltage is applied to an IMPATT diode, it results in:

a) avalanche multiplication

b) break down of depletion region

c) high reverse saturation current

d) none of the mentioned

View Answer

Answer: a

Explanation: A reverse bias voltage exceeding the breakdown voltage is applied to an IMPATT diode, a high electric field appears across the n+ p junction. This high field imparts sufficient energy to the holes and also to valence electrons to raise themselves to the conduction band. This results in avalanche multiplication of electron hole pair.

3. To prevent an IMPATT diode from burning, a constant bias source is used to maintain _______ at safe limit.

a) average current

b) average voltage

c) average bias voltage

d) average resistance

View Answer

Answer: a

Explanation: Avalanche multiplication is a cumulative process resulting in rapid increase of carrier density. To prevent the diode from burning due to this increased carrier density, a constant bias source is used to maintain average current at safe limit.

4. The number of semiconductor layers in IMPATT diode is:

a) two

b) three

c) four

d) none of the mentioned

View Answer

Answer: c

Explanation: IMPATT diode consists of 4 layers according to the construction. It consists of a p+ region and n+ layers at the two ends. In between these layers, a p type layer and an intrinsic region is sandwiched.

5. The resonant frequency of an IMPATT diode is given by:

a) Vd/2l

b) Vd/l

c) Vd/2πl

d) Vdd/4πl

View Answer

Answer: a

Explanation: The resonant frequency of an IMPATT diode is given by the expression Vd/2l. Here VD is the carrier drift velocity; L is the length of the intrinsic region in the IMPATT diode.

6. If the length of the intrinsic region in IMPATT diode is 2 µm and the carrier drift velocity are 107 cm/s, then the drift time of the carrier is:

a) 10-11 seconds

b) 2×10-11 seconds

c) 2.5×10-11 seconds

d) none of the mentioned

View Answer

Answer: b

Explanation: The drift time of the carrier is defined as the ratio of length of the intrinsic region to the carrier drift velocity. Substituting the given values in this relation, the drift time of the carrier is 2×10-11 seconds.

7. If the length of the intrinsic region in IMPATT diode is 2 µm and the carrier drift velocity are 107 cm/s, then the nominal frequency of the diode is:

a) 12 GHz

b) 25 GHz

c) 30 GHz

d) 24 GHz

View Answer

Answer: b

Explanation: Nominal frequency is defined as the ratio of the carrier drift velocity to twice the length of the intrinsic region. Substituting the given values in the above equation, the nominal frequency is 25 GHz.

8. IMPATT diodes employ impact ionization technique which is a noisy mechanism of generating charge carriers.

a) true

b) false

View Answer

Answer: a

Explanation: IMPATT devices employ impact ionization techniques which is too noisy. Hence in order to achieve low noise figure, impact ionization is avoided in BARITT diodes. The minority injection is provided by punch through of the intermediate region.

9. An essential requirement for the BARITT diode is that the intermediate drift region be completely filled to cause the punch through to occur.

a) true

b) false

View Answer

Answer: b

Explanation: An essential requirement for the BARITT diode is that the intermediate drift region be completely filled to cause the punch through to the emitter-base junction without causing avalanche breakdown of the base collector junction.

10. If the RMS peak current in an IMPATT diode is 700 mA and if DC input power is 6 watt, with the load resistance being equal to 2.5 Ω, the efficiency of the diode is:

a) 10.1 %

b) 10.21 %

c) 12 %

d) 15.2 %

View Answer

Answer: b

Explanation: Efficiency of IMPATT diode is defined as the ratio of output RMS power to the input DC power. Calculating the RMS output power from the given RMS current and substituting in the equation of efficiency, the efficiency is 10.21%.

11. If the critical field in a Gunn diode oscillator is 3.2 KV/cm and effective length is 20 microns, then the critical voltage is:

a) 3.2 V

b) 6.4 V

c) 2.4 V

d) 6.5 V

View Answer

Answer: b

Explanation: Critical voltage of a Gunn diode oscillator is given by the expression lEc where l is the effective length and Ec is the critical field. Substituting the given values in the above equation, critical voltage is 6.4 volts.

complete set of 1000+ Multiple Choice Questions and Answers.

Microwave Engineering Questions and Answers – Applications of RF Diodes

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Applications of RF Diodes”.

1. Classical p-n junction diode cannot be used for high frequency applications because of:

a) High bias voltage

b) High junction capacitance

c) Frequency sensitive

d) High forward biased current

View Answer

Answer: b

Explanation: p-n junction diodes have high junction capacitance that makes them not suitable for high frequency applications. A Schottky barrier diode relies on a semiconductor metal junction and hence making them suitable for high frequency application.

2. Schottky barrier diode is a sophisticated version of the point contact ______________

a) Germanium diode

b) Silicon crystal diode

c) GaAs diode

d) None of the mentioned

View Answer

Answer: b

Explanation: Schottky barrier diode is a sophisticated version of the point contact silicon crystal diode, wherein the metal-semiconductor junction so formed is a surface rather than a point contact as it is in point contact silicon crystal diode.

3. Advantage of Schottky diode over silicon crystal diode is the presence minority charge carriers.

a) True

b) False

View Answer

Answer: b

Explanation: The advantage of Schottky diode over point contact silicon crystal diode is the elimination of minority carrier flow in the reverse biased condition of the diode. Due to the elimination of holes, there is no delay due to hole-electron recombination and hence operation is faster.

4. As the area of rectifying contact goes on increasing, the forward resistance of the Schottky diode:

a) Increases

b) Decreases

c) Remains unchanged

d) None of the mentioned

View Answer

Answer: b

Explanation: The noise and forward resistance of Schottky diode is small as compared to the noise figure and forward resistance of point contact silicon crystal diode. Hence, as the area of rectifying contact goes on increasing, the forward resistance decreases.

5. The number of semiconductor layers in a TRAPATT diode is:

a) Two

b) Three

c) Four

d) One

View Answer

Answer: b

Explanation: Silicon is usually used for the manufacture of TRAPATT diodes and they have a configuration p+ nn+ the p-N junction is reverse biased beyond the breakdown region, so that the current density is larger.

6. In order to achieve high current density, a compromise in _______is made in a TRAPATT diode.

a) Gain

b) Size

c) Operating frequency

d) No compromise is made on any of the parameter

View Answer

Answer: c

Explanation: When a high current density achieved, it decreases the electric field in the space charge region and increases the carrier transit time. Due to this, the frequency of operation gets lowered to less than 10 GHz. But efficiency is increased due to low power dissipation.

7. TRAPATT diode is normally mounted at a point inside a coaxial resonator where there is minimum RF voltage swing.

a) True

b) False

View Answer

Answer: b

Explanation: Inside a coaxial resonator, the TRAPATT diode is normally mounted at a point where maximum RF voltage swing is obtained. When the combined DC bias and RF voltage exceeds breakdown voltage, avalanche occurs and a plasma of holes and electrons are generated which get trapped.

8. A major disadvantage of TRAPATT diode is:

a) Fabrication is costly

b) Low operational bandwidth

c) Low gain

d) High noise figure

View Answer

Answer: d

Explanation: The disadvantages of TRAPATT diode are high noise figure and generation of strong harmonics due to the short duration of the current pulse. Since short duration of current pulses are used, they find application in S band pulse transmitters.

9. _________ gives a frequency domain representation of a signal, displaying the average power density versus frequency.

a) CRO

b) Oscilloscope

c) Spectrum analyzer

d) Network analyzer

View Answer

Answer: c

Explanation: Spectrum analyzer gives a frequency domain representation of a signal, displaying the average power density versus frequency. Thus, its function is dual to that of oscilloscope, which displays the time domain representation of a signal.

10. The most important functional unit of a spectrum analyzer is:

a) Mixer

b) IF amplifier

c) Sensitive receiver

d) None of the mentioned

View Answer

Answer: c

Explanation: A spectrum analyzer basically consists of a sensitive receiver that tunes over a specified frequency band and gives out a video output that is proportional to the signal power in a narrow bandwidth.

11. A tunnel diode is a p-n junction diode with a doping profile that allows electron tunneling through a narrow energy band gap.

a) True

b) False

View Answer

Answer: a

Explanation: A tunnel diode is a pn junction diode with a doping profile that allows electron tunneling through a narrow energy band gap leading to negative resistance at high frequencies. Tunnel diode can be used for both oscillators and amplifiers.

Microwave Engineering Questions and Answers – Heterojunction BJT – 1

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Heterojunction BJT – 1”.

1. BJTs are bipolar junction transistors. The name bipolar is given because:

a) they are made of n type and p type semiconductor

b) they have holes as charge carriers

c) they have electrons as charge carriers

d) none of the mentioned

View Answer

Answer: d

Explanation: In bipolar junction transistors, both electrons and holes are charge carriers and both of them together constitute current flow in transistors. Since both carriers result in current, they are called bipolar devices.

2. BJTs are suitable for RF applications because:

a) good performance in terms of frequency

b) power capacity

c) noise characteristics

d) all of the mentioned

View Answer

Answer: d

Explanation: BJTs designed to operate at certain frequency can be operated over a wide range of frequencies hence offering higher bandwidth. Also they have high power handling capacity and very good noise characteristics.

3. Bipolar junction transistors have _______ 1/f characteristics hence making them suitable for oscillators.

a) high

b) low

c) constant

d) decreasing exponential

View Answer

Answer: b

Explanation: Bipolar junction transistors have very low 1/f noise. 1/f noise is nothing but thermal noise. Hence BJTs are not very temperature and can be used at high temperature applications as well.

4. Silicon junction transistors are used as amplifiers at frequency range of about:

a) 5-10 MHz

b) 2-10 GHz

c) 40-50 MHz

d) 12-45 GHz

View Answer

Answer: b

Explanation: Silicon junction transistors have unconditional stability as a two port device at a wide range of frequencies. They are more suitable as amplifiers in the frequency range of about 2-10 GHz. Junction transistors when used as oscillators are used in the frequency range of about 20 GHz.

5. At frequency range of about 2-4 GHz, BJTs are preferred over FETs.

a) true

b) false

View Answer

Answer: a

Explanation: At about 2-4 GHz frequency range, BJTs have higher gain as compared to FETs, power capacity is high and biasing can be done using a single power supply. Because of these advantages, BJTs are preferred over FETs.

6. One major disadvantage of BJTs over FETs is that:

a) they have low gain

b) they do not have a good noise figure

c) low bandwidth

d) none of the mentioned

View Answer

Answer: b

Explanation: Bipolar junction transistors are subject to shot noise as well as thermal noise effects, so their noise figure is not as good as that of FET. Noise figure can pose serious problems at high operating frequencies.

7. Bipolar junction transistor is a ________ driven device.

a) current

b) voltage

c) power

d) none of the mentioned

View Answer

Answer: a

Explanation: Bipolar junction transistor is a current driven device where the collector output current directly depends on the input base current. Base current modulates the collector current of the device.

8. The upper frequency limit of BJT depends on the:

a) collector length in the transistor

b) base length

c) emitter length

d) driving voltage

View Answer

Answer: b

Explanation: The upper operating frequency limit of a BJT depends on the base length of the transistor. Typical base length of a transistor is in the range of a 0.1 µm. the operating frequency is a few GHz for this base length.

9. In the hybrid –π model of a BJT, the capacitance Cc between the base and collector in the hybrid –π model is ignored.

a) true

b) false

View Answer

Answer: a

Explanation: The capacitance Cc in the hybrid –π model is small and can be neglected. This has the effect of making the S12 parameter of the BJT equal to zero, implying that the power flows only in one direction through the device.

10. with the increase in the operating frequency of a BJT, the S22 parameter of the transistor:

a) increases

b) decreases

c) remains constant

d) none of the mentioned

View Answer

Answer: b

Explanation: With increase in the operating frequency of the transistor, S22 parameter of the transistor decreases. S22 parameter signifies the voltage reflected back to port 2. S22 parameter has a value of about 0.93 at 0.1 GHz frequency and 0.33 at 4 GHz frequency.

Microwave Engineering Questions and Answers – Heterojunction BJT – 2

This set of Microwave Engineering Multiple Choice Questions & Answers focuses on “Heterojunction BJT – 2”.

1. The hybrid-π model of a BJT is useful for analysis at all frequency ranges and variation of other transistor parameters.

a) true

b) false

View Answer

Answer: a

Explanation: The element values of the hybrid-π model are fairly constant over a wide range of operating points, bias conditions, load conditions and frequency. Otherwise, the element values become frequency, bias or load dependent in which case the hybrid –π model becomes less useful.

2. If the S11 and S22 parameters of a common emitter operated BJT is high:

a) then the output and input ports are matched well

b) there is mismatch in the ports

c) the gain of the amplifier is high

d) none of the mentioned

View Answer

Answer: b

Explanation: S11 and S22 parameters of a two port network signify the amount of signal to the same port that is excited by the source, a high value of these values imply that these ports are not impedance matched properly.

3. If a common emitter configuration BJT is treated as a two port network, the gain of the amplifier is roughly given by the S parameter:

a) S11

b) S12

c) S21

d) S22

View Answer

Answer: c

Explanation: When a BJT is represented as a two port network, where the base is port 1 and collector is port 2. Gain of the amplifier is given by the parameter S21. This parameter drops quickly with an increase in the operating frequency.

4. Short circuit current gain of BJT is given by the expression:

a) gm/ωC

b) ωC/ gm

c) gm/C

d) none of the mentioned

View Answer

Answer: a

Explanation: Short circuit current gain of BJT is defined as the ratio of output collector current to the input base current assuming the base resistance to be zero. The frequency at which the short circuit current gain of the amplifier is unity is called upper frequency limit.

5. The output collector to emitter current of a BJT amplifier is independent of the input base current of the amplifier.

a) true

b) false

View Answer

Answer: b

Explanation: BJT is a current controlled device. Output collector current is controlled by the input base current. If the input base current is increased, the collector current also increases.

6. The current gain of a BJT ________ with frequency.

a) increases

b) decreases

c) remains constant

d) none of the mentioned

View Answer

Answer: a

Explanation: The short circuit current gain of a BJT amplifier is given by the expression gm/ωC. From the equation, it is seen that gain is inversely proportional to frequency. As the frequency of operation of BJT increases, current gain of the transistor reduces.

7. If a transistor has a short circuit current gain of 25 and the capacitance measured in the hybrid-π model of the transistor was 60 pF. Then the threshold frequency of operation of the transistor is:

a) 60 MHz

b) 45.6 GHz

c) 66.3 GHz

d) 34.8 GHz

View Answer

Answer: c

Explanation: The threshold frequency for a BJT is given by gm/2πC. substituting the given values; the threshold operating frequency of the transistor is 66.3 GHz.

8. Hetero junction bipolar transistors have the same working principle and operation as that of a BJT.

a) true

b) false

View Answer

Answer: b

Explanation: The operation of hetero junction BJT is same as that of BJT, but an HBT has a base emitter junction made from a compound semiconductor material such as GaAs in junction with thin layers of other materials.

9. Advantage of HJT over BJT is that it has:

a) higher gain

b) high frequency of operation

c) sophisticated construction

d) none of the mentioned

View Answer

Answer: b

Explanation: The gain of BJT reduces with the increase in the operating frequency. This problem is overcome in HJT which gives sustained gain even at frequencies of about 100 GHz. They are also suitable for application in low power circuits.

10. The S21 parameter of a HJT increases with increase in the operating frequency of the transistor.

a) true

b) false

View Answer

Answer: b

Explanation: S21 parameter of a transistor signifies the gain of the transistor. With the increase in the operating frequency of the transistor, the gain of the transistor reduces, but it is to be noted that the fall in gain of a HJT is not as rapid as fall in gain in a BJT.

Microwave Engineering Questions and Answers – Field Effect Transistors

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Field Effect Transistors”.

