Signals & Systems Pune University MCQs
Signals & Systems Pune University MCQs
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “System Classification and Properties – 1”.
1. The type of systems which are characterized by input and the output quantized at certain levels are called as
a) analog
b) discrete
c) continuous
d) digital
Answer: b
Explanation: Discrete systems have their input and output values restricted to enter some quantised/discretized levels.
2. The type of systems which are characterized by input and the output capable of taking any value in a particular set of values are called as
a) analog
b) discrete
c) digital
d) continuous
Answer: d
Explanation: Continuous systems have a restriction on the basis of the upper bound and lower bound, but within this set, the input and output can assume any value. Thus, there are infinite values attainable in this system
3. An example of a discrete set of information/system is
a) the trajectory of the Sun
b) data on a CD
c) universe time scale
d) movement of water through a pipe
Answer: b
Explanation: The rest of the parameters are continuous in nature. Data is stored in the form of discretized bits on CDs.
4. A system which is linear is said to obey the rules of
a) scaling
b) additivity
c) both scaling and additivity
d) homogeneity
Answer: c
Explanation: A system is said to be additive and scalable in order to be classified as a linear system.
5. A time invariant system is a system whose output
a) increases with a delay in input
b) decreases with a delay in input
c) remains same with a delay in input
d) vanishes with a delay in input
Answer: c
Explanation: A time invariant system’s output should be directly related to the time of the output. There should be no scaling, i.e. y = f).
6. Should real time instruments like oscilloscopes be time invariant?
a) Yes
b) Sometimes
c) Never
d) They have no relation with time variance
Answer: a
Explanation: Oscilloscopes should be time invariant, i.e they should work the same way everyday, and the output should not change with the time at which it is operated.
7. All real time systems concerned with the concept of causality are
a) non causal
b) causal
c) neither causal nor non causal
d) memoryless
Answer: b
Explanation: All real time systems are causal, since they cannot have perception of the future, and only depend on their memory.
8. A system is said to be defined as non causal, when
a) the output at the present depends on the input at an earlier time
b) the output at the present does not depend on the factor of time at all
c) the output at the present depends on the input at the current time
d) the output at the present depends on the input at a time instant in the future
Answer: d
Explanation: A non causal system’s output is said to depend on the input at a time in the future.
9. When we take up design of systems, ideally how do we define the stability of a system?
a) A system is stable, if a bounded input gives a bounded output, for some values of the input
b) A system is unstable, if a bounded input gives a bounded output, for all values of the input
c) A system is stable, if a bounded input gives a bounded output, for all values of the input
d) A system is unstable, if a bounded input gives a bounded output, for some values of the input
Answer: c
Explanation: For designing a system, it should be kept in mind that the system does not blow out for a finite input. Thus, every finite input should give a finite output.
10. All causal systems must have the component of
a) memory
b) time invariance
c) stability
d) linearity
Answer: a
Explanation: Causal systems depend on the functional value at an earlier time, compelling the system to possess memory.
This set of Signals & Systems Interview Questions and Answers focuses on “System Classification and Properties – 2”.
1. Amplifiers, motors, filters etc. are examples for which type of system?
a) Distributed parameter systems
b) Unstable systems
c) Discrete time systems
d) Continuous time systems
Answer: d
Explanation: Amplifiers, motors, filters etc. are examples of continuous time systems as these systems operate on a continuous time input signal and produce a continuous time output signal. Whereas discrete time systems operate on discrete time signals, distributed parameter systems have signals which are functions of space as well as time and unstable systems produce unbounded output from bounded or unbounded input.
2. Which among the following systems are described by partial differential functions?
a) Causal Systems and Dynamic systems
b) Distributed parameter systems and linear systems
c) Distributed parameter systems and Dynamic systems
d) Causal systems and linear systems
Answer: c
Explanation: In distributed parameter systems, signals are functions of space as well as time. In dynamic systems the output depends on past, present and future values of input, hence, both of these systems are described by differential functions.
3. Which one of the following systems is causal?
a) y=x+x+x(t 2 )
b) y=x
c) y=x+x
d) y=x(2n 2 )
Answer: c
Explanation: A causal system is one in which the output depends on the present or past values of the input, not future. If it depends on future values then it is non-causal. For y=x+x+x(t 2 ), y=x, and y=x(2n 2 ), the output depends on future values i.e., x (t 2 ), x and x (2n 2 ) respectively. Whereas in y=x+x, the output y depends on past values only i.e., x and x.
4. Which among the following is not a linear system?
signals-systems-interview-questions-answers-q4
Answer: a
Explanation: Here is the Explanation.
signals-systems-interview-questions-answers-q4exp
5. signals-systems-questions-answers-system-classification-properties-q5
a) Static, linear, causal and time variant
b) Dynamic, non – linear, causal and time invariant
c) Static, non – linear, causal and time variant
d) Dynamic, non – linear, causal and time variant
Answer: b
Explanation: Here is the Explanation.
signals-systems-interview-questions-answers-q5exp
6. Which one of the following is an example of a bounded signal?
a) e t coswt
b) e t sinw
c) e -t coswt
d) e t cosw
Answer: c
Explanation: A bounded signal is the one which satisfies the condition |x|< M < ∞ for all t.
Clearly, the signals e t coswt, e t sinw and e t cosw are exponentially growing signals as the power of the function is positive i.e., the signals will grow beyond infinity. Whereas the signal e -t coswt is an exponentially decaying signal, hence it will decay to zero and will always be less than infinity. Therefore, it is bounded.
7. A system produces zero output for one input and same gives the same output for several other inputs. What is the system called?
a) Non – invertible System
b) Invertible system
c) Non – causal system
d) Causal system
Answer: a
Explanation: A system is said to be invertible if the input fed to the system can be retrieved from the output of the system. Otherwise the system is non-invertible. Also, if a system gives zero output for any input and gives the same output for many inputs, then the system is non-invertible.
8. Which among the following is a LTI system?
a) dy/dt+ty=x
b) y=xcosπt
c) y=x+nx
d) y=x 3
Answer: d
Explanation: A system is said to be linear time invariant if the input-output characteristics do not change with time.
This expression has a coefficient which is a function of time. ∴ the system is time variant.
Output when input is delayed by T, y=xcosπt
If the output is delayed by T, y=xcosπ
Clearly, both expressions are not equal ∴ The system is time variant.
Output when input is delayed by N, y=x+nx
If the output is delayed by N, y=x+x
Clearly, both expressions are not equal ∴ The system is time variant.
Output when input is delayed by N, y=x 3
If the output is delayed by N, y= x 3
Clearly, both expressions are equal. ∴ The system is time invariant.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Signal Classification and Properties – 1”.
1. Which of the following signals are monotonic in nature?
a) 1-exp
b) 1-exp)
c) log)
d) cos
Answer: a
Explanation: All of the other functions have a periodic element in them, which means the function attains the same value after a period of time, which should not occur for a monotonic function.
2. What is the period of the following signal, x = sin?
a) 1 ⁄ 9
b) 2 ⁄ 9
c) 1 ⁄ 3
d) 4 ⁄ 9
Answer: b
Explanation: The signal can be expressed as sin, where the time period = 2*pi/w.
3. Which of the following signals is monotonic?
a) x = t 3 – 2t
b) x = sin
c) x = sin 2 2 + cos 2 2 – 2t
d) x = log)
Answer: c
Explanation: c) reduces to 1 – 2t, which is a strictly decreasing function.
4. For the signal, x = log) for a = 50 Hz, what is the time period of the signal, if periodic?
a) 0.16s
b) 0.08s
c) 0.12s
d) 0.04s
Answer: d
Explanation: Time period = 2*pi/pi = 1/25 = 0.04s
5. What are the steady state values of the signals, 1-exp, and 1-k*exp?
a) 1, k
b) 1, 1/k
c) k, k
d) 1, 1
Answer: d
Explanation: Consider limit at t tending to infinity, we obtain 1 for both cases.
6. For a bounded function, is the integral of the function from -infinity to +infinity defined and finite?
a) Yes
b) Never
c) Not always
d) None of the mentioned
Answer: c
Explanation: If the bounded function, is say y = 2, then the integral ceases to hold. Similarly, if it is just the block square function, it is finite. Hence, it depends upon the spread of the signal on either side. If the spread is finite, the integral will be finite.
7. For the signal x = a – b*exp, what is the steady state value, and the initial value?
a) c, b
b) c, c-a
c) a, a-b
d) b, a-b
Answer: c
Explanation: Put the limits as t tends to infinity and as t tends to zero.
8. For a double sided function, which is odd, what will be the integral of the function from -infinity to +infinity equal to?
a) Non-zero Finite
b) Zero
c) Infinite
d) None of the mentioned
Answer: b
Explanation: For an odd function, f = -f, thus the integrals will cancel each other, giving zero.
9. Find where the signal x = 1/(t 2 – 3t + 2) finds its maximum value between :
a) 1.40
b) 1.45
c) 1.55
d) 1.50
Answer: d
Explanation: Differentiate the function for an optima, put it to zero, we will obtain t = 1.5 as the required instant.
10. Is the signal x = exp*sin periodic in nature?
a) Yes
b) No
Answer: b
Explanation: Though sin is a periodic function, exp is not a periodic function, thus leading to non-periodicity.
This set of Signals & Systems Questions and Answers for Freshers focuses on “Signal Classification and Properties – 2”.
1. A signal is a physical quantity which does not vary with ____________
a) Time
b) Space
c) Independent Variables
d) Dependent Variables
Answer: d
Explanation: A signal is a physical quantity which varies with time, space or any other independent variables. Therefore, it does not vary with dependent variables.
2. Most of the signals found in nature are _________
a) Continuous-time and discrete-time
b) Continuous-time and digital
c) Digital and Analog
d) Analog and Continuous-time
Answer: d
Explanation: Signals naturally are continuous-time signals. These are also known as analog signals. Continuous-time or analog signals are defined for all values of time t.
3. Which one of the following is not a characteristic of a deterministic signal?
a) Exhibits no uncertainty
b) Instantaneous value can be accurately predicted
c) Exhibits uncertainty
d) Can be represented by a mathematical equation
Answer: c
Explanation: Deterministic signal is one which exhibits no uncertainty and its instantaneous value can be accurately predicted from its mathematical equation. Therefore, a deterministic signal doesn’t exhibit uncertainty. However, a random is always uncertain.
4. Determine the fundamental period of the following signal:sin60t.
a) 1/60 sec
b) 1/30 sec
c) 1/20 sec
d) 1/10 sec
Answer: b
Explanation: Consider the equation: sinΩ 0 t. Comparing this equation with the one given in the question: sin60t
⇒ Ω 0 =60π
signals-systems-questions-answers-freshers-q4
5. Sum of two periodic signals is a periodic signal when the ratio of their time periods is ____________
a) A rational number
b) An irrational number
c) A complex number
d) An integer
Answer: a
Explanation: Sum of two periodic signals is a periodic signal only when the ratio of their time periods is a rational number or it is the ratio of two integers. For e.g., T1/T2 = 5/7 → Periodic; T1/T2 = 5 → Aperiodic.
6. Determine the Time period of: x=3 cos+sin.
a) 1/10 sec
b) 1/20 sec
c) 2/5 sec
d 2/4 sec
Answer: c
Explanation: Here is the explanation.
signals-systems-questions-answers-freshers-q6
7. What is the even component of a discrete-time signal?
signals-systems-questions-answers-signal-classification-properties-q7
Answer: b
Explanation: Here is the explanation.
signals-systems-questions-answers-freshers-q7exp
8. Determine the odd component of the signal: x=cost+sint.
a) sint
b) 2sint
c) cost
d) 2cost
Answer: c
Explanation: Here is the explanation.
signals-systems-questions-answers-freshers-q8
9. Is the signal sin anti-symmetric?
a) YES
b) NO
Answer: a
Explanation: A signal is said to be anti-symmetric or odd signal when it satisfies the following condition:
⇒ x = – x
Now, here, x = sin ⇒ x = sin = – sin
∴ Sin is an anti-symmetric signal or an odd signal.
10. For an energy signal __________
a) E=0
b) P= ∞
c) E= ∞
d) P=0
Answer: d
Explanation: A signal is called an energy signal if the energy satisfies 0<E< ∞ and power P=0.
11. Determine the power of the signal: x = cos.
a) 1/2
b) 1
c) 3/2
d) 2
Answer: a
Explanation: Here is the explanation.
signals-systems-questions-answers-freshers-q11
12. Is the following signal an energy signal?
x = u – u
a) YES
b) NO
Answer: a
Explanation: Here is the explanation.
signals-systems-questions-answers-freshers-q12
13. A signal is anti-causal if ______________
a) x = 0 for t = 0
b) x = 1 for t < 0
c) x = 1 for t > 0
d) x = 0 for t > 0
Answer: d
Explanation: A signal is said to be anti-causal when x = 0 for t > 0.
14. Is the signal x= e at u causal?
a) YES
b) NO
Answer: a
Explanation: A signal is said to be causal if it is 0 for t < 0.
Now, we know, u = 1 for t ≥ 0.
∴ x= e at for t ≥0 .
∴ The signal is 0 for t < 0.
∴ The signal is causal.
15. Is the signal x = u – u causal?
a) YES
b) NO
Answer: b
Explanation: A signal is said to be causal if it is 0 for t < 0.
⇒ u = 1 for n ≥ – 4
⇒ u = 1 for n ≥ 4
∴ The signal x is defined for the interval [-4, 4] and it is zero for all other n values.
∵ The signal is defined for both n > 0 and n < 0
∴ The signal is non-causal.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Classification of Signals”.
1. What is single-valued function?
a) Single value for all instants of time
b) Unique value for every instant of time
c) A single pattern is followed by after ‘t’ intervals
d) Different pattern of values is followed by after ‘t’ intervals of time
Answer: b
Explanation: Single-valued function means “for every instant of time there exists unique value of the function”.
2. In real valued function and complex valued function, time is _______________
a) Real
b) Complex
c) Imaginary
d) Not predictable
Answer: a
Explanation: Time is an independent variable and it is real valued irrespective of real valued or complex valued function. And time is always real.
3. Discrete time signal is derived from continuous time signal by _____________ process.
a) Addition
b) Multiplying
c) Sampling
d) Addition and multiplication
Answer: c
Explanation: Sampling is a process wherein continuous time signal is converted to its equivalent discrete time signal. It is given by t = N*t.
4. Even signals are symmetric about the vertical axis.
a) True
b) False
Answer: a
Explanation: Signals are classified as even if it has symmetry about its vertical axis. It is given by the equation x = x .
5. If x = -x then the signal is said to be _____________
a) Even signal
b) Odd signal
c) Periodic signal
d) Non periodic signal
Answer: a
Explanation: Signals is said to be odd if it is anti- symmetry over the time origin. And it is given by the equation x = -x .
6. Which of the following is true for complex-valued function?
a) X = x*
b) X = x
c) X = – x
d) X = x*
Answer: a
Explanation: Complex-valued function is said to be conjugate symmetry if its real part is even and imaginary part is odd and it is shown by the equation x = x*.
7. When x is said to be non periodic signal?
a) If the equation x = x is satisfied for all values of T
b) If the equation x = x is satisfied for only one value of T
c) If the equation x = x is satisfied for no values of T
d) If the equation x = x is satisfied for only odd values of T
Answer: c
Explanation: A signal x is said to be non periodic signal if it does not satisfy the equation x = x. And it is periodic if it satisfies the equation for all values of T = T0, 2T0, 3T0…
8. Fundamental frequency x[n] is given by ___________
a) Omega = 2*pi /N
b) Omega = 2*pi*N
c) Omega = 4*pi *2N
d) Omega = pi / N
Answer: a
Explanation: Fundamental frequency is the smallest value of N which satisfies the equation
Omega = 2*pi/ N, Where N is a positive integer.
9. Noise generated by an amplifier of radio is an example for?
a) Discrete signal
b) Deterministic signal
c) Random signal
d) Periodic signal
Answer: c
Explanation: Random signal is the one which there is uncertainty before its actual occurrence. Noise is a best example for random signal.
10. Energy signal has zero average power and power signal has zero energy.
a) True
b) False
Answer: b
Explanation: Energy and power signals are mutually exclusive. Energy signal has zero average power and power signal has infinite energy.
signals-systems-questions-answers-classification-signals-q11
11. What is the fundamental frequency of discrete –time wave shown in fig a?
a) π/6
b) π/3
c) 2π/8
d) π
Answer: b
Explanation: Omega = 2* π / N. In the given example the number of samples in one period is N = 6. By substituting the value of N =6 in the above equation then we get fundamental frequency as π/3.
12. Calculate the average power of the discrete-time wave shown in fig a?
a) 1
b) 6
c) 0
d) -1
Answer: a
Explanation: The given formula is signals-systems-questions-answers-classification-signals-q12 used to calculate average power for Periodic -discrete signal. By substituting the value of N and x 2[n] in the given then we get the required answer.
signals-systems-questions-answers-classification-signals-q13
13. What is the total energy of rectangular pulse shown in fig b?
a) 8A 2
b) 4A
c) 2A
d) 4A 2
Answer: a
Explanation: signals-systems-questions-answers-classification-signals-q13a The total energy of the rectangular pulse can be found by integrating the square of the signal. Basically energy is given by area under the curve.
14. What is the total power of the rectangular pulse shown in fig b?
a) 0
b) 8A 2
c) ∞
d) 2A
Answer: a
Explanation: Energy signals have zero power and finite energy. Figure b is an example of an energy signal. This is one of the definition/ properties of the energy signal.
15. What is the total energy of the signal shown in fig c?
signals-systems-questions-answers-classification-signals-q15
a) 6
b) 0
c) 3
d) 1
Answer: a
Explanation: The given figure is an example of an energy signal hence the energy of a discrete-time signal is given by the equation E = ∑ x 2 [n].
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Basic Operations on Signals – 1”.
1. Which of the following is an example of amplitude scaling?
a) Electronic amplifier
b) Electronic attenuator
c) Both amplifier and attenuator
d) Adder
Answer: c
Explanation: Amplitude scaling refers to multiplication of a constant with the given signal.
It is given by y = a x . It can be both increase in amplitude or decrease in amplitude.
2. Resistor performs amplitude scaling when x is voltage, a is resistance and y is output current.
a) True
b) False
Answer: b
Explanation: The given statement is not true. The relation between voltage, current and resistance is given by V = IR. Comparing with equation y = a x , we can see that y is the output voltage for given current x with resistance R.
3. Which of the following is an example of physical device which adds the signals?
a) Radio
b) Audio mixer
c) Frequency divider
d) Subtractor
Answer: b
Explanation: Audio mixer is a device which combines music and voice signals. It is given by
Y = x1 + x2 .
4. AM radio signal is an example for __________
a) y = a x
b) y = x1 + x2
c) y = x1 * x2
d) y = -x
Answer: c
Explanation: AM radio signal is an example for y = x1 * x2 where, x1 consists of an audio signal plus a dc component and x2 is a sinusoidal signal called carrier wave.
5. Which of the passive component performs differentiation operation?
a) Resistor
b) Capacitor
c) Inductor
d) Amplifier
Answer: c
Explanation: Inductor performs differentiation. It is given by y = L d/dt i where, I denotes current flowing through an inductor of inductance L.
6. Which of the component performs integration operation?
a) Resistor
d) Diode
c) Capacitor
d) Inductor
Answer: c
Explanation: Capacitor performs integration. V developed across capacitor is given by
v = * ∫ t -∞ i .d∂, I is the current flowing through a capacitor of capacitance C.
7. Time scaling is an operation performed on _______
a) Dependent variable
b) Independent variable
c) Both dependent and independent variable
d) Neither dependent nor independent variable
Answer: b
Explanation: Time scaling is an example for operations performed on independent variable time.
It is given by y = x .
8. Y = x is ________
a) Compressed signal
b) Expanded signal
c) Shifted signal
d) Amplitude scaled signal by a factor of 2
Answer: a
Explanation: By comparing the given equation with y = x we get a=2. If a>1 then it is compressed version of x .
9. Y = x is _______
a) Compressed signal
b) Expanded signal
c) Time shifted signal
d) Amplitude scaled signal by factor 1/5
Answer: b
Explanation: y = x , comparing this with the given expression we get a = 1/5. If 0<a<1 then it is expanded version of x .
10. In discrete signal, if y [n] = x [k*n] and k>1 then ______
a) Some samples are lost from x [n]
b) Some samples are added to x [n]
c) It has no effect on samples
d) Samples will be increased with factor k
Answer: a
Explanation: For discrete time signal y [n] = x [k*n] and k>1, it will be compressed signal and some samples will be lost. The samples lost will not violate the rules of sampling theorem.
This set of Signals & Systems Interview Questions and Answers for freshers focuses on “Basic Operations on Signals – 2”.
signals-systems-interview-questions-answers-freshers-q1
1. Considering Figure 1, sketch y= 2* x .
signals-systems-interview-questions-answers-freshers-q1a
Answer: a
Explanation: Y = 2*x is an example for amplitude scaling. Here amplitude is scaled by a factor 2.
2. Considering Figure 1, sketch y= -3* x .
a) signals-systems-interview-questions-answers-freshers-q2a
b) signals-systems-interview-questions-answers-freshers-q2b
c) signals-systems-interview-questions-answers-freshers-q2c
d) signals-systems-interview-questions-answers-freshers-q2d
Answer: a
Explanation: Y = -3*x is an example for amplitude scaling. Here amplitude is scaled by a factor -3.
3. In the following diagram, X [n] and y [n] are related by ______
signals-systems-interview-questions-answers-freshers-q3
a) Y [n] = 2*x [n]
b) Y [n] = -2*x [n]
c) Y [n] = x [2n]
d) Y [n] = x [-2n]
Answer: a
Explanation: Y [n] = 2*x [n] is an example for amplitude scaling of discrete time signal. The given figure is an example for 2*x [n] hence Y [n] = 2*x [n] is correct.
4. X [n] and y [n] is as shown below, the relationship between x [n] and y [n] is given by ______
signals-systems-interview-questions-answers-freshers-q4
a) X [n] = y [n]/3
b) X [n] = 3* y [n]
c) Y [n] = x [n]/3
d) Y [n] = 3*x [n]
Answer: c
Explanation: The given y [n] is amplitude scaling of a discrete time signal by a factor 1/3.
Hence the amplitude is reduced by 1/3.
signals-systems-interview-questions-answers-freshers-q5
5. Considering figure 3 above, is the following figure true for y [n] = x [2n]?
signals-systems-interview-questions-answers-freshers-q5a
a) True
b) False
Answer: a
Explanation: X [2n] is an example of time scaling. For discrete time signal x [k*n], k>1 the samples will be lost.
6. Considering figure 3 above, is the following figure true for y [n] = x [n/2]?
signals-systems-interview-questions-answers-freshers-q6
a) True
b) False
Answer: b
Explanation: X [n/2] is an example for time scaling by factor ½ and it will be a stretched signal. The discrete time signal should extend from -10 to 10.
signals-systems-interview-questions-answers-freshers-q7
7. Consider figure 4, is the given y an integration of x ?
a) Y = ∫x .dt
b) Y = ∫x 2 .dt
c) Y = 3* ∫x .dt
d) Y = 3* ∫x 2 .dt
Answer: a
Explanation: The given y is integral of x and amplitude 3 remains constant for t>1.
It is because of the properties of integration.
8. . Consider figure 4, is the given y a differentiation of x ?
signals-systems-interview-questions-answers-freshers-q8
signals-systems-interview-questions-answers-freshers-q8a
Answer: a
Explanation: The given y is differentiation of x and hence we have impulses at -1, 0 and 1.
9. The given pair x and y is _______
signals-systems-interview-questions-answers-freshers-q9
a) Y = d/dt )
b) Y = ∫x .dt
c) Y = x -1
d) Y = x /2
Answer: a
Explanation: The given pair x and y is related by y = d/dt ). From -2 to 2 we have Y is zero because differentiation of constant is zero.
10. The given pair x and y is related by _______
signals-systems-interview-questions-answers-freshers-q10
a) Y = d/dt )
b) Y = x + 1
c) Y = ∫x .dt
d) Not related
Answer: c
Explanation: The given pair x and y is related by Y = ∫x .dt. The integral of x gives the Y . Y = 0 for t > 1.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Elementary Signals – 1”.
1. The general form of real exponential signal is________
a) X = be at
b) X = e at
c) X = b
d) X = be t
Answer: a
Explanation: X = be at is the most general way of representing the exponential signals where both b and a are real parameters.
2. In the equation x = be at if a < 0, then it is called ______
a) Growing exponential
b) Decaying exponential
c) Complex exponential
d) Both Growing and Decaying exponential
Answer: b
Explanation: If a > 0 in x = be at it is called growing exponential and if <0 it is called decaying exponential. Hence Decaying exponential is correct.
3. In the below figure if R value is increased then which of the following is true?
signals-systems-questions-answers-elementary-signals-1-q3
a) Slower the rate of decay of v
b) Greater the rate of decay of v
c) Decay rate is independent of R
d) Decay rate depends only on the capacitor value
Answer: a
Explanation: In the circuit shown voltage across capacitor decays exponentially with time at a rate determined by time constant RC. Hence the larger the resistor, the slower will be the rate of decay of v with time.
4. The time period of continuous-time sinusoidal signal is given by _____
a) T = 2π / w
b) T = 2π / 3w
c) T = π / w
d) T = π / 2w
Answer: a
Explanation: X = A cos is the continuous-time sinusoidal signal and its period is given by
T = 2π / w where w is the frequency in radians per second.
5. The natural angular frequency of the parallel LC circuit is?
signals-systems-questions-answers-elementary-signals-1-q5
Answer: a
Explanation: Wo is the natural angular frequency and for parallel LC circuit it is given by wo= 1 ⁄ √LC where, L is value of inductor and C is value of capacitor.
6. X [n] = 2 cos is periodic or not?
a) Periodic with period 2n
b) Periodic with period 2π
c) Periodic with period 2
d) Non periodic
Answer: d
Explanation: The given signal x [n] is non periodic as it doesn’t satisfy the equation w= 2πm ⁄ N where, N is fundamental period and m is an integer.
7. Check whether x [n] = 7 sin is periodic and if it is period calculate its fundamental period?
a) Periodic with fundamental period 6π
b) Periodic with fundamental period 3
c) Periodic with fundamental period 1
d) Non periodic
Answer: c
Explanation: X [n] = 7 sin is a periodic discrete time signal with period 1. By substituting w = 6π and m=3 in w= 2πm ⁄ N we get N =1.
8. Find the smallest angular frequency for which the discrete time signal with fundamental period N=8 would be periodic?
a) π ⁄ 4
b) π ⁄ 2
c) 3π ⁄ 4
d) π ⁄ 16
Answer: a
Explanation: By substituting N=8 and m=1 in the equation w= 2πm ⁄ N we get the smallest angular frequency as π ⁄ 4 .
9. Euler’s identity e jθ is expanded as _____
a) cos θ + j sin θ
b) cos θ – j sin θ
c) cos θ + j sin 2θ
d) cos 2θ+j sinθ
Answer: a
Explanation: The complex exponential e jθ is expanded as cos θ + j sin θ and is called Euler’s identity with cos θ as real part sin θ as imaginary part.
10. Exponentially damped sinusoidal signal is ______
a) Periodic
b) Non periodic
c) Insufficient information
d) Maybe periodic
Answer: b
Explanation: Exponentially damped sinusoidal signal of any kind is not periodic as it does not satisfy the periodicity condition.
This set of Signals & Systems Questions and Answers for Experienced people focuses on “Elementary Signals – 2”.
1. Mathematical representation of given rectangular pulse is ______
signals-systems-questions-answers-experienced-q1
a) X = {2A, t/2 < 0 < -t/2
b) X = {2A, -t/2 < 0 < t/2
c) X = {2A, 0 <= |t| <= t/2
{0, |t| > t/2
d) X = {2A, 0 <|t| < t/2
{0, |t| > t/2
Answer: c
Explanation: The given rectangular pulse is of amplitude 2A for the time interval –t/2 to t/2 and zero otherwise.
2. If signals-systems-questions-answers-experienced-q2 describe x [n] as superposition of two step functions.
a) X [n] = u [n] – u [n-5].
b) X [n] = u [n] + u [n-5].
c) X [n] = u [n-5] – u [n].
d) X [n] = u [n-5] + u [n].
Answer: a
Explanation: X [n] will be of amplitude for the interval 0 to 4 and zero otherwise. It can be obtained by the equation x [n] = u [n] – u [n-5].
3. Discrete-time version of unit impulse is defined as ______
signals-systems-questions-answers-experienced-q3
Answer: a
Explanation: Unit impulse is an elementary signal with zero amplitude everywhere except at n = 0.
4. Which of the following is not true about unit impulse function?
signals-systems-questions-answers-experienced-q4
Answer: d
Explanation: One option gives the definition of discrete-time version of impulse function, other options gives continuous-time representation of impulse function.
5. The step function u is integral of _______ with respect to time t.
a) Ramp function
b) Impulse function
c) Sinusoidal function
d) Exponential function
Answer: b
Explanation: Step function is an integral of impulse function and conversely, impulse is the derivative of step function u .
6. The area under the pulse defines _____ of the impulse.
a) Strength
b) Energy
c) Power
d) Duration
Answer: a
Explanation: The area under the pulse defines strength of the impulse and the strength of the impulse is denoted by the label next to the arrow.
7. Unit impulse ∂ is _____ of time t.
a) Odd function
b) Even function
c) Neither even nor odd function
d) Odd function of even amplitude
Answer: b
Explanation: For an impulse function, ∂= ∂. Hence unit impulse is an even function of time t.
8. Shifting property of impulse ∂ is given by ______
signals-systems-questions-answers-experienced-q8
Answer: a
Explanation: X be a function and the product of x with time shifted delta function ∂ gives x, this is referred to as shifting property of impulse function.
9. ∂ = 1 ⁄ a ∂, this property of unit impulse is called ______
a) Time shifting property
b) Time scaling property
c) Amplitude scaling property
d) Time reversal property
Answer: b
Explanation: Impulse function exhibits shifting property, time scaling property. And time scaling property is given by∂ = 1 ⁄ a ∂.
10. Which of the following is not true about the ramp function?
a) signals-systems-questions-answers-experienced-q10
b) r = t u
c) Ramp function with unit slope is integral of unit step
d) Integral of unit step is a ramp function of unit slope
Answer: d
Explanation: The impulse function is derivative of the step function. In the same way the integral of step function is a ramp function of unit slope.