1. Field effect transistors are different from BJTs in that they are _________

a) monopolar devices

b) bipolar devices

c) bidirectional device

d) none of the mentioned

View Answer

Answer: a

Explanation: FETs are called monopolar devices, with only one carrier type, either electrons or holes providing current flow through the device. N-channel FETs employ electrons while p-channel FETs employ holes as source of current.

2. GaAs MESFET –metal semiconductor field effect transistor are one of the widely used categories of FETs.

a) true

b) false

View Answer

Answer: a

Explanation: One of the most important developments in microwave technology has been the GaAs metal semiconductor field effect transistor, as this device permitted the first practical solid-state implementation of amplifiers and oscillators.

3. At frequencies above 10GHz, MESFET are not suitable for microwave applications due to parasitic effects.

a) true

b) false

View Answer

Answer: b

Explanation: GaAs MESFETs can be used at frequencies well into the millimeter wave range, with high gain and low noise figure often making them the device of choice in monolithic microwave integrated circuits above frequencies of 10 GHz.

4. Advantage of using GaAs in MESFET as compared to use of silicon is:

a) GaAs are cost effective

b) they have higher mobility

c) they have high resistance for flow of current in the reverse direction

d) none of the mentioned

View Answer

Answer: b

Explanation: The desired high gain and noise features of this transistor are a result of high electron mobility of GaAs compared to silicon and the absence of shot noise in them.

5. In MESFET, an applied signal at the gate modulates the electron carriers; this produces _______ in the FET.

a) voltage amplification

b) voltage attenuation

c) electron multiplication

d) electron recombination

View Answer

Answer: a

Explanation: In operation, the electrons are drawn from the source to drain by a positive voltage applied to the source and drain. These carriers are modulated by the voltage applied to the gate hence resulting in voltage amplification.

6. The frequency of operation of an FET is limited by:

a) drain to source voltage

b) gate to source voltage

c) gate length

d) effective area of an FET

View Answer

Answer: c

Explanation: The frequency of operation of an FET is given by the gate length. Present FETs have a gate length of the order of 0.2-0.6 µm, with corresponding upper frequency limits of 100-50 GHz.

7. The S21 parameter for a MESFET is lesser than 1.

a) true

b) false

View Answer

Answer: b

Explanation: From the analysis of the small signal equivalent circuit of MESFE, S21 parameter of the transistor was found to be greater than one under normal operating conditions. Here port 1 is at the gate and port 2 is at the drain.

8. The expression for short circuit current gain of an FET is given by:

a) gm/ ωCgs

b) Ig/gmVc

c) ωCgs/ gm

d) none of the mentioned

View Answer

Answer: a

Explanation: Short circuit current gain of an FET is defined as the ratio of drain current to gate current when the output is short circuited. This is expressed as ID/IG. This ratio in simplified form is given as gm/ ωCgs.

9. The upper threshold frequency of an FET, where short circuit gain is unity is given by:

a) gm/2πCgs

b) gm/Cgs

c) gm/ 2π

d) none of the mentioned

View Answer

Answer: a

Explanation: The upper threshold frequency is dependent on the factor gm, associated with the current generator of the small signal equivalent circuit. Cgs is the capacitance measured between the gate and source terminals.

10. The scattering parameter S11 for an FET __________ with increase in the frequency of operation of the transistor.

a) increases

b) decreases

c) remains constant

d) none of the mentioned

View Answer

Answer: b

Explanation: S11 parameter of an FET decreases with the increase in the frequency of operation of an FET. The measured values are 0.97 at 1 GHz, and 0.49 at 12 GHz.

11. The curve of IDS v/s VDS of an FET does not vary with the gate to source voltage applied.

a) true

b) false

View Answer

Answer: b

Explanation: Curve of IDS v/s VDS of an FET varies with the gate to source voltage applied. As the gate to source voltage applied becomes more positive, the drain to source current goes on increasing for an applied constant gate to source voltage.

12. High-power circuits generally use higher values of:

a) gate to source current

b) drain to source current

c) drain current

d) gate to source voltage

View Answer

Answer: c

Explanation: In order to achieve high drain current for high power applications, DC bias voltage must be applied to both gate and the drain, without disturbing the RF signal paths.

13. High drain current at RF levels is achieved with the biasing and decoupling circuitry for a dual polarity supply.

a) true

b) false

View Answer

Answer: a

Explanation: High drain current at RF levels is achieved with the biasing and decoupling circuitry for a dual polarity supply. The RF chokes provide a very low DC resistance for biasing, and a very high impedance at RF frequencies to isolate the signal from the bias supply.

Microwave Engineering Questions and Answers – Metal Oxide Semiconductor FET

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Metal Oxide Semiconductor FET”.

1. There exists no difference between the construction of GaAs MESFET and silicon MOSFET except for the material used in their construction.

a) True

b) False

View Answer

Answer: b

Explanation: There exists a difference between the construction of MESFET and MOSFET. There is a thin insulating layer of silicon dioxide between the gate contact and the channel region. Because the gate is insulated, it does not conduct DC bias current.

2. MOSFETs can provide a power of several hundred watts when the devices are packaged in:

a) Series

b) Parallel

c) Diagonal

d) None of the mentioned

View Answer

Answer: b

Explanation: MOSFETs can be used at frequencies into the UHF range and can provide powers of several hundred watts when devices are packaged in parallel. Laterally diffused MOSFETs have direct grounding of the source and can operate at low microwave frequencies with high power.

3. High electron mobility transistors can be constructed with the use of single semiconductor material like GaAs that have high electron mobility.

a) True

b) False

View Answer

Answer: b

Explanation: High electron mobility transistor is a hetero junction FET, meaning that it does not use a single semiconductor material, but instead is constructed with several layers of compound semiconductor materials.

4. The curve of IDS v/s VDS of an FET does not vary with the gate to source voltage applied.

a) True

b) False

View Answer

Answer: b

5. High-power circuits generally use higher values of:

a) Gate to source current

b) Drain to source current

c) Drain current

c) Gate to source voltage

View Answer

Answer: c

Explanation: In order to achieve high drain current for high power applications, DC bias voltage must be applied to both gate and the drain, without disturbing the RF signal paths.

6. High drain current at RF levels is achieved with the biasing and decoupling circuitry for a dual polarity supply.

a) True

b) False

View Answer

Answer: a

Explanation: High drain current at RF levels is achieved with the biasing and decoupling circuitry for a dual polarity supply. The RF chokes provide a very low DC resistance for biasing, and very high impedance at RF frequencies to isolate the signal from the bias supply.

7. Since multiple layers of semiconductor materials is used in high electron mobility transistors, this results in:

a) High gain

b) Power loss

c) Temperature sensitivity

d) Thermal stress

View Answer

Answer: d

Explanation: The multiple layers in the high electron mobility transistor result in the thermal and mechanical stress in the layers. To avoid this, the layers usually have matched crystal lattice.

8. A major disadvantage of high electron mobility transistor is that:

a) They have low gain

b) High manufacturing cost

c) Temperature sensitive

d) High driving voltage is required

View Answer

Answer: b

Explanation: High electron mobility transistors are devices containing multiple layers of different semiconductor materials. This complicated structure of HEMT requires sophisticated fabrication techniques leading to relatively high cost.

9. HEMT fabricated using GaN and aluminum gallium nitride on a silicon substrate can be used in :

a) High power transmitters

b) High power receivers

c) RADAR

d) Smart antennas

View Answer

Answer: a

Explanation: GaN HEMT operate with drain voltages in the range of 20-40 V and can deliver power up to 100 W at frequencies in the low microwave range, making these devices popular for high power transmitters.

10. The scattering parameter S11 for GaN HELMT increases with increase in frequency of operation

a) True

b) False

View Answer

Answer: b

Explanation: For GaN, the S11 parameter of the amplifier decreases with increase in frequency of operation. Experimental results have shown that S11 parameter was 0.96 at 0.5 GHz of frequency and 0.88 at 4 GHz of frequency.

Microwave Engineering Questions and Answers – Hybrid Microwave integrated Circuits

This set of Microwave Engineering Questions and Answers for Campus interviews focuses on “Hybrid Microwave integrated Circuits”.

1. In the course of development of microwave circuits, two distinct types of microwave integrated circuits have been developed according to the application requirements.

a) true

b) false

View Answer

Answer: a

Explanation: There are two distinct types of microwave integrated circuits fabricated. They are hybrid microwave integrated circuits and monolithic microwave integrated circuits. They differ in the method of fabrication in the layers of metallization done.

2. __________ is an important consideration for a hybrid integrated circuit.

a) material selection

b) processing units

c) design complexity

d) active sources

View Answer

Answer: a

Explanation: Material selection is an important consideration for a hybrid integrated circuit. Characteristics such as electrical conductivity, dielectric constant, loss tangent, thermal transfer and manufacturing compatibility of the material to be used for hybrid microwave circuits are evaluated first.

3. To fabricate a low frequency circuit using the hybrid microwave IC methodology, the material with _______ is preferred.

a) high dielectric constant

b) low dielectric constant

c) high resistivity

d) low resistivity

View Answer

Answer: a

Explanation: At low frequency applications, a high dielectric constant is desirable because it results in smaller circuit size. At higher frequencies, however the substrate thickness must be decreased to prevent radiation loss and other spurious effects.

4. The mask in a hybrid microwave circuit is made of:

a) rubylith

b) silicon

c) quartz

d) arsenic

View Answer

Answer: a

Explanation: The mask in hybrid microwave integrated circuits is made of Rubylith, a soft mylar film usually at a magnified scale for high accuracy. Then an actual size mask is made on a thin sheet of glass or quartz.

5. The metalized substrate is coated with __________ covered with the mast and exposed to light source.

a) photoresist

b) GaAs

c) germanium liquid

d) none of the mentioned

View Answer

Answer: a

Explanation: The metalized substrate is coated with photoresist, covered with the mast and exposed to light source. The substrate can be etched to remove the unwanted areas of the metal.

6. Commonly used software packages for CAD of hybrid microwave integrated circuits are:

a) CADENCE

b) ADS

c) DESIGNER

d) all of the mentioned

View Answer

Answer: d

Explanation: Before any microwave circuit design is implemented on the hardware, it is economical to simulate the same designs in software and check for the expected theoretical results. A few such software that provide such an environment is CADENCE, ADS, DESIGNER to name a few.

7. In hybrid microwave integrated circuits, the various components of the circuit are etched in the substrate.

a) true

b) false

View Answer

Answer: b

Explanation: In hybrid integrated circuit design, after all initial design steps are completed, the discrete components are soldered or wire bonded to the conductors This can be done manually or through automated computer-controlled pick and place machines.

8. Once the circuit components are designed and fabricated for certain specific values, they cannot be changed as per the requirement later.

a) true

b) false

View Answer

Answer: b

Explanation: IN HIC, provision is made for variations in component values and other circuit tolerances by providing tuning or trimming stubs that can be manually trimmed for each circuit. This increase circuit yield but also increases the cost of manufacture.

9.________ is a micromachining technique where suspended structures are formed on silicon substrates.

a) MMIC

b) HIC

c) RF MEMS

d) none of the mentioned

View Answer

Answer: c

Explanation: RF MEMS switch technology is a micro machining technique where suspended structures are formed in silicon substrates. These can be used in microwave resonators, antennas and switches.

10. Depending on the single path (capacitive or direct contact) and the attenuation mechanism MEMS switch can be used for various configurations for various devices.

a) true

b) false

View Answer

Answer: a

Explanation: A MEMS switch can be used in several different configurations depending on the single path, actuation mechanism, pull-back mechanism and the type of structure. One such example is switching the capacitance of a single path between high and low values by moving a flexible conductive membrane through the application of DC controlled voltage.

Microwave Engineering Questions and Answers – Monolithic Microwave Integrated Circuits

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Monolithic Microwave Integrated Circuits”.

1. Progress in ________ and other related semiconductors material processing led to the feasibility of monolithic microwave integrated circuits.

a) GaAs

b) Silicon

c) Germanium

d) GaAlAs

View Answer

Answer: a

Explanation: Progress in GaAs and other related semiconductor material processing led to the feasibility of MMIC, where all the passive and active components required for a given circuit can be grown or implanted in a substrate.

2. MMICs are high cost devices that involve complex fabrication methods and contain multiple layers to contain even small circuits.

a) True

b) False

View Answer

Answer: b

Explanation: Min MIC can be made at low cost because the labor involved with fabricating hybrid MIC’s is reduced. In addition, a single wafer can contain a large number of circuits, all of which can be processed and fabricated simultaneously.

3. The substrate of an MMIC must be a _____________ to accommodate the fabrication of all the type of devices.

a) Semiconductor

b) Insulator

c) Partial conductors

d) Metals operable at high frequencies

View Answer

Answer: a

Explanation: Substrate of MMIC must be a semiconductor material to accommodate the fabrication of active devices. The type of devices and the frequency range dictate the type of substrate material. One such material is GaAs MESFET.

4. GaAs MESFETs are versatile device because it finds application in:

a) Low-noise amplifiers

b) High gain amplifiers

c) Mixers

d) All of the mentioned

View Answer

Answer: d

Explanation: GaAs MESFET find application in low noise amplifiers, high gain amplifiers, broadband amplifiers, mixers, oscillators, phase shifters, and switches. These are the mostly used and cost effective substrates.

5. Transmission lines and other conductors in microwave devices are usually made with ___________

a) Gold metallization

b) Silver metallization

c) Copper metallization

d) Zinc metallization

View Answer

Answer: a

Explanation: Transmission lines and conductors at microwave operation are usually made with gold metallization. To improve the adhesion of gold to the substrate, a thin layer of chromium or titanium may be deposited first since these metals are relatively lossy.

6. For the capacitors used in MMICs, the insulating dielectric films used are:

a) Air

b) SiO

c) Titanium

d) GaAs

View Answer

Answer: b

Explanation: Capacitor and overlaying lines require insulating dielectric films, such as SiO, SiO2, SiN4 and Ta2O5. These materials have high dielectric constants and low loss and are compatible with integrated circuit processing.

7. Resistors used at normal operating frequencies can be directly used at microwave frequencies in MMIc.

a) True

b) False

View Answer

Answer: b

Explanation: Resistors used at normal operating frequencies cannot be used directly in MMICs. Resistors require the deposition of the lossy films like NiCr, Ta, Ti, and doped GaAs commonly used.

8. Processing in MMICs is done by __________

a) Ion implantation

b) Net list generation

c) Floor planning

d) None of the mentioned

View Answer

Answer: a

Explanation: Processing begins by forming an active layer in the semiconductor substrate for the necessary active devices. This is done by ion implantation or epitaxial techniques.

9. MMICs are the best microwave integrated circuit fabrication methodologies without any drawbacks in it.

a) True

b) False

View Answer

Answer: b

Explanation: Major drawback of MMIC is that they tend to waste large area of relatively expensive semiconductor substrate for components such as transmission lines and hybrids.

10. MMICs have higher circuit flexibility as compared to other microwave integrated fabrication methods.

a) True

b) False

View Answer

Answer: a

Explanation: Since the fabrication of additional FETs in an MMIC design is much easy, the circuit flexibility and performance can be enhanced with only little additional cost and the requirement for the fabrication of the entire device is prevented.

Microwave Engineering Questions and Answers – Micro-wave Tubes

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Micro-wave Tubes”.

1. The production of power at higher frequencies is much simpler than production of power at low frequencies.

a) True

b) False

View Answer

Answer: b

Explanation: As frequency increases to the millimeter and sub millimeter ranges, it becomes increasingly more difficult to produce even moderate power with solid state devices, so microwave tubes become more useful at these higher frequencies.

2. Microwave tubes are power sources themselves at higher frequencies and can be used independently without any other devices.

a) True

b) False

View Answer

Answer: b

Explanation: Microwave tubes are not actually sources by themselves, but are high power amplifiers. These tubes are in conjunction with low power sources and this combination is referred to as microwave power module.

3. Microwave tubes are grouped into two categories depending on the type of:

a) Electron beam field interaction

b) Amplification method

c) Power gain achieved

d) Construction methods

d) None of the mentioned

View Answer

Answer: a

Explanation: Microwave tubes are grouped into two categories depending on the type of electron beam field interaction. They are linear or ‘O’ beam and crossed field or the m type tube. Microwave tubes can also be classified as oscillators and amplifiers.

4. The klystron tube used in a klystron amplifier is a _________ type beam amplifier.

a) Linear beam

b) Crossed field

c) Parallel field

d) None of the mentioned

View Answer

Answer: a

Explanation: In klystron amplifier, the electron beam passes through two or more resonant cavities. The first cavity accepts an RF input and modulates the electron beam by bunching it into high density and low density regions.

5. In crossed field tubes, the electron beam traverses the length of the tube and is parallel to the electric field.

a) True

b) False

View Answer

Answer: b

Explanation: In a crossed field or ‘m’ type tubes, the focusing field is perpendicular to the accelerating electric field. Since the focusing field and accelerating fields are perpendicular to each other, they are called crossed field tubes.

6. ________ is a single cavity klystron tube that operates as on oscillator by using a reflector electrode after the cavity.

a) Backward wave oscillator

b) Reflex klystron

c) Travelling wave tube

d) Magnetrons

View Answer

Answer: b

Explanation: Reflex klystron is a single cavity klystron tube that operates as on oscillator by using a reflector electrode after the cavity to provide positive feedback via the electron beam. It can be tuned by mechanically adjusting the cavity size.

7. A major disadvantage of klystron amplifier is:

a) Low power gain

b) Low bandwidth

c) High source power

d) Design complexity

View Answer

Answer: b

Explanation: Klystron amplifier offers a very narrow operating bandwidth. This is overcome in travelling wave tube (TWT). TWT is a linear beam amplifier that uses an electron gun and a focusing magnet to accelerate beam of electrons through an interaction region.

8. In a _________ oscillator, the RF wave travels along the helix from the collector towards the electron gun.

a) Interaction oscillator

b) Backward wave oscillator

c) Magnetrons

d) None o the mentioned

View Answer

Answer: b

Explanation: In a backward wave oscillator, the RF wave travels along the helix from the collector towards the electron gun. Thus the signal for oscillation is provided by the bunched electron beam itself and oscillation occurs.

9. Extended interaction oscillator is a ________ beam oscillator that is similar to klystron.

a) Linear beam

b) Crossed beam

c) Parallel beam

d) M beam

View Answer

Answer: a

Explanation: Extended interaction oscillator is a linear beam oscillator that uses an interaction region consisting of several cavities coupled together, with positive feedback to support oscillation.

10. Magnetrons are microwave devices that offer very high efficiencies of about 80%.

a) True

b) False

View Answer

Answer: a

Explanation: Magnetrons are capable of very high power outputs, on the order of several kilowatts, and with efficiencies of 80% or more. But disadvantage of magnetron is that they are very noisy and cannot maintain frequency or phase coherence when operated in pulse mode.

11. Klystron amplifiers have high noise output as compared to crossed field amplifiers.

a) True

b) False

View Answer

Answer: b

Explanation: Crossed filed amplifiers have very good efficiencies – up to 80%, but the gain is limited to 10-15 db) In addition, the CFA has a noisier output than either a klystron amplifier or TWT. Its bandwidth can be up to 40%.

12. ____________ is a microwave device in which the frequency of operation is determined by the biasing field strength.

a) VTM

b) Gyratron

c) Helix BWO

d) None of the mentioned

View Answer

Answer: b

Explanation: Gyratron is a microwave device in which the frequency of operation is determined by the biasing field strength and the electron velocity, as opposed to the dimensions of the tube itself. This makes the gyrator especially useful for microwave frequencies.

Microwave Engineering Questions and Answers – Two Port Power Gains

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Two Port Power Gains”.

1. The power gain G of a two port network is independent of the source impedance of the two port network.

a) True

b) False

View Answer

Answer: a

Explanation: Power gain G is the ratio of power dissipated in the load ZL to the power delivered to the input of the two port network. This gain is independent of ZS although the characteristic of some active devices is dependent on ZS.

2. __________ is defined as the ratio of power available from the two port network to the power available from the source.

a) Transducer power gain

b) Available power gain

c) Power gain

d) None of the mentioned

View Answer

Answer: b

Explanation: Available power gain is defined as the ratio of power available from the two port network to the power available from the source. This assumes conjugate matching of both source and the load and depends on ZS, not ZL.

3. Transducer power gain of a two port network is dependent on :

a) ZS and ZL

b) ZS

c) ZL

d) Independent of both the impedances

View Answer

Answer: a

Explanation: Transducer power gain of a two port network is the ratio of the power delivered to the load to the power available from the source. This depends on both ZS and ZL.

4. For a two port network the voltage reflection coefficient seen looking towards the load, ГS is:

a) (ZS –Z0)/ (ZS –Z0)

b) (ZS +Z0)/ (Z0 – Z0)

c) ZS / (ZS –Z0)

d) Z0/ (ZS –Z0)

View Answer

Answer: a

Explanation: For a two port network, the reflection coefficient ГS seen looking towards the load is (ZS –Z0)/ (ZS –Z0). Here ZS is the input impedance of the transmission line and Z0 is the characteristic impedance of the transmission line.

5. In a two port network, the source impedance was measured to be 25 Ω and the characteristic impedance of the transmission line was measured to be 50 Ω. Then the reflection coefficient at the source end is:

a) -0.33333

b) -0.1111

c) 0.678

d) 0.2345

View Answer

Answer: a

Explanation: For a two port network, the reflection coefficient ГS seen looking towards the load is (ZS –Z0)/ (ZS –Z0). Substituting the given values in the above equation, reflection coefficient at the source end is -0.3333.

6. For a unilateral transistor, the S parameter that is zero is:

a) S11

b) S12

c) S21

d) S22

View Answer

Answer: b

Explanation: In a unilateral transistor power flow occurs only in one direction and hence S12 is sufficiently small and can be ignored. Also for a unilateral transistor the reflection coefficients reduce to Гin=S11 and Гout=S22.

7. Gain of an amplifier is independent of the operating frequency.

a) True

b) False

View Answer

Answer: b

Explanation: Gain of an amplifier depends on the operating frequency. Gain of a conjugate matched FET amplifier drops off as 1/f2 or 6dB per octave.

8. Gain of a conjugate matched FET amplifier is given by the relation:

a) Rds (fT)2/ 4Ri (f)2

b) 4Ri (f)2/Rds (fT)2

c) Rds/ Ri

d) None of the mentioned

View Answer

Answer: a

Explanation: Gain of FET amplifier is given by the relation Rds (fT)2/ 4Ri (f)2. Gain depends on the drain to source resistance, input resistance and also on the frequency of operation of the amplifier.

9. When both input and output of an amplifier are matched to zero reflection (in contrast to conjugate matching), the transducer power gain is:

a) │S21│2

b) │S22│2

c) │S12│2

d) |S11│2

View Answer

Answer: a

Explanation: When both input and output of an amplifier are matched to zero reflection, ГL=0 and ГS=0. This reduces the complex transducer gain equation to the s parameter of the amplifier S21. S21 signifies the power at port 2 due to input applied at port 1.

10. If the load impedance of a two port network is 40 Ω and the characteristic impedance is 50 Ω, then the reflection coefficient of the two port network at the load end is:

a) -0.111

b) -0.333

c) -0.987

d) None of the mentioned

View Answer

Answer: a

Explanation: Reflection coefficient at the load end of a two port network is given by the ratio (ZL-Z0)/ (ZL+Z0). ZL is the load impedance and Z0 is the characteristic impedance. Substituting, reflection at load end is -0.1111.

Microwave Engineering Questions and Answers – Stability Circles

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Stability Circles”.

1. For a transistor amplifier to be stable, either the input or the output impedance must have a real negative part.

a) True

b) False

View Answer

Answer: a

Explanation: For a transistor amplifier to be stable, either the input or the output impedance must have a real negative part. This would imply that │Гin│>1 or │Гout│>1, because these reflection coefficients depend on the source and load matching network.

2. ____________ condition, if met then the transistor can be impedance matched for any load.

a) Conditional stability

b) Unconditional stability

c) Infinite gain

d) Infinite input impedance

View Answer

Answer: b

Explanation: A network is said to be unconditionally stable if │Гin│<1 and │Гout│<1 for all passive source and load impedance. Transistors that are unconditionally stable can be easily matched.

3. A network is said to be conditionally stable if:

a) │Гin│<1, │Гout│<1.

b) │Гin│>1, │Гout│>1

c) │Гin│>1, │Гout│<1

d) │Гin│<1, │Гout│>1

View Answer

Answer: a

Explanation: For conditional stability, the condition to be satisfied is │Гin│<1, │Гout│<1. But this condition will be valid only for a certain range of passive source and load Impedance. His condition is also called potentially unstable.

4. Stability condition of an amplifier is frequency independent and hence can be operated at any frequency.

a) True

b) False

View Answer

Answer: a

Explanation: Stability condition of an amplifier is frequency dependent since the input and output matching networks generally depend on frequency. Hence it is possible for an amplifier to be stable at the designed frequency and unstable at other frequencies.

5. For a unilateral device condition for unconditional stability in terms of S parameters is:

a) │S11│<1, │S22│<1

b) │S11│>1, │S22│>1

c) │S11│>1, │S22│<1

d) │S11│<1, │S22│>1

View Answer

Answer: a

Explanation: For a unilateral device, the condition for unconditional stability is │S11│<1, │S22│<1. S11 parameter signifies the amount of power reflected back to port 1, which is the input port of the transistor. If this S parameter is greater is than 1, more amount of power is reflected back implying the amplifier is unstable.

6. If │S11│>1 or │S22│>1, the amplifier cannot be unconditionally stable.

a) True

b) False

View Answer

Answer: a

Explanation: If │S11│>1 or │S22│>1, the amplifier cannot be unconditionally stable because we can have a source or load impedance of Zₒ leading to Гs=0 or ГL=0, thus causing output and input reflection coefficients greater than 1.

7. For any passive source termination ГS, Unconditional stability implies that:

a) │Гout│<1

b) │Гout│>1

c) │Гin│<1

d) │Гin│>1

View Answer

Answer: a

Explanation: Unconditional stability implies that │Гout│<1 for any passive source termination, Гs. The reflection coefficient for passive source impedance must lie within the unit circle of the smith chart, nd the other boundary of the circle is written as Гs=ejφ.

8. The condition for unconditional stability of a transistor as per the K-∆ test is │∆│> 1 and K<1.

a) True

b) False

View Answer

Answer: b

Explanation: The condition for unconditional stability of a transistor is │∆│< 1 and K>1. Here, │∆│ and K are defined in terms of the s parameters of the transistor by defining the S matrix. To determine the unconditional stability of a transistor in K-∆ method, the S matrix of the transistor must be known.

9. If the S parameters of a transistor given are

S11=-0.811-j0.311

S12= 0.0306+j0.0048

S21=2.06+j3.717

S22=-0.230-j0.4517

Then ∆ for the given transistor is:

a) 0.336

b) 0.383

c) 0.456

d) None of the mentioned

View Answer

Answer: a

Explanation: Given the S parameters of a transistor, the ∆ value of the transistor is given by │S11S22-S12S21│. Substituting the given values in the above equation, the ∆ of the transistor is 0.336.

10. By performing the K-∆ test for a given transistor the values of K and ∆ were found to be equal to 0.383 and 0.334 respectively. The transistor with these parameters has unconditional stability.

a) True

b) False

View Answer

Answer: b

Explanation: The condition for unconditional stability of a transistor is │∆│< 1 and K>1. Here, │∆│ and K are defined in terms of the s parameters of the transistor by defining the S matrix. Here │∆│< 1 but the second condition is not satisfied. Hence they are not unconditionally stable.

Microwave Engineering Questions and Answers – Single Stage Transistor Amplifier Design

This set of Microwave Engineering online test focuses on “Single Stage Transistor Amplifier Design”.

1. The overall gain of a transistor is always a fixed value and cannot be changed as per design requirements.

a) True

b) False

View Answer

Answer: b

Explanation: For a given transistor gain G0 is a fixed value and cannot be changed. But the overall transducer gain of the amplifier will be controlled by GS and GL, of the matching section used with the transistor.

2. The frequency response of an amplifier is _______

a) Wide band

b) Narrow band

c) Pass band

d) None of the mentioned

View Answer

Answer: b

Explanation: Most transistors exhibit a significant impedance mismatch (large S11 and S22). This results in a frequency response of the transistor that being narrow band.

3. Maximum power transfer from the input matching port to the transistor will occur when:

a) Гin=Г*S

b) Гin=ГS

c) Гin=ГS. ejω

d) None of the mentioned

View Answer

Answer: a

Explanation: For a transistor, ГS is the reflection co-efficient of the amplifier looking towards the source. Гin is the reflection coefficient of the amplifier looking towards the input terminals of an amplifier. For maximum power transfer, the above mentioned condition must be satisfied.

4. The condition for maximum power transfer from the transistor to the output matching network will occur when:

a) Гout=ГL*

b) Гout=ГL

c) Гout=1/ ГL

d) Гout=1/ ГL*2

View Answer

Answer: a

Explanation: The condition for maximum power transfer from the transistor to the output matching network will occur when Гout=ГL*. ГL is the reflection coefficient seen looking towards the load. Гout is the looking towards the output ports of the transistor.

5. The input and output ports of an amplifier are always matched to the impedance Z0 .

a) True

b) False

View Answer

Answer: a

Explanation: With two different matching techniques called conjugate matching and lossless matching sections, the input and output ports of a transistor are matched to the characteristic impedance Z0 of the feed line used.

6. Unconditionally stable devices can always be ____________ for maximum gain.

a) Lossless matched

b) Conjugate matched

c) Forward biased

d) Driven with high current

View Answer

Answer: b

Explanation: Unconditionally stable devices can always be conjugate matched for maximum power gain and potentially unstable devices can be conjugate matched if K>1 and │∆│<1.

7. The maximum transducer gain occurs when the source and the load are matched to the impedance Z of the transistor by lossless method.

a) True

b) False

View Answer

Answer: b

Explanation: The maximum transducer power gain occurs when the source and load are conjugated matched to the transistor. This matched condition can be verified using the relation between the reflection coefficients.

8. Maximum transducer gain for an amplifier is the same as the maximum gain for an amplifier.

a) True

b) False

View Answer

Answer: a

Explanation: Maximum transducer gain is also referred to as matched gain. Maximum gain does not give a meaningful result when the device is only conditionally stable, since the simultaneous conjugate match of the load and source is not possible simultaneously.

9. In terms of S parameters for a transistor, the transducer gain is given by the relation:

a) │S21│/│S12│

b) │S12│/│S21│

c) │S22│/│S11│

d) │S11│/│S22│

View Answer

Answer: a

Explanation: Transducer power gain is defined as the ratio of power measured at the port 2 (output port) to the ratio of the power at the input port. This is redefined in terms of the S parameter of the network and can be written as │S21│/│S12│.

10. In the S matrix of a transistor, if the parameter S21 is 2.6 then the gain G0 of the transistor has the value

a) 6.2 dB

b) 8.3 dB

c) 2.22 dB

d) None of the mentioned

View Answer

Answer: b

Explanation: Gain G0 of a transistor amplifier is given as 10 log S12. Substituting for S12 in the equation, then the gain G0 of the amplifier is given by 8.3 dB.