∫u = r.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Properties of Systems – I”.
1. Is the system y = Rx, where R is a arbitrary constant, a memoryless system?
a) Yes
b) No
Answer: a
Explanation: The output of the system depends on the input of the system at the same time instant. Hence, the system has to be memoryless.
2. Does the following discrete system have the parameter of memory, y[n] = x[n-1] + x[n] ?
a) Yes
b) No
Answer: a
Explanation: y[n] depends upon x[n-1], i.e at the earlier time instant, thus forcing the system to have memory.
3. y[t]= ∫x[t],t ranges from 0 to t. Is the system a memoryless one?
a) Yes
b) No
c) Both memoryless and having memory
d) None of the Mentioned
Answer: b
Explanation: While evaluating the integral, it becomes imperative to know the values of x[t] from 0 to t, thus making the system requiring memory.
4. y = sin) : Comment on its memory aspects.
a) Having memory
b) Needn’t have memory
c) Memoryless system
d) Time invariant system
Answer: a
Explanation: The output at any time t = A, requires knowing the input at an earlier time, t = A – 1, hence making the system require memory aspects.
5. Construct the inverse system of y = 2x
a) y = 0.5x
b) y = 2x
c) y = x
d) y = x
Answer: a
Explanation: Now, y = 2x => x = 0.5*y
Thus, reversing x <-> y, we obtain the inverse system: y = 0.5x
6. y = x 2 . Is y = sqrt) the inverse of the first system?
a) Yes
b) No
c) Inverse doesn’t exist
d) Inverse exist
Answer: b
Explanation: We cannot determine the sign of the input from the second function, thus, the output doesn’t replicate the input. Thus, the second function is not an inverse of the first one.
7. Comment on the causality of y[n] = x[-n].
a) Time invariant
b) Causal
c) Non causal
d) Time varying
Answer: c
Explanation: For positive time, the system may seem to be causal. However, for negative time, the output depends on time at a positive sign, thus being in the future, enforcing non causality.
8. y = x + x. Comment on its causality:
a) Causal
b) Time variant
c) Non causal
d) All of the mentioned
Answer: c
Explanation: For a time instant existing between 0 and 1, it would depend on the input at a time in the future as well, hence being non causal.
9. Comment on the causality of y[n] = n*x[n].
a) Time invariant
b) Time varying
c) Non causal
d) Causal
Answer: d
Explanation: For positive time, the system may seem to be causal. For negative time, the output depends on the same time instant, thus making it causal.
10. Comment on the linearity of y[n] = n*x[n].
a) Linear
b) Only additive
c) Not scalable
d) Non linear
Answer: d
Explanation: The function obeys the scaling/homogeneity property, but doesn’t obey the additivity property, thus not being linear.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Properties of Systems – II”.
1. What is the following type of system called? y[n] = x[n] + y[n-1].
a) Subtractor system
b) Adder system
c) Product System
d) Divisor System
Answer: b
Explanation: If we write for n-1, n-2, .. we will obtain y[n] = x[n] + x[n-1] + x[n-2] …,
thus obtaining an adder system.
2. Which of the following systems is linear?
a) y = sin)
b) y = log)
c) y = cos)
d) y = dx/dt
Answer: d
Explanation: Only d satisfies both the scaling and the additivity properties.
3. Which of the following systems is stable?
a) y = log)
b) y = exp)
c) y = sin)
d) y = tx + 1
Answer: c
Explanation: Stability implies that a bounded input should give a bounded output. In a, b, d there are regions of x, for which y reaches infinity/negative infinity. Thus the sin function always stays between -1 and 1, and is hence stable.
4. Which of the following systems is time invariant?
a) y = x + x
b) y = x + x
c) y = -x + x
d) y = x + x
Answer: d
Explanation: In each of a, b and c there is a negative sign of t involved, which means a backward shift of t-0 in time, would mean a forward shift in each of them. However, only in d, the backward shift will remain as backward, and undiminished.
5. State whether the differentiator system is causal or not.
a) True
b) False
Answer: b
Explanation: The derivative of a function can be written in forward difference and in backward difference form, hence the derivative would depend on a slightly forward value of the function, thus making it non causal.
6. State whether the differentiator system is a stable system or not.
a) True
b) False
Answer: b
Explanation: The derivative of a function can be unbounded at some bounded inputs, like tan at x=pi/2, hence the differentiator system is unstable in general, when the input is not specified.
7. Which of the following systems is memoryless?
a) y = x + x
b) y = x + 2x
c) y = -x + x
d) y = x + 2x
Answer: b
Explanation: A system possessing no memory has its output depending upon the input at the same time instant, which is prevalent only in option b.
8. For what value of k, will the following system be time invariant?
y = x + x – x + x
a) 1
b) 2
c) 3
d) 2.5
Answer: b
Explanation: A system possessing no memory has its output depending upon the input at the same time instant, which is prevalent only in option b.
9. State if the following system is periodic or not. y = sin*x)
a) No
b) Yes
Answer: a
Explanation: The function y = sin is periodic only for rational ‘n’.
10. State whether the following system is periodic or not. y = log).
a) Yes
b) No
Answer: b
Explanation: Sin x is a periodic function, but log x is not a periodic function. Thus y is log t, where t= sin x, thus y is not periodic.
This set of Signals & Systems Interview Questions and Answers for Experienced people focuses on “Properties of Systems – 3”.
1. Which one of the following is an example of a system with memory?
a) Identity System
b) Resistor
c) y=x-2x
d) Accumulator
Answer: d
Explanation: An identity system gives the output same as input hence it totally depends on the present state of the input. Therefore, it is memory less. Similarly, a resistor and the expression in option c are memory less systems as they depend upon the present state of the input. An accumulator sums up the values of all past and present states of input. Therefore, it is a system with memory.
2. Which among the following is a memory less system?
a) Delay
b) Summer
c) Resistor
d) Capacitor
Answer: c
Explanation: Options Delay, Summer and Capacitor are all systems with memory as they depend upon past, past and present, past and present values of input respectively. Whereas, a resistor is a memory less system as its relationship with output always depends upon the current or present state of the input.
3. In a continuous-time physical system, memory is directly associated with _____________
a) Storage registers
b) Time
c) Storage of energy
d) Number of components in the system
Answer: c
Explanation: Memory is directly associated with storage of energy such as electric charge in the capacitor or kinetic energy in an automobile. Storage registers are for discrete time systems such as microprocessor etc. Time and number of components of a system have got nothing to do with memory.
4. A system with memory which anticipates future values of input is called _________
a) Non-causal System
b) Non-anticipative System
c) Causal System
d) Static System
Answer: a
Explanation: A system which anticipates the future values of input is called a non-causal system. A causal depends only on the past and present values of input. Non-anticipative is another name for the causal system. A static system is memory less system.
5. Determine the nature of the system: y=x.
a) Causal
b) Non-causal
c) Causal for all positive values of n
d) Non-causal for negative values of n
Answer: b
Explanation: The given system gives negative values of input i.e., past values of input when we feed positive integers to LHS. However, it gives positive values for negative values of n i.e., future values. Therefore, the system depends upon past values for some integers and future values for some other. A system cannot be called partially causal or non-causal, therefore, the system is non-causal.
6. Which among the following is an application of non-causal system?
a) Image processing
b) RC circuit
c) Stock market Analysis
d) Automobile
Answer: c
Explanation: Image processing, RC circuit, and an automobile are all causal systems as they do not anticipate the future values of an image, RC circuit and future actions of a driver respectively. Instead, they function upon either the stored information or on the current values of the input. Whereas, in the stock market, analysts try to figure out a trend in the future based upon the stored information. Therefore, it is non-causal.
7. Determine the nature of the given system: y=x
a) Causal, Non-linear
b) Causal, Linear
c) Non-Causal, Non-linear
d) Non-causal, Linear
Answer: d
Explanation: The system is non-causal as it gives future values for some inputs.
E.g. y = x ) = x
For linearity, it needs to satisfy superposition principle,
⇒ y 1 = x 1
⇒ y 2 = x 2
⇒ ay 1 + by 2 = ax 1 + bx 1 Equation 1
Now, y 3 = x 3 = (ax 1 + bx 2 ) = ax 1 + bx 1 Equation 2
Clearly, Equation 1 and 2 are equal, hence the system is linear.
8. Is the system y[n]=2x[n]+2 linear?
a) YES
b) NO
Answer: b
Explanation: The system needs to satisfy superposition principle for linearity:
For input x1[n], y 1 [n] = 2x 1 [n] + 2
For input x2[n], y 2 [n] = 2x 2 [n] + 2
⇒ ay 1 [n]+ by 2 [n] = 2(ax 1 [n]+ bx 2 [n]) + 2 Equation 1
For, x3[n], y 3 [n]=2x 3 [n]+2 = 2(ax 1 [n]+ bx 2 [n]) + 2 Equation 2
Clearly, Equation 1 is not equal to equation
∴ The system does not satisfy superposition principle ⇒ The system is not linear.
9. An inverse system with the original system gives an output equal to the input. How is the inverse system connected to the original system?
a) Series
b) Cascaded
c) parallel
d) No connection
Answer: c
Explanation: An inverse system when cascaded with the original system gives an output equal to the input.
10. Which among the following is an invertible system?
a) y[n] = 0
b) y[n] = 2x[n]
c) y = x 2
d) y = dx/dt
Answer: b
Explanation: A system is said to be invertible if it’s input can be found out from its output. Implying, if a system has same outputs for several inputs then it is impossible to find the correct input as output is same for many. Therefore, a system is invertible if it gives distinct outputs to distinct inputs. It is non-invertible if it gives same outputs for many inputs.
Option a produces 0 output for any input → Non-invertible
Option b produces different outputs for different inputs and also it’s inverse system is y[n] → Invertible
Option c, we get same output for both positive and negative values → Non-invertible
Option d, we get 0 for all constant input values → Non-invertible.
11. Is the system time invariant: y = x?
a) YES
b) NO
Answer: b
Explanation: A system is said to be time invariant if a change input causes the same change in output.
For change in input by T
⇒ y = x) = x Equation 1
For the same change in output
⇒ y = x Equation 2
Equation 1 is not equal to equation 2.
∴ The system is not time invariant or is time variant.
12. Determine the nature of the system: y[n] = x[n]x[n – 1] with unit impulse function as an input.
a) Dynamic, output always zero, non-invertible
b) Static, output always zero, non-invertible
c) Dynamic, output always 1, invertible
d) Dynamic, output always 1, invertible
Answer: a
Explanation: Since the system depends on present and past values, therefore, it is not memory less.
Now, input is a unit impulse function. Unit impulse function = 1 at n = 0, otherwise it is equal to 0.
For, y[0] = x[0]x[-1] = 1 × 0 = 0
For, y[1] = x[1]x[0] = 0 × 1 = 0
For, y[2] = x[2]x[1] = 0 × 0 = 0
∴ For any time, output is always zero.
Since, the output is always same, the system is non-invertible.
13. Determine the nature of the system: y= t 2 x
a) Linear, time invariant
b) Linear, time variant
c) Non-linear, time invariant
d) Non-linear, time variant
Answer: b
Explanation: For linearity:
For input x1: y 1 = t 2 x 1
For input x2: y 2 = t 2 x 2
⇒ ay 1 +by 2 = t 2 [x 1 + bx 2 ] Equation 1
For input x3: y 3 = t 2 x 3 = t 2 [ax 1 + bx 2 ] Equation 2
∴ The system is linear.
For time invariancy: Shift in input:
⇒y= t 2 x
Shift in output: y= 2 x
∵ The shift in output is not equal to the shift in input, therefore, the system is time variant.
14. y[n]=r n x[n] is ________ system.
a) LTI
b) Time varying
c) Linear and time invariant
d) Causal and time invariant
Answer: b
Explanation: The input-output relationship of the given system shows it does not satisfy the condition of time-invariant system. Hence it is time varying system.
15. A system is said to be linear if _______
a) It satisfies only the principle of superposition theorem
b) It satisfies only amplitude scaling
c) It satisfies both amplitude scaling and principle of superposition theorem
d) It satisfies amplitude scaling but not the principle of superposition theorem
Answer: c
Explanation: By the definition of linearity a system is said to be linear if it satisfies the condition y1 + y2 = ax1 + bx2.
16. If the input-output relationship is given by y = 2x d ⁄ dx x. What kind of system it represents?
a) Linear system
b) Non linear system
c) LTI system
d) Linear but time-invariant system
Answer: b
Explanation: The given input-output relationship of the system does not satisfy the principle of superposition theorem hence it is an example for non linear system.
This set of Signals & Systems test focuses on “Properties of Systems – 4”.
1. What is a stable system?
a) If every bounded input results in the bounded output
b) If every bounded input results in an unbounded output
c) If every unbounded input results in a bounded output
d) If unbounded input results in bounded as well as unbounded output
Answer: a
Explanation: The system is said to bounded input bounded output stable if every bounded input results in bounded output and also the output of such a system does not diverge if the input does not diverge.
2. If signals-systems-questions-answers-test-q2 This is an example for _______ system.
a) Stable system
b) Unstable system
c) Bounded input unbounded output system
d) Unbounded input system
Answer: a
Explanation: In the above example, the input is finite and the output is also finite as y is absolute integrable. Hence the system is stable.
3. If x= ∂ and y= e -t . This is an example for ______ system.
a) Stable
b) BIBO
c) Bounded input
d) Unstable
Answer: d
Explanation: In this example, the input is finite and output is not finite. Hence the given system is unstable.
3. If x=e t , y= e -2t this is a _____system.
a) Unstable
b) Stable
c) BIBO
d) Cannot classify the system
Answer: d
Explanation: In this example, the input is infinite and hence this input cannot be used to classify the system. Here the output is not considered.
4. Which of the following is not true about systems having memory?
a) It is also called dynamic systems
b) The output signal depends on the past values of the input signal
c) It is also called static system
d) Resistive circuit
Answer: c
Explanation: The system is said to have memory if its output signal depends on the past values of the input signal and also it is called dynamic system.
5. How far does the memory of the given system y[n]=1/2{x[n]+ x[n-1]} extend into past?
a) Two time units
b) One time unit
c) Three time units
d) Not predictable
Answer: b
Explanation: The given memory system extends into past by one time unit. It is determined by the term x [n-1].
6. The input- output relation of a device is represented asi=ao+a1v1+a2v 2 +⋯. Does this device have memory?
a) Has memory
b) Does not have memory
c) It is dynamic
d) Insufficient information
Answer: b
Explanation: In the given equation, v is the voltage applied I is the current flowing through the device and a0, a1, a2 are the constants. It does not involve any past value of the input signal and hence memory less.
7. Which is not an example for memory system?
a) Capacitive circuit
b) Inductive circuit
c) Resistive circuit
d) Parallel RC circuit
Answer: c
Explanation: Resistive circuit is memory less since the current I flowing through it in response to the applied voltage v is defined by i = 1 ⁄ R v.
8. What is the memory of the system if its input-output relation is given by signals-systems-questions-answers-test-q8 ?
a) Memory extends from time t to the infinite future
b) Memory extends from time t to the infinite past
c) Does not have memory
d) Insufficient information
Answer: b
Explanation: Given system has inductor involved in it. Hence it has memory. Since integral is from -∞ to time t, its memory extends from time t to infinite past.
9. Which of the following systems is memory less?
a) y = 2x + d ⁄ dx x
b) y = 2x 2 + d ⁄ dx x
c) y = ∫xdt
d) y = 2x 2
Answer: d
Explanation: A differentiator or integrator maybe realized with capacitors and inductors and cannot be realized using resistors. Hence differentiators and integrators can be considered as systems with memory.
10. An example for non-causal system is ________
a) Amplifier
b) Oscillator
c) Rectifiers
d) Does not exists
Answer: d
Explanation: Non-causal system is the one which results in output even without the application of input. Since all systems are real, non-causal systems do not exists.
11. Is Ideal low pass filter is an example for Non –causal system?
a) True
b) False
Answer: a
Explanation: Ideal low pass filter has sharp transitions which cannot be physically realized. Hence non – causal.
12. Can impulse response be measured?
a) Impulse cannot be generated
b) Impulse can be generated
c) Can be measured
d) Cannot be measured
Answer: c
Explanation: Impulse response can be measured but in an indirect manner. Hence by giving step response to a suitable differentiator impulse response is measured. Impulse response is derivative of step response.
13. Which of the following is an example for non- causal system?
a) y[n] = 1 ⁄ 3 {x[n-1] + x[n] + x[n-2]}
b) y[n] = 1 ⁄ 3 {x[n-1] + x[n] + x[n+1]}
c) y[n] = 1 ⁄ 2 {x[n-1] + x[n]}
d) y[n] = 1 ⁄ 2 {x[n] + x[n-2]}
Answer: b
Explanation: y[n] = 1 ⁄ 3 {x[n-1] + x[n] + x[n+1]} is an example for non- causal system since the output y [n] depends on the future value of the input namely x [n+1].
14. Which of the following is not true about invertible systems?
a) H -1 H=I
b) There must be one-to-one mapping between input and output signals for a system to be invertible
c) Input of the invertible system can be recovered from the system output
d) Input of the invertible system cannot be recovered from the system output
Answer: d
Explanation: By the definition of invertible system we can say that input of the invertible system can be recovered from the system output.
15. Is y= x 2 is an example for invertible system?
a) True
b) False
Answer: b
Explanation: In this example we don’t have a unique inverse hence the input of the system is not recovered from the system output. Hence it is considered as non-invertible system.
16. y = 2x + 3t d ⁄ dx x Is an example for _____
a) Time invariant system
b) Time varying system
c) LTI system
d) Time invariant and linear system
Answer: b
Explanation: The given system does not satisfy the condition {R{x}=y} hence the system is time varying.
17. y = 5x + 6 d ⁄ dx x Is an example for _____ system.
a) Time varying
b) Time invariant
c) Time varying and linear
d) Time varying and non linear
Answer: b
Explanation: The given system satisfies the condition {R{x}=y} hence the system is time invariant.
This set of Signals & Systems Questions & Answers for Exams focuses on “Properties of Systems – 5”.
1. For the system y = x 2 , which of the following holds true?
a) Invertible
b) Non-Invertible
c) Invertible as well as Non-Invertible in its respective domains
d) Cannot be determined
Answer: b
Explanation: When we pass the signal x through the system y , the system squares the input. Hence, inverse system should take square root, i.e.
W = \
Thus, two outputs are possible, + x or – x
This means there is no unique output for unique input. Hence, this system is Non-Invertible.
2. For the system \ = ∑_{k=-∞}^n x\), which of the following holds true?
a) Invertible
b) Non-Invertible
c) Invertible as well as Non-Invertible in its respective domains
d) Cannot be determined
Answer: a
Explanation: \ = ∑_{k=-∞}^n x\)
Or, y = x + x + x + …….
Or, y = x + x +
Or, y = x + y
Or, y = x + y
Or, y = x + y
This is an alternate form of given system equation.
∴ x = y – y
Taking z-transform on both sides,
X = Y =
\ = \frac{Y}{X} = \frac{1}{1-z-1}\)
Or, \(\frac{w}{y} \)= H -1 =
∴ w = y – y
Hence, the system is invertible.
3. For the system, y = u{x } which of the following holds true?
a) System is Linear, time-invariant, causal and stable
b) System is time-invariant, causal and stable
c) System is causal and stable
d) System is stable
Answer: b
Explanation: Let x 1 = v , then y 1 = u {v }
Let x 2 = k v , then y 2 = u {k v } = k y 1
Hence, non-linear
y 1 = u {v }
y 2 = u {v (t-t 0 )} = y 1 (t-t 0 )
Hence, time-invariant
Since the response at any time depends only on the excitation at time t=t 0 , and not on any further values, hence causal.
4. For the system, y = x – x which of the following holds true?
a) System is Linear, time-invariant, causal and stable
b) System is time-invariant, causal and stable
c) System is Linear, time-invariant and stable
d) System is Linear, time-invariant and causal
Answer: c
Explanation: y 1 = v – v
y 2 = k v – k v = ky 1
Let x 1 = v , then y 1 = v – v
Let x 2 = 2w , then y 2 = w -w
Let x 3 = x + w
Then, y 3 = y 1 + y 2
Hence it is linear.
Again, y 1 = v – v
∴ y 2 = y 1 (t-t 0 )
Hence, system is time-invariant
If x is bounded, then, x and x are also bounded, so stable system.
At t=0, y = x -x , therefore, the response at t=0 depends on the excitation at a later time t=3.
Therefore Non-Causal.
5. For the system, y = x
, which of the following holds true?
a) System is Linear, time-invariant, causal and stable
b) System is Linear and time-invariant
c) System is Linear and causal
d) System is Linear and stable
Answer: d
Explanation: y 1 = v
And y 2 = k v
= k y 1
If x 3 = v + w ,
Then, y 3 = v
+ w
= y 1 + y 2
Since, it satisfies both law of homogeneity and additivity, it is also linear.
Again, y 1 = v
, y 2 (\(\frac{t}{2}\)-t 0 ) ≠ y (t-t 0 ) = v
\)
∴ The system is time variant.
If x is bounded then y is bounded, so a stable system.
At time t=-2, y = x , therefore, the response at time t=-2 depends on the excitation at a later time, t=-1, so a Non-causal system.
6. For the system, y = cos 2πt x , which of the following holds true?
a) System is Linear, time-invariant, causal and stable
b) System is time-invariant, causal and stable
c) System is Linear, causal and stable
d) System is Linear, time-invariant and stable
Answer: c
Explanation: y 2 k cos 2πt v = k y 1
Again, x 3 = v + w
So, y 3 = cos 2πt [v + w ] = y 1 + y 2
Since, the system is both homogeneous and additive, therefore it is linear.
Again, y 1 = cos 2πt v
And y 2 = cos 2πt (t-t 0 ) ≠ y (t-t 0 )
= cos [2π (t-t 0 )] v (t-t 0 )
∴ The system is time variant.
The response at any time t=t 0 , depends only on the excitation at that time and not on the excitation at any later time, so Causal system.
If x is bounded then y is also bounded, so a stable system.
7. For the system, y = |x |, which of the following holds true?
a) System is Linear, time-invariant, causal and stable
b) System is Linear, time-invariant and causal
c) System is Linear, time-invariant and stable
d) System is Linear, causal and stable
Answer: c
Explanation: y 1 = |v |, y 2 = |k v |= |k|y 1
If k is negative, |k| y 1 ≠k y 1
Since it is not homogeneous, so non-linear system.
Again, y 1 = |v |, y 2 = |y (t-t 0 )| = y 1 (t-t 0 )
∴ System is time invariant.
The response at any time t=t 0 , depends only on the excitation at that time and not on the excitation at any later time, so causal system.
If x is bounded then y is also bounded, so stable system.
8. For the system, \
= x , which of the following holds true?
a) System is Linear, time-invariant, causal and stable
b) System is Linear, time-invariant and causal
c) System is time-invariant, causal and stable
d) System is Linear, causal and stable
Answer: c
Explanation: All options are linear. Hence linearity is not required to be checked.
Let x 1 = v , then \(t\frac{dy_1 }{dt}\) – 8 y 1 = v
Let x 2 = v (t-t 0 )
Then, \(t\frac{dy_2 }{dt}\) – 8 y 2 = v (t-t 0 )
The first equation can be written as (t-t 0 ) \(t\frac{dy
}{dt}\) – 8 y (t-t 0 ) = x (t-t 0 )
This equation is not satisfied if y 2 = y 1 (t-t 0 ). Therefore y 2 ≠ y 1 (t-t 0 )
∴ System is time variant.
The response at any time t=t 0 , depends only on the excitation at that time and not on the excitation at any later time, so Causal system.
The response will increase without bound as time increases, so unstable system.
9. For the system, \ = \int_{-∞}^{t+3} x \,dt\), which of the following holds true?
a) System is Linear, time-invariant and causal
b) System is time-invariant and causal
c) System is Linear and time-invariant
d) System is Linear and stable
Answer: c
Explanation: \
= \int_{-∞}^{t+3} v \,dt\)
And, \
= \int_{-∞}^{t+3} kv \,dt\)
= \
\,dt\) = k y 1
Now, x 3 = v + w
And, y 3 = \
+ w]dt
= \
dt + \
dt
= y 1 + y 2
Since, it is both homogeneous and additive, so linear system.
Again, y 1 = \
dt
And, y 2 = \(\int_{-∞}^{t+3}\) v(t-t 0 ) dt
= y 1 (t-t 0 )
∴ System is time invariant.
The response at any time t=t 0 , depends partially on the excitation at time t 0 < t < (t 0 + 3), which are in future, so non-causal system.
The response will increase without bound as time increases, so unstable system.
10. The impulse response of a continuous time LTI system is \ =
\,u \). The system is ____________
a) Causal and stable
b) Causal but not stable
c) Stable but not causal
d) Neither causal nor stable
Answer: b
Explanation: For t<0, h = 0.
Therefore from the definition of causality, we can infer that the system is Causal.
Now, \
| \,dt = ∞\)
∴ From the definition of stability, we can infer that the system is unstable.
Hence, the given system is causal but not stable.
11. The impulse response of a continuous time LTI system is H = e -|t| . The system is ___________
a) Causal and stable
b) Causal but not stable
c) Stable but not causal
d) Neither causal nor stable
Answer: c
Explanation: For t<0,
H ≠ 0
Therefore the system is not causal
Again, \
| \,dt\) = \(\frac{1}{3}\) < ∞
∴ The system is stable.
12. The impulse response of a continuous time LTI system is H = e -t u . The system is __________
a) Causal and stable
b) Causal but not stable
c) Stable but not causal
d) Neither causal nor stable
Answer: d
Explanation: For t<0, h ≠ 0
Therefore the system is not causal.
Again, \
| \,dt = \int_{-∞}^∞ e-t \,u \,dt = ∞ \)
∴ System is unstable.
13. The impulse response of a continuous time LTI system is H = e -t u . The system is __________
a) Causal and stable
b) Causal but not stable
c) Stable but not causal
d) Neither causal nor stable
Answer: a
Explanation: Since, h = 0 for t<0, so the system is causal.
Again, \
| \,dt = \int_{-∞}^∞ e-t \,u \,dt\) < ∞
∴ The system is stable.
14. The continuous time convolution integral y = cos πt [u – u * u] is __________
a) \
– u]
b) \
c) \
d) \
Answer: a
Explanation: For t<-1, y = 0
For t<1, y = \
= \
= \
– u].
15. The continuous time convolution integral y = e-3tu * u is ___________
a) \(\frac{1}{3}\)[1 – e -3 ] u
b) \(\frac{1}{3}\)[1 – e -3 ] u
c) \(\frac{1}{3}\)[1 – e -3t ] u
d) \(\frac{1}{3}\)[1 – e -3t ] u
Answer: a
Explanation: For t+3<0 or t<-3, y=0
For t≥-3, y = \(\int_{-∞}^3 e^{-3t} \,dt\)
= \(\frac{1}{3}\)[1 – e -3 ]
∴ y = \(\frac{1}{3}\)[1 – e -3 ] u.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Discrete Time Signals”.
1. Is the function y[n] = sin periodic or not?
a) True
b) False
Answer: b
Explanation: ‘y’ will be periodic only if x attains the same value after some time, T. However, if x is a one-one discrete function, it may not be possible for some x[n].
2. What is the time period of the function x[n] = exp?
a) pi/2w
b) pi/w
c) 2pi/w
d) 4pi/w
Answer: c
Explanation: Using Euler’s rule, exp = 1 for all integer n. Thus, the answer can be derived.
3. What is the nature of the following function: y[n] = y[n-1] + x[n]?
a) Integrator
b) Differentiator
c) Subtractor
d) Accumulator
Answer: d
Explanation: If the above recursive definition is repeated for all n, starting from 1,2.. then y[n] will be the sum of all x[n] ranging from 1 to n, making it an accumulator system.
4. Is the above function defined, causal in nature?
a) True
b) False
Answer: a
Explanation: As the value of the function depends solely on the value of the input at a time presently and/or in the past, it is a causal system.
5. Is the function y[n] = x[n-1] – x[n-4] memoryless?
a) True
b) False
Answer: b
Explanation: Since the function needs to store what it was at a time 4 units and 1 unit before the present time, it needs memory.
6. Is the function y[n] = x[n-1] – x[n-56] causal?
a) The system is non causal
b) The system is causal
c) Both causal and non causal
d) None of the mentioned
Answer: b
Explanation: As the value of the function depends solely on the value of the input at a time presently and/or in the past, it is a causal system.
7. Is the function y[n] = y[n-1] + x[n] stable in nature?
a) It is stable
b) It is unstable
c) Both stable and unstable
d) None of the mentioned
Answer: a
Explanation: It is BIBO stable in nature, i.e. bounded input-bounded output stable.
8. If n tends to infinity, is the accumulator function a stable one?
a) The function is marginally stable
b) The function is stable
c) The function is unstable
d) None of the mentioned
Answer: c
Explanation: The system would be unstable, as the output will grow out of bound at the maximally worst possible case.
9. We define y[n] = nx[n] – x[n]. Now, z[n] = z[n-1] + y[n], is z[n] stable?
a) Yes
b) No
Answer: a
Explanation: As we take the sum of y[n], terms cancel out and deem z[n] to be BIBO stable.
10. We define y[n] = nx[n] – x[n]. Now, z[n] = z[n-1] + y[n]. Is z[n] a causal system?
a) No
b) Yes
Answer: b
Explanation: As the value of the function depends solely on the value of the input at a time presently and/or in the past, it is a causal system.
11. Discrete-time signals are _________________
a) Continuous in amplitude and continuous in time
b) Continuous in amplitude and discrete in time
c) Discrete in amplitude and discrete in time
d) Discrete in amplitude and continuous in time
Answer: b
Explanation: A discrete-time signal is continuous in amplitude and discrete in time. It can either be present in nature or is sampled from an analog signal. A digital signal is discrete in amplitude and time.
12. Determine the discrete-time signal: x=1 for n≥0 and x=0 for n<0
a) Unit ramp sequence
b) Unit impulse sequence
c) Exponential sequence
d) Unit step sequence
Answer: d
Explanation: Unit step is defined by: x=1 for n≥0 and x=0 for n<0.
13. Determine the value of the summation: ∑ ∞ n= -∞ δsin2n.
a) 1
b) 0
c) sin2
d) sin4
Answer: c
Explanation: ∑ ∞ n= -∞ δsin2n
⇒ We know, δ is impulse function which means δ=1 when n=0
⇒ δ=1 when n=1 otherwise it is 0.