Microwave Engineering Questions and Answers – Broadband Transistor Amplifier Design

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Broadband Transistor Amplifier Design”.

1. High gain is not achievable at microwave frequencies using BJT amplifiers because:

a) device construction

b) complex architecture

c) ports are not matched at high frequencies

d) none of the mentioned

View Answer

Answer: c

Explanation: At higher frequencies, if higher bandwidth is desired, a compromise on maximum achievable gain is made. But at these higher frequencies, the ports of the amplifier are not matched to 50 Ω.

2. To flatten the gain response of a transistor:

a) biasing current has to be increased

b) input signal level has to increased

c) increase the operational bandwidth

d) give negative feedback to the amplifier

View Answer

Answer: d

Explanation: Negative feedback can be used to increase the gain response of the transistor, improve the input and output match, and increase the stability of the device.

3. In conventional amplifiers, a flat gain response is achieved at the cost of reduced gain. But this drawback can be overcome by using:

a) balanced amplifiers

b) distributed amplifiers

c) differential amplifiers

d) none of the mentioned

View Answer

Answer: a

Explanation: In conventional amplifiers, a flat gain response is achieved at the cost of reduced gain. But this drawback can be overcome by using balanced amplifiers. This is overcome by using two 900 couplers to cancel input and output reflections from two identical amplifiers.

4. Bandwidth of balanced amplifier can be an octave or more, but is limited by the bandwidth of the coupler.

a) true

b) false

View Answer

Answer: a

Explanation: In order to achieve flat gain response, balanced amplifiers use couplers to minimize reflections. But this in turn reduces the bandwidth of the amplifier to the coupler bandwidth.

5. Coupler that is mostly used in balanced amplifiers to achieve the required performance is:

a) branch line coupler

b) wilkinson coupler

c) lange coupler

d) waveguide coupler

View Answer

Answer: c

Explanation: Lange couplers are broadband couplers and are compact in size. Since the bandwidth of a balanced amplifiers depends on the bandwidth of the coupler used. Lange coupler is thus preferred over couplers.

6. Distributed amplifiers offer very high _________

a) gain

b) bandwidth

c) attenuation

d) none of the mentioned

View Answer

Answer: b

Explanation: Distributed amplifiers offer very high bandwidth of about 10 decade. But higher gain cannot be achieved using distributed amplifiers and matching at the ports is very important to achieve higher bandwidth.

7. In distributed amplifiers, all the FET stages in the amplifier are connected in series to one another.

a) true

b) false

View Answer

Answer: b

Explanation: In distributed amplifiers, cascade of N identical FETs have their gates connected to a transmission line having a characteristic impedance of Zg with a spacing of lg while the drains are connected to a transmission line of characteristic impedance Zd, with a spacing ld.

8. ____________ uses balanced input and output, meaning that there are 2 signal lines, with opposite polarity at each port.

a) differential amplifier

b) distributed amplifier

c) balanced amplifier

d) none of the mentioned

View Answer

Answer: a

Explanation: Differential amplifier uses balanced input and outputs, meaning that there are 2 signal lines, with opposite polarity at each port. It has two input ports and one output port. The difference of the 2 input signals is amplified.

9. A major advantage of differential amplifiers is:

a) high gain

b) low input impedance

c) higher output voltage swing

d) none of the mentioned

View Answer

Answer: c

Explanation: Differential amplifiers can provide higher voltage swings that are approximately double that obtained with single ended amplifier.

10. Along with a differential amplifier, 1800 hybrid is used both at the input and output.

a) true

b) false

View Answer

Answer: a

Explanation: A differential amplifier can be constructed using two single-ended amplifiers and 1800 hybrids at the input and output to split and then recombine the signals.

Microwave Engineering Questions and Answers – Power Amplifiers

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Power Amplifiers”.

1. _______________ are used in the final stages of radar and radio transmitters to increase the radiated power level.

a) Power amplifiers

b) Oscillators

c) Transistors

d) Attenuators

View Answer

Answer: a

Explanation: Power amplifiers are used in the final stages of radar and radio transmitters to increase the radiated power level. Output of power amplifiers are in the range of 100-500 mW.

2. Important factors to be considered for power amplifier design are:

a) Efficiency

b) Gain

c) Thermal effect

d) All of the mentioned

View Answer

Answer: d

Explanation: As per the application requirement and considering various aspects of an amplifier like efficiency, gain, thermal efficiency and inter modulation distortion, amplifiers need to be designed.

3. Amplifier efficiency is the ratio of RF output power to DC input power. This parameter determines the performance of an amplifier.

a) True

b) False

View Answer

Answer: a

Explanation: Power amplifier is the primary consumer of DC power in most hand-held wireless devices, so amplifier efficiency is an important consideration. Amplifier efficiency is the ratio of RF output power to DC input power.

4. Gain of power amplifiers __________ with increase in operating frequency.

a) Increases

b) Decreases

c) Increases exponentially

d) Decreases exponentially

View Answer

Answer: b

Explanation: Silicon bipolar junction transistor amplifiers in the cellular telephone band of 800-900 MHz band have power added efficiencies of about 80%. But this efficiency drops quickly with increase in the operating frequency.

5. ___________ amplifiers are linear circuits, where the transistor is biased to conduct over the entire range of the input signal cycle.

a) Class A amplifiers

b) Class B amplifiers

c) Class C amplifiers

d) None of the mentioned

View Answer

Answer: a

Explanation: Class A amplifiers are linear circuits, where the transistor is biased to conduct over the entire range of the input signal cycle. Because of this, class A amplifiers theoretically have a maximum efficiency of 50%.

6. A class B amplifier consists of _______ transistors in order to conduct the input signal over the entire cycle.

a) 1

b) 2

c) 4

d) 6

View Answer

Answer: b

Explanation: Class B amplifier is biased to conduct only during one-half of the input signal cycle. 2 complementary transistors are operated in a class B push pull amplifier to provide amplification over the entire cycle.

7. Power amplifiers in the increasing order of efficiency is:

a) Class A, B, C

b) Class C, A, B

c) Class B, A, C

d) Efficiency of all the 3 amplifiers is the same

View Answer

Answer: a

Explanation: Class A amplifiers have an efficiency of about 50%. Class B amplifiers have an efficiency of about 78%, class C amplifiers can achieve efficiencies up to 100%. In the increasing order of efficiency, C > B> a)

8. Behavior of a transistor in power amplifiers is unpredictable at all input signal levels.

a) True

b) False

View Answer

Answer: b

Explanation: A transistor behaves linearly for signal powers below 1dB compression point and so, the small –signal scattering parameters should not depend either on the input power level or the output termination impedance.

9. If the output power of an amplifier is 10 V, and the input power supplied to the amplifier is 0.229 V given that the DC voltage used is 38.5 V, efficiency of the power amplifier is:

a) 25%

b) 50%

c) 75%

d) 35%

View Answer

Answer: a

Explanation: Efficiency of a power amplifier is (Pout- Pin)/ PDc Substituting the given values in the above expression, efficiency of the power amplifier is 25%.

10. If a power amplifier has an output power of 10 W, and an amplifier gain of 16.4 dB, then the input drive power is:

a) 400 mW

b) 225 mW

c) 229 mW

d) 240 mW

View Answer

Answer: c

Explanation: Input drive power required to get an output of 10 W is Pout (dBm)- G (dB). G is the gain of the amplifier. Substituting the given values in the above equation, 229 mW.

Microwave Engineering Questions and Answers – RF Oscillators

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “RF Oscillators”.

1. _________ is a non linear circuit that converts DC power to an AC waveform of desired frequency based on the oscillator design.

a) Attenuator

b) Amplifier

c) Oscillator

d) None of the mentioned

View Answer

Answer: c

Explanation: Oscillator is a non linear circuit that converts DC power to an AC waveform. Most RC oscillators provide sinusoidal outputs, which minimizes undesired harmonics and noise sidebands.

2. The transfer function of an RF oscillator is given by:

a) A/ (1-AH (ω))

b) A/ (1+AH (ω))

c) A/ (-1+AH (ω))

d) 1/ (1-AH (ω))

View Answer

Answer: a

Explanation: Transfer function of an RF oscillator is given by A/ (1-AH (ω)). Here, A is the gain of the transistor multiplier used. H(ω) is the function representing the feedback network. In an oscillator, positive feedback is used.

3. The criterion on which oscillations are produced in the oscillator circuit is called:

a) Shannon’s criteria

b) Barkhausen criteria

c) Colpitts criteria

d) None of the mentioned

View Answer

Answer: b

Explanation: When the condition 1-AH (ω) =0 occurs, it is possible to achieve non zero output voltage for zero input voltage, thus forming an oscillator. This is called Barkhausen criteria.

4. The necessary condition for oscillation in a Colpitts oscillator is:

a) C2/C1=gm/Gi

b) C1/C2=gm/Gi

c) C2/C1= gm*Gi

d) None of the mentioned

View Answer

Answer: a

Explanation: The condition for sustained oscillation in a Colpitts oscillator is C2/C1 = gm/Gi. Here C1 and C2 are the capacitance in the feedback network, gm is the transconductance of the transistor and Gi is the input admittance.

5. Colpitts oscillator operating at 50 MHz has an inductor in the feedback section of value 0.10µH. then the values of the capacitors in the feedback section is:

a) 100 pF, 100 pF

b) 100 pF, 50 pF

c) 70 pF, 130 pF

d) 80 pF, 60 pF

View Answer

Answer: a

Explanation: The equivalent value of series combination of the capacitors is given by 1/ω2L. This gives the equivalent capacitance value of 200 pF. C1C2/ (C1+C2) =200 pF. C1 and C2 values can be chosen in several ways. One of the way is C1=C2=100 pF.

6. The necessary condition for oscillation in a Hartley oscillator is:

a) L1/L2 = gm/Gi

b) L1/L2 =Gi /gm

c) L2L/L1 = gm/Gi

d) None of the mentioned

View Answer

Answer: a

Explanation: Necessary condition for oscillation in a Hartley oscillator is L1/L2 = gm/Gi. Here, L1 and L2 are the inductances in the feedback network and gm is the transconductance of the transistor and Gi is the input admittance.

7. An inductor is operating at frequency of 50 MHz. Its inductance is 0.1 µH, and then the series resistance associated with the inductor is: (Qo=100)

a) 0.31 Ω

b) 1.32 Ω

c) 1 Ω

d) 1.561 Ω

View Answer

Answer: a

Explanation: Series resistance associated with an inductor is given by ωL/Qₒ. Substituting in this equation, the series of an inductor is given by 0.31.

8. Hartley oscillator has inductance values of 12 mH and 4 mH in the feedback section and a capacitor of 4 nF. Then the resonant frequency of the circuit is:

a) 19.89 kHz

b) 25 kHz

c) 45 kHz

d) 12 kHz

View Answer

Answer: a

Explanation: Resonant frequency of Hartley oscillator is given by 1/ 2π√(C1 (L1 + L2)). Substituting the given values in the above equation, cut-off frequency is 19.89 kHz.

9. Colpitts oscillator in the feedback section has an inductance of 4 mH and capacitors of 12 nH and 4 nH. Then the resonant frequency of Colpitts oscillator is:

a) 50.4 kHz

b) 35.1 kHz

c) 45.9 kHz

d) None of the mentioned

View Answer

Answer: c

Explanation: Resonant frequency of Colpitts oscillator is given by 1/2π√LCₒ, where C0 is the equivalent capacitance given by C1C2/ (C1+C2). Substituting and solving the equation, resonant frequency is 45.9 kHz.

10. For Colpitts oscillator, the capacitors C1 and C2 in the feedback network are 1 µF and 25 µF respectively. Then the β value of the transistor is:

a) 35

b) 000.76

c) 25

d) 0.0025

View Answer

Answer: c

Explanation: β for a transistor is defined as the ratio of transconductance of the transistor to the input admittance, which is equal to the ratio of C2/C1. Substituting the given values, β of the transistor is 25.

Microwave Engineering Questions and Answers – Crystal Oscillators

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Crystal Oscillators”.

1. One condition to be satisfied in an oscillator circuit so that stable oscillations are produced is:

a) positive feedback is to be achieved

b) negative feedback is to be achieved

c) 1800 phase shift is required between the transistor input and output.

d) none of the mentioned

View Answer

Answer: c

Explanation: In an oscillator a total of 3600 of phase shift is to be achieved in the entire circuit to produce oscillations. The transistor used in the oscillator circuit must produce a phase shift of 1800 to achieve stable oscillations. Hence this condition has to be satisfied by the oscillator.

2. In an oscillator, the resonant feedback circuit must have must have a low Q in order to achieve stable oscillation.

a) true

b) false

View Answer

Answer: b

Explanation: If the resonant feedback circuit has a high Q, so that there is random phase shift with frequency, the oscillator will have good frequency stability.

3. Quartz crystals are more efficient as a feedback network because:

a) less circuit complexity

b) cost effective

c) crystals operate at high voltage levels

d) LC circuits have unloaded Q of a few hundreds

View Answer

Answer: d

Explanation: At frequencies below a few hundred MHz, where LC resonators seldom have unloaded Qs greater than a few hundred. Quartz crystals have unloaded Q of about 10000 and have a temperature drift of 0.001%/C0.

4. Quartz crystal and tourmaline used in oscillators work on the principle of:

a) photo electric effect

b) piezo electric effect

c) Raman effect

d) black body radiation

View Answer

Answer: b

Explanation: Quartz crystals work on the principle of piezo electric effect. When electrical energy is applied to these crystals, they vibrate in a direction perpendicular to the application of energy producing oscillations.

5. A quartz crystals equivalent circuit is a series LCR circuit and has a series resonant frequency.

a) true

b) false

View Answer

Answer: b

Explanation: A quartz crystal has an equivalent circuit such that a series LCR network is in parallel with a capacitor. A quartz crystal thus has both series and parallel resonant frequencies.

6. Quartz crystal is used in the _______region, where the operating point of the crystal is fixed.

a) resistive

b) inductive reactance

c) capacitive reactance

d) none of the mentioned

View Answer

Answer: b

Explanation: Quartz crystal is always operated in the inductive reactance region so that the crystal is used in place of an inductor in a Colpitts or pierce oscillator.

7. In the plot of reactance v/s frequency of a crystal oscillator, the reactance between series resonant frequency and parallel resonant frequency is:

a) capacitive

b) inductive

c) both capacitive and inductive

d) none of the mentioned

View Answer

Answer: b

Explanation: In the plot of reactance v/s frequency of a crystal oscillator, the reactance between series resonant frequency and parallel resonant frequency is inductive. In this region between the series and parallel and series resonant frequencies, the operating point of the crystal is fixed and hence can be used as part of other circuits.

8. In the equivalent circuit of a quartz crystal, LCR arm has an inductance of 4 mH and capacitor has a value of 4nF, then the series resonant frequency of the oscillator is:

a) 0.25 MHz

b) 2.5 MHz

c) 25 MHz

d) 5 MHz

View Answer

Answer: a

Explanation: The series resonant frequency of a crystal oscillator is given by 1/√LC. Substituting the given values of L and C in the expression, the series resonant frequency is 0.25 MHz.

9. Parallel resonant frequency of quartz crystal is given by:

a) 1/ √(LCₒC/(Cₒ+C))

b) 1/√LC

c) 1/√LCₒ

d) 1/ √(L(Cₒ+C) )

View Answer

Answer: a

Explanation: Parallel resonant frequency of an oscillator is given by√(LCₒC/(Cₒ+C)). Here L and C are the inductance and capacitance in the LCR arm of the equivalent circuit of the crystal. Co is the capacitance existing in parallel to this LCR arm.