Therefore, the summation’s limit reduces to n=1
⇒ ∑ ∞ n= -∞ δsin2n = sin2n| n=1 = sin2.
14. Determine the value of the summation: ∑ ∞ n= -∞ δ(n 2 +n).
a) 3
b) 6
c) 9
d) 12
Answer: b
Explanation: ∑ ∞ n= -∞ δ(n 2 +n)
⇒ δ=1 when n= -3 otherwise 0.
Therefore, the limit reduces to n = -3
⇒ ∑ ∞ n = -∞ δ(n 2 +n) = (n 2 +n)| n = -3 = 2 -3 = 9 – 3 = 6.
15. Determine the product of two signals: x 1 = {2,1,1.5,3}; x 2 = { 1,1.5,0,2}.
a) {2,1.5,0,6}
b) {2,1.5,6,0}
c) {2,0,1.5,6}
d) {2,1.5,0,3}
Answer: a
Explanation: Product of discrete-time signals is computed element by element.
⇒ x = x 1 * x 2 = {2×1, 1×1.5, 1.5×0, 3×2} = {2,1.5,0,6}.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Useful Signals”.
1. What is the value of d[0], such that d[n] is the unit impulse function?
a) 0
b) 0.5
c) 1.5
d) 1
Answer: d
Explanation: The unit impulse function has value 1 at n = 0 and zero everywhere else.
2. What is the value of u[1], where u[n] is the unit step function?
a) 1
b) 0.5
c) 0
d) -1
Answer: a
Explanation: The unit step function u[n] = 1 for all n>=0, hence u[1] = 1.
3. Evaluate the following function in terms of t: {sum from -1 to infinity:d[n]}/{Integral from 0 to t: u}
a) t
b) 1 ⁄ t
c) t 2
d) 1 ⁄ t 2
Answer: b
Explanation: The numerator evaluates to 1, and the denominator is t, hence the answer is 1/t.
4. Evaluate the following function in terms of t: {integral from 0 to t}{Integral from -inf to inf}d
a) 1 ⁄ t
b) 1 ⁄ t 2
c) t
d) t 2
Answer: c
Explanation: The first integral is 1, and the overall integral evaluates to t.
5. The fundamental period of exp is
a) pi/w
b) 2pi/w
c) 3pi/w
d) 4pi/w
Answer: b
Explanation: The function assumes the same value after t+2pi/w, hence the period would be 2pi/w.
6. Find the magnitude of exp. Find the boundness of sin and cos.
a) 1, [-1,2], [-1,2]
b) 0.5, [-1,1], [-1,1]
c) 1, [-1,1], [-1,2]
d) 1, [-1,1], [-1,1]
Answer: d
Explanation: The sinand cos can be found using Euler’s rule.
7. Find the value of {sum from -inf to inf} exp*d[n].
a) 0
b) 1
c) 2
d) 3
Answer: b
Explanation: The sum will exist only for n = 0, for which the product will be 1.
8. Compute d[n]d[n-1] + d[n-1]d[n-2] for n = 0, 1, 2.
a) 0, 1, 2
b) 0, 0, 1
c) 1, 0, 0
d) 0, 0, 0
Answer: d
Explanation: Only one of the values can be one at a time, others will be forced to zero, due to the delta function.
9. Defining u, r and s in their standard ways, are their derivatives defined at t = 0?
a) Yes, Yes, No
b) No, Yes, No
c) No, No, Yes
d) No, No, No
Answer: d
Explanation: None of the derivatives are defined at t=0.
10. Which is the correct Euler expression?
a) exp = cos + jsin
b) exp = cos + jsin
c) exp = cos + sin
d) exp = jcos + jsin
Answer: b
Explanation: Euler rule: exp = cos + jsin.
11. The range for unit step function for u, is ________
a) t < a
b) t ≤ a
c) t = a
d) t ≥ a
Answer: d
Explanation: A unit step signal u = 1 when t ≥ 0 and 0 when t < 0
∴ u = 1 when t – a ≥ 0 ⇒ t ≥ a
12. Which one of the following is not a ramp function?
a) r = t when t ≥ 0
b) r = 0 when t < 0
c) r = ∫udt when t < 0
d) r = du ⁄ dt
Answer: d
Explanation: Ramp function r = t when t ≥ 0 and r = 0 when t < 0
Also, r= ∫udt = ∫dt = t = 1 for t≥0)
⇒ du ⁄ dt = d ⁄ dt = 0 which is not a ramp function.
13. Which one of the following is not a unit step function?
signals-systems-questions-answers-useful-signals-q13
Answer: d
Explanation: Unit step function, u = 1 for t ≥ 0 and u = 0 for t < 0. Also,
signals-systems-questions-answers-useful-signals-q13a
14. Unit Impulse function is obtained by using the limiting process on which among the following functions?
a) Triangular Function
b) Rectangular Function
c) Signum Function
d) Sinc Function
Answer: b
Explanation: Unit impulse function can be obtained by using a limiting process on the rectangular pulse function. Area under the rectangular pulse is equal to unity.
15. Evaluate: signals-systems-questions-answers-useful-signals-q15
a) {2,1.5,0,6}
b) {2,1.5,6,0}
c) {2,0,1.5,6}
d) {2,1.5,0,3}
Answer: a
Explanation: From the impulse function property, signals-systems-questions-answers-useful-signals-q15a
16. When is a complex exponential signal pure DC?
a) σ = 0 and Ω < 0
b) σ < 0 and Ω = 0
c) σ = 0 and Ω = 0
d) σ < 0 and Ω < 0
Answer: c
Explanation: A complex exponential signal is represented as x= e st
Where, s = σ + jΩ
⇒ x = e σt [cosΩt + jsinΩt] When, σ = 0 and Ω = 0 ⇒ x = e 0 [cos0 + jsin0] = 1 × 1 = 1 which is pure DC.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “The Complex Exponential”.
1. What is exp equal to, where j is the square root of unity?
a) cos ja + jsin a
b) sin a + jcos a
c) cos j + a sin j
d) cos a + jsin a
Answer: d
Explanation: This is the corollary of DeMoivre/Euler’s Theorem.
2. What is the magnitude of exp?
a) exp
b) exp
c) exp
d) exp
Answer: c
Explanation: exp =exp * exp, and |exp| = 1.
3. What is the fundamental frequency of exp?
a) 1pi*w
b) 2pi*w
c) w
d) 2w
Answer: c
Explanation: Fundamental period = 2pi/w, hence fundamental frequency will be w.
4. Total energy possessed by a signal exp is?
a) 2pi/w
b) pi/w
c) pi/2w
d) 2pi/3w
Answer: a
Explanation: Energy possessed by a periodic signal is the integral of the square of the magnitude of the signal over a time period.
5. Sinusoidal signals multiplied by decaying exponentials are referred to as
a) Amplified sinusoids
b) Neutralized sinusoids
c) Buffered sinusoids
d) Damped sinusoids
Answer: d
Explanation: The decaying exponentials dampen the amplitudes of sinusoids. Hence, the term damped sinusoids.
6. What is the power possessed by a signal exp?
a) 1
b) 0.5
c) 1 ⁄ w
d) w
Answer: a
Explanation: The power = Energy/Time period for a periodic signal. Hence, Power = 1.
7. What is the period of expt?
a) 4
b) 8
c) 16
d) 20
Answer: b
Explanation: The fundamental period = 2pi/ = 8.
8. exp is periodic
a) for any w
b) for any t
c) for no w
d) for no t
Answer: a
Explanation: Any two instants, t and t + 2pi will be equal, hence the signal will be periodic with period 2pi.
9. Define the fundamental period of the following signal x[n] = exp + exp?
a) 8
b) 12
c) 18
d) 24
Answer: d
Explanation: The first signal, will repeat itself after 3 cycles. The second will repeat itself after 8 cycles. Thus, both of them together will repeat themselves only after LCM = 24 cycles.
10. exp[jwn] is periodic
a) for any w
b) for any t
c) for w=2pi*M/n
d) for t = 1/w
Answer: c
Explanation: Discrete exponentials are periodic only for a particular choice of the fundamental frequency.
11. The most general form of complex exponential function is:
a) e σt
b) e Ωt
c) e st
d) e at
Answer: c
Explanation: The general form of complex exponential function is: x = e st where s = σ + jΩ.
12. A complex exponential signal is a decaying exponential signal when
a) Ω = 0 and σ > 0
b) Ω = 0 and σ = 0
c) Ω ≠ 0 and σ < 0
d) Ω = 0 and σ < 0
Answer: d
Explanation: Let x be the complex exponential signal
⇒ x = e st = e t = e σt e jΩt
Now, when Ω = 0 ⇒ x = e σt which will be an exponentially decaying signal if σ < 0.
13. When is a complex exponential signal sinusoidal?
a) σ =0 and Ω = 0
b) σ < 0 and Ω = 0
c) σ = 0 and Ω ≠ 0
d) σ ≠ 0 and Ω ≠ 0
Answer: c
Explanation: A signal is sinusoidal when σ = 0 and Ω ≠ 0
⇒ x = e st = e t = e σt e jΩt = e jΩt = cosΩt + jsinΩt which is sinusoidal.
14. An exponentially growing sinusoidal signal is:
a) σ = 0 and Ω = 0
b) σ > 0 and Ω ≠ 0
c) σ < 0 and Ω ≠ 0
d) σ = 0 and Ω ≠ 0
Answer: b
Explanation: A complex exponential signal is sinusoidal when Ω has a definite value i.e., Ω ≠ 0. It can either be growing exponential or decaying exponential based on the value of σ.
∴ A signal is sinusoidal growing exponential when σ > 0 and Ω ≠ 0.
15. Determine the nature of the signal: x = e -0.2t [cosΩt + jsinΩt].
a) Exponentially decaying sinusoidal signal
b) Exponentially growing sinusoidal signal
c) Sinusoidal signal
d) Exponential signal
Answer: a
Explanation: Clearly the signal has negative exponential ⇒ Decaying exponential signal.
The signal also has sinusoidal component.
∴ The signal is exponentially decaying sinusoidal signal.
1. Is the function y[n] = cos periodic or not?
a) True
b) False
Answer: a
Explanation: ‘y’ will be periodic only if x attains the same value after some time, T. However, if x is a one-one discrete function, it may not be possible for some x[n].
2. If n tends to infinity, is the accumulator function an unstable one?
a) The function is marginally stable
b) The function is unstable
c) The function is stable
d) None of the mentioned
Answer: b
Explanation: The system would be unstable, as the output will grow out of bound at the maximally worst possible case.
3. Comment on the causality of the following discrete time system: y[n] = x[-n].
a) Causal
b) Non causal
c) Both Casual and Non casual
d) None of the mentioned
Answer: b
Explanation: For positive time, the output depends on the input at an earlier time, giving causality for this portion. However, at a negative time, the output depends on the input at a positive time, i.e. at a time in the future, rendering it non causal.
4. Comment on the causality of the discrete time system: y[n] = x[n+3].
a) Causal
b) Non Causal
c) Anti Causal
d) None of the mentioned
Answer: c
Explanation: The output always depends on the input at a time in the future, rendering it anti-causal.
5. Consider the system y[n] = 2x[n] + 5. Is the function linear?
a) Yes
b) No
Answer: b
Explanation: As we give two inputs, x1 and x2, and give an added input x1 and x2, we do not get the corresponding y1 and y2. Thus, additive rule is disturbed and hence the system is not linear.
6.Comment on the time invariance of the following discrete system: y[n] = x[2n+4].
a) Time invariant
b) Time variant
c) Both Time variant and Time invariant
d) None of the mentioned
Answer: b
Explanation: A time shift in the input scale gives double the time shift in the output scale, and hence is time variant.
7. Is the function y[2n] = x[2n] linear in nature?
a) Yes
b) No
Answer: a
Explanation: The function obeys both additivity and homogeneity properties. Hence, the function is linear.
8. How is a linear function described as?
a) Zero in Finite out
b) Zero in infinite out
c) Zero in zero out
d) Zero in Negative out
Answer: c
Explanation: The system needs to give a zero output for a zero input so as to conserve the law of additivity, to ensure linearity.
9. Is the system y[n] = x 2 [n-2] linear?
a) Yes
b) No
Answer: b
Explanation: The system is not linear, as x1 2 + x2 2 is not equal to 2 .
10. Is the above system, i.e y[n] = x 2 [n-2] time invariant?
a) Yes
b) No
Answer: a
Explanation: A time shift of t0 will still result in an equivalent time shift of t0 in the output, and hence will be time invariant.
This set of Signals & Systems Quiz focuses on “Discrete-Time Systems in the Time-Domain – 2”.
1. The difference equation for an N th order discrete-time system is ___________
signals-systems-questions-answers-quiz-q1
Answer: c
Explanation: The difference equation for an Nth order discrete-time system is:
signals-systems-questions-answers-discrete-time-systems-time-domain-q1-1
2. The response of any discrete time system can be decomposed as _____________
a) Total Response=Impulse+step
b) Total Response=Impulse+Ramp
c) Total Response=zero-output response
d) Total Response=zero-state response+zero-input response
Answer: d
Explanation: There are two approaches to analyzing response of a system:
Direct solution of difference solution
Decomposing in terms of impulse signals
In the first method, the response of the system can be decomposed as:
Total Response = zero-state response + zero-input response.
3. Zero-state response of the system is _____________
a) Response of the system when initial state of the system is zero
b) Response of the system due to input alone
c) Response of the system due to input alone when initial state of the system is zero
d) Response of the system due to input alone when initial state is neglected
Answer: c
Explanation: Zero-state response of the system is the response of the system due to input alone when the initial state of the system is zero. That is the system is relaxed at time n = 0.
4. Zero-input response is also known as ____________
a) zero-state response
b) Natural response
c) state-input response
d) Forced response
Answer: b
Explanation: Natural response of the system is when the input x = 0.
5. The general solution of natural response is of the form of _________
a) y h = c 1 λ 1 n +c 2 λ 2 n +⋯+c N λ N n
b) y h = c 1 λ 1 n +c 2 λ 2 n +⋯+c N λ N n
c) y h = c 1 λ 1 2 +c 2 λ 2 2 +⋯+c N λ N 2
d) y h = c 1 λ 1 n -c 2 λ 2 n +⋯+c N λ N n
Answer: a
Explanation: The general solution of natural response is of the form:
y h = c 1 λ 1 n +c 2 λ 2 n +⋯+c N λ N n
The form will vary if the roots are repeating or complex.
6. Determine the natural response of the system: Difference equation is
y-y-2y=x and y = 1; y = 0
signals-systems-questions-answers-discrete-time-systems-time-domain-q6
Answer: c
Explanation: Natural Response of the system:
Homogenous equation ⇒ y-y-2y=0
The homogenous solution: y h = λ n
⇒ λ n – λ -2λ =0
⇒ λ [λ 2 – λ 1 -2]=0
⇒ λ 2 – λ-2=0
⇒ λ 2 -2λ+λ-2=0
⇒ λ+1=0
⇒ =0
⇒ λ 1 =2,λ 2 =-1
General form of homogenous solution is
y h = c 1 n +c 2 n
⇒ y= c 1 +c 2
⇒ y=2c 1 – c 2
⇒ y-y-2y=0
Given, y = 1 and y = 0
⇒ y-1=0⇒y=1
Similarly, y-y-2y=0⇒y=1+2=3
∴ y = 1 and y = 3
Comparing the above values with equations and
⇒ c 1 +c 2 =1 and 2c 1 – c 2 =3
Solving the two equations we get, c1 = 4/3 and c2 = -1/3
signals-systems-questions-answers-discrete-time-systems-time-domain-q6-1
7. Forced Response is solution of difference equation when ____________
a) Input is zero
b) Input is given and initial conditions are zero
c) Natural Response
d) Input is given and initial conditions are non-zero
Answer: b
Explanation: Forced response is solution of difference equation when input is given and initial conditions are zero. Also known as zero-state response.
8. Forced response consists of _________
a) Homogenous solution and general solution
b) General solution alone
c) Homogenous solution and particular solution
d) Particular solution alone
Answer: c
Explanation: Forced response consists of homogenous solution and particular solution.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Periodic and Non-Periodic Signals”.
1. Given the signal
X = cos t, if t<0
Sin t, if t≥0
The correct statement among the following is?
a) Periodic with fundamental period 2π
b) Periodic but with no fundamental period
c) Non-periodic and discontinuous
d) Non-periodic but continuous
Answer: c
Explanation: From the graphs of cos and sin, we can infer that at t=0, the function becomes discontinuous.
Since, cos 0 = 1, but sin 0 = 0
As 1 ≠ 0, so, the function X is discontinuous and therefore Non-periodic.
2. The fundamental period of the signal X = 10 cos 2 is __________
a) 0.2
b) 0.1
c) 0.5
d) No fundamental period exists
Answer: b
Explanation: X = 10 cos 2
Since, cos 2t = 2cos 2 t – 1
Or, cos 2 t = \
= 5 + 5 cos 20πt
Now, Y = cos 20πt
Fundamental period of the signal is = \(\frac{2π}{20π} = \frac{1}{10}\) = 0.1.
3. The even component of the signal X = e jt is _________________
a) Sin t
b) Cos t
c) Sinh t
d) Cosh t
Answer: b
Explanation: Let X e represents the even component of X
Now, X e = \
+ X ]
= \(\frac{1}{2}\)[e jt + e -jt ]
= cos t.
4. The odd component of the signal X = e jt is _______________
a) Sin t
b) Cos t
c) Sinh t
d) Cosh t
Answer: a
Explanation: Let X o represents the odd component of X
Now, X o = \
– X ]
= \(\frac{1}{2}\)[e jt + e -jt ]
= sin t.
5. The period of the signal X = 24 + 50 cos 60πt is _______________
a) \
60 π s
c) \
Non-periodic
Answer: a
Explanation: Period of cos t = 2π
Period of cos at = \(\frac{2π}{a}\)
Here, a = 60π
So, period of cos 60πt = \(\frac{2π}{60π}\)
= \(\frac{1}{30}\) s.
6. The period of the signal X = 10 sin 5t – 4 cos 9t is _______________
a) \
\
2π
d) Non-periodic
Answer: c
Explanation: Period of cos t = 2π
Period of cos at = \
= LCM [Period of X 1 , Period of X 2 ]
∴ Period of X = LCM
= 2π.
7. The period of the signal X = 5t – 2 cos 6000 πt is ________________
a) 0.96 ms
b) 1.4 ms
c) 0.4 ms
d) Non-periodic
Answer: d
Explanation: Period of cos t = 2π
Period of cos at = \
= LCM [Period of X 1 , Period of X 2 ]
∴ Period of X = LCM
= Indefinite.
8. The period of the signal X = 4 sin 6t + 3 sin \Missing open brace for subscript \
\
2π s
d) Non-periodic
Answer: d
Explanation: Period of sin t = 2π
Period of sin at = \
= LCM [Period of X 1 , Period of X 2 ]
∴ Period of X = LCM
= Indefinite.
9. The period of the signal Z = sin3t + cos 4t is _______________
a) periodic without a definite period
b) periodic with a definite period
c) non- periodic over an interval
d) non-periodic throughout
Answer: b
Explanation: Period of cos t = 2π
Period of cos at = \
= LCM [Period of X 1 , Period of X 2 ]
∴ Period of X = LCM
= definite
Hence Z is periodic with a definite period.
10. The signal X = e -4t u is _______________
a) Power signal with P ∞ = \
Power signal with P ∞ = 0
c) Energy signal with E ∞ = \
Energy signal with E ∞ = 0
Answer: c
Explanation: If a signal has E∞ as ∞ and P∞ as a finite value, then the signal is a power signal. If a signal has E∞ as a finite value and P ∞ as ∞, then the signal is an energy signal.
|x | < ∞, E ∞ = \
|^2 \,dt\)
= \
\,dt \)
= \
|^2 \,dt = ∞.\)
11. The signal X = \Missing open brace for subscript Power signal with P ∞ = 1
b) Power signal with P ∞ = 2
c) Energy signal with E ∞ = 2
d) Energy signal with E ∞ = 1
Answer: a
Explanation: If a signal has E ∞ as ∞ and P ∞ as a finite value, then the signal is a power signal. If a signal has E ∞ as a finite value and P ∞ as ∞, then the signal is an energy signal.
|x | = 1, E ∞ = \
|^2 \,dt = ∞\)
So, this is a power signal not an energy signal.
\
|^2 \,dt = 1.\).
12. Signal X is as shown in the figure below.
signals-systems-questions-answers-periodic-non-periodic-signals-q12
The total energy of X is _______________
a) 0
b) 13
c) \
\(\frac{26}{3}\)
Answer: d
Explanation: E = 2\
\,dt\)
= 2 \
\,dt\)
= 8 + \(\frac{2}{3} = \frac{26}{3}\).
13. A discrete time signal is as given below
\
\)
The period of the signal X [n] is ______________
a) 126
b) 32
c) 252
d) Non-periodic
Answer: a
Explanation: Given that, N 1 = 18, N 2 = 14
We know that period of X [n] = LCM (N 1 , N 2 )
∴ Period of X [n] = LCM = 126.
14. A discrete time signal is as given below
\
cos
\)
The period of the signal X [n] is _____________
a) 16 π
b) 16
c) 8
d) Non-periodic
Answer: d
Explanation: We know that for X [n] = X 1 [n] × X 2 [n] to be periodic, both X 1 [n] and X 2 [n] should be periodic with finite periods.
Here X 2 [n] = cos
, is periodic with fundamental period as \(\frac{8}{n}\)
But X 1 [n] = cos
is non periodic.
∴ X [n] is a non-periodic signal.
15. A discrete time signal is as given below
\
– sin
+ 3 cos
\)
The period of the signal X [n] is _____________
a) 16
b) 4
c) 2
d) Non-periodic
Answer: a
Explanation: Given that, N 1 = 4, N 2 = 16, N 3 = 8
We know that period of X [n] = LCM (N 1 , N 2 , N 3 )
∴ Period of X [n] = LCM = 16.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “The Impulse Function”.
1. How is the discrete time impulse function defined in terms of the step function?
a) d[n] = u[n+1] – u[n].
b) d[n] = u[n] – u[n-2].
c) d[n] = u[n] – u[n-1].
d) d[n] = u[n+1] – u[n-1].
Answer: c
Explanation: Using the definition of the Heaviside function, we can come to this conclusion.
2. What is the definition of the delta function in time space intuitively?
a) Defines that there is a point 1 at t=0, and zero everywhere else
b) Defines that there is a point 0 at t=0, and 1 everywhere else
c) Defines 1 for all t > 0, and 0 else
d) Defines an impulse of area 1 at t=0, zero everywhere else
Answer: d
Explanation: Arises from the definition of the delta function. There is a clear difference between just the functional value and the impulse area of the delta function.
3. Is it practically possible for us to provide a perfect impulse to a system?
a) Certainly possible
b) Impossible
c) Possible
d) None of the mentioned
Answer: b
Explanation: The spread of the impulse can never be restricted to a single point in time, and thus, we cannot achieve a perfect impulse.
4. The convolution of a discrete time system with a delta function gives
a) the square of the system
b) the system itself
c) the derivative of the system
d) the integral of the system
Answer: b
Explanation: The integral reduces to the the integral calculated at a single point, determined by the centre of the delta function.
5. Find the value of 2sgnd[0] + d[1] + d[45], where sgn is the signum function.
a) 2
b) -2
c) 1
d) 0
Answer: d
Explanation: sgn=0, and d[n] = 0 for all n not equal to zero. Hence the sum reduces to zero.
6. Where h*x denotes h convolved with x, x[n]*d[n-90] reduces to
a) x[n-89].
b) x[n-91].
c) x[n=90].
d) x[n].
Answer: c
Explanation: The function gets shifted by the center of the delta function during convolution.
7. Where h*x denotes h convolved with x, find the value of d[n]*d[n-1].
a) d[n].
b) d[n-1].
c) d 2 [n].
d) d 2 [n-1].
Answer: b
Explanation: Using the corollary, if we take d[n] to be the ‘x’ function, it will be shifted by -1 when convolved with d[n-1], thus rendering d[n-1].
8. How is the continuous time impulse function defined in terms of the step function?
a) u = d)/dt
b) u = d
c) d = du/dt
d) d = u 2
Answer: c
Explanation: Using the definition of the Heaviside function, we can come to this conclusion.
9. In which of the following useful signals, is the bilateral Laplace Transform different from the unilateral Laplace Transform?
a) d
b) s
c) u
d) all of the mentioned
Answer: c
Explanation: The bilateral LT is different from the aspect that the integral is applied for the entire time axis, but the unilateral LT is applied only for the positive time axis. Hence, the u [unit step function] differs in that aspect and hence can be used to differentiate the same.
10. What is the relation between the unit impulse function and the unit ramp function?
a) r = dd/dt
b) d = dr/dt
c) d = d 2 /dt 2
d) r = d 2 /dt 2
Answer: c
Explanation: Now, d = du/dt and u = dr/dt. Hence, we obtain the above answer.
This set of Signals & Systems MCQs focuses on “The Impulse Function – 2”.
1. What is the other name of a Continuous Time Unit Impulse Function?
a) Dirac delta function
b) Unit function
c) Area function
d ) Direct delta function
Answer: a
Explanation: The continuous time unit impulse function is also known as the Dirac delta function. This because it was first defined by Paul Adrein Maurice Dirac as ∂=0.
2. What is the area of a Unit Impulse function?
a) Zero
b) Half of Unity
c) Depends on the function
d) Unity
Answer: d
Explanation: The area under an impulse function is unity. It is defined between limits negative infinity to positive infinity with ∂dt=1, i.e ∫∂dt=1. It can be seen as a rectangular pulse with width that is negligible and the height that is infinitely large and area as one.
3. Why is the impulse duration important?
a) It is zero
b) It changes with time
c) It approaches zero
d) It depends on the situation
Answer: c
Explanation: One of the most interesting features of the impulse function, is not its shape, but the fact that its effective duration approaches zero, while the area remains unity. Hence, ∫∂dt=1.
4. What are the singularity functions?
a) Derivatives and integrals of unit impulse functions
b) Derivatives of a unit impulse function
c) Integrals of an impulse function
d) Sum of successive impulse function
Answer: a
Explanation: All the function derived from an impulse function are called singularity functions. Here, impulse function is taken as a generalized function than an ordinary function.
5. What properties does a Continuous time unit Impulse function follow?
a) Shifting, sampling, differentiation, multiplication
b) Multiplication, sampling, shifting
c) Shifting, multiplication, differentiation
d) Sampling only
Answer: a
Explanation: Continuous time impulse functions follows all the properties like shifting, scaling, sampling or multiplication property, differential.
6. Impulse function is an odd function.
a) True
b) False
Answer: b
Explanation: The Impulse Function is an even function. By scaling property of an Impulse function we can see, ∂=1/|a|∂
So, substituting, ∂=1/|-1|∂ we get ∂, hence, it is an even function. .
7. Multiplication of a signal with a Unit Impulse function gives the value of the signal at which the impulse is located.
a) True
b) False
Answer: a
Explanation: Multiplying the signal by a unit impulse samples the value of the signal at the point at which the impulse is located. That is x*∂=x|t=0=x∂.
8. What is a doublet function?
a) Branch of an impulse function
b) The output of an impulse function
c) The first derivative of an impulse function
d) Any continuous time impulse function has another name that is doublet function
Answer: c
Explanation: The first derivative of d∂/∂=∂’ is referred to as a doublet function. The derivatives of all orders of the impulse functions are also singularity functions. It is defined as d∂/dt=∂’=0.
9. What is the area under a doublet function?
a) Unity
b) Negative
c) Zero
d) Positive
Answer: c
Explanation: We can explain by-
Integration -infinity to +infinity x∂’dt= negative of Integration -infinity to +infinity x’∂’dt=-x’|t=0=-X’, where x is any continuous function having a continuous derivative at t=0. This is ∫∂’=0.
10. How are discrete unit impulse functions and discrete time unit step functions related?
a) They are inverse of each other
b) ∂=u-u
c) ∂=u*2∂
d) Integration of unit step function gives unit step function.
Answer: b
Explanation: From definition of u and u,
u – u=∂+sigma k=1 to infinity∂- sigma k=1 to infinity ∂ = ∂. In continuous time, ∂=du/dt.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “BIBO Stability”.
1. Which of the following systems is stable?
a) y = log)
b) y = sin)
c) y = exp)
d) y = tx + 1
Answer: b
Explanation: Stability implies that a bounded input should give a bounded output. In a,b,d there are regions of x, for which y reaches infinity/negative infinity. Thus the sin function always stays between -1 and 1, and is hence stable.
2. State whether the integrator system is stable or not.
a) Unstable
b) Stable
c) Partially Stable
d) All of the mentioned
Answer: a
Explanation: The integrator system keep accumulating values and hence may become unbounded even for a bounded input in case of an impulse.
3. For what values of k is the following system stable, y = (k 2 – 3k -4)log + sin?
a) k=1,4
b) k=2,3
c) k=5,4
d) k =4,-1
Answer: d
Explanation: The values of k for which the logarithmic function ceases to exist, gives the condition for a stable system.
4. For a bounded function, is the integral of the odd function from -infinity to +infinity defined and finite?
a) Yes
b) Never
c) Not always
d) None of the mentioned
Answer: a
Explanation: The odd function will have zero area over all real time space.
5. When a system is such that the square sum of its impulse response tends to infinity when summed over all real time space,
a) System is marginally stable
b) System is unstable
c) System is stable
d) None of the mentioned
Answer: b
Explanation: The system turns out to be unstable. Only if it is zero/finite it is stable.
6. Is the system h = exp stable?
a) Yes
b) No
c) Can’t say
d) None of the mentioned
Answer: c
Explanation: If w is a complex number with Im < 0, we could have an unstable situation as well. Hence, we cannot conclude [no constraints on w given].
7. Is the system h = exp stable?
a) Yes
b) No
c) Can’t say
d) None of the mentioned
Answer: a
Explanation: The integral of the system from -inf to +inf equals to a finite quantity, hence it will be a stable system.
8. Comment on the stability of the following system, y[n] = n*x[n-1].
a) Stable
b) Unstable
c) Partially Stable
d) All of the mentioned
Answer: b
Explanation: Even if we have a bounded input as n tends to inf, we will have an unbounded output. Hence, the system resolves to be an unstable one.
9. Comment on the stability of the following system, y[n] = n .
a) Stable
b) Unstable
c) Partially Stable
d) All of the mentioned
Answer: a
Explanation: Even if we have a bounded input as n tends to inf, we will have an bounded output. Hence, the system resolves to be a stable one.
10. What is the consequence of marginally stable systems?
a) The system will turn out to be critically damped
b) The system will be an overdamped system
c) It will be a damped system
d) Purely oscillatory system
Answer: d
Explanation: The system will be a purely oscillatory system with no damping involved.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “BIBO Stability & Systems in the Time Domain”.
1. What is the full form of BIBO?
a) Boundary input Boundary Output
b) Boundary Input Bounded Output
c) Bonded Input Bonded Output
d) Bounded Input, Bounded Output
Answer: d
Explanation: BIBO stands for Bounded input, Bounded Output. It gives the stability of a system through a simple explanation that a system will be stable if it’s both input and output are bounded i.e it is not infinity.
2. When is a system said to be BIBO stable?
a) When the boundary conditions of the system are stable
b) When there is stability in the overall system
c) Every Bounded input results in a bounded output
d) When the input and output conditions are stable
Answer: c
Explanation: A system is said to be stable if, for any bounded input x, the response y is also Bounded.
i.e |x|≤Bx<∞ implies |y≤Bx<∞.
3. When does a signal say to be bounded?
a) When it is stable
b) When it gives slow responses
c) Magnitude does not grow without bound
d) When it has small inputs
Answer: c
Explanation: A signal x is said to be bounded if its magnitude does not grow without bound.
i.e |x|≤Bx<∞.
4. How do you describe a stable system informally?
a) When small inputs lead to responses that do not diverge
b) When small inputs lead to responses that diverge
c) When large inputs lead to diverging outputs
d) All inputs lead to outputs that converge
Answer: a
Explanation: When small inputs lead to output responses that do not tend to infinity. The output of such systems does not diverge if the input does not diverge.
5. The system is stable when y= tx.
a) True
b) False
Answer: b
Explanation: Here we can see,
|x|≤Bx<∞, for all t.
Using the input output relation we have y= tx. And so we may write, |Y|=|tx|=|t||x|=|t|Bx.
As tends to infinity, the output also tends to infinity. That is it is unbounded. So it is unstable.
6. How is a time domain system analyzed?
a) Study of a system in accordance to changes in its inputs over time
b) Study of a system in accordance to changes in its over time
c) Study of a system in accordance to changes in its overall structure over time
d) Study of a system in accordance to how a system change itself overall in a time
Answer: d
Explanation: Analysis in the time domain by done in how signals behave over time. That is, a system or a signal is studied in accordance to how it changes itself overall in time.
7. What is the frequency domain?
a) Analysis of signals in a frequency range
b) Analysis of signals in their bandwidth
c) Analysis of a signal with respect to its frequency
d) Study of a system in accordance to changes in its overall frequency
Answer: c
Explanation: Though this answer is a bit confusing frequency domain is defined as an analysis of a signal or a system with respect to its frequency. This concept has emerged from the transformations and the ‘spectrum’ concept.
8. Time domain is easier for mathematical operation than frequency domain.
a) True
b) False
Answer: b
Explanation: Time domain analysis is much tedious and difficult to perform when it comes to lengthy solvable problems. Whereas, in a frequency domain, it is very quick to perform. Even stability is easily attained in frequency domain analysis.
9. What are the mathematical tools to convert a system from a time domain to frequency domain?
a) Fourier series, Fourier transform, Laplace transform, Z-transform
b) Fourier series only
c) Fourier series and Laplace transform only
d) Fourier series, Fourier transform and Laplace transform only
Answer: a
Explanation: Fourier series, Fourier transform, Laplace transform, z-transform are some tools to convert a system from a time domain to frequency domain analysis to make it simpler. In fact, the concept of frequency domain has emerged from these transformations. It was first given by Joseph Fourier.
10. One of the main limitations of time domain analysis is the noise and frequency.
a) True
b) False
Answer: a
Explanation: True, one of the main limitations of time domain analysis is the noise and frequency. This is because it is easier in the frequency domain to read it and detect it and solve it. Time domain analysis is much tedious and difficult to perform when it comes to lengthy solvable problems.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Applications of Signals on Circuits”.
1. A circuit tuned to a frequency of 1.5 MHz and having an effective capacitance of 150 pF. In this circuit, the current falls to 70.7 % of its resonant value. The frequency deviates from the resonant frequency by 5 kHz. Q factor is?
a) 50
b) 100
c) 150
d) 200
Answer: c
Explanation: \(Q = \frac{ω}{ω1 – ω2} = \frac{f}{f2-f1} \)
Here, f = 1.5 × 10 6 Hz
f1 = (1.5 × 10 6 – 5 × 10 3 )
f2 = (1.5 × 10 6 + 5 × 10 3 )
So, f2 – f1 = 10 × 10 3 Hz
\(∴ Q = \frac{1.5 × 10^6}{10 × 10^3}\) = 150.
2. A 440 V, 50 HZ AC source supplies a series LCR circuit with a capacitor and a coil. If the coil has 100 mΩ resistance and 15 mH inductance, then at a resonance frequency of 50 Hz the half power frequencies of the circuit are ______________
a) 50.53 Hz, 49.57 Hz
b) 52.12 HZ, 49.8 Hz
c) 55.02 Hz, 48.95 Hz
d) 50 HZ, 49 Hz
Answer: a
Explanation: Bandwidth, BW = \(\frac{f_o}{Q} = \frac{50}{47.115}\) = 1.061 Hz
f 2 , higher half power frequency = f 0 + \(\frac{BW}{2}\)
∴ \(f_2 = 50 + \frac{1.061}{2}\) = 50.53 Hz
f 1 , lower half power frequency = f 0 – \(\frac{BW}{2}\)
∴ f 1 = 100 – \(\frac{1.59}{2}\) = 49.47 Hz.
3. The even component of the signal Y = a jt is _________________
a) Sin t
b) Cos t
c) Sinh t
d) Cosh t
Answer: b
Explanation: Let Y a represents the even component of Y
Now, Y a = \
+ Y ]
= \(\frac{1}{2}\)[a jt + a -jt ]
= cos t.
4. The odd component of the signal Y = a jt is _______________
a) Sin t
b) Cos t
c) Sinh t
d) Cosh t
Answer: a
Explanation: Let Y o represents the odd component of Y
Now, Y o = \
– Y ]
= \(\frac{1}{2}\)[a jt + a -jt ]
= sin t.
5. The signal Y = e -2t u is _______________
a) Power signal with P ∞ = \
Power signal with P ∞ = \
Energy signal with E ∞ = \
Energy signal with E ∞ = 0
Answer: b
Explanation: If a signal has E∞ as ∞ and P∞ as a finite value, then the signal is a power signal. If a signal has E∞ as a finite value and P ∞ as ∞, then the signal is an energy signal.
|Y | < ∞, E ∞ = \
|^2 \,dt\)
= \
\,dt \)
= \
|^2 \,dt = ∞.\)
6. The signal X = \Missing open brace for subscript Energy signal with E ∞ = 2
b) Power signal with P ∞ = 2
c) Power signal with P ∞ = 1
d) Energy signal with E ∞ = 1
Answer: c
Explanation: If a signal has E ∞ as ∞ and P ∞ as a finite value, then the signal is a power signal. If a signal has E ∞ as a finite value and P ∞ as ∞, then the signal is an energy signal.
|x | = 1, E ∞ = \
|^2 \,dt = ∞\)
So, this is a power signal not an energy signal.
\
|^2 \,dt = 1.\).
7. Given the signal
Y = cos t, if t>0
Sin t, if t≥0
The correct statement among the following is?
a) Periodic with fundamental period 2π
b) Non-periodic and discontinuous
c) Periodic but with no fundamental period
d) Non-periodic but continuous
Answer: b
Explanation: From the graphs of cos and sin, we can infer that at t=0, the function becomes discontinuous.
Since, cos 0 = 1, but sin 0 = 0
As 1 ≠ 0, so, the function X is discontinuous and therefore Non-periodic.
8. Two series resonant filters are shown below. Let the cut-off bandwidth of filter 1 be B 1 and that of filter 2 be B 2 . The value of \(\frac{B_1}{B_2}\) is ____________
signals-systems-questions-answers-applications-signals-circuits-q8
a) 0.25
b) 1
c) 0.5
d) 0.75
Answer: a
Explanation: For series resonant circuit, 3dB bandwidth is \(\frac{R}{L}\)
B 1 = \(\frac{R}{L_1}\)
B 2 = \(\frac{R}{L_2} = \frac{4R}{L_1}\)
Hence, \(\frac{B_1}{B_2}\) = 0.25.
9. In a series RLC circuit for lower frequency and for higher frequency, power factors are respectively ______________
a) Leading, Lagging
b) Lagging, Leading
c) Independent of Frequency
d) Same in both cases
Answer: a
Explanation: A Leading power factor means that the current in the circuit leads the applied voltage. This condition occurs in capacitive circuits. On the other hand, a lagging power factor indicates that current lags the voltage and this condition happens in an inductive circuit.
10. A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q = 50. If R and L are doubled and C is kept same, the new Q of the circuit is ____________
a) 25.52
b) 35.35
c) 45.45
d) 20.02
Answer: b
Explanation: Quality factor Q of the series RLC circuit is given by, \(Q = \frac{1}{R} \sqrt{\frac{L}{C}}\)
Given that Q = 50
Q new = \(\frac{1}{2R} \sqrt{\frac{2L}{C}}\)
= \(\frac{1}{2} × \frac{1}{R} \sqrt{\frac{2L}{C}}\)
= \(\frac{1}{2} × \sqrt{2} × Q \)
= \(\frac{1}{2} × \sqrt{2}\) × 50 = 35.35.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Continuous Time Convolution – 1”.
For all the following questions, ‘*’ indicates convolution. $ indicates integral
1. Find the value of h[n]*d[n-1], d[n] being the delta function.
a) h[n-2].
b) h[n].
c) h[n-1].
d) h[n+1].
Answer: c
Explanation: Convolution of a function with a delta function shifts accordingly.
2. Evaluate u)*u, u being the heaviside function.
a) ) u/a
b) ) u/a
c) ) u/a
d) ) u/a
Answer: c
Explanation: Use the convolution formula.
3. Find the value of h[n]*d[n-5], d[n] being the delta function.
a) h[n-2].
b) h[n-5].
c) h[n-4].
d) h[n+5].
Answer: b
Explanation: Convolution of a function with a delta function shifts accordingly.
4. Evaluate u)*u, u being the heaviside function.
a) ) u/a
b) ) u/a
c) ) u/a
d) ) u/a
Answer: b
Explanation: Use the convolution formula.
5. Find the value of h[n-1]*d[n-1], d[n] being the delta function.
a) h[n-2].
b) h[n].
c) h[n-1].
d) h[n+1].
Answer: a
Explanation: Convolution of a function with a delta function shifts accordingly.
6. Find the convolution of x = expu, and h = u
a) 0.5exp u + 0.5u
b) 0.5exp u + 0.8u
c) 0.5exp u + 0.5u
d) 0.5exp u + 0.8u
Answer: a
Explanation: Divide it into 2 sectors and apply the convolution formula.
7. Find the value of h[n]*d[n+1], d[n] being the delta function.
a) h[n-2].
b) h[n].
c) h[n-1].
d) h[n+1].
Answer: d
Explanation: Convolution of a function with a delta function shifts accordingly.
8. Find the convolution of x = expu, and h = u
a) 0.33exp u + 0.5u
b) 0.5exp u + 0.8u
c) 0.33exp u + 0.5u
d) 0.33exp u + 0.33u
Answer: d
Explanation: Divide it into 2 sectors and apply the convolution formula.
9. Find the value of d*x, d being the delta function.
a) x
b) x
c) x
d) x
Answer: c
Explanation: Convolution of a function with a delta function shifts accordingly.
10. Find x*u
a) tx
b) t 2 x
c) $x(t 2 )
d) $x
Answer: d
Explanation: Apply the convolution formula. The above corollary exists for any x [not impulsive].
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Continuous Time Convolution – 2”.
For all the following problems, h*x denotes h convolved with x. $ indicates integral.
1. Find the value of [d – d] * -x[t+1].
a) x – x
b) x – x
c) x – x
d) x – x
Answer: b
Explanation: The delta function convolved with another function results in the shifted function.
2. If h1, h2 and h3 are cascaded, find the overall impulse response
a) h1 * h2 * h3
b) h1 + h2 + h3
c) h3
d) all of the mentioned
Answer: a
Explanation: The resultant impulse response will be the convolution of all the subsequent impulse responses.
3. Find the value of [d – d] * x[t+3].
a) x – x
b) x – x
c) x – x
d) x – x
Answer: c
Explanation: The delta function convolved with another function results in the shifted function.
4. If h1, h2 and h3 are cascaded, and h1 = u, h2 = d and h3 = d, find the overall impulse response
a) s
b) d
c) u
d) all of the mentioned
Answer: c
Explanation: The resultant impulse response will be the convolution of all the subsequent impulse responses.
5. Find the value of [d – u] * x[t+1].
a) x – x – x
c) x – x – x
Answer: a
Explanation: The delta function convolved with another function results in the shifted function.
6. If h1, h2 and h3 are cascaded, and h1 = u, h2 = d and h3 = d, find the overall impulse response
a) u
b) u
c) u
d) all of the mentioned
Answer: a
Explanation: The resultant impulse response will be the convolution of all the subsequent impulse responses.
7. Find the value of [u – d] * -x[t+1].
a) x
c) x – x
Answer: b
Explanation: The delta function convolved with another function results in the shifted function.
8. If h1, h2 and h3 are parallelly summed, find the overall impulse response
a) h1 + h2 + h3
b) h1 – h2 + h3
c) h1*h2*h3
d) all of the mentioned
Answer: a
Explanation: The resultant impulse response will be the convolution of all the subsequent impulse responses.
9. Find the value of [u – u] * x[t+1].
a) x
b) x
c) x
d) x
Answer: d
Explanation: The delta function convolved with another function results in the shifted function.
10. If h1, h2 and h3 are cascaded, and h1 = u, h2 = exp and h3 = sin, find the overall impulse response
a) sin*exp*u
b) sin + exp + u
c) u*sin
d) all of the mentioned
Answer: a
Explanation: The resultant impulse response will be the convolution of all the subsequent impulse responses.
11. Who started the Convolution theorem?
a) Sylvestre François Lacroix
b) Vito Volterra
c) Pierre Simon Laplace
d) D’Alembert
Answer: d
Explanation: One of the earliest uses of the convolution integral appeared in D’Alembert’s derivation of Taylor’s theorem, 1754.Sylvestre François Lacroix, has also used convolution on page 505 of his book entitled Treatise on differences and series.
12. What is periodic convolution?
a) Continuous type superposition
b) Periodic type summation
c) Discrete type addition
d) Summation of both continuous and periodic type
Answer: b
Explanation: When a function g is periodic, with period T, then for functions, f, such that f∗g exists, the convolution is also periodic. This is called a periodic convolution.
Example: f*g=∫∑[f]g.
13. What is a circular or cyclic convolution?
a) Convolution of a periodic and continuous time function
b) Convolution of a periodic and discrete time function
c) Superposition of periodic and periodic function
d) Summation of continuous time and a convolution of a periodic function convolution
Answer: d
Explanation: The circular convolution is done of two aperiodic functions happens when one of them is convolved in the normal way with a periodic summation of the other given function.
[n]=∑h[m].xn[n-m], for discrete sequences n.
This set of Signals & Systems online test focuses on “Continuous Time Convolution – 3”.
1. What is the full form of the LTI system?
a) Linear time inverse system
b) Late time inverse system
c) Linearity times invariant system
d) Linear Time Invariant system
Answer: d
Explanation: LTI system stands for linear time Invariant system. It investigates the response of a linear and time variant system to an arbitrary input signal.
2. What is a unit impulse response?
a) The output of a linear system
b) The response of an invariant system
c) The output of an LTI system due to unit Impulse signal
d) The output of an input response signal
Answer: c
Explanation: The impulse response is defined as the output of an LTI System due to a unit impulse signal input applied at time t=0 or n=0.
x–>y
∂–>h
Where ∂ is the unit impulse function and h is the unit impulse response of a continuous time LTI system.
3. How are the convolution integral of signals represented?
a) x+h
b) x-h
c) x*h
d)x**h
Answer: c
Explanation: We obtain the system output y to an arbitrary input x in terms of the input response h.
y= ∫xhdα=x*h.
4. How do you define convolution?
a) Weighted superposition of time shifted responses
b) Addition of responses of an input signal
c) Multiplication or various shifted responses of a stable system
d) Superposition of various outputs
Answer: a
Explanation: This is defined as-
y= ∫xhdα=x*h, output y to an arbitrary input x in terms of the input response h.
This is defined as a weighted superposition of time shifted responses where the whole of the signals is taken into account i.e its full limits.
5. The Convolution of the continuous functions f=e-t2 and g=3t2 is 5.312t2.
a) True
b) False
Answer: b
Explanation: f=e-t2 and g=3t2
f*g= ∫fgdα
=∫e-α2-32 dα
=3t2√π-0+3√π/2
=5.31t2+2.659.
6. What is the difference between convolution and multiplication?
a) Convolution leads to addition and multiplication leads to the multiplication
b) Convolution leads to a superposition of signals while multiplication does not consider all the signals
c) Convolution is multiplication but of signals
d) Convolution is a multiplication of added signals.
Answer: b
Explanation: Convolution is defined as weighted superposition of time shifted responses where the whole of the signals is taken into account. But multiplication leads to loss of those signals which are after the limits.
7. Convolution leads to loss of signals.
a) True
b) False
Answer: b
Explanation: False, convolution is superimposition hence it does not lead to loss of signals. But multiplication does. It keeps the signal intact while superimposing it.
8. Convolution is considered in case of ________
a) Discrete time systems only
b) Continuous time only
c) In both continuous time and discrete time
d) Superposition of various outputs
Answer: c
Explanation: Convolution is considered in case of both continuous time and discrete time systems. In continuous time it is represented by x*h and in discrete time as x[n]*h[n], x is input and h is the response in both cases.
9. Choose the properties which are very important in case of LTI signals and systems?
a) Linearity and time invariance
b) Linearity and stability
c) Stability and invariance
d) Linearity and causality
Answer: a
Explanation: Linearity and time invariance are the most important properties which are very important in case of LTI signals and systems as they even derive their name Linear time invariance from them. It is also because many physical properties possess these properties.
10. Why is a linear time invariant systems important?
a) They can be structured as wanted
b) They can be molded in any domain
c) They are easy to define
d) They can be represented as a linear combination of signals
Answer: d
Explanation: A Linear time invariant system is important because they can be represented as linear combination of delayed impulses. This is in case of both continuous and discrete time signals. So, output can be easily calculated through superposition that is convolution.
11. What is a dummy variable?
a) Unused variable
b) Extra variable
c) Free variable
d) Something that is used to store extra numbers
Answer: c
Explanation: A free variable or a dummy variable which are mostly used in LTI systems are notation that specifies places in an expression where substitution may take place. They are very useful for calculation of convolution.
12. When are dummy variables used in continuous time convolution?
a) To change the limits of integration
b) To change the domain of integration
c) To substitute time analysis
d) To substitute frequency analysis
Answer: b
Explanation: Dummy variables are used in continuous time convolution to change the domain of integration i.e the input and response are first changed into a dummy variable domain before convolution.
Suppose x is the input and h is the impulse response so to find the convolution, the domain is changed to a dummy variable e.
So, x*h=x*h.
13. After converting the input and output to a dummy variable, the next step of convolution is ________
a) Shifting any one of the signals to left side i.e towards the negative direction
b) Changing the dummy variables
c) Shift the impulse response
d) Shift the input
Answer: a
Explanation: The next step of convolution after conversion to dummy variable is shifting any one of the input or response to the negative side. This is done for superimposition purposes.
14. Continuous time convolution is done from negative infinity to positive infinity.
a) True
b) False
Answer: a
Explanation: Convolution is a superposition theorem hence we have to consider the signals from negative to positive infinity. We start at t, at−∞ and slide it all the way to +∞. Wherever the two functions intersect, we find the integral of their product.
15. It does not matter which one we shift, the input signal or the unit impulse response of a system during linear convolution in an integral.
a) True
b) False
Answer: a
Explanation: It does not matter which one we shift input or output. We start at t at −∞ and slide it all the way to +∞. Wherever the two functions intersect, we find the integral of their product.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Properties of LTI Systems – 1”.
1. What is the rule h* = *h called?
a) Commutativity rule
b) Associativity rule
c) Distributive rule
d) Transitive rule
Answer: a
Explanation: By definition, the commutative rule h*x=x*h.
2. Does the system h = exp correspond to a stable system?
a) Yes
b) No
c) Marginally Stable
d) None of the mentioned
Answer: c
Explanation: The system corresponds to an oscillatory system, this resolving to a marginally stable system.
3. What is the rule h* = *c called?
a) Commutativity rule
b) Associativity rule
c) Distributive rule
d) Associativity and Commutativity rule
Answer: d
Explanation: By definition, the commutative rule i h*x=x*h and associativity rule = h* = *c.
4. Is y[n] = n*cosu[n] a stable system?
a) Yes
b) No
c) Marginally stable
d) None of the mentioned
Answer: b
Explanation: The ‘n’ term in the y[n] will dominate as it reaches to infinity, and hence could reach infinite values.
5. What is the rule *c = h* called?
a) Commutativity rule
b) Associativity rule
c) Distributive rule
d) Transitive rule
Answer: b
Explanation: By definition, the associativity rule = h* = *c.
4. Is y[n] = n*sinu[-n] a stable system?
a) Yes
b) No
c) Marginally stable
d) None of the mentioned
Answer: b
Explanation: The ‘n’ term in the y[n] will dominate as it reaches to negative infinity, and hence could reach infinite values. Eventhough + infinity would not be a problem, still the resultant system would be unstable.
7. What is the following expression equal to: h*)), d is the delta function
a) h*c + h*b
b) h*c*b + b
c) h*c*b + h*c
d) h*c*b + h
Answer: c
Explanation: Apply commutative and associative rules
8. Does the system h = exp correspond to a stable system?
a) Yes
b) No
c) Marginally Stable
d) None of the mentioned
Answer: b
Explanation: The system corresponds to an unstable system, as the Re term is a positive quantity.
9. The system transfer function and the input if exchanged will still give the same response.
a) True
b) False
Answer: a
Explanation: By definition, the commutative rule i h*x=x*h=y. Thus, the response will be the same.
10. For an LTI discrete system to be stable, the square sum of the impulse response should be
a) Integral multiple of 2pi
b) Infinity
c) Finite
d) Zero
Answer: c
Explanation: If the square sum is infinite, the system is an unstable system. If it is zero, it means h = 0 for all t. However, this cannot be possible. Thus, it has to be finite.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Properties of LTI Systems – 2”.
1. What is the rule h*x = x*h called?
a) Commutativity rule
b) Associativity rule
c) Distributive rule
d) Transitive rule
Answer: a
Explanation: By definition, the commutative rule h*x=x*h.
2. For an LTI discrete system to be stable, the square sum of the impulse response should be
a) Integral multiple of 2pi
b) Infinity
c) Finite
d) Zero
Answer: c
Explanation: If the square sum is infinite, the system is an unstable system. If it is zero, it means h = 0 for all t. However, this cannot be possible. Thus, it has to be finite.
3. What is the rule *c = h* called?
a) Commutativity rule
b) Associativity rule
c) Distributive rule
d) Transitive rule
Answer: b
Explanation: By definition, the associativity rule = h* = *c.
4. Does the system h = exp correspond to a stable system?
a) Yes
b) No
c) Marginally Stable
d) None of the mentioned
Answer: a
Explanation: The system corresponds to a stable system, as the Re term is negative, and hence will die down as t tends to infinity.
5. What is the following expression equal to: h*, d is the delta function
a) h + d
b) b + d
c) d
d) h + b
Answer: c
Explanation: Apply commutative and associative rules and the convolution formula for a delta function
6. Does the system h = exp correspond to a stable system?
a) Yes
b) No
c) Marginally Stable
d) None of the mentioned
Answer: c
Explanation: The system corresponds to an oscillatory system, this resolving to a marginally stable system.
7. What is the rule c* = *c called?
a) Commutativity rule
b) Associativity rule
c) Distributive rule
d) Associativity and Commutativity rule
Answer: d
Explanation: By definition, the commutative rule i h*x=x*h and associativity rule = h* = *c.
8. Is y[n] = n*sinu[-n] a causal system?
a) Yes
b) No
c) Marginally causal
d) None of the mentioned
Answer: b
Explanation: The anti causal u[-n] term makes the system non causal.
9. The system transfer function and the input if exchanged will still give the same response.
a) True
b) False
Answer: a
Explanation: By definition, the commutative rule i h*x=x*h=y. Thus, the response will be the same.
10. Is y[n] = nu[n] a linear system?
a) Yes
b) No
Answer: a
Explanation: The system is linear s it obeys both homogeneity and the additive rules.
This set of Signals & Systems online quiz focuses on “Properties of LTI Systems – 3”.
1. Which is special the property listed below only holds good by an LTI system?
a) Memory
b) Stability
c) Causality
d) Distributive property
Answer: d
Explanation: An LTI system holds a unique property of Associativity, Commutativity and Distributive Property which are not held by other systems. They have very special representations in terms of convolution and integrals.
2. What are the three special properties that only LTI systems follow?
a) Commutative property, Associative property, Causality
b) Associative property, Distributive property, Causality
c) Commutative property, Distributive property, Associative property
d) Distributive property, Stability, Causality
Answer: c
Explanation: Commutative property, Distributive property, Associative property are the unique properties of LTI systems which are special representations in terms of convolution and integrals.
3. Which is the commutative property of the LTI System in case of discrete time system?
a) x[n]+h[n]=h[n]+x[n]
b) x[n]+h[n]=h[n]*x[n]
c) x[n]*h[n]=h[n]*x[n]
d) x[t]*h[t]=h[n]*x[n]
Answer: c
Explanation: We represent commutative property as: x[t]*h[t]=h[t]*x[t] because it proves that convolution of two signals in either order will be same, with x[n] being the input and h[n] being the impulse response.
4. Does the commutative property holds good for both continuous and discrete signal?
a) Yes
b) No
Answer: a
Explanation: Yes, the commutative property is followed by both continuous time and discrete time LTI system. In this system, convolution of one side of the equation is equal to the other side.
5. Which is the correct representation of Commutative Continuous time LTI system?
a) ∫h + x = ∫x + h
b) ∫hx = ∫xh
c) ∫h – x = ∫xh
d) ∫h * x = ∫x * h
Answer: b
Explanation: According to the commutative rule, x*h = h*x= ∫hx = ∫xh, with x being the input and h being the impulse response.
6. What is the Distributive property of the LTI system?
a) x[n] + h1[n] + h2[n] = h1[n] + h2[n] + x[n]
b) x[n]* = x[n]**
c) x[n]* = x[n]*h1[n] + x[n]*h2[n]
d) x[n]* = * + x[n]*h2[n]
Answer: c
Explanation: x[n]* = x[n]*h1[n] + x[n]*h2[n], with x[n] being the input and h1[n] and h2[n] being the impulse responses.
7. What does the Distributive property signify?
a) The sum of signals in both the sides in any number must be equal
b) The responses must be equal in any side of an LTI system
c) The sum of two inputs must be equal to responses to these signals
d) The Multiplication of two signals in the inputs side is equal to multiplication of the responses
Answer: c
Explanation: The sum of two inputs must be equal to responses to these signals individually.
x[n]*=x[n]*h1[n]+x[n]*h2[n], with x[n] being the input and h1[n] and h2[2] being the impulse responses.
8. Which is the associative property of the LTI system?
a) x[n]*=x[n]*h1[n]+x[n]*h2[n]
b) x[n]*=h1[n]+x[n]+h2[n]
c) x[n]*h[n]=h[n]*x[n]
d) x[n]*=*h2[n]
Answer: d
Explanation: x[n]*=*h2[n], with x[n] being the input and h1[n] and h2[n] being the impulse responses which is the associative property.
9. What does the associative property apply?
a) Parallel connection of two systems is equivalent to a single system only in case of a continuous system
b) Series and parallel interconnection of two systems is equivalent to a single system only in case of discrete system
c) Series interconnection of two systems is equivalent to a single system in case of both continuous and discrete system
d) Series interconnection of two systems is equivalent to a single system only in case of discrete
Answer: c
Explanation: Series interconnection of two systems is equivalent to a single system in case of both continuous and discrete system. This can be generalized to an arbitrary number of LTI systems in cascade.
10. The order of the Cascade system doesn’t depend on the output. Which is the property?
a) Commutative
b) Associative
c) Commutative and distributive
d) Distributive
Answer: b
Explanation: Series interconnection of two systems is equivalent to a single system in case of both continuous and discrete system, which is the associative property. This can be generalized to an arbitrary number of LTI systems in cascade.
This set of Signals & Systems Question Bank focuses on “Properties of LTI Systems – 4”.
1. What are the properties of an LTI system posse other than Associative, Commutative and Distributive properties?
a) Memory, invertibility, causality, stability
b) Memory and non-causality
c) Invertibility and stability
d) Causality only
Answer: a
Explanation: A LTI System follows most of the properties that a normal system follows. This includes memory and memory-less property, invertibility, causality and stability.
2. An LTI system is memoryless only if ____________
a) It does not store the previous value of the input
b) It does not depend on any previous value of the input
c) It does not depend on stored values of the system
d) It does not depend on the present value of the input
Answer: b
Explanation: A LTI system is said to be memoryless only if it does not depend on any previous value of the input. That is we can say that if its output at any time depends only on the value of the input at the same time.
3. A continuous time LTI system has memory only when __________
a) It does not depend on the present value of the input
b) It only depends on the past values of the input
c) Its output always depends both on the previous and past values of the input
d) Its output might depend on the present value as well as the previous value of the input
Answer: d
Explanation: An LTI system is said to have a memory when its output at any time depends on the previous value of the input. This does not mean its value does not depend on present values. It depends both on past and present values according to the situation.
4. Which of the following system is memoryless?
a) h=0,t ≠0
b) h=x
c) h=0, t=0
d) h=kx
Answer: a
Explanation: A continuous-time LTI system is memoryless when h=0,t ≠0. Such memoryless system has the form h=kx, for some constant k has the impulse response h = k∂.