10. The equivalent circuit of a quartz crystal has LCR arm capacitance of 12nF and inductance of 3mH and parallel arm capacitance of 4nF. Parallel resonant frequency for the circuit is:

a) 3 MHz

b) 0.3 MHz

c) 6 MHz

d) 9 MHz

View Answer

Answer: a

Explanation: The parallel resonant frequency of a crystal oscillator is given by 1/√(LCₒC/(Cₒ+C)). Substituting the given values in the equation, the parallel resonant frequency is found to be 3 MHz.

Microwave Engineering Questions and Answers – Microwave Oscillators

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Microwave Oscillators”.

1. In microwave oscillators, negative resistance transistors and diodes are used in order to generate oscillations in the circuit.

a) True

b) False

View Answer

Answer: a

Explanation: In microwave oscillator, for a current to flow in the circuit the negative impedance of the device must be matched with positive impedance. This results in current being non-zero and generates oscillation.

2. Any device with negative impedance as its characteristic property can be called:

a) Energy source

b) Energy sink

c) Oscillator

d) None of the mentioned

View Answer

Answer: a

Explanation: A positive resistance implies energy dissipation while a negative resistance implies an energy source. The negative resistance device used in the microwave oscillator, thus acts as a source. The condition Xin+ XL=0 controls the frequency of oscillation. Xin is the impedance of the negative resistance device.

3. In a microwave oscillator, a load of 50+50j is connected across a negative resistance device of impedance -50-50j. Steady state oscillation is not achieved in the oscillator.

a) True

b) False

View Answer

Answer: b

Explanation: The condition for steady state oscillation in a microwave oscillator is Zin=-ZL. Since this condition is satisfied in the above case, steady state oscillation is achieved.

4. For achieving steady state oscillation, the condition to be satisfied in terms of reflection coefficients is:

a) Гin=ГL

b) Гin=-ГL

c) Гin=1/ГL

d) None of the mentioned

View Answer

Answer: c

Explanation: The condition for steady state oscillation to be achieved in terms of reflection coefficient is Гin=1/ГL. Here Гin is the reflection coefficient towards the reflection coefficient device and ГL is the reflection coefficient towards the load.

5. A one port oscillator uses a negative resistance diode having Гin=0.9575+j0.8034 (Z0=50Ω) at its desired frequency point. Then the input impedance of the diode is:

a) -44+j123

b) 50+j100

c) -44+j145

d) None of the mentioned

View Answer

Answer: a

Explanation: The input impedance of the diode given reflection coefficient and characteristic impedance is Z0 (1+Гin)/ (1-Гin). Substituting in the given equation, the input impedance is -44 +j123 Ω.

6. If the input impedance of a diode used in the microwave oscillator is 45-j23 Ω, then the load impedance is to achieve stable oscillation is:

a) 45-j23 Ω

b) -45+j23 Ω

c) 50 Ω

d) 23-j45 Ω

View Answer

Answer: b

Explanation: The condition for stabilized oscillation is Zin=-ZL. According to this equation, the load impedance required for stabilized oscillation is – (45-j23) Ω. The load impedance is thus -45+j23 Ω.

7. To achieve stable oscillation, Zin + ZL=0 is the only necessary and sufficient condition to be satisfied by the microwave oscillator.

a) True

b) False

View Answer

Answer: b

Explanation: The condition Zin + ZL=0 is only a necessary condition for stable oscillation and not sufficient. Stability requires that any perturbation in current or frequency is damped out, allowing the oscillator to return to its original state.

8. In transistor oscillators, the requirement of a negative resistance device is satisfied using a varactor diode.

a) True

b) False

View Answer

Answer: b

Explanation: In a transistor oscillator, a negative resistance one port network is created by terminating a potentially unstable transistor with impedance designed to drive the device in an unstable region.

9. In transistor oscillators, FET and BJT are used. Instability is achieved by:

a) Giving a negative feedback

b) Giving a positive feedback

c) Using a tank circuit

d) None of the mentioned

View Answer

Answer: b

Explanation: Oscillators require a device that has high instability. To achieve this condition, transistors are used with a positive feedback to increase instability.

10. In a transistor amplifier, if the input impedance is -84-j1.9 Ω, then the terminating impedance required to create enough instability is:

a) -84-j1.9 Ω

b) 28+j1.9 Ω

c) – (28+j1.9) Ω

d) None of the mentioned

View Answer

Answer: b

Explanation: Relation between terminating impedance and input impedance is Zs=-Rin/3. Zs is the terminating impedance. Substituting in the given equation, the terminated impedance is 28+j1.9 Ω.

Microwave Engineering Questions and Answers – Dielectric Resonator Oscillators

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Dielectric Resonator Oscillators”.

1. The stability of an oscillator is enhanced with the use of:

a) high Q tuning network

b) passive elements

c) appropriate feedback methods

d) none of the mentioned

View Answer

Answer: a

Explanation: The stability of an oscillator is enhanced with the use of high Q tuning circuits. At microwave frequencies, lumped elements cannot provide high Q factor. Waveguide tuning circuits cannot be integrated with small microwave circuits easily. Hence dielectric resonators are used to provide high Q factor.

2. In oscillator tuning circuits, dielectric resonators are preferred over waveguide resonators because:

a) they have high Q factor

b) compact size

c) they are easily integrated with microwave integrated circuits

d) all of the mentioned

View Answer

Answer: d

Explanation: Dielectric resonators have all the above properties mentioned. Also, they are made of ceramic materials which also increases the temperature stability of resonator which further increases the fields of application.

3. A dielectric resonator coupled with an oscillator operates in:

a) TE10δ

b) TE01δ

c) TM10δ

d) TM01δ

View Answer

Answer: a

Explanation: A dielectric resonator is coupled to an oscillator by positioning it in close proximity to a microstrip line. The resonator operates in TE10δ mode, and couples to the fringing field magnetic field of the microstrip line.

4. A dielectric resonator is modeled as __________ when it is used as a tuning circuit with a oscillator.

a) series RLC circuit

b) parallel RLC circuit

c) LC circuit

d) tank circuit

View Answer

Answer: b

Explanation: The strength of coupling between a microstrip line and the resonator is determined by the spacing between them. The resonator appears as a series load on the microstrip line. The resonator is modeled a s a parallel RLC circuit and the coupling to the feed line is modeled by the turns ratio of the transformer.

5. The coupling factor between the resonator and the microstrip line is the ratio of external Q to the unloaded Q.

a) true

b) false

View Answer

Answer: b

Explanation: The coupling factor the resonator and the microstrip line is the ratio of unloaded Q to external Q. The simplified expression for coupling factor is N2R/2Z0.

6. If the reflection coefficient seen on the terminated microstrip line looking towards the resonator is 0.5, then the coupling coefficient is:

a) 0.5

b) 0.25

c) 0.234

d) 1

View Answer

Answer: d

Explanation: Coupling coefficient for a line in terms of reflection coefficient is Г/ (1-Г). Substituting the given reflection coefficient in this expression, the coupling coefficient is 1.

7. A dielectric resonator can be incorporated into a circuit to provide _________ using either parallel or series arrangement.

a) frequency stability

b) oscillations

c) high gain

d) optimized reflection coefficient

View Answer

Answer: a

Explanation: Oscillators can be obtained in various transistor configurations, their instability can be enhanced by using series or shunt elements. A dielectric resonator can be incorporated into a circuit to provide frequency stability using either parallel or series arrangement.

8. It is desired to design a frequency oscillator at 2.4 GHz and the reflection coefficient desired is 0.6, then the coupling coefficient between the feed line and the dielectric resonator is:

a) 1.5

b) 1

c) 0.5

d) none of the mentioned

View Answer

Answer: a

Explanation: Coupling coefficient for a line in terms of reflection coefficient is Г/ (1-Г). Substituting the given reflection coefficient in this expression, the coupling coefficient is 1.5.

9. If the reflection coefficient between the feed line and the resonator is -0.6, then the equivalent impedance of the resonator at resonance given that the characteristic impedance of the microstrip line is:

a) 50 Ω

b) 12.5 Ω

c) 25 Ω

d) none of the mentioned

View Answer

Answer: b

Explanation: The equivalent impedance of the resonator at resonance is Z0 (1 + Г)/(1- Г). Substituting the given values in this expression, equivalent impedance of the resonator is 12.5 Ω.

10. If the equivalent impedance of the resonator at resonance is 12.5 Ω and the characteristic impedance of the feed line is 50 Ω, then the coupling coefficient is:

a) 0.25

b) 0.5

c) 0.75

d) 1

View Answer

Answer: a

Explanation: Coupling coefficient is defined as the ratio of the equivalent impedance of the resonator to the characteristic impedance of the feed line. Substituting accordingly, coupling coefficient is 0.25.

Microwave Engineering Questions and Answers – Oscillator Phase Noise

This set of Microwave Engineering online quiz focuses on “Oscillator Phase Noise”.

1. A practical oscillator has a frequency spectrum consisting of a single delta function at its operating frequency.

a) True

b) False

View Answer

Answer: b

Explanation: An ideal oscillator has a frequency spectrum consisting of a single delta function at its operating frequency, but a practical oscillator has a spectrum in the form of a Gaussian curve consisting of some noise component.

2. ____________ due to random fluctuations caused by thermal and other noise sources appear as broad continuous distribution localized about the output signal.

a) Phase noise

b) White noise

c) Thermal noise

d) Shot noise

View Answer

Answer: a

Explanation: Phase noise is defined as the ratio of power in one phase modulation sideband to the total signal power per unit bandwidth at a particular offset fm. phase noise due to random fluctuations caused by thermal and other noise sources appear as broad continuous distribution localized about the output signal.

3. The phase variation for an oscillator or synthesizer is given by:

a) ∆f*sin ωmt/ fm

b) ∆f / fm

c) Sin ωmt/ fm

d) None of the mentioned

View Answer

Answer: a

Explanation: The phase variation for an oscillator or synthesizer is given by ∆f*sin ωmt/ fm. here, fm is the modulating signals frequency, ∆f is the change in the frequency.

4. The expression for phase noise in an oscillator is given by:

a) θrms2

b) θrms2/√2

c) θrms2/2

d) θrms2/ 3

View Answer

Answer: c

Explanation: The expression for phase noise in an oscillator is given by θrms2/2. θrms is the rms value of the phase deviation. Phase noise is directly proportional to the square of the RMS value of the phase deviation. Greater the deviation, higher is the phase noise.

5. Phase noise at the output of an oscillator is given by:

a) kBFGT0

b) kT0F/Pc

c) kT0F/Pc

d) None of the mentioned

View Answer

Answer: b

Explanation: Phase noise at the output of an oscillator is given by kT0F/Pc. here k is the Boltzmann’s constant. B is the operating bandwidth of the system, here the equation is considered for a bandwidth of 1 Hz as per the definition of phase noise. F is the figure of merit of system.

6. Noise power versus frequency for an amplifier has spikes at the operating frequency without the application of an input voltage.

a) True

b) False

View Answer

Answer: b

Explanation: Noise power versus frequency for an amplifier has spikes at the operating frequency with the application of an input voltage. When an input voltage is applied to the amplifier, noise component also is added. Along with the signal, noise is also amplified and peaks at the operating frequency.

7. An idealized power spectral density of amplifier has a straight line parallel to X axis and the noise is constant at all frequencies.

a) True

b) False

View Answer

Answer: b

Explanation: The curve has a negative slope up to a frequency called fα due to the thermal noise also called as 1/f noise. Above this frequency, the graph is a straight line parallel to X axis.

8. At higher frequencies of operation of an oscillator, induced noise is mostly:

a) Thermal noise

b) White noise

c) Shot noise

d) Flicker noise

View Answer

Answer: a

Explanation: At higher frequencies of operation of an oscillator, induced noise is mostly thermal, and constant with frequency. This noise is also proportional to the noise figure of the amplifier.

9. A GSM cellular telephone standard requires a minimum of 9 dB rejection of interfering signal levels of -23 dBm at 3 MHz from the carrier, -33 dBm at 1.6 MHz from the carrier, and -43 dBm at 0.6 MHz from the carrier, for a carrier level of -99 dBm. Determine the required local oscillator phase noise at 3 MHz carrier frequency offset.

a) -138 dBc/Hz

b) -128 dBc/Hz

c) -118 dBc/Hz

d) None of the mentioned

View Answer

Answer: a

Explanation: Phase noise is given by the expression C (dBm)-S (dB) –I (dBm)-10log (B). Substituting the given values in the above expression, the oscillator phase noise is -138.

10. The most affected parameter of a receiver by the phase noise is signal to noise ratio.

a) True

b) False

View Answer

Answer: b

Explanation: The effect of phase noise in a receiver is to degrade both the signal to noise ratio and the selectivity. Of these, the most severely affected is the selectivity. Phase noise degrades the receiver selectivity by causing down conversion of signals located near by the desired signal frequency.

Microwave Engineering Questions and Answers – Frequency Multipliers

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Frequency Multipliers”.

1. Oscillators operating at millimeter wavelength are difficult to realize and are also less efficient.

a) true

b) false

View Answer

Answer: a

Explanation: As frequency increases to the millimeter wave range, it becomes increasingly difficult to build fundamental frequency oscillators with good power, stability and noise characteristics. An alternative approach is to produce a harmonic of a low frequency oscillator through the use of frequency multiplier.

2. __________ is an example for a frequency multiplier.

a) resistor

b) inductor

c) capacitor

d) transistor

View Answer

Answer: d

Explanation: A non linear device has the ability to generate the harmonics of the input sinusoidal signal. Transistor and diodes are non linear devices and hence can be used as a frequency multiplier.

3. The major drawback of frequency multipliers is that they have:

a) higher attenuation

b) complex construction methods

c) complex design

d) none of the mentioned

View Answer

Answer: c

Explanation: Designing a good quality frequency multiplier is more difficult since it non-linear analysis, matching at multiple frequencies, stability analysis and thermal considerations. Considering all these issues for designing a multiplier makes it very complex.

4. A reactive diode multiplier uses _______ as the key electronic component for frequency multiplication.

a) zener diode

b) light emitting diode

c) varactor diode

d) Gunn diode

View Answer

Answer: c

Explanation: Reactive diode multipliers use either a varactor diode or step recovery diode biased to present a non linear junction capacitance. Since losses in these diodes are low, the fraction of RF power converted to the desired harmonic is relatively high.

5. A major disadvantage of frequency multipliers is that they multiply the noise factor along with frequency.

a) true

b) false

View Answer

Answer: a

Explanation: A disadvantage of frequency multipliers is that noise levels are also increased by the multiplication factor. Frequency multiplication process is a phase multiplication process as well, so phase noise variations get multiplied by the same factor as the frequency gets multiplied.

6. If a frequency multiplier has a multiplication factor of 10, then the increase in noise level due to frequency multiplication is:

a) 10 dB

b) 20 dB

c) 25 db

d) 15 dB

View Answer

Answer: b

Explanation: For a frequency multiplier, the increase in noise power is given by 20 log n, where n is the multiplication factor of the multiplier. Substituting in the below equation, increase in noise level is 20 dB.

7. In a diode frequency multiplier, an input signal of frequency fo applied to the diode is terminated with_________ at all frequencies other than required harmonic.

a) real impedances

b) reactive impedance

c) complex impedance

d) none of the mentioned

View Answer

Answer: b

Explanation: In a diode frequency multiplier, an input signal of frequency fo applied to the diode is terminated with reactive impedance at all frequencies other than required harmonic nfo. if the diode junction capacitance has a square –law I-V characteristic , it is necessary to terminate unwanted harmonics with short circuit.

8. Resistive multipliers are more efficient as compared to reactive multipliers.

a) true

b) false

View Answer

Answer: b

Explanation: Resistive multipliers generally use forward biased Schottky-barrier diodes to provide non linear characteristic. Resistive multipliers have low efficiency but have better bandwidth.

9. Reactive multipliers have a disadvantage that they cannot be used at very high frequencies and they become less efficient.

a) true

b) false

View Answer

Answer: a

Explanation: At millimeter frequencies, varactor diode exhibits resistive property. Hence, at high frequency the multiplier becomes lossy and also does not offer high bandwidth, which is a major disadvantage.

10. For a resistive frequency multiplier of multiplication factor 2, the maximum theoretical conversion efficiency is:

a) 50 %

b) 25 %

c) 75 %

d) 12.5 %

View Answer

Answer: b

Explanation: For a resistive frequency multiplier of multiplication factor 2, the maximum theoretical conversion efficiency is given by 1/m2 where m is the multiplication factor. For a factor 2 multiplier, maximum theoretical conversion efficiency is 25 %.

Microwave Engineering Questions and Answers – Transistor Multipliers

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Transistor Multipliers”.

1. Transistor multipliers are more efficient compared to diode multipliers from all operational aspects.

a) True

b) False

View Answer

Answer: a

Explanation: Transistor multipliers offer better bandwidth and the possibility of conversion efficiency greater than 100% (conversion gain). FET multipliers also require less input and DC power than diode multipliers.

2. A major characteristic property required by frequency multipliers for frequency multiplication to happen is:

a) High gain

b) High conversion efficiency

c) Non linearity

d) None of the mentioned

View Answer

Answer: c

Explanation: Non linearity property of devices like transistors and diodes is exploited in frequency multipliers. In transistor amplifiers, FET are used since several nonlinearities exist in FET can be used for harmonic generation.

3. If the input power for a frequency doubler is 10.7 mW and the output measured after the frequency doubling process is 21 mW, then the conversion gain for the frequency doubler is:

a) 4.5 dB

b) 8.4 dB

c) 9.8 dB

d) 2.9 dB

View Answer

Answer: d

Explanation: The conversion gain for a frequency doubler is given by the expression, P2/Pavail. Here P2 is the power measured at the output of the frequency doubler and Pavail is the power input. Substituting the given values in the equation, the conversion gain is 2.9 dB.

4. An ideal _______ produces an output consisting of the sum and difference frequencies of the two input signals.

a) Mixer

b) Amplifier

c) Product modulator

d) Phase modulator

View Answer

Answer: a

Explanation: A mixer is a three port device that uses a mixer or a time varying element to achieve frequency conversion. An ideal mixer produces an output consisting of the sum and difference frequencies of the two input signals.

5. A mixer consists of a non-linear device that produces various harmonics of the input frequency.

a) True

b) False

View Answer

Answer: a

Explanation: A non linear device can generate a wide variety of harmonics and other products of the input frequencies. A filter is used to extract only the desired frequency components.

6. A mixer can be used for both up conversion and down conversion at the transmitter and receiver respectively.

a) True

b) False

View Answer

Answer: a

Explanation: At the transmitter, mixer is used to convert the baseband signal to a broadband signal with use of a high frequency local oscillator. At the receiver, a mixer is used to convert the received broadband signal to baseband signal using a local oscillator. This is called down conversion.

7. A mixer having high conversion loss is said to have very high:

a) Gain

b) Loss

c) Bandwidth

d) None of the mentioned

View Answer

Answer: b

Explanation: Conversion loss for a mixer is defined as the ratio of available RF input power to the available IF output power. A higher value of conversion loss implies that large amount of power is lost in down converting the frequency from RF to IF range. This makes them less efficient.

8. The IS-54 digital cellular telephone system uses a receive frequency band of 869-894 MHz, with a first IF frequency range of 87 MHz, one possible range of local oscillator frequency is:

a) 956 to 981 MHz

b) 750 to 784 MHz

c) 869 to 894 MHz

d) None of the mentioned

View Answer

Answer: a

Explanation: The two possible local oscillator frequency range is given by fLO = fRF ± fIF.. fLO is the local oscillator frequency, fRF is the received frequency and fIF is the intermediate frequency range. Substituting the given values in the above equation, one possible frequency range is 956 to 981 MHz.

9. The curve of FET transconductance v/s gate-to-source voltage is a straight line through origin.

a) True

b) False

View Answer

Answer: b

Explanation: The gate to source voltage of an FET is slowly increased from negative voltage towards zero. As the voltage applied becomes more positive, the transconductance increases up to a certain level and then remains a constant. This property of a FET is used in single-ended FET mixer.

10. RF input matching and RF-LO isolation can be improved through the use of:

a) Balanced mixer

b) Single-ender diode mixer

c) Single ended FET mixer

d) Image reject mixer

View Answer

Answer: a

Explanation: RF input matching and RF-LO isolation can be improved through the use of balanced mixer. It consists of two single ended mixers combined with a hybrid junction.

Microwave Engineering Questions and Answers – System Aspects of Antennas

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “System Aspects of Antennas”.

1. If an antenna has a directivity of 16 and radiation efficiency of 0.9, then the gain of the antenna is:

a) 16.2

b) 14.8

c) 12.5

d) 19.3

View Answer

Answer: a

Explanation: Gain of an antenna is given by the product of radiation efficiency of the antenna and the directivity of the antenna. Product of directivity and efficiency thus gives the gain of the antenna to be 16.2.

2. Gain of an antenna is always greater than the directivity of the antenna.

a) True

b) False

View Answer

Answer: b

Explanation: Gain of an antenna is always smaller than the directivity of an antenna. Gain is given by the product of directivity and radiation efficiency. Radiation efficiency can never be greater than one. So gain is always less than or equal to directivity.

3. A rectangular horn antenna has an aperture area of 3λ × 2λ. Then the maximum directivity that can be achieved by this rectangular horn antenna is:

a) 24 dB

b) 4 dB

c) 19 dB

d) Insufficient data

View Answer

Answer: c

Explanation: Given the aperture dimensions of an antenna, the maximum directivity that can be achieved is 4π A/λ2, where A is the aperture area and λ is the operating wavelength. Substituting the given values in the above equation, the maximum directivity achieved is 19 dB.

4. A rectangular horn antenna has an aperture area of 3λ × 2λ. If the aperture efficiency of an antenna is 90%, then the directivity of the antenna is:

a) 19 dB

b) 17.1 dB

c) 13 dB

d) 21.1 dB

View Answer

Answer: b

Explanation: Given the aperture dimensions of an antenna, the directivity that can be achieved is ap4π A/λ2, where A is the aperture area and λ is the operating wavelength, ap is the aperture efficiency. Substituting the given values in the above equation, the directivity achieved is 17.1 dB.

5. If an antenna has a directivity of 16 and is operating at a wavelength of λ, then the maximum effective aperture efficiency is:

a) 1.27λ2

b) 2.56λ2

c) 0.87λ2

d) None of the mentioned

View Answer

Answer: a

Explanation: Maximum effective aperture efficiency of an antenna is given by D λ2/4π, D is the directivity of the antenna. Substituting in the equation the given values, the maximum effective aperture is 1.27λ2.

6. A resistor is operated at a temperature of 300 K, with a system bandwidth of 1 MHz then the noise power produced by the resistor is:

a) 3.13×10-23 watts

b) 4.14×10-15 watts

c) 6.14×10-15 watts

d) None of the mentioned

View Answer

Answer: b

Explanation: For a resistor noise power produced is given by kTB, where T is the system temperature and B is the bandwidth. Substituting in the above expression, the noise power produced is 4.14×10-15 watts.

7. With an increase in operating frequency, the background noise temperature:

a) Increases

b) Decreases

c) Remains constant

d) Remains unaffected

View Answer

Answer: a

Explanation: The plot of frequency v/s background noise temperature shows that with the increase of the signal frequency, the background noise temperature increases. Also, with the increase of the elevation angle from the horizon, background noise temperature increases.

8. The noise temperature of an antenna is given by the expression:

a) radTb + (1-rad) Tp

b) (1-rad) TP

c) radTb

d) None of the mentioned

View Answer

Answer: a

Explanation: The noise temperature of an antenna is given by the expression radTb + (1-rad) Tp. here, Tb is the brightness temperature and Tp is the physical temperature of the system. rad is the radiation efficiency. Noise temperature of a system depends on these factors.

9. Low is the G/T ratio of an antenna, higher is its efficiency.

a) True

b) False

View Answer

Answer: b

Explanation: In the G/T ratio of an antenna, G is the gain of an antenna and T is the antenna noise temperature. Higher the G/T ratio of an antenna better is the performance of the antenna.

10._________ has a constant power spectral density.

a) White noise

b) Gaussian noise

c) Thermal noise

d) Shot noise

View Answer

Answer: a

Explanation: Thermal noise has a power spectral density for a wide range of frequencies. Its plot of frequency v/s noise power is a straight line parallel to Y axis.

Microwave Engineering Questions and Answers – Antenna Gain and Efficiency

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Antenna Gain and Efficiency”.

1. A __________ is a device that converts a guided electromagnetic wave on a transmission line into a plane wave propagating in free space.

a) Transmitting antenna

b) Receiving antenna

c) Radar

d) Mixer

View Answer

Answer: a

Explanation: A transmitting antenna is a device that converts a guided electromagnetic wave on a transmission line into a plane wave propagating in free space. It appears as an electrical circuit on one side, provides an interface with a propagating plane wave.

2. Antennas are bidirectional devices.

a) True

b) False

View Answer

Answer: a

Explanation: Antennas can be used both as transmitters and receivers. As transmitters they radiate energy to free space and as receivers they receive signal from free space. Hence, they are called bidirectional devices as they are used at both transmitting end and receiving end.

3. Dipole antennas are an example for:

a) Wire antennas

b) Aperture antennas

c) Array antennas

d) None of the mentioned

View Answer

Answer: a

Explanation: Dipoles, monopoles, oops, Yagi-Uda arrays are all examples for wire antennas. These antennas have low gains, and are mostly used at lower frequencies.

4. _________ antennas consist of a regular arrangement of antenna elements with a feed network

a) Aperture antennas

b) Array antennas

c) Printed antennas

d) Wire antennas

View Answer

Answer: b

Explanation: Array antennas consist of a regular arrangement of antenna elements with a feed network. Pattern characteristics such as beam pointing angle and side lobe levels can be controlled by adjusting the amplitude and phase excitation of array elements.

5. A parabolic reflector used for reception with the direct broadcast system is 18 inches in diameter and operates at 12.4 GHz. The far-field distance for this antenna is:

a) 18 m

b) 13 m

c) 16.4 m

d) 17.3 m

View Answer

Answer: d

Explanation: Far field distance for a reflector antenna is given by 2D2/λ. D is the diameter and λ is the operating signal wavelength. Substituting in the above expression, far field distance is 17.3 m.

6._________ of an antenna is a plot of the magnitude of the far field strength versus position around the antenna.

a) Radiation pattern

b) Directivity

c) Beam width

d) None of the mentioned

View Answer

Answer: a

Explanation: Radiation pattern of an antenna is a plot of the magnitude of the far field strength versus position around the antenna. This plot gives the detail regarding the region where most of the energy of antenna is radiated, side lobes and beam width of an antenna.

7. Antennas having a constant pattern in the azimuthal plane are called _____________

a) High gain antenna

b) Omni directional antenna

c) Unidirectional antenna

d) Low gain antenna

View Answer

Answer: b

Explanation: Omni directional antennas radiate EM waves in all direction. If the radiation pattern for this type of antenna is plotted, the pattern is a constant signifying that the radiated power is constant measured at any point around the antenna.

8. Beamwidth and directivity are both measures of the focusing ability of an antenna.

a) True

b) False

View Answer

Answer: a

Explanation: Beamwidth and directivity are both measures of the focusing ability of an antenna. An antenna with a narrow main beam will have high directivity, while a pattern with low beam will have low directivity.

9. If the beam width of an antenna in two orthogonal planes are 300 and 600. Then the directivity of the antenna is:

a) 24

b) 18

c) 36

d) 12

View Answer

Answer: b

Explanation: Given the beam width of the antenna in 2 planes, the directivity is given by 32400/θ*∅, where θ,∅ are the beam widths in the two orthogonal planes. Substituting in the equation, directivity of the antenna is 18.

10. If the power input to an antenna is 100 mW and if the radiated power is measured to be 90 mW, then the efficiency of the antenna is:

a) 75 %

b) 80 %

c) 90 %

d) Insufficient data

View Answer

Answer: c

Explanation: Antenna efficiency is defined as the ratio of radiated power to the input power to the antenna. Substituting the given data in the efficiency equation, the efficiency of the antenna is 90%.

Microwave Engineering Questions and Answers – Wireless Communication

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Wireless Communication”.

1. Most of the wireless systems today operate at a frequency of about:

a) 800 MHz

b) 100 MHz

c) 80 MHz

d) None of the mentioned

View Answer

Answer: a

Explanation: With all advancement in wireless communication today, the need of the hour is higher data rates of transmission and reception. These higher data rates can be achieved only at microwave frequency range and in giga hertz frequency range.

2. Point to point communication systems use low gain antennas for communication.

a) True

b) False

View Answer

Answer: b

Explanation: In point to point communication a single transmitter communicates with a single receiver. Such systems use high gain antennas to maximize received power and minimize interference with other radios.

3. In this method of wireless communication, communication happens only in one direction:

a) Simplex

b) Duplex

c) Half duplex

d) None of the mentioned

View Answer

Answer: a

Explanation: In simplex systems, communication happens only in one direction that is from the transmitter to the receiver. Examples for this type of communication include radio, television and paging systems.

4. The power density radiated by an isotropic antenna is given by the relation:

a) Pt/4πR2

b) Pt/4R2

c) Pt/R2

d) None of the mentioned

View Answer

Answer: a

Explanation: An isotropic antenna radiates energy equally in all the directions. Hence, the power density radiated at a distance R is given by the relation Pt/4πR2.

5. The power received by a receiving antenna given that Pt is the transmitted power is:

a) GrGtλ2pt/ (4πR)2

b) Gtλ2pt/ (4πR)2

c) Grλ2pt/ (4πR)2

d) None of the mentioned

View Answer

Answer: a

Explanation: The power received by a receiving antenna given that Pt is the transmitted power is GrGtλ2pt/ (4πR)2. Here Gr is the gain of the receiving antenna; Gt is the gain of the transmitting antenna. R is the distance between the transmitting and receiving antenna.

6. If the distance between a transmitting station and receiving station is 1 Km and if the antennas are operating at a wavelength of 5 cm, then the path loss is:

a) 108 dB

b) 12 dB

c) 45 dB

d) 48 dB

View Answer

Answer: a

Explanation: Path loss is given by the expression 20 log (4πR/λ) in db. Substituting the given values in the above expression, the path loss is 108 dB.

7. The amount of power by which the received power must be greater than the threshold level required to maintain a minimum quality of service is called _______

a) Line loss

b) Link budget

c) Link margin

d) None of the mentioned

View Answer

Answer: c

Explanation: Link margin is the amount of power by which the received power must be greater than the threshold level required to maintain a minimum quality of service. Link margin signifies the minimum amount of power required to sustain communication maintaining a minimum quality of service.

8. Link margin that is used to account for fading effects is called fade margin.

a) True

b) False

View Answer

Answer: a

Explanation: Link margin that is used to account for fading effects is called fade margin. Satellite links operating at frequencies of above 10 GHz require a fading margin of about 20dB or more to account for attenuation during heavy rain.

9. One of the most important requirements of a radio receiver is high gain.

a) True

b) False

View Answer

Answer: a

Explanation: Radio receivers must have very high gain of about 100 dB in order to detect the very low power level of the received signal to a level near its original baseband value.

10. A radio receiver operating at microwave frequencies must have very high selectivity.

a) True

b) False

View Answer

Answer: a

Explanation: Today, most of the applications use wireless communication at microwave frequency. Hence space is a sea of EM waves. In order to receive only the desired signal in the desired range of frequencies, the radio receiver must have high sensitivity.