5. A continuous time LTI system is invertible only when its inverse exists.
a) True
b) False
Answer: a
Explanation: Yes, a continuous time LTI system is invertible only when its inverse exists that, when connected in series with the original system produces an output equal to the input to the first system. Furthermore, if a system is invertible we can say its inverse exists.
6. Invertibility is only followed by continuous time LTI systems.
a) True
b) False
Answer: b
Explanation: False, discrete time LTI System also follows invertibility properties. It can be shown by Impulse response h1[n] of the inverse system for an LTI system for an impulse response h[n] must satisfy
h[n]*h1[n]=∂[n].
7. Which property of an LTI system does the following equation prove h[n]*h1[n]=∂[n]?
a) Invertibility
b) Stability
c) Associativilty
d) Commutative
Answer: a
Explanation: This equation proves that the condition that h1[n] must satisfy to be the impulse response of the inverse system in case of discrete time LTI system. Thus this gives the necessary condition for the invertibility property of an LTI system.
8. A continuous time LTI system is causal only when __________
a) It depends on the present value of the input
b) It depends on the past values of the input
c) Its output always depends on future values of the input
d) Its output might depend only on the past and present values of the system
Answer: d
Explanation: An LTI system is said to be causal when its output at any time depends on the previous and present value of the input. That is its value does not depend only on past values.
9. An important property for causality of the system is __________
a) Initial rest
b) Final rest
c) It is memoryless
d) It is unstable
Answer: a
Explanation: A causal system follows what is called initial rest concept. That is if the input of the system is 0 upto some point in time than the output of the system should also be zero upto that time.
10. When a discrete time LTI system is said to be causal?
a) Output y[n] must not depend on x[k] for k>n
b) Output y[n] must not depend on x[k] for k=n
c) Output y[n] must not depend on x[k] for k<n
d) Output y[n] must depend on x[k] for k>n
Answer: a
Explanation: A causal system cannot depend on the future values of the input. It can only depend on the past values or present values.
11. Which of the following system is causal?
a) y[n] = 2[n] – 3[n+1]
b) y[n] = 2[n] + 3
c) y[n] = 2[n-7] – 3[n+1]
d) y[n] = 2[n]*3[n+1]
Answer: b
Explanation: An LTI system is said to be causal only when its output at any time depends on the previous or present value of the input. A causal system cannot depend on the future values of the input.
12. When are LTI systems stable?
a) Only when bounded input produces a bounded output
b) Only when bounded input produces a unbounded output
c) Only when unbounded input produces a bounded output
d) Only when unbounded input produces an unbounded output
Answer: a
Explanation: It follows from previous BIBO stability factor. It is the same for an LTI system too. Hence, a continuous or discrete time LTI system is said to be stable only when bounded input produces a bounded output.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Discrete Time Convolution – 1”.
1. Is discrete time convolution possible?
a) True
b) False
Answer: a
Explanation: Yes, like continuous time convolution discrete time convolution is also possible with the same phenomena except that it is discrete and superimposition occurs only in those time interval in which signal is present.
2. How is discrete time convolution represented?
a) x[n] + h[n]
b) x[n] – h[n]
c) x[n] * h[n]
d) x[n] + h[n]
Answer: c
Explanation: Discrete time convolution is represented by x[n]*h[n]. Here x[n] is the input and h[n] is the impulse response.
3. What are the tools used in a graphical method of finding convolution of discrete time signals?
a) Plotting, shifting, folding, multiplication, and addition in order
b) Scaling, shifting, multiplication, and addition in order
c) Scaling, multiplication and addition in order
d) Scaling, plotting, shifting, multiplication and addition in order
Answer: a
Explanation: The tools used in a graphical method of finding convolution of discrete time signals are basically plotting, shifting, folding, multiplication and addition. These are taken in the order in the graphs. Both the signals are plotted, one of them is shifted, folded and both are again multiplied and added.
4. Choose the correct equation for finding the output of a discrete time convolution?
a) y[n] = ∑x[k]h[n-k], k from 0 to ∞
b) y[n] = ∑x[k]h[n-k], k from -∞ to +∞
c) y[n] = ∑x[k]h[k], k from 0 to ∞
d) y[n] = ∑x[k]h[n], k from -∞ to +∞
Answer: b
Explanation: y[n]=∑x[k]h[n-k], k from -∞ to +∞
Is the correct equation, where x[n] is the input and h[n] is the impulse response of the ∂[n] input of an LTI system. This is referred to as the convolution sum.
5. What is a convolution sum?
a) ∑x[k]h[n-k], k from -∞ to +∞
b) ∑x[k]*∑h[n-k], k from -∞ to +∞
c) ∑x[k]+∑h[n-k], k from -∞ to +∞
d) ∑∑x[k]h[n-k], k from -∞ to +∞
Answer: a
Explanation: y[n]=∑x[k]h[n-k], k from -∞ to +∞, y[n] is the output of the summation of the components on the right hand side. Where x[n] is the input of an LTI system and h[n] is the impulse response of the ∂[n] input of an LTI system. The response is due to superposition, in short.
6. What is the convolution of x[n]= e-n2 and h[n]=n 2 ?
a) 5.318n 2 + .123
b) 6.318n 2 + .123
c) 5.318n+.88
d) 5.318n 2 +.8846
Answer: d
Explanation: x[n]*h[n]=∑x[k]h[n-k] =∑e -k2 [ 2 ] =3n 2 ∑e -k2 + ∑k 2 e -k2
=5.318n 2 +.8846.
7. Choose the properties which are followed by a discrete time convolution?
a) Associative, commutative, distributive
b) Associative
c) Commutative and distributive
d) Distributive and associative
Answer: a
Explanation: The properties which are followed by a discrete time convolution are same as continuous time convolution. These are – associative, commutative, distributive property.
8. What is the convolution of a signal with an impulse?
a) Signal itself
b) Impulse
c) A new signal
d) Signal multiplied by impulse
Answer: a
Explanation: The convolution of a signal x with a unit impulse function ∂ results in the signal x itself:
x* ∂=x.
9. What is the commutative property?
a) x*h=h*x
b) x+h=h+x
c) x**h=h**x
d) xh=hx
Answer: a
Explanation: The commutative property is x*h=h*x, where x is the input and h is the impulse response of the ∂ input of an LTI system.
∑x[k]h[n-k], when we change the variables to n-k to k-n makes it equal to LHS and RHS.
10. What is the associative property of discrete time convolution?
a) [x 1 * x 2 ]*h = x 1 * [x 2 *h]
b) [x 1 * x 2 ]+h = x 1 + [x 2 *h]
c) [x 1 + x 2 ]*h = x 1 * [x 2 +h]
d) [x 1 * x 2 ]h = x 1 [x 2 *h]
Answer: a
Explanation: [x 1 * x 2 ]*h= x 1 * [x 2 *h], x 1 and x 2 are inputs and h is the impulse response.
This can be proved by considering two x 1 * x 2 as one output and then using the commutative property proof.
This set of Signals & Systems Questions and Answers for Entrance exams focuses on “Discrete Time Convolution – 2”.
1. What is the distributive property of a discrete time convolution?
a) [x 1 + x 2 ]*h = x 1 * [x 2 + h]
b) [x 1 + x 2 ] = x 1 * [x 2 + h]
c) [x 1 + x 2 ]*h = x 1 * h+ x 2 * h
d) [x 1 + x 2 ]*h = x 1 * h* x 2 * h
Answer: c
Explanation: x 1 + x 2 ]*h = x 1 * h + x 2 * h,)], x 1 and x 2 are inputs and h is the impulse response of discrete time system.
2. What is this property of discrete time convolution?
x[n]*h[n]=y[n], then x[n]*h[n-n 0 ] = x[n-n 0 ]*h[n] = y[n-n 0 ]
a) Distributive
b) Commutative
c) Sym property
d) Shifting property
Answer: d
Explanation: x[n]*h[n]=y[n], then x[n]*h[n-n 0 ]= x[n-n 0 ]*h[n] = y[n-n 0 ] This gives x[n-n 1 ]*h[n-n 0 ] = y[n-n 0 -n 1 ] Is the shifting property of discrete time convolution.
3. What is the sum of impulses in a convolution sum of two discrete time sequences?
a) S y = S x S h , S x =∑x and S h = ∑h
b) S y = S x +S h , S x =∑x and S h = ∑h
c) S y = S x -S h , S x =∑x and S h = ∑h
d) S y = S x *S h , S x =∑x and S h = ∑h
Answer: a
Explanation: S y =S x +S h , , S x = ∑x and S h = ∑h, the sum of impulses in a convolution sum of two discrete time sequences is the product of the sums of the impulses in the two individual sequences. Here, y=x*h.
4. How can a cascade connected discrete time system respresented?
a) y[n] = x[n] + t[n] + r[n]
b) y[n] = x[n] * t[n] * r[n]
c) y[n] = x[n] * t[n] + r[n]
d) y[n] = x[n] + t[n] * r[n]
Answer: b
Explanation: y[n] = x[n]*t[n]*r[n], is how we can represent a cascade connected discrete time system.
Proof:
If Y1[n]=x[n]*t[n]
y[n]=Y1*r[n], using properties.
5. How can a parallel connected discrete time system respresented?
a) y[n] = x[n] + t[n] + r[n]
b) y[n] = x[n] * t[n] * r[n]
c) y[n] = x[n] *
d) y[n] = x[n] + t[n] * r[n]
Answer: b
Explanation: y[n] = x[n]* is how we can represent a parallel connected discrete time system.
Proof:
If Y1[n]=t[n]+r[n]
y[n]=Y1*r[n], using properties.
6. How can we solve discrete time convolution problems?
a) The graphical method only
b) Graphical method and tabular method
c) Graphical method, tabular method and matrix method
d) Graphical method, tabular method, matrix method and summation method
Answer: c
Explanation: Discrete time convolution problems are mostly solved by a graphical method, tabular method and matrix method. Even if the graphical method is very popular, the tabular and matrix method is more easy to calculate.
7. Which method uses sum of diagonal elements for discrete time convolution?
a) Matrix method only
b) Graphical method and tabular method
c) Graphical method, tabular method and matrix method
d) Graphical method, tabular method, matrix method and summation method
Answer: a
Explanation: Even if the graphical method is very popular, the tabular and matrix method is more easy to calculate. And matrix method uses the sum of diagonal elements for discrete time convolution.
8. Which method is close to a graphical method for discrete time convolution?
a) Matrix method only
b) Tabular method
c) Tabular method and matrix method
d) Summation method
Answer: b
Explanation: Tabular method is close to graphical method for discrete time convolution except that tabular representation of sequences is employed instead of graphical representation. Here every input is folded and shifted ad represented by a row.
9. The sample of x={1,2,3,1} and h={1,2,1,-1}, origin at 2, is 7.
a) True
b) False
Answer: a
Explanation: The input starts at n=0 and impulse at n=-1. So, output starts at n=0+=-1.
Output at is =4+4=1=7 samples. So, its true.
10. The convolution of x={1,2,3,1} and h={1,2,1,-1}, origin at 2, is?
a) {1,4,8,8,3,-2,-1}, origin at 4
b) {1,4,8,8,3,-2,1}, origin at 4
c) {1,3,8,8,3,-2,-1}, origin at 4
d) {1,4,8,3,-2,-1}, origin at 4
Answer: a
Explanation: Using tabular method:
signals-systems-questions-answers-entrance-exams-q10
Hence, result is {1,4,8,8,3,-2,-1}, origin at 4.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Convolution : Impulse Response Representation for LTI Systems – 1”.
1. Impulse response is the output of ______ system due to impulse input applied at time=0?
a) Linear
b) Time varying
c) Time invariant
d) Linear and time invariant
Answer: d
Explanation: Impulse response is the output of LTI system due to impulse input applied at time = 0 or n=0. Behaviour of an LTI system is characterised by the impulse response.
2. Which of the following is correct regarding to impulse signal?
a) x[n]δ[n] = x[0]δ[n]
b) x[n]δ[n] = δ[n]
c) x[n]δ[n] = x[n]
d) x[n]δ[n] = x[0]
Answer: a
Explanation: When the input x[n] is multiplied with an impulse signal, the result will be impulse signal with magnitude of x[n] at that time.
3. Weighted superposition of time-shifted impulse responses is termed as _______ for discrete-time signals.
a) Convolution integral
b) Convolution multiple
c) Convolution sum
d) Convolution
Answer: c
Explanation: Weighted superposition of time-shifted impulse responses is called convolution sum for discrete-time signals and convolution integral for continuous-time signals.
4. Which of the following is a correct expression for Impulse response?
signals-systems-questions-answers-convolution-impulse-response-representation-lti-systems-1-q4
Answer: a
Explanation: X [n] is represented as the weighted sum of time shifted impulses and concisely written as x[n] = ∑ ∞ k=-∞ x[k]δ[n-k]..
5. Determine the output of a LTI system, signals-systems-questions-answers-convolution-impulse-response-representation-lti-systems-1-q5
a) signals-systems-questions-answers-convolution-impulse-response-representation-lti-systems-1-q5a
b) signals-systems-questions-answers-convolution-impulse-response-representation-lti-systems-1-q5b
c) signals-systems-questions-answers-convolution-impulse-response-representation-lti-systems-1-q5c
d) signals-systems-questions-answers-convolution-impulse-response-representation-lti-systems-1-q5d
Answer: a
Explanation: X [n] is written as x[n]=δ[n]+4δ[n-1]-5δ[n-2] and y[n]=h[n]+4h[n-1]-5h[n-2]. By substituting the values for n = ….-2,-1, 0, 1, 2…. And corresponding h [n] we get y [n].
6. Determine the output of LTI system to signals-systems-questions-answers-convolution-impulse-response-representation-lti-systems-1-q6
a) signals-systems-questions-answers-convolution-impulse-response-representation-lti-systems-1-q6a
b) signals-systems-questions-answers-convolution-impulse-response-representation-lti-systems-1-q6b
c) signals-systems-questions-answers-convolution-impulse-response-representation-lti-systems-1-q6c
d) signals-systems-questions-answers-convolution-impulse-response-representation-lti-systems-1-q6d
Answer: b
Explanation: x[n]=5δ[n]+6δ[n-1] andy[n]=5h[n]+6h[n-1]. By substituting the values for n = ….-2,-1, 0, 1, 2…. And corresponding h[n] we get the output sequence y[n].
7. Find the convolution sum of sequences x1[n] = and x2[n] = .
a) {2, 5, 12, 11, 12}
b) {2, 12, 5, 11, 12}
c) {2, 11, 5, 12, 12}
d) {-2, 5,-12, 11, 12}
Answer: a
Explanation: x1[n] = δ+2δ+3δ and x2[n] = 2δ+δ+4δ
Y[n] = x1[n]*x2[n] by performing convolution operation on x1[n] and x2[n] we get the sequence as {2, 5, 12, 11, 12}.
8. Find the convolution sum of sequence x1[n] = {1, 2, 3} and signals-systems-questions-answers-convolution-impulse-response-representation-lti-systems-1-q8a
signals-systems-questions-answers-convolution-impulse-response-representation-lti-systems-1-q8
Answer: a
Explanation: x1[n] and x2[n] can be written as x1[n] = δ+2δ+3δ and
x2[n] = δ + 2 2δ + 3δ + 4δ and convolving x1[n], x2[n] we get
Y [n] = {1, 4, 10, 16, 17, 12}.
↑
9. Determine the output of LTI with input x[n] =2δ-δ and h[n] given as below.
signals-systems-questions-answers-convolution-impulse-response-representation-lti-systems-1-q9
signals-systems-questions-answers-convolution-impulse-response-representation-lti-systems-1-q9a
Answer: b
Explanation: The h[n] can be written as
{1, 3, 2,-1, 1}.
↑
By convolving x[n] with the given h[n] we get the following sequence as result
Y [n] = {2, 5, 1, -4, 3, -1}.
↑
10. Consider h [n] as in figure 1 and x[n] = u[n] – u [n-3], determine the output y [n] of the LTI system?
signals-systems-questions-answers-convolution-impulse-response-representation-lti-systems-1-q9
signals-systems-questions-answers-convolution-impulse-response-representation-lti-systems-1-q10
Answer: a
Explanation: The h[n] can be written as
{1, 3, 2,-1, 1}.
↑
By convolving x [n] with the given h [n] we get the output
y [n] = {1,4,6,4,2,0,1}.
↑
This set of Signals & Systems Questions and Answers for Campus interviews focuses on “Convolution : Impulse Response Representation for LTI Systems – 2”.
1. The convolution sum is given by _____ equation.
signals-systems-questions-answers-campus-interviews-q1
Answer: a
Explanation: By the definition of convolution sum we can write the equation as
x[n]*h[n] = ∑ ∞ k=-∞ x[k]h[n-k].
2. When the sequences x1 [n] = u [n] and x2 [n] = u [n-3], the output of LTI system is given as _____
a) y[n] = n-2, n>3
b) y[n] = n-2, n≥3
c) y[n] = n+2, n>3
d) y[n] = n-2, n≤3
Answer: b
Explanation: The output y[n] =∑ ∞ k=-∞ uu, by solving the above summation either by graphically or by direct summation we get signals-systems-questions-answers-campus-interviews-q2 .
3. The impulse response h of an LTI system is given by e -2t .u . What is the step response?
a) y = 1 ⁄ 2 (1 – e -2t ) u
b) y = 1 ⁄ 2 (1 – e -2t )
c) y = (1- e -2t ) u
d) y = 1 ⁄ 2 (e -2t ) u
Answer: a
Explanation: Given x = u and h = e -2t .u. By using convolution integral
signals-systems-questions-answers-campus-interviews-q3
We get output y as y = 1 ⁄ 2 (1 – e -2t ) u .
4. Is *h = h*x?
a) True
b) False
Answer: a
Explanation: By the properties of convolution we say that x*h = h*x
It can be proved using the convolution integral
signals-systems-questions-answers-campus-interviews-q3 .
5. Compute u convolved with itself?
a) y=t.u
b) y=u
c) y=t 2 .u
d) y=t.u
Answer: a
Explanation: By taking x = u and h = u and substituting in the integral
signals-systems-questions-answers-campus-interviews-q3 On solving the given integral we get y = t u .
6. Convolve the signals e -2t u, e -3t u. Determine the output?
a) y = (e -2t – e -3t )u
b) y = (e -2t – e -3t )
c) y = (e -3t – e -2t )u
d) y = (e -t – e -3t )u
Answer: a
Explanation: By solving the convolution integral
signals-systems-questions-answers-campus-interviews-q3 , we get output asy = (e -2t – e -3t )u.
7. Convolve graphically.
signals-systems-questions-answers-campus-interviews-q7
a) signals-systems-questions-answers-campus-interviews-q7a
b) signals-systems-questions-answers-campus-interviews-q7b
c) signals-systems-questions-answers-campus-interviews-q7c
d) signals-systems-questions-answers-campus-interviews-q7d
Answer: a
Explanation: Step 1: sketch x and h
Step 2: Obtain the product x h and the area under this product will give y
Step 3: sketch h and compute y and so on
Step 4: similarly sketch h and compute y and so on.
Hence we get the output as signals-systems-questions-answers-campus-interviews-q7a .
8. Convolve graphically the below given signals, and determine the correct sequence?
signals-systems-questions-answers-campus-interviews-q8
a) Y = 0, y = 2, y = 2
b) Y = 2, y = 2, y = 2
c) Y = 0, y = 0, y = 2
d) Y = 0, y = 3, y = 2
Answer: a
Explanation: Step 1: sketch x and h
Step 2: Obtain the product x h and the area under this product will give y
Step 3: sketch h and compute y and so on
Step 4: similarly sketch h and compute y and so on.
By following above steps we get the output as signals-systems-questions-answers-campus-interviews-q8a
9. Convolve given x with itself and choose the correct output.
signals-systems-questions-answers-campus-interviews-q9
a) signals-systems-questions-answers-campus-interviews-q9a
b) signals-systems-questions-answers-campus-interviews-q9b
c) signals-systems-questions-answers-campus-interviews-q9c
d) signals-systems-questions-answers-campus-interviews-q9d
Answer: a
Explanation: Step 1: sketch x and h
Step 2: Obtain the product x h and the area under this product will give y
Step 3: sketch h and compute y and so on
Step 4: similarly sketch h and compute y and so on.
By computing the above steps we get output y .
10. Find the convolution of x1[n] = {1, 2, 3, 4} and x2[n] = {2, 1, 2, 1}.
a) Y[n] = {14, 10, 14, 10}
b) Y [n] = {14, 16, 14, 16}
c) Y [n] = {14, 16,-14,-16}
d) Y [n] = {14,-16,-14, 16}
Answer: b
Explanation: By using convolution sum we get x1[n]*x2[n] = {14, 16, 14, 16}. This can be verified using tabular method of convolution.
This set of Signals & Systems Questions and Answers for Aptitude test focuses on “Properties of the Impulse Response Representation for LTI Systems”.
[NOTE:’*’ is reffered as convolution operation]
1. If two LTI systems with impulse response h1 and h2 and are connected in parallel then output is given by ______
a) y = x * + h2)
b) y = x + + h2)
c) y = x * h2)
d) y = * h1) + h2
Answer: a
Explanation: The equivalent impulse response of two systems connected in parallel is the sum of individual impulse responses. It is represented as
y = x * h1 + x * h2 = x * + h2).
2. When two LTI systems with impulse responses ha and hb are cascaded then equivalent response is given by ______
a) h = ha + hb
b) h = ha – hb
c) h = ha hb
d) h = ha * hb
Answer: d
Explanation: The equivalent impulse response of two systems connected in series is given by convolution of individual impulse responses.
3. What is this property of impulse response is called ___________
h1 * h2 = h2 * h1
a) Associative property
b) Commutative property
c) Distributive property
d) Closure law
Answer: b
Explanation: Impulse response exhibits commutative property and it is given mathematically by the equation
h1 * h2 = h2 * h1.
4. The overall impulse response of the system is given by ______
signals-systems-questions-answers-aptitude-test-q4
a) h = + h2 * h3) – h4
b) y = x * + h2*h3) – h4
c) h = + h2 * h3) + h4 * x
d) h = h2 * h3) – h4
Answer: a
Explanation: In the above given system h1 and h2 are connected in parallel hence it is
h1 +h2 which is cascaded to h3 and its equivalent is connected in parallel with h4 . Hence the equivalent impulse response is given by h = + h2 * h3) – h4.
5. The overall impulse response of the system is given by ______
signals-systems-questions-answers-aptitude-test-q5
a) h[n] = *h3[n]+h5[n]*h4[n]
b) h[n] = *h3[n])+h5[n])*h4[n]
c) h[n] = *h3[n])-h5[n])*h4[n]
d) h[n] = *-h3[n])-h5[n])*h4[n]
Answer: c
Explanation: Here in the above system h1 [n] and h2 [n] are connected in parallel and given by h1 [n] – h2 [n], this is cascaded with h3 [n] and given by * h3 [n], this is again connected in parallel with h5 [n] and its equivalent is cascaded with h4 [n]. The equivalent response is given by h[n] = *h3[n])-h5[n])*h4[n].
6. The condition for memory-less system is given by _____
a) h[k] = cδ[k]
b) h[k] = cδ[n-k]
c) h[k] = ch[k]δ[k]
d) h[k] = ch[n-k]δ[k]
Answer: a
Explanation: The LTI discrete-time system is said to be memory-less if and only if it satisfies the condition h[k]=cδ[k]. All memory-less LTI systems perform scalar multiplication on the input.
7. The causal continuous system with impulse response should satisfy ____ equation.
a) h=0,t<0
b) h=0,t>0
c) h≠0,t<0
d) h≠0,t≤0
Answer: a
Explanation: To the continuous system to be causal, the impulse response should satisfy the equation h=0,t<0 and convolution integral is given by y=∫ 0 ∞ hxdτ.
8. Causal systems are ______
a) Anticipative
b) Non anticipative
c) For certain cases anticipative
d) For certain cases anticipative and non anticipative
Answer: b
Explanation: Causal systems are non anticipative. They cannot generate an output before the input is applied. Which indicates the impulse response is zero for negative time.
9. Which of the following is true for discrete-time stable systems?
signals-systems-questions-answers-aptitude-test-q9
Answer: a
Explanation: If the condition ∑ ∞ k=-∞ |h[k]|<∞ is satisfied by an LTI system then it is said to be stable and for continuous time signal the condition is given by integral
∫ ∞ -∞ |h|dτ<∞.
10. The impulse response of discrete-time signal is given by h [n] = u [n+3]. Whether the system is causal or not?
a) Causal
b) Non-causal
c) Insufficient information
d) The system cannot be classified
Answer: b
Explanation: The given impulse response h [n] = u [n+3] is not causal because of the term u [n+3] which implies it is non zero for n= -1, -2, -3.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Basics of Linear Algebra”.
1. Find the values of x, y, z and w from the below condition.
\
x=1, y=3, z=4, w=0
b) x=2, y=3, z=8, w=1
c) x=1, y=2, z=3, w=1
d) x=1, y=2, z=4, w=1
Answer: d
Explanation: 5z=10+5 => 5z=15 => z=3
5x=2+z => 5x=5 => x=1
5y=3+7 => 5y=10 => y=2
5w=2+2+w => 4w=4 => w=1.
2. The matrix A is represented as \
\
\
\
\(
\)
Answer: c
Explanation: Given matrix is a 3×2 matrix and the transpose of the matrix is 3×2 matrix.
The values of matrix are not changed but, the elements are interchanged, as row elements of a given matrix to the column elements of the transpose matrix and vice versa but the polarities of the elements remains same.
3. Find the inverse of the matrix \
\
\
\
\(\frac{1}{13}*
\)
Answer: d
Explanation: The inverse of matrix A = \(\frac{adjA}{|A|}\),
adjA=AA -1 ,
adjA = \(\frac{1}{13}*
\), |A|=13.
4. Given the equations are 4x+2y+z=8, x+ y+ z=3, 3x+y+3z=9. Find the values of x, y and z.
a) 5/3, 0, 2/3
b) 1, 2, 3
c) 4/3, 1/3, 5/3
d) 2, 3, 4
Answer: a
Explanation: The matrix from the equations is represented as M=\(
\)
The another matrix is X = \(
\)
Then |M| = 6
For x=\(
\) = 5/3
Similarly, y=0, z=-2/3.
5. Find the adjacent A as A=\
\
\
\
\(
\)
Answer: b
Explanation: The adjacency of A is given by AA T
A T = \(
\),
AA T = \(
×
\)
adjA=\(
\).
6. Find the rank of the matrix A=\
3
b) 2
c) 1
d) 0
Answer: a
Explanation: To find out the rank of the matrix first find the |A|
If the value of the|A| = 0 then the matrix is said to be reduced
But, as the determinant of A has some finite value then, the rank of the matrix is 3.
7. The rank of the matrix where m<n cannot be more than?
a) m
b) n
c) m*n
d) m-n
Answer: a
Explanation: let us consider a 2×3 matrix \(
\)
Where R 1 ≠R 2 rank is 2
Another 2×3 matrix \(
\)
Here, R 1 =R 2 rank is 1
And the rank of these two matrices is 1, 2
So rank is cannot be more than m.
8. Given A=\
\
\
\
\(\frac{5}{20}\)
Answer: b
Explanation: AA -1 = I = \(
=
\)
Therefore, a = \(\frac{1}{60}\) and b = \(\frac{1}{3}\) and a + b = \(\frac{7}{20}\).
9. If a square matrix B is skew symmetric then.
a) B T = -B
b) B T = B
c) B -1 = B
d) B -1 = B T
Answer: a
Explanation: The transpose of a skew symmetric matrix should be equal to the negative of the matrix
Example: let us consider a matrix B = \(
\), B T = \(
\).
10. For the following set of simultaneous equations 1.5x-0.5y=2, 4x+2y+3z=9, 7x+y+5=10.
a) The solution is unique
b) Infinitely many solutions exist
c) The equations are incompatible
d) Finite number of multiple solutions exist
Answer: a
Explanation: The equations can be written as \(
\)
It can also be written as A = \(
\), |A|=19
Hence, it has a unique solution.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Eigenvalues”.
1. Find the Eigen values of matrix \
2 + \ 2, 1, 2
c) 2, 2, 0
d) 2, 2, 2
Answer: a
Explanation: To find the Eigen values it satisfy the condition, |A-λI|=0
|A-λI| = \(
– λ
\)
|A-λI| = \(
\)
= 2 – (λ 2 -4λ+3) –
By solving the above equation, we get,
λ = 2 + \(\sqrt{2}\), 2-\(\sqrt{2}\), 2.
2. Find the product of Eigen values of a matrix \
60
b) 45
c) -60
d) 40
Answer: c
Explanation: According to the property of Eigen values, the product of the Eigen values of a given matrix is equal to the determinant of the matrix |A| = 1 – 2 + 4
= -60.
3. Let us consider a square matrix A of order n with Eigen values of a, b, c then the Eigen values of the matrix A T could be.
a) a, b, c
b) -a, -b, -c
c) a-b, b-a, c-a
d) a -1 , b -1 , c -1
Answer: a
Explanation: According to the property of the Eigen values, any square matrix A and its transpose A T have the same Eigen values.
4. What is Eigen value?
a) A vector obtained from the coordinates
b) A matrix determined from the algebraic equations
c) A scalar associated with a given linear transformation
d) It is the inverse of the transform
Answer: c
Explanation: Eigen values is a scalar associated with a given linear transformation of a vector space and having the property that there is some nonzero vector which is when multiplied by the scalar is equal to the vector obtained by letting the transformation operate on the vector.
5. Find the sum of the Eigen values of the matrix \
7
b) 8
c) 9
d) 10
Answer: b
Explanation: According to the property of the Eigen values, the sum of the Eigen values of a matrix is its trace that is the sum of the elements of the principal diagonal.
Therefore, the sum of the Eigen values = 3 + 4 + 1 = 8.
6. Let the matrix A be the idempotent matrix then the Eigen values of the idempotent matrix are ________
a) 0, 1
b) 0
c) 1
d) -1
Answer: a
Explanation: According to the property of the Eigen values, the Eigen values of the idempotent matrix are either zero or unity.
So, the answer is 0 or 1.
7. Let us consider a 3×3 matrix A with Eigen values of λ 1 , λ 2 , λ 3 and the Eigen values of A -1 are?
a) λ 1 , λ 2 , λ 3
b) \
-λ 1 , -λ 2 , -λ 3
d) λ 1 , 0, 0
Answer: b
Explanation: According to the property of the Eigen values, if is the Eigen value of A, then \( \frac{1}{λ}\) is the Eigen value of A -1 .
So the Eigen values of A -1 are \( \frac{1}{λ_1}, \frac{1}{λ_2}, \frac{1}{λ_3}\).
8. The Eigen values of a 3×3 matrix are λ 1 , λ 2 , λ 3 then the Eigen values of a matrix A 3 are __________
a) λ 1 , λ 2 , λ 3
b) \
\
1, 1, 1
Answer: c
Explanation: If λ 1 , λ 2 , λ 3 ……… λ n are the Eigen values of matrix A then the Eigen values of matrix A m are said to be \(λ_1^m, λ_2^m, λ_3^m,………λ_n^m\).
So, the answer is \(λ_1^3, λ_2^3, λ_3^3\).
9. Find the Eigen values of matrix A=\
3, -3
b) -3, -5
c) 3, 5
d) 5, 0
Answer: c
Explanation: According to the property of the Eigen value, the eigen values are determined as follows:
4 + 4 = 8
3 + 5 = 8
The sum of the Eigen values is equal to the sum of the principal diagonal elements of the matrix.
10. Where do we use Eigen values?
a) Fashion or cosmetics
b) Communication systems
c) Operations
d) Natural herbals
Answer: b
Explanation: Eigen values are used in communication systems, designing bridges, designing car stereo system, electrical engineering, mechanical companies.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Periodic Signals – 1”.
1. What are periodic signals?
a) The signals which change with time
b) The signals which change with frequency
c) The signal that repeats itself in time
d) The signals that repeat itself over a fixed frequency
Answer: c
Explanation: Those signals which repeat themselves in a fixed interval of time are called periodic signals. The continuous-time signal x is periodic if and only if
x= x.
2. Periodic signals are different in case of continuous time and discrete time signals.
a) True
b) False
Answer: b
Explanation: Periodic signals are same in case of continuous time and discrete time signals.
In case of continuous time signal, x=x, for all t>0
In case of discrete time signal,
x=x, for all n>0.
3. What is the time period of a periodic signal in actual terms?
a) The signals which start at t=-∞ and end at t=+∞
b) The signals which have a finite interval of occurrence
c) The signals which start at t= -∞ and ends at a finite time period
d) The signals which have a short period of occurrence
Answer: a
Explanation: The periodic signals have actually a time period between t=-∞ and at t= + ∞. These signals have an infinite time period, that is periodic signals are actually continued forever. But this is not possible in case of real time signals.
4. Periodic signals actually exist according to a definition.
a) True
b) False
Answer: b
Explanation: Periodic signals are defined as signals having time period in between t=-∞ and t=+ ∞. These signals have an infinite time period that is periodic signals are continued forever. But real time signals always cease at some time due to distortion and resistance.
5. What is a fundamental period?
a) Every interval of a periodic signal
b) Every interval of an aperiodic signal
c) The first interval of a periodic signal
d) The last interval of a periodic signal
Answer: c
Explanation: The first time interval of a periodic signal after which it repeats itself is called a fundamental period. It should be noted that the fundamental period is the first positive value of frequency for which the signal repeats itself.
6. Comment on the periodicity of a constant signal?
a) It is periodic
b) It is not periodic
c) It is a mixture of period and aperiodic signal
d) It depends on the signal
Answer: b
Explanation: A constant signal is not periodic. It is because it does not repeat itself over in time. It is constant at any time, it is aperiodic.
7. A discrete time periodic signal is defined as x= x
How is the N defined here?
a) Samples/ cycle
b) Samples/ twice cycle
c) Fundamental period
d) Rate of change of the period
Answer: a
Explanation: The value of N is a positive integer and it represents the period of any discrete time periodic signal measured in terms of number of sample spacing . The smallest value of N is a fundamental period.
8. What is the general range of a period of a signal?
a) It can have of any value from positive to negative
b) It can be negative
c) It can be positive
d) It is always positive
Answer: d
Explanation: The period of a periodic signal is always positive. The smallest positive value of a periodic interval is called a fundamental period in case of both discrete and continuous time signal.
9. What is the area of a periodic signal in a periodic interval?
a) It depends on the situation
b) It is same as the area in the previous interval
c) It is different in different situations
d) It is the square of the fundamental period
Answer: b
Explanation: The area of any periodic signal in any interval is the same. Hence it is same as the previous interval. This results from the fact that a periodic signal takes same values at the intervals of T.
10. When is the sum of M periodic signals periodic?
a) T/T i = 1
b) T/T i = 4
c) T/T i = n i
d) T/T i = m+n
Answer: c
Explanation: The sum of M periodic signal is not necessarily periodic. It is periodic only with the condition that
T/T i = n i , 1≤i≤M,
where T i is the period of the signal and in the sum of n i is an integer.
11. How is the period of the sum signal computed as?
a) T*n
b) T*T
c) T*N+M
d) T *
Answer: a
Explanation: If a signal is periodic then we have to convert each of the periods to the ratio of integers. We have to take the ratio of greatest common divisor from the numerator to the gcd of denominator. The LCM of the denominators of the resulting ratios is the value of n the period of the sum signal is T*n.
12. What is the necessary and sufficient condition for a sum of a periodic continuous time signal to be periodic?
a) Ratio of period of the first signal to period of other signals should be constant
b) Ratio of period of the first signal to period of other signals should be finite
c) Ratio of period of the first signal to period of other signals should be real
d) Ratio of period of first signal to period of other signal should be rational
Answer: d
Explanation: The necessary and sufficient condition for a sum of a periodic continuous time signal to be periodic is that the ratio of a period of the first signal to the period of other signals should be rational.
I.e T/T i = a rational number.
13. Under what conditions the three signals x, y and z with period t1 t2 and t3 respectively are periodic?
a) t1/t2= t2/t3
b) t1/t2 is rational
c) All the ratios of the three periods in any order is rational
d) t1/t2/t3= rational
Answer: c
Explanation: if x , y and z are to be periodic then,
t1/t2 should be rational and simultaneously
t1/t3 should be rational and
t2/t3 should be rational. Hence, all the ratios of the three periods in any order is rational.
14. What is the fundamental period of the signal : e jwt ?
a) 2π/w
b) 2π/w2
c) 2π/w3
d) 4π/w
Answer: a
Explanation: The complex exponential signal can be represented as
e jwt = e jwt+jwT
Hence, wt=2 π,
T= 2π/w.
15. What is the period of the signal :je jw11t ?
a) 2π/10
b) 2π/11
c) 4π/10
d) 4π/11
Answer: b
Explanation: From the definition of periodic signal, we express a periodic exponential signal as :
e jw11t = e jwt+jwT
Hence, 11wt=2 π,
which gives the fundamental period as
2π/11.
This set of Signals & Systems Assessment Questions and Answers focuses on “Periodic Signals – 2”.
1. Is the sum of discrete time periodic signals periodic?
a) No, they are not
b) Yes they are
c) Depends on the signal
d) Not periodic if their ratio is not rational
Answer: b
Explanation: The sum of discrete time periodic signals always periodic because the period ratios N/N are always rational.
For the continuous time, it depends on the ratio.
2. How can we generate a periodic signal from a periodic signal itself?
a) By extending a signal with duration T
b) Cannot be extended
c) By extending the periodic signal’s amplitude
d) By extending the sugar with duration 2π
Answer: a
Explanation: A periodic signal x can be generated by a periodic extension of any segment of x of duration T.
As a result, we can generate x from any segment x having a duration of one period by replacing this segment and reproduction thereof end to end ad infinitum on either side.
3. Is a non periodic signal same as aperiodic signal?
a) No, it is not same as an aperiodic signal
b) Yes it is the other name of aperiodic signal
c) It is a branch of aperiodic signal
d) Aperiodic signal is a branch of non periodic signal
Answer: b
Explanation: A signal which does not satisfy the condition:
x = x is called an aperiodic signal.
Non periodic is another name of an aperiodic signal. Hence it is exactly the same.
4. What is the period of the signal: 2cost/6?
a) 8π
b) 16π
c) 12π
d) 10π
Answer: c
Explanation: Comparing the above signal with the standard form Acos2πFt, where A is the amplitude and F is the frequency,
We get, 2πF=⅙
So, F= 1/12π
Hence, t= 12π.
5. When a continuous signal is a mixture of two continuous periodic signals, what is its periodicity?
a) LCM of the periods of the two signals, provide their ratio is unity
b) LCM of the periods of the two signals, provide their ratio is rational
c) HCF of the periods of the two signals, provide their ratio is rational
d) LCM of the periods of the two signals provide their ratio is real
Answer: b
Explanation: When a continuous signal is a mixture of two continuous periodic signals if their time periods are T1 and T2, and their ratio is rational number, then, the periodicity of the continuous time signal will be the LCM of T1 and T2.
6. Is the signal e αt periodic?
a) Not periodic
b) Yes periodic
c) Depends on the value of
d) Semi- periodic
Answer: c
Explanation: Using the definition of x,
x = e αt
e jwt = e jwt+jwαT
For any value of α, if alpha is positive, it has a remaining term e jwαT
Hence it is not periodic.
7. What is a fundamental angular frequency?
a) The inverse of the fundamental time period
b) The inverse of fundamental frequency
c) Fundamental frequency in radians
d) Fundamental frequency in degree
Answer: c
Explanation: The inverse of the fundamental time period is called fundamental frequency. If it is F, then 2πF is called the fundamental angular frequency ie it is a fundamental frequency in radians.
8. What is the period of cos3t + sin14t?
a) 4π
b) π
c) 2π
d) 3π
Answer: b
Explanation: We know, T1= 2 π/3 and T2= 2 π/14
Now, T1/T2=14/3.
So, LCM gives the time period as π.
9. What is the periodicity of a discrete time signal?
a) 2πm/w
b) 2πm/w
c) 2πm/w
d) 2πm/w
Answer: b
Explanation: Using exponential function, we can show that
2π/N= w/m
Which when rearranged gets us 2πm/w.
10. What is the condition of a periodicity of exponential signal e αt ?
a) α=1
b) α=2
c) α=3
d) Depends on equation
Answer: a
Explanation: From, x= e αt+T = e αt e αt . For any value of α, e αt ≠1 so x ≠x. So only if α=1, the signal will be periodic.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Fourier Series”.
1. What is Fourier series?
a) The representation of periodic signals in a mathematical manner is called a Fourier series
b) The representation of non periodic signals in a mathematical manner is called a Fourier series
c) The representation of non periodic signals in terms of complex exponentials or sinusoids is called a Fourier series
d) The representation of periodic signals in terms of complex exponentials or sinusoids is called a Fourier series
Answer: d
Explanation: The Fourier series is the representation of non periodic signals in terms of complex exponentials, or equivalently in terms of sine and cosine waveform leads to Fourier series. In other words, Fourier series is a mathematical tool that allows representation of any periodic wave as a sum of harmonically related sinusoids.
2. Who discovered Fourier series?
a) Jean Baptiste de Fourier
b) Jean Baptiste Joseph Fourier
c) Fourier Joseph
d) Jean Fourier
Answer: b
Explanation: The Fourier series is the representation of non periodic signals in terms of complex exponentials or sine or cosine waveform. This was discovered by Jean Baptiste Joseph Fourier in 18th century.
3. Fourier series representation can be used in case of Non-periodic signals too. True or false?
a) True
b) False
Answer: b
Explanation: False. The Fourier series is the representation of periodic signals in terms of complex exponentials, or equivalently in terms of sine and cosine waveform leads to Fourier series. In other words, Fourier series is a mathematical tool that allows representation of any periodic wave as a sum of harmonically related sinusoids. They are for periodic signals only.
4. What are the conditions called which are required for a signal to fulfil to be represented as Fourier series?
a) Dirichlet’s conditions
b) Gibbs phenomenon
c) Fourier conditions
d) Fourier phenomenon
Answer: a
Explanation: When the Dirichlet’s conditions are satisfied, then only for a signal, the fourier series exist. Fourier series is of two types- trigonometric series and exponential series.
5. Choose the condition from below that is not a part of Dirichlet’s conditions?
a) If it is continuous then there are a finite number of discontinuities in the period T1
b) It has a finite average value over the period T
c) It has a finite number of positive and negative maxima in the period T
d) It is a periodic signal
Answer: d
Explanation: Even if the Fourier series demands periodicity as the major necessity for its formation still it is not a part of Dirichlet’s condition. It is the basic necessity for Fourier series.
6. What are the two types of Fourier series?
a) Trigonometric and exponential
b) Trigonometric and logarithmic
c) Exponential and logarithmic
d) Trigonometric only
Answer: a
Explanation: The two types of Fourier series are- Trigonometric and exponential. The exponential is more convenient for Fourier series calculations.
7. How is a trigonometric Fourier series represented?
a) A 0 +∑[ancos(w 0 t)+ ansin(w 0 t)]
b) ∑[ancos(w 0 t)+ ansin(w 0 t)]
c) A 0 *∑[ancos(w 0 t)+ ansin(w 0 t)]
d) A 0 +∑[ancos(w 0 t)+ ansin(w 0 t)] + sinwt
Answer: a
Explanation: A 0 + ∑[ancos(w 0 t)+ ansin(w 0 t)] is the correct representation of a trigonometric Fourier series. Here A 0 = 1/T∫xdt and an =2/T∫xcos(w 0 t)dt and bn= 2/T∫xsin(w 0 t)dt.
8. How is the exponential Fourier series represented?
a) X = ∑X n e jnwt + wt
b) X = 1/T∑Xne jnwt
c) X = ∑X n e jnwt
d) X = T*∑X n e jnwt
Answer: c
Explanation: The exponential Fourier series is represented as – X=∑X n e jnwt . Here, the X is the signal and X n =1/T∫xe -jnwt .
9. What is the equation – X=∑X n e jnwt called?
a) Synthesis equation
b) Analysis equation
c) Frequency domain equation
d) Discrete equation
Answer: a
Explanation: The equation – X = ∑X n e jnwt called the synthesis equation of an exponential Fourier series. It is because it is used to synthesize the Fourier series.
10. What is the equation – X n =1/T∫x e jwtn called?
a) Synthesis equation
b) Analysis equation
c) Frequency domain equation
d) Discrete equation
Answer: b
Explanation: The equation – X n =1/T∫xe -jwtn called the analysis equation of an exponential Fourier series. It is because it is used to synthesize the Fourier series.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Fourier Series & Coefficients”.
1. What are fourier coefficients?
a) The terms that are present in a fourier series
b) The terms that are obtained through fourier series
c) The terms which consist of the fourier series along with their sine or cosine values
d) The terms which are of resemblance to fourier transform in a fourier series are called fourier series coefficients
Answer: c
Explanation: The terms which consist of the fourier series along with their sine or cosine values are called fourier coefficients. Fourier coefficients are present in both exponential and trigonometric fourier series.
2. Which are the fourier coefficients in the following?
a) a 0 , a n and b n
b) a n
c) b n
d) a n and b n
Answer: a
Explanation: These are the fourier coefficients in a trigonometric fourier series.
a 0 = 1/T∫xdt
a n = 2/T∫xcosdt
b n = 2/T∫xsindt
3. Do exponential fourier series also have fourier coefficients to be evaluated.
a) True
b) False
Answer: a
Explanation: The fourier coefficient is : X n = 1/T∫xe -njwt dt.
4. The fourier series coefficients of the signal are carried from –T/2 to T/2.
a) True
b) False
Answer: a
Explanation: Yes, the coefficients evaluation can be done from –T/2 to T/2. It is done for the simplification of the signal.
5. What is the polar form of the fourier series?
a) x = c 0 + ∑cncos(nwt+ϕ n )
b) x = c 0 + ∑cncos(ϕ n )
c) x = ∑cncos(nwt+ϕ n )
d) x = c 0 + ∑cos(nwt+ϕ n )
Answer: a
Explanation: x = c 0 + ∑cncos(nwt+ϕ n ), is the polar form of the fourier series.
C 0 =a 0 and c n = √a2 n + b2 n for n≥1
And ϕ n = tan -1 b n /a n .
6. What is a line spectrum?
a) Plot showing magnitudes of waveforms are called line spectrum
b) Plot showing each of harmonic amplitudes in the wave is called line spectrum
c) Plot showing each of harmonic amplitudes in the wave is called line spectrum
d) Plot showing each of harmonic amplitudes called line spectrum
Answer: b
Explanation: The plot showing each of harmonic amplitudes in the wave is called line spectrum. The line rapidly decreases for waves with rapidly convergent series.
7. Fourier series is not true in case of discrete time signals.
a) True
b) False
Answer: b
Explanation: Fourier series is also true in case of discrete time signals. They just need to follow the dirichlet’s conditions.
8. What is the disadvantage of exponential Fourier series?
a) It is tough to calculate
b) It is not easily visualized
c) It cannot be easily visualized as sinusoids
d) It is hard for manipulation
Answer: c
Explanation: The major disadvantage of exponential Fourier series is that it cannot be easily visualized as sinusoids. Moreover, it is easier to calculate and easy for manipulation leave aside the disadvantage.
9. Fourier series uses which domain representation of signals?
a) Time domain representation
b) Frequency domain representation
c) Both combined
d) Neither depends on the situation
Answer: b
Explanation: Fourier series uses frequency domain representation of signals. X=1/T∑X n e jnwt . Here, the X is the signal and X n = 1/T∫xe -jwtn .
10. How does Fourier series make it easier to represent periodic signals?
a) Harmonically related
b) Periodically related
c) Sinusoidally related
d) Exponentially related
Answer: a
Explanation: Fourier series makes it easier to represent periodic signals as it is a mathematical tool that allows the representation of any periodic signals as the sum of harmonically related sinusoids.
This set of Signals & Systems Objective Questions & Answers focuses on “Fourier Series Coefficients – 2”.
1. The Fourier series coefficient for the signal 10δ is ___________
a) 1
b) Cos
c) sin
d) 2
Answer: d
Explanation: \
e^{-jkωt} \,dt\)
= \(\frac{A}{2}\)
Here, A=10, T=5
∴ X[k] = 2.
2. The Fourier series coefficient for the periodic rectangular pulses of height 2A is ____________
a) \
\
\
\(\frac{2A}{kπ} \,cos\, \frac{π}{2} k\)
Answer: c
Explanation: \
e^{-jkωt} \,dt\)
\
\(=\frac{2A}{kπ} \,sin\, \frac{π}{2} k\).
3. The Fourier series coefficient for the periodic signal x = sin 2 t is _____________
a) –\
–\
–\
–\(\frac{1}{2}\) δ[k-2] + δ[k] – \(\frac{1}{2}\) δ[k+2]
Answer: a
Explanation: sin 2 t = \
^2\)
= –\(\frac{1}{4}\) (e 2jt – 2 + e -2jt )
The fundamental period of sin2t is π and ω = \(\frac{2π}{π}\) = 2
∴ X[k] = –\(\frac{1}{4}\) δ[k-1] + \(\frac{1}{2}\) δ[k] – \(\frac{1}{4}\) δ[k+1].
4. The Fourier series coefficient of time domain signal x is X[k] = jδ[k-1] – jδ[k+1] + δ[k+3] + δ[k-3], the fundamental frequency of the signal is ω=2π. The signal is ___________
a) 2
b) -2
c) 2
d) -2
Answer: c
Explanation: \ = ∑_{k=-∞}^∞ X[k]e^{j2πkt}\)
= je j2πt – je -j2πt + e j6πt + e -j6πt
= 2.
5. The Fourier series coefficient of time domain signal x is X[k] = \
^{|k|}\). The fundamental frequency of signal is ω=1. The signal is _____________
a) \
\
\
\(\frac{4}{5 + 3 sint}\)
Answer: d
Explanation: \ = ∑_{k=-∞}^∞ X[k]e^{jkt}\)
Or, x = \
^{-k} e^{jk} + ∑_{k=0}^∞
^k e^{jkt}\)
= \(\frac{\frac{-1}{3} e^{-jt}}{1+\frac{1}{3} e{-jt}} + \frac{1}{1 + \frac{1}{3} e^{jt}}\)
= \(\frac{4}{5 + 3 sint}\).
6. The Fourier series coefficient of the signal y = x(t-t 0 ) + x(t+t 0 ) is _____________
a) 2 cos (\(\frac{2π}{t}\) kt 0 ) X[k]
b) 2 sin (\(\frac{2π}{t}\) kt 0 ) X[k]
c) 2 cos (\(\frac{2π}{t}\) kt 0 )
d) 2 sin (\(\frac{2π}{t}\) kt 0 )
Answer: a
Explanation: x (t-t 0 ) is periodic with period T. the Fourier series coefficient of x (t-t 0 ) is X 1 [k] = \(\frac{1}{T}\) ∫ x (t-t 0 )e -jkωt dt
= e -jkωt 0 X[k]
Similarly, the Fourier series coefficient of x (t+t 0 ) is X 2 [k] = e jkωt 0 X[k]
The Fourier series coefficient of x (t-t 0 ) + x (t+t 0 ) is
Y[k] = X 1 [k] + X 2 [k]
= e -jkωt 0 X[k] + e jkωt 0 X[k]
= 2 cos (\(\frac{2π}{t}\) kt 0 ) X[k].
7. The Fourier series coefficient of the signal y = Even{x} is ___________
a) \
\
\
\(\frac{X[k]-X^* [-k]}{2}\)
Answer: a
Explanation: even {x } = \
is
X 1 [k] = \
e -jkωt dt
= \
e jkωα dα
= X [-k] ∴ The Fourier coefficient of Even{x} = Y[k] = \(\frac{X[k]+X[-k]}{2}\).
8. The Fourier series coefficient of the signal y = Re{x} is ____________
a) \
\
\
\(\frac{X[k]-X^* [-k]}{2}\)
Answer: c
Explanation: Re{x } = \
is
X 1 [k] = \
e -jkωt dt = \
e jkωt dt = X [-k]
So, X 1 [k] = \(X_1^*\) [-k]
∴ Y[k] = \(\frac{X[k]+X^* [-k]}{2}\).
9. The Fourier series coefficient of the signal y = \Missing open brace for subscript \
^2 X[k]\)
b) –\
^2 X[k]\)
c) j\
^2 X[k]\)
d) -j\
^2 X[k]\)
Answer: b
Explanation: \ = ∑_{k=-∞}^∞ X[k]e^{j \frac{2π}{T} kt}\)
Now, \
k ∑_{k=-∞}^∞ X[k]e^{j \frac{2π}{T} kt}\)
And, \
^2 k^2 ∑_{k=-∞}^∞ X[k]e^{j \frac{2π}{T} kt}\)
∴ Y[k] = – \
^2 X[k]\).
10. The Fourier series coefficient of the signal y = x is ______________
a) \
\
\
\(e^{jk \frac{8π}{T}} X[k]\)
Answer: c
Explanation: The period of x is a fourth of the period of x . The Fourier series coefficient of x is still X[k]. Hence, the coefficient of x is \(e^{-jk \frac{8π}{T}} X[k]\).
11. The discrete time Fourier coefficients of \Missing open brace for subscript –\
\
–\
\(\frac{1}{2}\) for all k
Answer: b
Explanation: N=4, ω = \(\frac{2π}{4} = \frac{π}{2}\)
\(X[k] = \frac{1}{4} ∑_{n=4}^3 x[n]e^{-j
nk}\)
= \(\frac{1}{4}\) x[0] = \(\frac{1}{4}\) for all k.
12. The discrete time Fourier coefficient of cos 2
is ______________
a) \
b) \
c) \
d) \
Answer: c
Explanation: N=8, ω = \(\frac{2π}{8} = \frac{π}{4}\)
X[n] = cos 2
= \
^2\)
\
^2\)
Or, X[k] = \
.
13. V = 5, 0≤t<1;
t, t≥1;
The Laplace transform of V is ___________
a) \
\
\
\(\frac{5}{s} – \frac{e^{-s}}{s^2} + \frac{4e^{-s}}{s}\)
Answer: b
Explanation: V = 5 + u
L {5 + u } = \
\)
= \(\frac{5}{s} + \frac{e^{-s}}{s^2} – \frac{4e^{-s}}{s}\).
14. W = 2, 0≤t<4;
t 2 , t≥4;
The Laplace transform of W is ___________
a) \
\)
b) \
\)
c) \
\)
d) \
\)
Answer: d
Explanation: W = 2 + u (t 2 -2)
L {2 + u (t 2 -2)} = \(\frac{2}{s}\) + L {u (t 2 -2)}
= \(\frac{2}{s} + e^{-4s}\) L { 2 -2}
= \(\frac{2}{s} + e^{-4s}\) L {t 2 + 8t + 14}
= \
\).
15. U = 0, 0≤t<7;
3 , t≥7;
The Laplace transform of U is ___________
a) \
\
\
\(\frac{3e^{-7s}}{s^3}\)
Answer: a
Explanation: U = u 3
L {u 3 } = e -7s L {t 3 }
= \(\frac{3!e^{-7s}}{s^4} = \frac{6e^{-7s}}{s^4}\).
This set of Advanced Signals & Systems Questions and Answers focuses on “Miscellaneous Examples on Fourier Series”.
1. A signal g = 10 sin is ___________
a) A periodic signal with period 6 s
b) A periodic signal with period \
A periodic signal with period \
An aperiodic signal
Answer: c
Explanation: 10 sin = A sin ωt
∴ω = 12π
∴ 2πf = 1
So, fundamental frequency = 6
Hence, fundamental period = \(\frac{1}{6}\) s.
2. A voltage having the waveform of a sine curve is applied across a capacitor. When the frequency of the voltage is increased, what happens to the current through the capacitor?
a) Increases
b) Decreases
c) Remains same
d) Is zero
Answer: a
Explanation: The current through the capacitor is given by,
I C = ωCV cos .
As the frequency is increased, I C also increases.
3. What is the steady state value of F , if it is known that \ = \frac{2}{s}\)?
a) \
\
\
Cannot be determined
Answer: b
Explanation: From the equation of F, we can infer that, a simple pole is at origin and all other poles are having negative real part.
∴ F = lim s→0 sF
= lim s→0 \(\frac{2s}{s}\)
= \(\frac{2}{s}\)
= \(\frac{2}{6} = \frac{1}{3}\).
4. What is the steady state value of F , if it is known that F = \
-5
b) 5
c) 10
d) Cannot be determined
Answer: d
Explanation: The steady state value of this Laplace transform is cannot be determined since; F is having two poles on the imaginary axis . Hence the answer is that it cannot be determined.
5. A periodic rectangular signal X has the waveform as shown below. The frequency of the fifth harmonic of its spectrum is ______________
advanced-signals-systems-questions-answers-q5
a) 40 Hz
b) 200 Hz
c) 250 Hz
d) 1250 Hz
Answer: d
Explanation: Periodic time = 4 ms = 4 × 10 -3
Fundamental frequency = \(\frac{10^3}{4}\) = 250 Hz
∴ Frequency of the fifth harmonic = 250 × 5 = 1250 Hz.
6. A CRO probe has an impedance of 500 kΩ in parallel with a capacitance of 10 pF. The probe is used to measure the voltage between P and Q as shown in the figure. The measured voltage will be?
advanced-signals-systems-questions-answers-q6
a) 3.53 V
b) 3.47 V
c) 5.54 V
d) 7.00 V
Answer: b
Explanation: \(X_C = \frac{1}{jCω} = \frac{-j}{2 × 3.14 × 100 × 10^3 × 10 × 10^{-12}}\)
Applying KCL at the node,
\(\frac{V_a-10}{100} + \frac{V_a}{100} + \frac{V_a}{500} + \frac{V_a}{-j159}\)
∴ V a = 4.37∠-15.95°.
7. The Fourier series coefficient of the signal z = Re{x} is ____________
a) \
\
\
\(\frac{X[k]-X^* [-k]}{2}\)
Answer: c
Explanation: Re{x } = \
is
X 1 [k] = \
e -jkωt dt = \
e jkωt dt = X [-k]
So, X 1 [k] = \(X_1^*\) [-k]
∴ Z[k] = \(\frac{X[k]+X^* [-k]}{2}\).
8. The Fourier series expansion of a real periodic signal with fundamental frequency f 0 is given by g p = \(∑_{n=-∞}^∞ C_n e^{j 2πnf_0 t}\). Given that C 3 = 3 + j5. The value of C -3 is ______________
a) 5 + 3j
b) -5 + 3j
c) -3 – j5
d) 3 – j5
Answer: d
Explanation: Given that C 3 = 3 + j5.
We know that for real periodic signal C -k = \(C_k^*\)
So, C -3 = \(C_3^*\) = 3 – j5.
9. A signal e -at sin is the input to a real linear time invariant system. Given K and ∅ are constants, the output of the system will be of the form Ke -bt sin . The correct statement among the following is __________
a) b need not be equal to a but v must be equal to ω
b) v need not be equal to ω but b must be equal to a
c) b must be equal to a and v must be equal to ω
d) b need not be equal to a and v need not be equal to ω
Answer: a
Explanation: For a system with input e -at sin and output Ke -bt sin , frequency to output must be equal to input frequency while b will depend on system parameters and need not be equal to a.
10. Let us suppose that the impulse response of a causal LTI system is given as h . Now, consider the following two statements:
Statement 1- Principle of superposition holds
Statement 2- h = 0
Which of the following is correct?
a) Statement 1 is correct and Statement 2 is wrong
b) Statement 2 is correct and Statement 1 is wrong
c) Both Statement 1 and Statement 2 are wrong
d) Both statement 1 and Statement 2 are correct
Answer: d
Explanation: As we know that linear system possesses superposition theorem.
Hence, Statement 1 satisfies the given equation. We also know that time invariant condition depend on time.
Hence, Statement 2 satisfies the given equation.
11. The Fourier series representation of an impulse train denoted by s = \
\) is given by _____________
a) \
\)
b) \
\)
c) \
\)
d) \
\)
Answer: b
Explanation: s = \(∑_{n=-∞}^∞ C_n e^{jnω_0 t}\), where ω 0 =(2T/T 0 )
And C n = \
e^{-jnω_0 t} \,dt\)
= \
\).
12. For which of the following a Fourier series cannot be defined?
a) 3 sin
b) 4 cos + 2 sin
c) exp sin
d) 1
Answer: c
Explanation: 3 sin = 25
4 cos + 2sin sum of two periodic function is also periodic function
For 1 which is a constant, Fourier series exists.
For exp sin , due to decaying exponential decaying function, it is not periodic. So Fourier series cannot be defined for it.