Microwave Engineering Questions and Answers – Digital Modulation and Bit Error Rate

This set of Microwave Engineering Question Bank focuses on “Digital Modulation and Bit Error Rate”.

1. All modern wireless communication systems rely on digital modulation methods due to:

a) superior performance

b) low power requirements

c) sustainability

d) all of the mentioned

View Answer

Answer: d

Explanation: Major advantages of digital modulation over analog modulation is that they have superior performance in the presence of noise and signal fading and the power requirement is very as compared to analog communication.

2. The probability of error depends on the ratio of bit energy to noise power density.

a) true

b) false

View Answer

Answer: a

Explanation: If the noise power is higher than the energy transmitted per bit, then the probability of error is high. Suppose signal 0 is transmitted, if the noise level is greater than the threshold to detect signal 1, bit 0 is interpreted as 1.

3. Probability of bit error is greater for ASK as compared to FSK.

a) true

b) false

View Answer

Answer: a

Explanation: Frequency shift keying method of modulation is more efficient than amplitude shift keying where the carrier is varied with respect to the amplitude of the message signal. Hence FSK has a lower bit error rate as compared to ASK.

4. Probability of bit error rate is greater for QPSK as compared to FSK.

a) true

b) false

View Answer

Answer: b

Explanation: In QPSK (Quadrature phase shift keying), two data bits are used to select one of the four possible phase states of the received data sequence. Hence, this has a low error probability as compared to FSK.

5. Global positioning system uses _____ satellites in medium earth orbits to provide accurate position information.

a) 12

b) 24

c) 36

d) 48

View Answer

Answer: b

Explanation: Global positioning system uses 24 satellites in medium earth orbits to provide accurate position information to users on land, air and sea. GPS has become one of the most pervasive applications of wireless technology to consumers throughout the world.

6. GPS operates at a single frequency band.

a) true

b) false

View Answer

Answer: b

Explanation: GPS operates at two frequency bands. L1 at 1575.42 MHz and L2 at 1227.60 MHz, transmitting spread spectrum signals with BPSK modulation.

7. WLAN is used for providing connection between a host computer and satellite for communication.

a) true

b) false

View Answer

Answer: b

Explanation: WLANs are used to provide connections between a computer and peripherals over a short distance. These types of networks are used in airports, college campuses and many more.

8. There is no standard to be formed by commercial WLAN products.

a) true

b) false

View Answer

Answer: b

Explanation: Most commercial WLANs follow IEEE 802.11 standards (Wi-Fi). They operate at either 2.4 or 5.7 GHz in the industrial, scientific and medical frequency bands and use either frequency hopping or direct sequence spread spectrum techniques.

9. Bluetooth devices operate at a frequency of :

a) 2.4 GHz

b) 4.5 GHz

c) 5.7 GHz

d) none of the mentioned

View Answer

Answer: a

Explanation: Bluetooth operates at a frequency range of 2.4 GHz and RF power range of 1-100 mW and corresponding operating ranges of 1-100 m.

10. The modulation technique in which both amplitude and phase of the carrier are varied simultaneously is:

a) ASK

b) BPSK

c) QAM

d) QPSK

View Answer

Answer: c

Explanation: In Quadrature amplitude modulation, both amplitude and phase of the carrier is varied with respect to the message signal. This is one of the most advanced modulation techniques which is used in advanced applications.

Microwave Engineering Questions and Answers – Noise Characteristics of Receivers

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Noise Characteristics of Receivers”.

1. The noise power will determine the maximum detectable signal level for a receiver.

a) True

b) False

View Answer

Answer: b

Explanation: The noise power will determine the minimum detectable signal level of the receiver for a given transmitter power, maximum range of a communication link. There is a limit on the maximum noise that can be associated with a signal in spite of which the signal can be recovered from the noise.

2. Equivalent noise temperature of a transmission line connecting the antenna to the receiver is:

a) TP (LP-1)

b) TP (LP + 1)

c) TP/ (LP-1)

d) TP / (LP + 1)

View Answer

Answer: a

Explanation: The transmission line connecting the antenna to the receiver has a loss of LT and is at a physical temperature TP. its noise equivalent temperature is given by TP (LP-1).

3. In a receiver, if the noise figure of the mixer stage in the receiver is 7 dB, then the equivalent noise temperature is given that the receiver is operating at 290 K:

a) 1163 K

b) 1789 K

c) 1000 K

d) 1234 K

View Answer

Answer: a

Explanation: Equivalent noise temperature for a given noise figure is given by To (FM-1). FM is the noise figure in dB. Substituting the given values for noise figure and temperature, noise equivalent temperature is 1163 K.

4. If a transmission line connecting the antennas to the receiver has a loss of 1.5 dB, given the physical temperature is 270C, noise equivalent temperature is:

a) 123 K

b) 145 K

c) 345 K

d) 234 K

View Answer

Answer: a

Explanation: The noise equivalent temperature of the transmission line is given by TP(LP-1). Converting the value from dB scale and substituting, noise equivalent temperature is 123 K.

5. Given that the antenna efficiency is 0.9, equivalent brightness temperature is 200 K; physical temperature is 300 K, noise temperature of an antenna is:

a) 220 K

b) 210 K

c) 240 K

d) None of the mentioned

View Answer

Answer: b

Explanation: Noise temperature of an antenna is given by rad Tb + (1- rad) TP. Tb is the equivalent brightness temperature and TP is the physical temperature. Substituting the given values, noise temperature of the antenna is 210 K.

6. If a receiver is operating at a bandwidth of 1 MHz and has antenna noise temperature of 210 K, then the input noise power is:

a) -90 dBm

b) -115 dBm

c) -56 dBm

d) -120 dBm

View Answer

Answer: b

Explanation: Input noise power is given the expression kBTa) Here k is the Boltzmann’s constant, B is the operational bandwidth of the antenna and TA is the antenna noise temperature. Substituting in the above expression, input noise power is -115 dBm.

7. Antenna noise temperature of a system is 210 K, noise temperature of transmission line is 123 K, loss of a transmission line connecting the antenna to receiver is 1.41 and noise temperature of the receiver cascade is 304 K. then the total system noise temperature is:

a) 840 K

b) 762 K

c) 678 K

d) 1236 K

View Answer

Answer: b

Explanation: The total system noise temperature is given by the expression TA+TTL+LTTREc) TA is the antenna noise temperature, TTL is the transmission line noise temperature, TREC is the noise temperature of receiver cascade. Substituting the given values, total system noise temperature is 762 K.

8. If the received power at antenna terminals is -80dBm, and if the input noise power is -115 dBm, then the input SNR is:

a) 45 dB

b) -195 dB

c) -35 dB

d) 35 dB

View Answer

Answer: d

Explanation: Input SNR of a system is (Si-Ni) in dB. Substituting the given signal power and noise power in dB, input SNR of the system is 35 dB.

9. A receiver system is operating at a bandwidth of 1 MHz and has a total system noise temperature of 762 K. then the output noise power is:

a) -110 dBm

b) -234 dBm

c) -145 dBm

d) -124 dBm

View Answer

Answer: a

Explanation: Output noise power of a receiver system is kBTsys. B is the operating bandwidth and Tsys is the total system noise temperature. Substituting the given values in the given equation, output noise power is -110 dBm.

10. If the received power at the antenna terminals is Si=-80 dBm and the output noise power is -110 dBm then the output signal to noise ratio is given by:

a) 30 dB

b) -30 dB

c) 35 dB

d) -35 dB

View Answer

Answer: a

Explanation: Output signal to noise ratio in dB is given by (So-No). Substituting the given values in the above equation, the output SNR is 30 dB.

Microwave Engineering Questions and Answers – Radar Systems

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Radar Systems”.

1. In a RADAR system the transmitter of the radar is more sensitive than the receiver.

a) True

b) False

View Answer

Answer: b

Explanation: The basic operation of RADAR is that the transmitter sends out a signal, which is partially reflected by the distant target, and then detected by a sensitive receiver. Because of the presence of noise in the received signal, the receiver has to be more sensitive.

2. For radar system, antennas with a large beam width are preferred over narrow beam antennas.

a) True

b) False

View Answer

Answer: b

Explanation: If a narrow beam width antenna is used in radar, the target’s direction can be accurately given by the angular position of the antenna. Hence narrow beam antennas give more accurate position of the objects.

3. The radar in which both transmission and reception is done using the same antenna are called:

a) Monostatic radar

b) Bistatic radar

c) Monopole radar

d) Dipole radar

View Answer

Answer: a

Explanation: Radar transmits electromagnetic waves and receives the waves that are reflected by objects. If a single antenna is used both for transmission and reception of the signals, they are called monostatic radar.

4. For applications like missile fire control, bistatic radars are used.

a) True

b) False

View Answer

Answer: a

Explanation: In missile fire control, the target is illuminated with one antenna and the reflected wave is received from another antenna in the radar. In situations where continuous transmission and reception of signals is required, bistatic radars are used.

5. When a power Pt is transmitted by an antenna, amount of energy incident on the target is given by the expression:

a) Pt×G/4πR2

b) Pt/4πR2

c) Pt×4 πR2/G

d) None of the mentioned

View Answer

Answer: a

Explanation: The amount of energy incident on the target is proportional to the energy radiated; gain of the antenna G, and R is the distance of the target from the radar system. As the distance from the radar system, the energy incident on the target reduces.

6. The term radar cross section defines the:

a) Scattering ability of the target

b) Power radiating ability of the radar

c) Amount of energy scattered by unwanted objects

d) Cross section of radar area through which energy is emitted

View Answer

Answer: a

Explanation: Radar cross section is defined as the ratio of scattered power in a given direction to the power incident on it. The power incident is the energy radiated by the transmitting antenna of the radar.

7. A ________ determines the target range by measuring the round trip time of a pulsed microwave signal.

a) Pulse radar

b) Doppler radar

c) Cross section radar

d) None of the mentioned

View Answer

Answer: a

Explanation: The working principle of pulse radar is that continuous pulses are transmitted and time is recorded until the pulse is received back by the radar. Based on this delay recorded, the range of target is estimated.

8. Construction of pulse radar is much simpler than a Doppler radar.

a) True

b) False

View Answer

Answer: b

Explanation: In Doppler radar the power / signal is continuously radiated by the transmitting antenna. In pulse radar, pulses are transmitted to the target. Generation and transmission of pulses is more complex as compared to continuous signal.

9. In military applications the radar cross sections of vehicles is minimized.

a) True

b) False

View Answer

Answer: a

Explanation: In military applications the radar cross sections of vehicles is minimized so that the military vehicles remain undetected. Lower the radar cross section, lower is the power scattered, and hence the object remains undetected.

10. Pulse radar operating at 10GHz frequency has an antenna with a gain of 28 dB and a transmitted power of 2kW. If it is desired to detect a target of cross section 12m2, and the minimum detectable signal is -90 dBm, the maximum range of the radar is:

a) 8114 m

b) 2348 m

c) 1256 m

d) 4563 m

View Answer

Answer: a

Explanation: The maximum range of a radar system is given the expression, [PtG2σλ22/ (4π) Pmin] 0.25. Pt is the transmitted power, σ is the radar cross section, G is the antenna gain. Substituting the given values in the above equation, the maximum range of the radar is 8114 m.

Microwave Engineering Questions and Answers – Radiometer Systems

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Radiometer Systems”.

1. __________ system obtains information about a target by transmitting a signal and receiving the echo from the target.

a) Radar

b) Sonar

c) Radiometer

d) None of the mentioned

View Answer

Answer: a

Explanation: Radar stands for radio detection and ranging. A radar system obtains information about a target by transmitting a signal and receiving the echo from the target. They are also called as active remote sensing systems.

2. Radiometry is a technique or the principle on which radar works.

a) True

b) False

View Answer

Answer: b

Explanation: Radiometry is a passive technique which develops information about a target solely from the microwave portion of the blackbody radiation that it either emits directly or reflects from the surrounding bodies.

3. According to Planck’s radiation law, a body can radiate energy to the medium surrounding it under all conditions invariably.

a) True

b) False

View Answer

Answer: b

Explanation: A body in thermal equilibrium only can radiate energy according to Planck’s radiation law. The radiating body has to be maintained in thermal equilibrium. In the microwave region this reduces to P=kTB, where k is the Boltzmann’s constant.

4. A major challenge in designing a radiometer is:

a) Design complexity

b) High cost

c) Requirement of highly sensitive receivers

d) None of the mentioned

View Answer

Answer: c

Explanation: The basic problem with building a radiometer is to build a receiver that can distinguish between the desired external radiometric noise and the inherent noise of the receiver. This application thus requires that are highly efficient in detecting the required signal.

5. The receiver model of a total power radiometer is based on the:

a) AM receiver

b) FM receiver

c) Super heterodyne receiver

d) None of the mentioned

View Answer

Answer: c

Explanation: The front end of a receiver is a standard super heterodyne circuit consisting of an RF amplifier, a mixer/ local oscillator, and an IF stage. These stages are the same as that used in a super heterodyne receiver.

6. The system bandwidth of a total power radiometer is determined by the:

a) RF amplifier

b) Local oscillator

c) IF filter

d) IF amplifier

View Answer

Answer: c

Explanation: The system bandwidth of a total power radiometer is determined by the IF filter section present in the receiver circuit of the radiometer. The upper and lower cutoff frequency of the IF filter specify the system bandwidth.

7. The integrator circuit after the detector in the receiver circuit is used to smooth out the short term variations in the signal power.

a) True

b) False

View Answer

Answer: b

Explanation: The integrator used is essentially a low pass filter with a fixed cutoff frequency, and serves to smooth out short term variations in the noise power.

8. The dominant factor affecting the accuracy of the total power radiometer is the variation of ___________

a) Gain in the overall system

b) The feedback circuit

c) Efficiency of the system

d) None of the mentioned

View Answer

Answer: a

Explanation: The dominant factor affecting the accuracy of the total power radiometer is the variation of the gain of the overall system. Since such variations have relatively a longer time constant, it is possible to eliminate this error by repeatedly calibrating the device.

9. A Dicke radiometer is identical to the total power radiometer in all aspects except that they have different receiving antenna.

a) True

b) False

View Answer

Answer: b

Explanation: In a Dicke radiometer, the input is periodically switched between the antenna and a variable power noise source. This switch is called Dicke switch. Repeatedly calibrating the device is the principle behind the operation of Dicke radiometer.

10. A typical radiometer would measure the brightness temperature over the range of about:

a) 50-300 K

b) 100-200 K

c) 400-500 K

d) None of the mentioned

View Answer

Answer: a

Explanation: Typical radiometer would measure the brightness temperature over the range of about 50-300K. This then implies that the reference noise source would have to cover the same range, which is difficult to achieve practically.

Microwave Engineering Questions and Answers – Antenna Basics

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Antenna Basics”.

1. The basic requirements of transmitting antennas are:

a) High efficiency

b) Low side lobes

c) Large signal to noise ratio

d) Lone of the mentioned

View Answer

Answer: a

Explanation: The basic requirements of a transmitting antenna are high gain and efficiency while requirements of receiving antennas are low side lobes and large signal to noise to ratio.

2. _________ is a device that converts electrons to photons or vice-versa.

a) Antenna

b) Electron gun

c) Photon amplifier

d) Microwave tube

View Answer

Answer: a

Explanation: Antenna is a device that converts electrons into photons or vice versa. A transmitting antenna converts electrons into photons while a receiving antenna converts photons into electrons.

3. The basic equation of radiation that is applied to any antenna irrespective of the type of the antenna is:

a) iL= Qv

b) iQ = Lv

c) i/L=Q/v

d) None of the mentioned

View Answer

Answer: a

Explanation: Basic equation of radiation is given by iL=Qv. i is the time change in current, l is the length of the current element, q is the charge v is the acceleration of the charge.

4. When the separation between two lines that carry the TEM wave approaches λ the wave tends to be radiated.

a) True

b) False

View Answer

Answer: a

Explanation: When the separation between two lines that chary the TEM wave approaches λ the wave tends to be radiated so that the opened – out line act as an antenna which lunches a free space wave.

5. The number of patterns radiation pattern required to specify the characteristic are :

a) Three

b) Four

c) Two

d) Five

View Answer

Answer: a

Explanation: The three patterns required are, θ component of the electric field as the function of the angles as θ and φ, the φ component of the electric field as the function of the angles θ and φ, the phase of these fields as a functions of the angle φ and θ .

6. The beam width of the antenna pattern measured at half power points is called:

a) Half power beam width

b) Full null beam width

c) Beam width

d) None of the mentioned

View Answer

Answer: a

Explanation: The beam width of an antenna measure at half of the maximum power received by an antenna or the 3 dB beam width of the antenna is termed as half null beam width.

7. An antenna has a field pattern of E (θ) = cos2 θ, θ varies between 0 and 900. Half power beam width of the antenna is:

a) 330

b) 660

c) 12000

d) None of the mentioned

View Answer

Answer: b

Explanation: Half power beam width of the antenna is obtained by equating the field pattern of the antenna to 0.707 (half power point) and finding θ. 2θ gives the value of beam width. Solving the given problem in the same flow, half power beam width of the antenna is 660.

8. An antenna has a field pattern E (θ) =cos θ. cos 2θ. The first null beam width of the antenna is:

a) 450

b) 900

c) 1800

d) 1200

View Answer

Answer: b

Explanation: Half power beam width of the antenna is obtained by equating the field pattern of the antenna to 0.707 (half power point) and finding θ. 2θ gives the value of beam width. Twice the half power beam width gives the first null beam width. With the same steps applied, the half power beam width of the antenna is 450. First null beam width is 900.

9. The solid area through which all the power radiated by the antenna is:

a) Beam area

b) Effective area

c) Aperture area

d) Beam efficiency

View Answer

Answer: a

Explanation: The beam area is the solid angle through which all of the power radiated by the antenna would stream if P (θ, φ) maintained its maximum value over beam area and zero elsewhere. This value is approximately equal to the angles subtended by the half power points of the main lobe in the two principal planes.

10. Power radiated from an antenna per unit solid angle is called radiation intensity.

a) True

b) False

View Answer

Answer: a

Explanation: Power radiated from an antenna per unit solid angle is called radiation intensity. Unit of radiation intensity is watts per steridian or per square degree.

Microwave Engineering Questions and Answers – Antenna Basics-2

This set of Microwave Engineering Questions and Answers for Entrance exams focuses on “Antenna Basics-2”.

1. As the beam area of an antenna decreases, the directivity of the antenna:

a) Increases

b) Decreases

c) Remains unchanged

d) Depends on the type of the antenna

View Answer

Answer: a

Explanation: Beam area of an antenna and the directivity of the antenna are inversely proportional. As the beam area is reduced, the directivity increases, meaning smaller the radiating area of the transmitting antenna, more directed is the emitted energy.

2. If an antenna radiates over half a sphere, directivity of the antenna is:

a) Two

b) Four

c) Three

d) One

View Answer

Answer: a

Explanation: Since the antenna radiates over half the sphere, beam area of the antenna is 2π, directivity of the antenna is given by 4π/ beam area. Substituting for beam area, the directivity of the antenna is two.

3. The half power beam width of an antenna in both θ and φ are 400 each. Then the gain of the antenna is:

a) 23

b) 25

c) 14

d) 27

View Answer

Answer: b

Explanation: Approximate gain of an antenna is given by the expression 40000/ (HPBW) 2. Substituting the given values in the above expression, the gain of the antenna is 25. In dB scale the gain of the antenna is 14 dB.

4. The number N of radio transmitters or point sources of radiation distributed uniformly over the sky which an antenna can resolve is given by:

a) 4π/ ΩA

b) 2π/ ΩA

c) π/ ΩA

d) None of the mentioned

View Answer

Answer: a

Explanation: Resolution may be defined as equal to half the beam width between first nulls. In the above expression the resolution N is given as 4π/ ΩA.. Here ΩA is the beam area.

5. Ideally, the number of point sources an antenna can resolve is numerically equal to:

a) Gain of the antenna

b) Directivity

c) Beam efficiency

d) Beam area

View Answer

Answer: b

Explanation: The number of point source an antenna can resolve is given by 4π/ ΩA Directivity of an antenna is mathematically given by the relation 4π/ ΩA . Numerically resolution and directivity are equal.

6. Effective aperture is a parameter of the antenna that gives the physical aperture of the antenna.

a) True

b) False

View Answer

Answer: b

Explanation: Effective aperture defines the amount of the total aperture of the antenna that is utilized for radiation of energy. Higher the effective aperture of an antenna, more is the aperture efficiency.

7. Effective aperture in terms of beam area and operating wavelength is given by the relation:

a) λ2/ ΩA

b) ΩA / λ2

c) λ2× ΩA

d) No such relationship exists

View Answer

Answer: a

Explanation: Effective aperture is given as λ2/ ΩA. Here ΩA is the beam area. If the beam area is specified in terms of the operating wavelength λ, then effective are of the antenna can be made operating wavelength independent.

8.________ of an antenna is defined as the ratio of the induced voltage to the incident electric field.

a) Effective height

b) Gain

c) Directivity

d) Loss

View Answer

Answer: a

Explanation: At the receiving end, effective height of an antenna is defined as the ratio of the induced voltage to the incident electric field. Otherwise, if the effective length of the receiving antenna is known and if the induced voltage is measured, then the field strength can be determined.

9. The directivity of an antenna in terms of the aperture efficiency and operating wavelength is given by:

a) 4πAe/λ2

b) 2πAe/λ2

c) πAe/λ2

d) None of the mentioned

View Answer

Answer: a

Explanation: The directivity of an antenna in terms of aperture efficiency is given by 4πAe/λ2. Here Ae is the aperture efficiency. λ is the operating frequency. With an increase in the effective aperture area of an antenna, directivity of the antenna can be increased making the radiated beam narrower.

10. A radio link has 15 W transmitter connected to an antenna of 2.5 m2 effective aperture at 5 GHz. The receiving antenna has an effective aperture of 0.5 m2 and is located at a 15 Km line of sight distance from transmitting antenna. Assuming lossless, matched antennas, the power delivered to the receiver is:

a) 20 µW

b) 15 µm

c) 23 µm

d) 25 µm

View Answer

Answer: c

Explanation: The power delivered to the receiving antenna is Pt (A1 A2/ r2λ2). Substituting the given values in the above equation, the power at the receiver is 23 µm.

Microwave Engineering Questions and Answers – Antenna Family

This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Antenna Family”.

1. The members of the antenna family which are made of wires of certain value in terms of operating wavelength are called:

a) Loop antennas

b) Wire antennas

c) Dipole antenna

d) Slot antennas

View Answer

Answer: c

Explanation: Wires of half wavelength are termed as dipoles. Their radiation resistance is about 73 Ω. If only half of this length is used, then it is called quarter-wave monopole with a radiation resistance of 36.5 Ω.

2. The antenna in which location of the feed determines the direction of the lobe are:

a) Wire antenna

b) Loop antenna

c) Helical antenna

d) Horn antenna

View Answer

Answer: a

Explanation: In a wire antenna, the location of the feed determines the direction of the lobe and the orientation of the wire determines the polarization. These wires can be thick or thin. Thickness of the wire determines the radiation resistance of the antenna.

3. Based on the size of the loops, loop antennas are classified as small and large loops. This is the only classification of loop antenna.

a) True

b) False

View Answer

Answer: b

Explanation: Loop antennas are classified based on various antenna parameters. To name a few, small and large loops, circular and square loops, loops having single or multi turns, loops with turns wound using a single wire or multiple wires.

4. Antenna that does not belong to the horn antenna family among the following are:

a) Pyramidal horn

b) Conical horn

c) bi-conical horn

d) None of the mentioned

View Answer

Answer: d

Explanation: All of the above mentioned antennas belong to the horn antenna family. Horn antennas may be made of pointed or rounded waveguides. The waveguides may contain disc at an end or some dielectric.

5. Patch antennas are the antennas of small size and are made of:

a) Strip line

b) Microstrip lines

c) Coaxial cables

d) Rectangular waveguide

View Answer

Answer: b

Explanation: Patch antennas are microstrip antennas that can be of any shape. Patch antennas can be aperture-coupled fed or proximity fed. For obtaining circular polarization, a patch may also be doubly fed.

6. Reflector antennas are widely used to modify radiation patterns of radiating elements.

a) True

b) False

View Answer

Answer: a

Explanation: Reflector antennas are used to modify radiation patterns of radiating elements. Reflector antennas are classified into two categories. They are passive reflectors and active reflectors. Based on the type of the radiating element and the modification in the radiation pattern required, accordingly either active or passive reflectors are chosen.

7. The pattern of the reflector in a reflector antenna is called:

a) Primary pattern

b) Secondary pattern

c) Reflector pattern

d) None of the mentioned

View Answer

Answer: b

Explanation: In a reflector antenna, the feed pattern is called primary pattern and the pattern of the reflector is called secondary pattern. These antennas are widely employed in RADARs and other types of point to point communication links.

8. ______ antennas have gain less than reflector antennas but have more lenient tolerance on surfaces.

a) Helical antennas

b) Lens antennas

c) Array antennas

d) Slot antennas

View Answer

Answer: b

Explanation: Lens antennas are complex in nature but are able to scale wider angles. In comparison to reflectors, their gain is 1 or 2 dB less, but these have more lenient tolerance on surfaces. These have less rearward reflection, relatively low loss and can be easily shaped to the desired contours.

9. Lens antennas are classified into two types. One being fast antenna, the other one is:

a) Slow antenna

b) Delay antenna

c) Dynamic antenna

d) None of the mentioned

View Answer

Answer: b

Explanation: In delay lenses, the electrical path length is increased or the wave is retarded by the lens medium. Dielectric lenses and H-plane metal lenses fall in this category.

10. The antennas which offer high operational bandwidth and the antenna parameters are maintained over a wide range of antennas are called:

a) Wide band antennas

b) Array antennas

c) Parabolic antennas

d) None of the mentioned

View Answer

Answer: a

Explanation: In this class of antennas, constancy of impedance and radiation characteristics is maintained over a wide range of frequencies. To be wide band or frequency independent, antennas should expand or contract in proportion to the wavelength.

11. High directivity required in RADAR communication is satisfied using this type of antennas:

a) Wide band antennas

b) Antenna arrays

c) Slot antennas

d) Patch antennas

View Answer

Answer: b

Explanation: Higher directivity is the requirement in point to point communication. This can be achieved by increasing the size of the antennas in terms of electrical length. When much high directivity is required, antenna arrays are used.

12. The terminal impedance of a dipole antenna is 710 Ω. The terminal impedance of the slot antenna given the intrinsic impedance of air is 377 Ω is:

a) 100 Ω

b) 50 Ω

c) 25 Ω

d) None of the mentioned

View Answer

Answer: b

Explanation: The terminal impedance ZS of the slot is given by the relation Z02/ 4Zd) Zₒ is the intrinsic impedance of the medium and ZD is the terminal impedance of the dipole. Substituting the given values in the above equation, the terminal impedance of sot is 50 Ω.

13. If the length of aperture in a pyramidal horn antenna is 10cm and δ for the design is 0.25. Then, the flaring angle of the pyramidal horn is:

a) 30⁰

b) 25.4⁰

c) 45⁰

d) 60⁰

View Answer

Answer: b

Explanation: The flaring angle of pyramidal horn is given by 2cos-1(L/L+δ). Substituting the values of L and δ, flaring angle is 25.4⁰.

14. If the directivity of a square corner receiving antenna is 20 and operating at a wavelength of 0.25m, the effective aperture of a square corner antenna is:

a) 0.4 m2

b) 0.2 m2

c) 0.1 m2

d) None of the mentioned

View Answer

Answer: a

Explanation: Given the directivity of the antenna, effective aperture of the antenna is given by Dλ2/4π. substituting the given values of the variables; the effective aperture of the antenna is 0.4 m2.

Microwave Engineering Questions and Answers – Antenna Radiation

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This set of Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Antenna Radiation”.

1. An antenna source that radiates energy uniformly in all the directions is called:

a) Isotropic source

b) Anisotropic source

c) Point source

d) None of the mentioned

View Answer

Answer: a

Explanation: Isotropic source radiates energy in all the direction uniformly. For such a source, the radial component Sr of the pointing vector is independent of θ and φ. The three dimensional power pattern of n isotropic source is a sphere.

2. Antennas that radiate energy only in a specified are called anisotropic antennas.

a) True

b) False

View Answer

Answer: a

Explanation: All physically realizable, simplest antennas also have directional properties. That is, they radiate energy in one direction than in any other direction. Such sources are called anisotropic point sources.

3. The expression for pointing vector of an isotropic point source at a distance ‘r’ from the source is given by:

a) P/ 4πR2

b) P/4π

c) P/ 4πR

d) P×4πR2

View Answer

Answer: a

Explanation: The pointing field vector for an isotropic source is given by the expression P/ 4πR2.P is the total power radiated y the source. As the distance of the point from the source increases, the magnitude of pointing vector decreases.

4. A source has a cosine radiation-intensity pattern given by U=UM cos (θ). The directivity of this source is:

a) 2

b) 4

c) 6

d) 8

View Answer

Answer: b

Explanation: To find the directivity of the given source, the power radiated by the given source is found out by the method of integration. Taking the ratio of the power radiated by the given source to the power radiated by an isotropic source gives the directivity. Following the above steps, the directivity of the given source is 4.

5. A source has a cosine power pattern that is bidirectional. Given that the directivity of a unidirectional source with cosine power pattern has a directivity of 4, then the directivity of the unidirectional source is:

a) 1

b) 2

c) 4

d) 8

View Answer

Answer: b

Explanation: Given the directivity of unidirectional power pattern, the directivity of bidirectional power pattern is half of it. Hence the directivity of the source is 2.

6. A source has a radiation intensity pattern given by U=UM sin θ. The directivity of the source with this power pattern is:

a) 1

b) 1.27

c) 2.4

d) 3.4

View Answer

Answer: b

7. A source has a sine squared radiation intensity power pattern. The directivity of the given source is:

a) 1.5

b) 3

c) 2.5

d) 3.5

View Answer

Answer: a

8. A source with a unidirectional cosine squared radiation intensity pattern is given by UMcos2 (θ). The directivity of the given source is:

a) 6

b) 8

c) 2

d) 7

View Answer

Answer: a

Explanation: To find the directivity of the given source, the power radiated by the given source is found out by the method of integration. Taking the ratio of the power radiated by the given source to the power radiated by anisotropic source gives the directivity. Following the above steps, the directivity of the given source is 6.

9. Considering distance as a parameter, two types of field zones can be defined around an antenna.

a) True

b) False

View Answer

Answer: a

Explanation: Considering distance as a parameter, two types of field zones can be defined around an antenna) .The field near the antenna is called near field or Fresnel region and the other region is the far field that is also called as Fraunhofer region.

10. If the field strength at receiving antenna is 1 µV/m, and the effective aperture area is 0.4 m2 and the intrinsic impedance of the medium is 377 Ω, then the power received by the antenna is:

a) 1.06 pW

b) 1.06 fW

c) 2 µW

d) None of the mentioned

View Answer

Answer: b

Explanation: The received power by the antenna is given by E2Ae/Zₒ. Substituting the known values in the above equation, the power received is 1.06×10-15 watts.

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