13. The RMS value of a rectangular wave of period T, having a value of +V for a duration T 1 and −V for the duration T − T 1 = T 2 , equals _____________
a) V
b) \
\
\(\frac{T_1}{T_2}\) V
Answer: a
Explanation: Period = T = T 1 + T 2
RMS value = \(\sqrt{\frac{1}{T} ∫_0^T x^2 dt}\)
= \(\sqrt{\frac{1}{T}[V^2.
+ V^2
]}\)
= \(\sqrt{V^2}\)
= V.
14. A periodic square wave is formed by rectangular pulses ranging from -1 to +1 and period = 2 units. The ratio of the power in the 7th harmonic to the power in the 5th harmonic for this waveform is equal to ____________
a) 1
b) 0.5
c) 2
d) 2.5
Answer: b
Explanation: For a periodic square wave nth harmonic component ∝ \(\frac{1}{n}\)
Thus the power in the nth harmonic component is ∝ \(\frac{1}{n^2}\)
∴ Ratio of power in 7th harmonic to 5th harmonic for the given wage form = \(\frac{\frac{1}{7^2}}{\frac{1}{5^2}}\)
= \(\frac{25}{4}\) ≅ 0.5.
15. A useful property of the unit impulse 6 is ________________
a) 6 = a 6
b) 6 = 6
c) 6 = \
d) 6 = [6] a
Answer: c
Explanation: Time-scaling property of 6
We know that by this property,
6 = \
, a>0
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Fourier Series Properties – 1”.
1. How do we represent a pairing of a periodic signal with its fourier series coefficients in case of continuous time fourier series?
a) x ↔ X n
b) x ↔ X n+1
c) x ↔ X
d) x ↔ X n
Answer: a
Explanation: In case of continuous time fourier series, for simplicity, we represent a pairing of a periodic signal with its fourier series coefficients as,
x ↔ X n
here, x is the signal and Xn is the fourier series coefficient.
2. What are the properties of continuous time fourier series?
a) Linearity, time shifting
b) Linearity, time shifting, frequency shifting
c) Linearity, time shifting, frequency shifting, time reversal, time scaling, periodic convolution
d) Linearity, time shifting, frequency shifting, time reversal, time scaling, periodic convolution, multiplication, differentiation
Answer: d
Explanation: Linearity, time shifting, frequency shifting, time reversal, time scaling, periodic convolution, multiplication, differentiation are some of the properties followed by continuous time fourier series. Integration and conjugation are also followed by continuous time fourier series.
3. Integration and conjugation are also followed by continuous time fourier series?
a) True
b) False
Answer: a
Explanation: Linearity, time shifting, frequency shifting, time reversal, time scaling, periodic convolution, multiplication, differentiation are some of the properties followed by continuous time fourier series. Integration and conjugation are also followed by continuous time fourier series.
4. If x and y are two periodic signals with coefficients X n and Y n then the linearity is represented as?
a) ax + by = aX n + bY n
b) ax + by = X n + bY n
c) ax + by = aX n + Y n
d) ax + by = X n + Y n
Answer: a
Explanation: ax + by = aX n + bY n , x and y are two periodic signals with coefficients X n and Y n .
5. How is time shifting represented in case of periodic signal?
a) If x is shifted to t 0 , X n is shifted to t 0
b) x(t-t 0 ), Y n = X n e -njwt0
c) X n = x(t-t 0 ), Y n = X n e -njwt0
d) X n = x(-t 0 ), Y n = X n e -njwt0
Answer: c
Explanation: If x and y are two periodic signals with coefficients X n and Y n , then if a signal is shifted to t 0 , then the property says,
X n = x(t-t 0 ), Y n = X n e -njwt0
6. What is the frequency shifting property of continuous time fourier series?
a) Multiplication in the time domain by a real sinusoid
b) Multiplication in the time domain by a complex sinusoid
c) Multiplication in the time domain by a sinusoid
d) Addition in the time domain by a complex sinusoid
Answer: b
Explanation: If x and y are two periodic signals with coefficients Xn and Yn,
Then y= e jmwt x↔Y n =X n-m .
Hence, we can see that a frequency shift corresponds to multiplication in the time domain by complex sinusoid whose frequency is equal to the time shift.
7. What is the time reversal property of fourier series coefficients?
a) Time reversal of the corresponding sequence of fourier series
b) Time reversal of the last term of fourier series
c) Time reversal of the corresponding term of fourier series
d) Time reversal of the corresponding sequence
Answer: a
Explanation: x↔ X n
Y = x↔Y n =X -n .
That is the time reversal property of fourier series coefficients is time reversal of the corresponding sequence of fourier series.
8. It does not depend whether the signal is odd or even, it is always reversal of the corresponding sequence of fourier series.
a) True
b) False
Answer: b
Explanation: It does depend whether the signal is odd or even.
If the signal is even, the reversal is positive and if the signal is odd, the reversal is negative.
9. Why does the signal change while time scaling?
a) Because the frequency changes
b) Time changes
c) Length changes
d) Both frequency and time changes
Answer: a
Explanation: x↔X n
Y = x↔Y n = X n
Hence, the fourier coefficients have not changed but the representation has changed because of changes in fundamental frequency.
10. What is the period of the signal when it is time shifted?
a) Changes according to the situation
b) Different in different situation
c) Remains the same
d) Takes the shifted value
Answer: c
Explanation: The period of the periodic signal does not change even if it is time shifted.
If x and y are two periodic signals with coefficients X n and Y n , then if a signal is shifted to t 0 , then the property says,
X n = x(t-t 0 ), Y n = X n e -njwt0 .
This set of Signals & Systems Problems focuses on “Fourier Series Properties – 2”.
1. Can continuous time fourier series undergo periodic convolution?
a) They cannot undergo periodic convoluion
b) They can undergo in certain situations
c) They undergo periodic convolution
d) Only even signals undergo periodic convolution
Answer: c
Explanation: Continuous time fourier series undergoes periodic convolution.
X*y=z ↔ X n Y n = Z n .
2. What is the outcome of a periodic convolution of signals in case of continuous time fourier series?
a) Division in frequency domain
b) Multiplication in frequency domain
c) Convolution is easier
d) Addition of signals in frequency domain
Answer: b
Explanation: This is a very important property of continuous time fourier series, it leads to the conclusion that the outcome of a periodic convolution is the multiplication of the signals in frequency domain representation.
X*y=z ↔ X n Y n =Z n .
3. What is the multiplication property of continuous time fourier series?
a) Convolution of the signals
b) Multiplication of the elements of the signal
c) Division of the frequency domain
d) Addition of the signals in frequency domain
Answer: a
Explanation: In the case of continuous time fourier series, the multiplication property leads to discrete time convolution of the signals.
z=xy ↔ Z n = X n Y n-k .
4. What is the differentiation property of continuous time fourier series?
a) Y n = jnwtX n
b) Y n = jntX n
c) Y n = jnwX n
d) X n = jnwtX n
Answer: c
Explanation: x ↔X n , x is the signal and X n is the coefficient.
Then, Y n = jnwX n .
5. What is the fourier series coefficient for n=0?
a) Zero
b) Unity
c) Depends on the situation
d) Non zero positive
Answer: a
Explanation: The differentiation property of the continuous time fourier series is,
Y = dx/dt ↔ Y n = jnwX n .
Hence, the differentiation property of time averaged value of the differentiated signal to be zero, hence, fourier series coefficient for n=0 is zero.
6. What is the integration property of the continuous time fourier series?
a) y ↔ Y n = 1/jnwX n
b) y ↔ Y n = 1/jwX n
c) y ↔ Y n = 1/jnX n
d) y ↔ Y n = 1/jnw
Answer: a
Explanation: y↔ Y n = 1/jnwX n , here x is the signal and y is the output.
This is the integration property of the signal.
7. What is the smoothing operation?
a) Differentiation property
b) Multiplication property
c) Integration property
d) Conjugation property
Answer: c
Explanation: The integration attenuates the magnitude of the high frequency components of the signal. High frequency contributors cause sharp details such as occurring at the points of discontinuity. Hence, integration smoothens the signal, hence it is called a smoothening operation.
8. What is the complex conjugate property of a fourier series?
a) It leads to convolution
b) It leads to time reversal
c) It leads to multiplication
d) It leads to addition of signals
Answer: x ↔ X n
Y = *x ↔Y n =*X -n
It leads to time reversal.
9. If the signal x is odd, what will be the fourier series soeffiients?
a) Real and even
b) Odd
c) Real only
d) Real and odd
Answer: a
Explanation: If the signal is real and odd, the fourier series coefficients are conjugate symmetric.
And its fourier series coefficients are real and even.
X n = X -n *= X n .
10. If the signal x is even, what will be the fourier series coefficients?
a) Real and even
b) Odd
c) Real only
d) Imaginary and odd
Answer: d
Explanation: If the signal is real and even, the fourier series coefficients are conjugate symmetric.
And its fourier series coefficients are imaginary and even.
X n = X -n *= -X n .
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Fourier Series and LTI Systems”.
1. Which system among the following is a time invariant system?
a) y = n x
b) y = x – x
c) y = x
d) y = x cos 2nf
Answer: b
Explanation: We know that, for any system y = k x , to be a time invariant system, it must satisfy the relation, y (n-n 1 ) = k x (n-n 1 ) [where k is a constant or a function of n].
For y = n x , y (n-n 1 ) = (n-n 1 ) x (n-n 1 )
This does not satisfy the criteria as stated above. Hence not time invariant.
For y = x , y (n-n 1 ) = x (-n+n 1 )
This also does not satisfy the criteria as stated above. Hence not time invariant.
For y = x cos 2nf, y (n-n 1 ) = x (n-n 1 ) cos 2(n-n 1 ) f
This also does not satisfy the criteria as stated above. Hence not time invariant.
For y = x – x , y (n-n 1 ) = x (n-n 1 ) – x (n-n 1 -1)
This satisfies the above criteria. Hence given system is time invariant.
2. Which of the following is a causal system?
a) y = 3x – 2x
b) y = 3x + 2x
c) y = 3x + 2x
d) y = 3x + 2x + x
Answer: a
Explanation: We know that, for a causal system, output must depend on present and past but not on future.
For y = 3x + 2x , we can observe that output depends on future because of the term x . Hence, not a causal system.
For y = 3x + 2x , we can observe that output depends on future because of the term x . Hence not a causal system.
For y = 3x + 2x + x , we can observe that output depends on future because of the term x . Hence not a causal system.
For y = 3x – 2x , we can observe that output depends on present and past but not on the future. Hence, it is a causal system.
3. Which of the following is a dynamic system?
a) y = y + y
b) y = y
c) y = x
d) y + y + y = 0
Answer: a
Explanation: We know that for a dynamic system, the present output of the system should depend only on the past output and the future output.
For y = y , we can observe that output depends only on the past but not on the future. Hence it is not a dynamic system.
For y = x , we can observe that output depends on the present. Hence it is not a dynamic system.
For y + y + y = 0, we can observe that output is a constant. Hence it is not a dynamic system.
For y = y + y , we can observe that output depends only on past and future outputs. Hence it is a dynamic system.
4. A series RC circuit excited by voltage V is __________
a) A memory less system
b) A causal system
c) A dynamic system
d) Static system
Answer: c
Explanation: Dynamic systems are those systems which consist of memory. In the series RC circuit excited by voltage V, the capacitor C is an energy storing element which acts as a memory for the circuit. Therefore since the system has memory it is not a memoryless system. Also, a causal system depends only on the past and present value. But since the future value of the charge is also under consideration in this type of circuit, so the system is not causal. Since charge moves about in the circuit due to the applied voltage V, hence the system is not a static system. Therefore the system is a dynamic system.
5. A LTI system is characterized by ___________
a) Unit impulse response
b) Time shifted impulses
c) Unit step response
d) Response to any signal
Answer: a
Explanation: The response of an LTI system to an arbitrary input is given by,
\(Y[n] = ∑_{k=-∞}^∞ x[k]h[n]\), where \(x[n] = ∑_{k=-∞}^∞ x[k]δ[n-k]\) represents sequences as a linear combination of shifted unit impulses δ[n-k]. Thus, any LTI system is completely characterized by its unit impulse response.
6. Given a stable and causal system with impulse response h and system function H. Let us suppose H is rational, contains a pole at s=-2 and does not have a zero at origin and location of all other zeroes is unknown, poles are present at some unknown location other than the origin. Then,
i) H = H
ii) \
i – true, ii – true
b) i – false, ii – true
c) i – true, ii – false
d) i – false, ii – false
Answer: b
Explanation: H = H
If it were true, then H has a pole at s=-2, it must also have a pole at s=2. This is contradicting the fact that all the poles of a causal and stable system must be in the left half of the s-plane.
For \
, but multiplication with s does not eliminate pole at s=-2.
7. From the given conditions, what are the Dirichlet conditions?
i. X should be absolutely integrable
ii. X should have finite discontinuities
iii. X should have a finite number of maxima as well as minima in its domain
a) i, ii and iii
b) i and ii
c) i and iii
d) ii and iii
Answer: a
Explanation: For both periodic and non-periodic signals to have their Laplace transforms, they must satisfy the Dirichlet conditions. Dirichlet conditions state that a system should be absolutely integrable, have finite discontinuities, and have a finite number of maxima and minima in its domain.
8. If y = e x , then the relation is _________
a) Dynamic
b) Static
c) Memory
d) Memoryless but not static
Answer: b
Explanation: Given relation, y = e x .
The system represented by the above relation is static since the output at time t is dependent on t only. Further, the input-output relation is not an integrodifferential relation. Hence it is a static system.
9. A signal e -at sin is the input to a real linear time-invariant system. Given K and ∅ are constants, the output of the system will be of the form Ke -bt sin . The correct statement among the following is __________
a) b need not be equal to a but v must be equal to ω
b) v need not be equal to ω but b must be equal to a
c) b must be equal to a and v must be equal to ω
d) b need not be equal to a and v need not be equal to ω
Answer: a
Explanation: For a system with input e -at sin and output Ke -bt sin , frequency to output must be equal to input frequency while b will depend on system parameters and need not be equal to a.
10. Which of the following is true about the bounded signal?
a) A finite signal is always bounded
b) A bounded signal always possess finite energy
c) A bounded signal is always zero outside a given interval
d) A bounded signal is always finite
Answer: b
Explanation: A bounded signal always possesses finite energy as a matter of fact. Mathematically on integrating it can be also shown that the energy of the bounded signal is finite while power is zero.
11. A system is said to be dynamic if the output of the system depends on ___________
a) The past Input
b) The Future Input
c) The Present Input
d) Both the Present and future Inputs
Answer: a
Explanation: A dynamic system is a system whose present output depends only on past inputs. Mathematically we can say that for a dynamic system y , the condition y = y should be always satisfied.
12. The system \ = 7
^k \,u – 6
^k \,u\) is ___________
a) Causal
b) Anti-causal
c) Non-causal
d) Cannot be determined
Answer: c
Explanation: Taking the z-transform, we get,
\ = 7\leftMissing or unrecognized delimiter for \right – 6\leftMissing or unrecognized delimiter for \right\)
∴ the ROC for given condition is as derived above.
∴ the bounded signal as a whole is non-causal.
13. The element which can weaken a signal system is ___________
a) Attenuation
b) Distortion
c) Noise
d) Attenuation, Distortion & Noise
Answer: d
Explanation: We know that distortion is the alteration of the original shape, attenuation is the reduction of signal strength and noise is the unwanted output of a signal or system. So, attenuation, distortion and noise all can impair a signal.
14. The difference between the highest and lowest frequencies of a signal is _________
a) Frequency
b) Period
c) Bandwidth
d) Amplitude
Answer: c
Explanation: Bandwidth is the name for that frequency range that a signal requires for transmission and is also a name for the frequency capacity of a particular transmission medium.
15. A system is said to be causal if the output of the system depends on ___________
a) The Past and Present Inputs
b) The Present Inputs
c) The Future Inputs
d) The Past and Future Inputs
Answer: a
Explanation: A system is causal if and only if the current output is only a function of present and past inputs. The current output does not depend on the future inputs y = f); u; u;).
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Symmetry Properties of the Fourier Series”.
1. How can fourier series calculations be made easy?
a) Using symmetry conditions
b) Using formula
c) Using integration
d) Calculations are easy anyways
Answer: a
Explanation: Fourier series calculations are made easy because the series consists of sine and cosine functions and if they are in symmetry they can be easily done. Some integration is always even or odd, hence, we can calculate.
2. What is the function of an even signal?
a) x = -x
b) x = x
c) x = -x
d) x = x
Answer: b
Explanation: An even signal is one in which the functional values of the signal in t and –t is same. Hence, even signal is one in which x and x is same.
3. What is the function of an odd signal?
a) x = -x
b) x = x
c) x = -x
d) x = x
Answer: a
Explanation: An even signal is one in which the functional values of the signal in t is negative of the value of -t . Hence, even signal is one in which x is –x.
4. What is the product of an even signal and odd signal?
a) Even signal
b) Odd signal
c) Mixture of even and odd
d) Odd signals sometimes
Answer: b
Explanation The product of an even and odd signal is an odd signal. But product or sum of two even signals is even signal.
5. What are the values of an and bn when the signal is even?
a) a n =0 and b n =0
b) a n =0 and b n =4/T∫xcosdt
c) a n =4/T∫xcosdt and b n =0
d) a n =4/T∫xsindt and b n =4/T∫xcosdt
Answer: c
Explanation: In an even signal the summation of 0 to T/2 is in sine series is zero.
Hence, 4/T∫xsindt=0, hence b n =0. So, a 0 = 2/T∫xdt
And a n =4/T∫xcosdt .
6. What are the values of a n and b n when the signal is even?
a) a n =0 and b n =0
b) a n =0 and b n =4/T∫xsindt
c) a n =4/T∫xcosdt and b n =0
d) a n =4/T∫xsindt and b n =4/T∫xcosdt
Answer: b
Explanation: In an odd signal the summation of 0 to T/2 is in cosine series is zero.
Hence, 4/T∫xcosdt=0, hence a n =0. So, a 0 = 0
And b n =4/T∫xsindt .
7. How can we define half wave symmetry?
a) x = -x
b) x = x
c) x = x
d) x = -x
Answer: d
Explanation: x = -x is how we define a half wave symmetry. In this case, the waveform is neither even nor odd, it must be both.
7. How can we define the coefficients half wave symmetry when n is even?
a) a n =0 and b n =0 and a 0 =0
b) a n =4/T∫xcosdt and b n =0= a 0
c) a n =4/T∫xsindt and b n =4/T∫xcosdt and a 0 =0
d) a n =4/T∫xsindt and b n =4/T and a 0 =0
Answer: a
Explanation: If it is half wave symmetry then we define the fourier coefficients as-
an=0 and b n =0 and a 0 =0 if n is an even number.
8. How can we define the coefficients a half wave symmetry when n is even?
a) a n =0 and b n =0 and a 0 =0
b) a n =4/T∫xcosdt and b n =0= a 0
c) a n =4/T∫xsindt and b n =4/T∫xcosdt and a 0 =0
d) a n =4/T∫xsindt and b n =4/T and a 0 =0
Answer: c
Explanation: If it is a half wave symmetry then we define the fourier coefficients as-
a n =0 and a n =4/T∫xcosdt and b n =4/T∫xsindt.
9. When does a wave posses a quarter wave symmetry?
a) It has either even or odd symmetry
b) It has half wave symmetry
c) even/odd symmetry and half wave symmetry
d) It is even in one quarter and odd in the other
Answer: c
Explanation: A signal is said to posses quarter wave symmetry when it is either odd or even symmetry and has half wave symmetry. These signals are shifted to left or right by T/4 with a half wave symmetry but the even symmetry of the original unshifted towards the odd symmetry it is said to be quarter wave symmetry.
10. What are the types of symmetry shown by signals?
a) Even symmetry and odd symmetry
b) Even, odd and quarter wave symmetry
c) Even, odd, half-wave and quarter wave symmetry
d) Half wave symmetry
Answer: c
Explanation: The types of symmetry shown by signals are- Even, odd, half-wave and quarter wave symmetry.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Dirichlet’s Conditions”.
1. How many dirichlet’s conditions are there?
a) One
b) Two
c) Three
d) Four
Answer: c
Explanation: There are three dirichlet’s conditions. These conditions are certain conditions that a signal must possess for its fourier series to converge at all points where the signal is continuous.
2. What are the Dirichlet’s conditions?
a) Conditions required for fourier series to diverge
b) Conditions required for fourier series to converge if continuous
c) Conditions required for fourier series to converge
d) Conditions required for fourier series to diverge if continuous
Answer: b
Explanation: Dirichlet’s conditions are Conditions required for fourier series to converge. That is there are certain conditions that a signal must posses for its fourier series to converge at all points where the signal is continuous.
3. What is the first Dirichlet’s condition?
a) Over any period, signal x must be integrable
b) Multiplication of the signals must be continuous
c) x should be continuous only
d) A signal can be integrable except break points
Answer: a
Explanation: In the case of Dirichlet’s conditions, the first property leads to the integration of signal. It states that over any period, signal x must be integrable.
That is ∫|x|dt<∞.
4. Is dirichlet’s condition possible in case of discrete signals?
a) True
b) False
Answer: b
Explanation: Dirichlet’s conditions is not possible in case of discrete signals. That is these are certain conditions that a signal must posses for its fourier series to converge at all points where the signal is continuous only.
5. What guarantees that coefficient is finite in a dirichlet’s condition?
a) First condition
b) Second condition
c) Third condition
d) Fourth condition
Answer: a
Explanation: The first property is:
That is ∫|x|dt<∞
Now, X n = 1/T∫|xe -jwt |dt ≤ 1/T∫|x|dt
So, X n <∞.
6. What is the second dirichlet’s condition?
a) In any finite interval, x is of bounded variation
b) In most of a finite interval, x is of bounded variation
c) In any finite interval, x is of unbounded variation
d) In majority finite interval, x is of unbounded variation
Answer: a
Explanation: In any finite interval, x is of bounded variation. That is there are no more than a finite number of maxima and minima during a single period of the signal.
7. There are maxima and minima not possible in dirichlet’s conditions.
a) True
b) False
Answer: b
Explanation: Maxima and minima are possible if they are infinite number as stated by the second dirichlet’s condition. In any finite interval, x is of bounded variation. That is there are no more than a finite number of maxima and minima during a single period of the signal.
8. What is the third dirichlet’s condition?
a) Finite discontinuities in the infinite interval
b) Finite discontinuities in the finite interval
c) Infiinite discontinuities in the infinite interval
d) Finite discontinuities in the all the intervals
Answer:b
Explanation: The third condition states that in any finite interval of time, there is an only a finite number of discontinuities. Hence, finite discontinuities in the finite interval are the correct option.
9. In the third condition, does each of the discontinuities need to be finite?
a) All the time
b) Sometimes
c) Never
d) Rarely
Answer: a
Explanation: The third condition states that in any finite interval of time, there is the only finite number of discontinuities. And furthermore, each of these discontinuities must be finite too.
10. What is the sum of the series at a point when the signal is discontinuous?
a) Average of the previous limits
b) Previous limits are considered
c) Future limits are added
d) Average of left hand limit and right hand limit are taken
Answer: d
Explanation: If the signal is discontinuous at a point t, then an average of left hand limit and right hand limit of the signal x are taken.
That is x = ½[ x+x].
This set of Basic Signals & Systems Questions and Answers focuses on “Gibb’s Phenomena, Convergence of Fourier Series”.
1. When is the gibbs phenomenon present in a signal x?
a) Only when there is a discontinuity in the signal
b) Only when the signal is discrete
c) Only when there is a jump discontinuity in the signal
d) Gibbs phenomenon is not possible in continuous signals
Answer: c
Explanation: The gibbs phenomenon present in a signal x , only when there is a jump discontinuity in the signal.
2. Where does the gibbs phenomenon occur?
a) Gibbs phenomenon occurs near points of discontinuity
b) Gibbs phenomenon occurs only near points of discontinuity
c) Gibbs phenomenon occurs only ahead of points of discontinuity
d) Gibbs phenomenon does not occur near points of discontinuity
Answer: b
Explanation: The gibbs phenomenon present in a signal x, only when there is a jump discontinuity in the signal. Gibbs phenomenon occurs only near points of discontinuity that is approximated by a fourier series in which only a finite number of terms are kept constant.
3. What causes the gibbs phenomenon?
a) Abruptly terminating the signals
b) Abruptly integrating the signals
c) x should be continuous only
d) Signal should be discontinuous
Answer: a
Explanation: In case gibbs phenomenon, When a continuous function is synthesized by using the first N terms of the fourier series, we are abruptly terminating the signal, giving weigtage to the first N terms and zero to the remaining. This abrupt termination causes it.
4. When a continuous function is synthesized by using the first N terms of the fourier series does the gibbs phenomenon occur?
a) True
b) False
Answer: b
Explanation: The gibbs phenomenon present in a signal x, only when there is a jump discontinuity in the signal. When a continuous function is synthesized by using the first N terms of the fourier series, the synthesized function approaches the signal for all t↔∞. No gibbs phenomenon occurs.
5. When is fourier convergence theorem applicable?
a) Infinite series limit
b) Continuous function limit
c) Discrete function limit
d) Break point limits
Answer: a
Explanation: According to fourier convergence theorem, near a point of discontinuity the fourier series approximation oscillates about the numerical value it should achieve, which is valid in an infinite series limit.
6. What is the fourier convergence theorem?
a) Fourier series approximation oscillates about the numerical value
b) Fourier coefficients converge near a discontinued point
c) In any finite interval, x is of unbounded variation
d) In majority finite interval, x is of unbounded variation
Answer: a
Explanation: According to fourier convergence theorem, near a point of discontinuity the fourier series approximation oscillates about the numerical value it should achieve. This is valid in an infinite series limit.
7. The overshoot near discontinuity vanishes as more and modes are retained.
a) True
b) False
Answer: b
Explanation: The overshoot near discontinuity does not vanish as more and modes are retained. Instead, the overshoot is finite no matter what finite numbers of modes N are retained.
8. What is the overshoot number?
a) Infinite
b) Finite
c) Zero
d) More than 10
Answer: b
Explanation: The overshoot near discontinuity does not vanish as more and modes are retained. Instead, the overshoot is finite no matter what finite numbers of modes N are retained. Even though the region of overshoot gets progressively smaller as N ↔∞.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Trigonometric Fourier Series”.
1. The Fourier transform of sin e-t u is ____________
a) \Missing or unrecognized delimiter for \right\)
b) \Missing or unrecognized delimiter for \right\)
c) \Missing or unrecognized delimiter for \right\)
d) \Missing or unrecognized delimiter for \right\)
Answer: b
Explanation: The Fourier Transform of e -t u = \(\frac{1}{1+jω}\)
∴ Fourier transform of e -3|t| = \
<–> X {j }
∴ X = \Missing or unrecognized delimiter for \right\).
2. The Fourier series coefficient of time domain signal x is X[k] = jδ[k-1] – jδ[k+1] + δ[k+3] + δ[k-3], the fundamental frequency of the signal is ω=2π. The signal is ___________
a) 2
b) -2
c) 2
d) -2
Answer: c
Explanation: x = \(∑_{k=-∞}^∞ X[k]e^{j2πkt}\)
= je j2πt – je -j2πt + e j6πt + e -j6πt
= 2.
3. The unit impulse response of the system c = -4e -t + 6e -2t for t>0. The step response of the same system for t ≥ 0 is _______________
a) -3e -2t – 4e -t + 1
b) -3e -2t + 4e -t – 1
c) -3e -2t – 4e -t – 1
d) 3e -2t + 4e -t – 1
Answer: b
Explanation: Impulse response = -4e -t + 6e -2t
And step response = \
\,dt\)
= \(\int_0^t -4e^{-t} \,dt + \int_0^t 6e^{-2t} \,dt\)
= 4e -t – 3e -2t -1.
4. Given a real valued function x with period T. Its trigonometric Fourier series expansion contains no term of frequency ω = 2π \
satisfies the equation ____________
a) x = x = -x
b) x = x = x
c) x = x = -x
d) x = x = x
Answer: d
Explanation: For an even symmetry, x = x
Thus no sine component will exist because b n =0 and by half wave symmetry condition odd harmonics will exist.
Now, x = x
Combining the two conditions, we get, x = x = x
.
5. Given a periodic rectangular waveform of frequency 1 kHz, symmetrical about t=0 and having a pulse width of 500 µs and amplitude 5 V. The Fourier series is ___________
a) 2.5 + 3.18 cos (2π × 10 3 t) – 1.06 cos (6π × 10 3 t)
b) 2.5 – 3.18 cos (2π × 10 3 t) – 1.06 cos (6π × 10 3 t)
c) 2.5 + 3.18 cos (2π × 10 3 t) + 1.06 cos (6π × 10 3 t)
d) 2.5 – 3.18 cos (2π × 10 3 t) + 1.06 cos (6π × 10 3 t)
Answer: a
Explanation: Period, T = \
= \frac{Aτ}{T} + 2 \frac{Aτ}{T} \frac{sin \frac{πτ}{T}}{\frac{πτ}{T}} cos[\frac{2πt}{τ}] + \frac{2Aτ}{T} \frac{sin \frac{2πτ}{T}}{\frac{2πτ}{T}} cos \frac{4πt}{T}\)
= 5 + 2 \(\frac{sin0.5π}{0.5π}\) cos(2π × 10 3 t) + 2 \(\frac{sinπ}{π}\) cos(4π × 10 3 t)
= 2.5 + 3.18 cos (2π × 10 3 t) – 1.06 cos (6π × 10 3 t).
6. The lengths of two discrete time sequence x 1 [n] and x 2 [n] are 5 and 7 respectively. The maximum length of a sequence x 1 [n] * x 2 [n] is ____________
a) 5
b) 6
c) 7
d) 11
Answer: d
Explanation: Maximum length of x 1 [n] * x 2 [n] is L .
Now, we know that, L = L 1 + L 2 -1 where L 1 is length of x 1 [n] and L 2 is length of x 2 [n] So, the maximum length of a sequence x 1 [n] * x 2 [n] = 5 + 7 – 1
= 11.
7. A waveform is given by v = 10 sin2π 100 t. The magnitude of the second harmonic in its Fourier series representation is ____________
a) 0 V
b) 20 V
c) 100 V
d) 200 V
Answer: a
Explanation: Given that v = 10 sin2π 100 t
It holds the half-wave symmetry i.e. x = -x
And for half wave symmetry, a 0 = 0 and a k = 0
So, second harmonic component a 2 = 0.
8. If the FT of x is \
, then the FT of e j5t x is ____________
a) \
b) \
\
d) \(\frac{2}{ω-5}\) sin{π}
Answer: d
Explanation: Given that the FT of x is \
Applying time shifting and scaling property, we get the FT of e j5t x as \(\frac{2}{ω-5}\) sin{π}.
9. A signal x has its FT as X . The inverse FT of X is _____________
a) \
e^{j3πt}\)
b) \
e^{-j4πt/3}\)
c) 3xe -j4πt
d) x
Answer: b
Explanation: Applying the time shifting and scaling property, we get,
X [3] = \
e^{-j4πt/3}\).
10. The impulse response of a continuous time system is given by h = δ + δ. The value of the step response at t=2 is _______________
a) 0
b) 1
c) 2
d) 3
Answer: b
Explanation: For step response, the impulse response can be integrated.
So, y = u + u
Hence, y = u + u
= 1 + 0 = 1.
11. The first two components of trigonometric Fourier series of the given signal is ____________
signals-systems-questions-answers-trigonometric-fourier-series-q11
a) 0, 4/π
b) 4/π, 0
c) 0, -4/π
d) -4/π, 1
Answer: a
Explanation: Signal is odd and ω = \(\frac{2π}{t}\) = π
Here, c 0 = a 0 = 0
So, c n = b n = \
.sinωt \,dt\)
∴ c 1 = b 1 = \(2\int_0^t sinωt \,dt\)
= \(\frac{2}{ω}[cosωt]_0^t\)
= \(-\frac{2}{π}\)[-1-1] = 4/π
So, the first two components are 0, 4/π.
12. The convolution of two continuous time signals x = e -t u and h = e -2t u is ________________
a) (e t – e 2t ) u
b) (e -2t – e -t ) u
c) (e -t – e -2t ) u
d) (e -t + e -2t ) u
Answer: c
Explanation: Given x = e -t u and h = e -2t u .
Now, y = \
hdτ\)
= e -2t (e t -1); t≥0
Or, y = e -t – e -2t ; t≥0
Since, y = 0 for t<0
Therefore, y = (e -t – e -2t ) u .
13. A linear phase channel with phase delay T p and group delay T g must have ____________
a) T p = T g = constant
b) T p ∝ f and T g ∝ f
c) T p = constant and T g ∝ f
d) T p ∝ f and T g = constant
Answer: a
Explanation: θ = -ωt 0
T p = \(-\frac{θ}{ω}\) = t 0
T g =\(-\frac{d θ}{dω}\) = t 0
∴ T p = T g = t 0 = constant.
14. The z-transform of cos
u[n] is __________
a) \
\
\
\(\frac{z}{2} \frac{}{
}\), |z|>1
Answer: b
Explanation: Performing z-transform on a n u[n], we get \
= 0.5 \Missing or unrecognized delimiter for \right\)
Hence, X = \(\frac{z}{2} \frac{}{
}\), |z|>1.
15. The Laplace transform of the function cosh 2 is ____________
a) \
\
\
\(\frac{s^2+2}{s
}\)
Answer: b
Explanation: L ((\(\frac{1}{2}\)(e t – e -t )) 2 )
= L \Missing or unrecognized delimiter for \right\)
= \(\frac{1}{4} \frac{1}{s-2} + \frac{1}{2} \frac{1}{s} + \frac{1}{4} \frac{1}{s+2}\)
= \(\frac{s^2-2}{s
}\).
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Average Power and Energy of a Signal”.
1. For the signal \ = δ – \frac{4}{3} e^{-t} \,u + \frac{1}{3} e^{2t} \,u\). It has how many poles and zeroes at infinity?
a) 1, 0
b) 2, 2
c) 2, 0
d) 0, 0
Answer: d
Explanation: Since the degree of the numerator and denominator of X are equal as shown below,
\ = 1 – \frac{4}{3} \frac{1}{s+1} + \frac{1}{3} \frac{1}{s-2}\)
\
has neither poles nor zeroes at infinity.
2. The resultant signal obtained after frequency convolution along with constant multiplier \
e -2t + e -t
b) e -2t
c) e -3t
d) e -t
Answer: c
Explanation: f 3 = f 1 . f 2
The Laplace transform of f 3 is \(\frac{1}{2πj}\) F 1 * F 2
Then, F 1 = \(\frac{1}{s+2}\)
∴ f 1 = e -2t u
F 2 = \(\frac{1}{s+1}\)
∴ f 2 = e -t u
∴ f 3 = e -3t .
3. Given signal given, \ = 2sin
+ 4sin
+ 6sin
+ 8sin
. The period of y is ________________
a) 12 π
b) 24 π
c) 8 π
d) 16 π
Answer: b
Explanation: ω 1 = \(\frac{2}{3}\)
Or, T 1 = \(\frac{2π}{ω_1}\) = 3π
Again, ω 2 = \(\frac{1}{4}\)
Or, T 2 = \(\frac{2π}{ω_2}\) = 8π
Again, ω 3 = \(\frac{1}{3}\)
Or, T 3 = \(\frac{2π}{ω_3}\) = 6π
And, ω 4 = \(\frac{1}{2}\)
Or, T 4 = \
∴ T = 24π.
4. The signal power of the periodic rectangular pulses of height 1 and width 1, is _______________
a) 0.25 W
b) 0.75 W
c) 0.5 W
d) 1 W
Answer: c
Explanation: The signal power in the given signal using Parsevals’s relation is
\
\,dt\)
\(= \frac{1}{2} \int_0^1 1.dt\)
= 0.5 W.
5. Power spectral density of the signal given below is _______________
signals-systems-questions-answers-average-power-energy-signal-q5
a) 0
b) 6
c) ∞
d) -6
Answer: b
Explanation: P = \
d\)
= \
+ 2 + 1
= 1 + 4 + 1 = 6.
6. A periodic signal has power \Missing open brace for subscript \
\
\
\(\frac{\sqrt{P}}{2}\)
Answer: d
Explanation: For an energy signal if the average energy per period is P, then the rms value which corresponds to a DC signal is equal to \
, then x rms –> \(\frac{\sqrt{P}}{2}\).
7. For a LTI system consisting of two integrators, the output of the first integrator is inverted and fed as input to the sigma at beginning and output of the second integrator is also inverted and fed as input to the sigma at the beginning. The input is x and output y , the transfer function is ______________
a) \
\
\
\(\frac{1}{s^2-4s+3}\)
Answer: b
Explanation: Transfer function of linear loop = \(\displaystyle\frac{\frac{1}{s}}{1 + \frac{4}{s}}\)
= \(\frac{1}{s+4}\)
∴ Transfer function of the overall system = \(\displaystyle\frac{\frac{1}{s}}{1 + \frac{3}{s}}\)
= \(\frac{1}{s^2+4s+3}\).
8. A system has a block G 1 which has input x and output y 1 . This output is fed as input to a second block G 2 which has output y 2 . For this system, G 1 = \(\frac{s^{-1}}{1 + 2s^{-1} + s^{-2}}\).
G 2 for the given system to be invertible is given by _______________
a) \
\
\
1+2s+s 2
Answer: b
Explanation: Given that G 2 should be the inverse of G 1 .
∴ G 2 = \
\)
Or, G 1 = \(\frac{s^{-1}}{1 + 2s^{-1} + s^{-2}}\)
∴ G 2 = \
= \frac{1+2s+s^2}{s}\).
9. The convolution y [n] = x [n] * h [n], where x [n] = {1,2,4} and h [n] = {1,1,1,1,1} is ________________
a) {1,3,7,7,7,6,4}
b) {1,3,3,7,7,6,4}
c) {1,2,4}
d) {1,3,7}
Answer: a
Explanation: Given x [n] = {1, 2, 4} and h [n] = {1, 1, 1, 1, 1}.
Now, y [n] is the convolution of x [n] and h [n].
Hence, y [n] = {1, 3, 7, 7, 7, 6, 4}.
10. Given a periodic function having half wave symmetry. Then the function is ________________
a) An even function
b) An odd function
c) Both odd and even functions
d) Neither odd nor even functions
Answer: c
Explanation: Let the time period be T then, f = f
The function has half wave symmetry i.e., f = -f \
\).
11. The signal power of the signal x = 2sin 2t + 4sin 4t + 6cos 4t + 2cos 2t with period 0.5 is ________________
a) 30 W
b) 36 W
c) 60 W
d) 12 W
Answer: a
Explanation: Signal power = 0.5(2 2 + 4 2 + 6 2 + 2 2 )
= 0.5
= 0.5 = 30 W.
12. A signal is a power signal if the signal has average power equal to __________
a) Infinite
b) Finite
c) Zero
d) Does not depend on the average power value
Answer: b
Explanation: A signal is said to be a power signal if and only if the average power of the signal is finite. In other words, we can say that a signal is a power signal if the energy of the signal is infinite, i.e., E = ∞.
13. A signal is an energy signal if the signal has average energy equal to __________
a) Infinite
b) Finite
c) Zero
d) Does not depend on the average energy value
Answer: b
Explanation: A signal is said to be an energy signal if and only if the average energy of the signal is finite. In other words, we can say that a signal is an energy signal if the average power of the signal is infinite, i.e., P = ∞.
14. The energy in the time-domain representation of a signal is the same as in the frequency domain representation normalized by ___________
a) 2π
b) π
c) \
\(\frac{π}{4}\)
Answer: a
Explanation: The solution lies in the basic definition of Parseval’s theorem.
We know that Parseval’s energy and power theorem states that, the energy in the time-domain representation of a signal is equal to the energy in the frequency domain representation normalized by 2π.
15. The system characterized by the equation y = ax + b is ____________
a) Linear for any value of b
b) Linear if b>0
c) Linear if b<0
d) Non-linear
Answer: d
Explanation: The system is non-linear. The principle of homogeneity states that if for any input signal X , i.e. scaling any input signal scales the output signal by the same factor, then the signal X is homogeneous. Because x does not lead to y = 0, which is a direct violation of the principle of homogeneity.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Power and Energy Signals”.
1. For any given signal, average power in its 6 harmonic components as 10 mW each and fundamental component also has 10 mV power. Then, average power in the periodic signal is _______________
a) 70
b) 60
c) 10
d) 5
Answer: b
Explanation: We know that according to Parseval’s relation, average power is equal to the sum of the average powers in all of its harmonic components.
∴ P avg = 10 × 6 = 60.
2. The property of Fourier Transform which states that the compression in time domain is equivalent to the expansion in the frequency domain is ____________
a) Duality
b) Scaling
c) Time scaling
d) Frequency shifting
Answer: c
Explanation: From the time scaling property, we can infer that, time scaling is the property which states that compression in the time domain is equivalent to the expansion in the frequency domain.
3. The impulse response h[n] of a linear time-invariant system is h[n] = u [n+3] + u [n-2] – 2u [n-7] where, u[n] is the step unit response. The above system is ________________
a) Stable but not causal
b) Stable and causal
c) Causal but unstable
d) Unstable and not causal
Answer: a
Explanation: \
= ∑_{k=-3}^∞ u + ∑_{k=2}^∞ u – 2∑_{k=7}^∞ u\)
= \
. So, the system is not causal.
4. The final value of X = \Missing open brace for subscript 1
b) 2
c) 3
d) Undetermined
Answer: d
Explanation: Final value theorem is not applicable for poles on imaginary axis.
Given axis has poles at s = +j2, -j2 and s = 0.
Therefore, x is undetermined.
5. The Laplace transform of a signal is given as Y = \Missing open brace for subscript -1
b) 0
c) 1
d) Unbounded
Answer: d
Explanation: Final value theorem is applicable only when all the poles of the system lie in the left half of the s-plane.
Here, s=1 lies on the right s-plane pole.
Hence the final value is unbounded.
6. Given that, F = \
is ___________
a) 0
b) 1
c) [-1,1]
d) ∞
Answer: c
Explanation: L -1 [F] = sin wt
So, f = sin wt
As sin lies between [-1, 1], so, final value of f lies in [-1, 1].
7. A voltage LT is given as \Missing open brace for subscript 0
b) 1/5 A
c) 2/7 A
d) 2/5 A
Answer: b
Explanation: We know that, i = lim s→0 sI
Now, V = Ls I
∴ i = lim s→0 \(\frac{V}{L}\) = lim s→0 \(\frac{1}{2}\Big[\frac{4s^2+3s+2}{7s^2+6s+5}\Big]\)
= 1/5 A.
8. A band-limited signal with a maximum frequency of 5 kHz is to be sampled. According to the sampling theorem, the sampling frequency which is not valid is _______________
a) 5 kHz
b) 12 kHz
c) 15 kHz
d) 20 kHz
Answer: a
Explanation: (f s ) min = 2 f m
Also, (f s ) min = 2 X 5 = 10 kHz
So, f s ≥ 10 kHz.
9. A current I given by I = – 8 + 6\) A is passed through three meters. The respective readings will be?
a) 8, 6 and 10
b) 8, 6 and 8
c) – 8, 10 and 10
d) -8, 2 and 2
Answer: c
Explanation: PMMC instrument reads only DC value and since it is a centre zero type, so it will give – 8 values.
So, rms = \(\sqrt{8^2 +
^2)}\) = 10 A
Moving iron also reads rms value, so its reading will also be 10 A.
10. Given, R = 10 Ω, L = 100 mH and C = 10 μF. Selectivity is ____________
signals-systems-questions-answers-power-energy-signals-q10
a) 10
b) 1.2
c) 0.15
d) 0.1
Answer: d
Explanation: Selectivity = \(\frac{1}{Q}\)
Q = \(\frac{ωL}{R} = \frac{1000×100×10^{-3}}{10}\)
∴ Q = 10
So, Selectivity = 0.1.
11. Given that X is the Laplace transform of the signal cos 2t u . The time signal corresponding to X is _________________
a) u
b)
u
c) u
d)
u
Answer: a
Explanation: We know that, s X + X is having Laplace transform \
Or, y = u .
12. Let x be a continuous time, real valued signal band limited to F Hz. The Nyquist sampling rate in Hz, for y = x + x – x is ______________
a) F
b) 2F
c) 4F
d) 8F
Answer: c
Explanation: Expansion in time domain is compression in frequency domain and vice versa. So, the maximum frequency component in given signal is 2F Hz. And according to sampling theorem,
Nyquist rate = 2 f m
= 2 X 2 F
= 4F Hz.
13. Increased pulse-width in the flat-top sampling , leads to ______________
a) Attenuation of high frequencies in reproduction
b) Attenuation of low frequencies in reproduction
c) Greater aliasing errors in reproduction
d) No harmful effects in reproduction
Answer: a
Explanation: As pulse width T is increased, the width 1/T of the first lobe of the spectrum is decreased.
Hence, increased pulse-width in the flat-top sampling, leads to attenuation of high frequencies in reproduction.
14. A relay coil having voltage of magnitude 210 V and frequency 50 Hz is connected to a 210 V, 50 Hz supply. If it has resistance of 50 Ω and an inductance of 0.2 H, the apparent power is _____________
a) 549.39 VA
b) 275.6 VA
c) 157 VA
d) 187 VA
Answer: a
Explanation: Z = 50 + j = 50 + 62.8
\(S = \frac{|V|^2}{Z} = \frac{210^2}{50-62.8}\)
Apparent Power |S| = \(\frac{210^2}{\sqrt{50^2 + 62.8^2}}\)
= 549.39 VA.
15. A band pass signal extends from 4-6 kHz. The smallest sampling frequency required to attain all the information in the signal is _________________
a) 10 kHz
b) 8 kHz
c) 6 kHz
d) 4 kHz
Answer: d
Explanation: Upper cut-off frequency f H = 6 kHz
Bandwidth = 6k – 4k
= 2 kHz
Now, f s = \(\frac{2f_H}{k}\)
Where, k = \(\frac{f_H}{B}\)
= \(\frac{6 kHz}{2 kHz}\)
= 3
Also, f s = \(\frac{2 X 6 kHz}{3}\) = 4 kHz.
This set of Signals & Systems Question Paper focuses on “Exponential Fourier Series and Fourier Transforms”.
1. The Fourier transform of u is B and the Laplace transform of u is A. Which of the following is correct?
a) B = A
b) A = \
≠ \
A ≠ \
≠ \
A ≠ \
= \(\frac{1}{jω}\)
Answer: b
Explanation: Laplace transform of u is given by u –> A = \
is given by, u = B =
+ π δ
Therefore A = \
≠ \(\frac{1}{jω}\) is satisfied.
2. Given, X (e jω ) = \Missing open brace for subscript b n u [n] + a n u [n-1]
b) b n u [n] – a n u [-n-1]
c) b n u [n] + a n u [-n-1]
d) b n u [n] – a n u [n+1]
Answer: c
Explanation: X (e jω ) = \( \frac{ e^{jω}}{e^{-j2ω}- e^{jω} + ab}\)
= \( \frac{ e^{jω}}{1-e^{-jω}+ab e^{-j2ω}}\)
= \(\frac{1}{1-be^{-jω}} + \frac{}{1-ae^{-jω}}\)
∴ x [n] = b n u [n] + a n u [-n-1].
3. The input and output of an LTI system are x = e -3t u and y = e -t u . The differential equation which characterizes the system is ___________
a) \
= \frac{dx}{dt} + 3x\)
b) \
= \frac{dx}{dt} + 3x\)
c) \
= \frac{dx}{dt} + 3x\)
d) \
= \frac{dx}{dt} + 3x\)
Answer: a
Explanation: X = \
= \
= \
+ Y = s X + 3 X
So, the differential equation together with the condition of initial rest that characterizes the system is \
= \frac{dx}{dt} + 3x\).
4. The Fourier transform of signal e -2t u is ___________
a) \
\
\
\(\frac{e^{3}}{2+jω}\)
Answer: b
Explanation: X = \
e^{-jωt} \,dt\)
Given u , hence value of this will be equal to 1 when t>=3
∴ X = \(\int_3^∞ e^{-2t} e^{-jωt} \,dt \)
= \(\frac{e^{-3}}{2+jω}\).
5. The Fourier transform of the signal e -4|t| is ____________
a) \
\
\
\(\frac{-4}{16+ω^2}\)
Answer: a
Explanation: X = \(\int_{-∞}^∞ e^{-4|t|} e^{-jωt} \,dt\)
Now, e -4|t| = e -4t , when t>0 and
e 4t , when t<0
∴X = \(\int_{-∞}^0 e^{4t} e^{-jωt} \,dt + \int_0^∞ e^{-4t} e^{-jωt} \,dt\)
= \(\frac{8}{16+ω^2}\).
6. The Inverse Fourier transform of the signal e -2|ω| is ____________
a) \
\
\
\(\frac{1}{
}\)
Answer: a
Explanation: e -2|ω| = e -2ω , ω > 0 and e 2ω , ω<0
Hence, x = \(\frac{1}{2π} \int_{-∞}^∞ e^{-2|ω|} e^{-jωt} \,dω\)
= \(\frac{1}{2π} \int_{-∞}^0 e^{2ω} e^{-jωt} \,dω + \frac{1}{2π} \int_0^∞ e^{-2ω} e^{-jωt} \,dω\)
= \(\frac{2}{π
}\).
7. The inverse Laplace transform of F = \Missing open brace for subscript 2e -k u
b) 2e -k u
c) 2e k u
d) 2e k u
Answer: b
Explanation: Let G = \
= L -1 {G} = 2e -ct
∴ F = L -1 {G e -bs }
= 2e -k u .
8. The Laplace transform of the function e -2t cos + 5e -2t sin is ____________
a) \
\
\
\(\frac{-15}{^2+9}\)
Answer: c
Explanation: L {e -2t cos + 5e -2t sin}
= \(\frac{}{^2+9} + \frac{53}{^2+9}\)
= \(\frac{+15}{^2+9}\).
9. A band pass signal extends from 1 KHz to 2 KHz. The minimum sampling frequency that is needed to retain all information of the sampled signal is ___________
a) 1 KHz
b) 2 KHz
c) 3 KHz
d) 4 KHz
Answer: b
Explanation: We know that the minimum sampling frequency is twice the maximum bandwidth.
Here, maximum bandwidth = 2-1 = 1 KHz
So, Minimum sampling frequency = 2 = 2 = 2 KHz.
10. The Laplace transform of the function 6e 5t cos – e 7t is ______________
a) \
\
\
\(\frac{6}{^2+4} + \frac{1}{s-7}\)
Answer: a
Explanation: We know that, Laplace transform of e at = \(\frac{1}{s-a}\)
Here, a=7, so L {e 7t } = \(\frac{1}{s-7}\)
And the Laplace transform of e at cos = \(\frac{}{^2+b^2}\)
Here, a=5 and b=2, so L {6e 5t cos } = \(\frac{6}{^2+4}\)
∴ L {6e 5t cos – e 7t } = \(\frac{6}{^2+4} – \frac{1}{s-7}\).
11. The inverse Laplace transform of F = \Missing open brace for subscript {0.5 + 0.5e -2 -e - } u
b) {0.5 + 0.5e -2 -e - } u
c) {0.5 – 0.5e -2 -e - } u
d) 0.5 + 0.5e -2t -e -t )
Answer: b
Explanation: Let G = \
= G e -3s
G = L -1 {G}
= L -1 {\Extra close brace or missing open brace = 0.5 + 0.5e -2t -e -t
The inverse Laplace transform is F = {0.5 + 0.5e -2 -e - } u .
12. The Fourier transform of the signal e -t+2 u is ___________
a) \
\
\
\(\frac{e^{2jω}}{1-2ω}\)
Answer: c
Explanation: Fourier transform of e -t u = \
= e -2jω X [Using Shifting property]
Hence, X = \(\frac{e^{-2jω}}{1+2ω}\).
13. The Fourier transform of the signal te -3|t-1| is _____________
a) \
\
–\
\(\frac{6e^{-jω}}{9+ω^2} + \frac{12jωe^{-jω}}{9+ω^2}\)
Answer: a
Explanation: e -3|t| = e -3t , when t>0 and
e 3t , when t<0
Now, Fourier transform of e -3|t| = \
↔ e -jω X
And t x ↔ \
∴ X = \(j\frac{d}{dω} [e^{-jω} \frac{6}{9+ω^2}]\)
= \(\frac{6e^{-jω}}{9+ω^2} – \frac{12jωe^{-jω}}{9+ω^2}\).
14. The Fourier transform of the signal te -t u is _____________
a) \
\
\
\(\frac{1}{^2}\)
Answer: c
Explanation: Fourier transform of te -at u is \
= \(\int_0^∞ te^{-t} e^{-jωt} \,dt\)
= \(\int_0^∞ te^{-t} \,dt\)
= \(\frac{1}{^2}\).
15. The Fourier transform of the signal \
\) is _____________
a) \
\
\
\(\frac{1}{1-ae^{-jω}}\)
Answer: d
Explanation: X = \
e^{-jωt} \,dt\)
= \
^m \)
= \(\frac{1}{1-ae^{-jω}}\).
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Fourier Analysis”.
1. The CTFT of a continuous time signal x = e -A|t| , A>0 is _________
a) \
\
\
\(\frac{A}{ω^2} \)
Answer: c
Explanation: CTFT {x } = X = \
e^{-jωt} \,dt\)
= \
= \(\frac{2A}{A^2+ω^2} \).
2. A signal x has the Fourier transform X having the following facts:
F -1 { X} = Ae -2t u and \
|^2 \,dω = 2π\)
The signal x is ___________
a) \(\sqrt{3}\) (e -t – e -2t )u
b) \(\sqrt{12}\) (e -t – e -2t )u
c) \(\sqrt{3}\) (e -2t – e -t )u
d) \(\sqrt{12}\) (e -2t – e -t )u
Answer: b
Explanation: F -1 { X} = Ae -2t u
Or, X = \
= \
\)
Or, x = Ae -t u – Ae -2t u
Given that, \
|^2 \,dω\) = 2π
Or, \
|^2 \,dt\) = 1
∴ \
dt = 1
Or, \
is non-negative.
3. The system characterized by the differential equation \
= x\) is _____________
a) Linear and stable
b) Linear and unstable
c) Nonlinear and unstable
d) Nonlinear and stable
Answer: b
Explanation: \
= x\)
Now, x –> h so, the system is linear.
Again, taking Laplace Transform with zero initial conditions, we get, s 2 Y – s Y – 2Y = X
Or, H = \(\frac{Y}{X} = \frac{1}{s^2-s-2} = \frac{1}{} \)
Since pole is at s = +2, the system is unstable.
4. An LTI system with impulse response h 1 [n] = -2
^n\) u[n] is connected in parallel with another causal LTI system with impulse response h 2 [n]. The resulting interconnection has frequency response H (e jω ) = \(\frac{-12+5e^{-jω}}{12+7e^{-jω}+e^{-j2ω}}\). Then h 2 [n] is ___________
a)
n u[n-1]
b)
n u[n]
c)
n u[n]
d)
n u[n-1]
Answer: b
Explanation: H (e jω ) = H 1 (e jω ) + H 2 (e jω )
Or, \(\frac{-12+5e^{-jω}}{12+7e^{-jω}+e^{-j2ω}} = \frac{1}{1-\frac{1}{3} e^{-jω}} + \frac{}{1-\frac{1}{4} e^{-jω}}\)
∴ H 2 (e jω ) = \(\frac{1}{1-\frac{1}{3} e^{-jω}}\)
So, h 2 [n] =
n u[n].
5. A pulse of unit amplitude and width a, is applied to a series RL circuit having R = 1 Ω, L = 1H. The current I at t = ∞ is __________
a) 0
b) Infinite
c) 2 A
d) 1 A
Answer: a
Explanation: Since, the circuit has a resistance; the current will eventually decrease and ultimately die down.
Now, I = \
[u
– u
]\)
But R = 1Ω and L = 1H
∴ I = [u(1-e -t ) – u(1-e - )]
This also leads to I = 0.
6. The rms value of a rectangular wave of period T, having value +V for a duration, T 1 and –V for the duration, T-T 1 = T 2 is __________
a) V
b) \ \
0
Answer: a
Explanation: \(\sqrt{V^2}\)n: Mean square value = \(\frac{1}{T}[∫_0^{T_1} V^2 \,dt + ∫_1^T V^2 \,dt] \)
Where, T 1 + T 2 = T
= \(\frac{1}{T}\)[V 2 T 1 + V 2 T – V 2 T 1 ] = V 2
∴ RMS = V.
7. X (e jω ) = \Missing open brace for subscript b n u [n] + a n u [n-1]
b) b n u [n] – a n u [-n-1]
c) b n u [n] + a n u [-n-1]
d) b n u [n] – a n u [n+1]
Answer: c
Explanation: X (e jω ) = \
\)
= \(\frac{1}{1-be^{-jω}} + \frac{}{1-ae^{-jω}}\)
∴ x [n] = b n u [n] + a n u [-n-1].
8. Frequency and time period are ____________
a) Proportional to each other
b) Inverse of each other
c) Same
d) equal
Answer: b
Explanation: Frequency is the number of occurrences of a repeating event per unit time.
Time period, denoted by T, is the duration of one cycle and is equal to the reciprocal of the frequency. Mathematically we can write T = 1/f.
9. The Fourier series for the function f = sin 2 x is ______________
a) 0.5 + 0.5 sin 2x
b) 0.5 – 0.5 sin 2x
c) 0.5 + 0.5 cos 2x
d) 0.5 – 0.5 cos 2x
Answer: d
Explanation: f = sin 2 x
Now, f = sin 2 x = \(\frac{1-cos2x}{2}\)
= 0.5 – 0.5 cos 2x.
10. The continuous time system described by the equation y = x(t 2 ) comes under the category of ____________
a) Causal, linear and time varying
b) Causal, non-linear and time varying
c) Non-causal, non-linear and time invariant
d) Non-causal, linear and time variant
Answer: d
Explanation: Let y = x (t 2 ). We can infer that y depends on x (t 2 ) i.e. on future values of input if t>1. Hence, the system is non-casual.
Again, α x 1 → y 1 = α x 1 (t 2 ) and β x 2 –> y 2 = β x 2 (t 2 )
Therefore α x 1 + β x 2 –> y = α x 1 (t 2 ) + β x 2 (t 2 ) = y 1 + y 2 , which implies that the system is linear.
Again, x = u – u –> y and X 1 = x –> y 1 .
So, we get, y 1 ≠ y , which implies that the system is time varying.
11. The running integrator, given by y = \
\,dt\) has ____________
a) No finite singularities in it’s double sided Laplace transform Y
b) Produces an abounded output for every causal bounded input
c) Produces a bounded output for every anti-causal bounded input
d) Has no finite zeroes in it’s double sided Laplace transform Y
Answer: b
Explanation: The running integrator \
\,dt = 0\) for every causal system. As causal systems have no memory and the initial value is zero, the output is followed by input. So, y will always be bounded if this function is a causal bounded system.
12. A signal x is given by
x = 1, -T/4<t≤3T/4
= -1, 3T/4<t≤7T/4
= -x
Which among the following gives the fundamental Fourier terms of x ?
a) \
\)
b) \
\)
c) \
\)
d) \
\)
Answer: c
Explanation: Given signal,
x = 1, -T/4<t≤3T/4
= -1, 3T/4<t≤7T/4
= -x
Now by property of symmetry of Fourier transform of x , we get the fundamental Fourier term as, \
\).
13. The type of systems which are characterized by input and the output capable of taking any value in a particular set of values are called as __________
a) Analog
b) Discrete
c) Digital
d) Continuous
Answer: d
Explanation: We know that continuous systems have a restriction on the basis of the upper bound and lower bound. However within this set, the input and output can assume any value. Hence, there are infinite values attainable in this system.
14. In Maxwell’s capacitance bridge for calculating unknown inductance, the various values at balance are, R 1 = 300 Ω, R 2 = 700 Ω, R 3 = 1500 Ω, C 4 = 0.8 μF. Calculate R 1 , L 1 and Q factor, if the frequency is 1100 Hz.
a) 240 Ω, 0.12 H, 3.14
b) 140 Ω, 0.168 H, 8.29
c) 140 Ω, 0.12 H, 5.92
d) 240 Ω, 0.36 H, 8.29
Answer: b
Explanation: From Maxwell’s capacitance, we have
R 1 = \(\frac{R_2 R_3}{R_4} = \frac{300 × 700}{1500}\) = 140 Ω
L 1 = R 2 R 3 C 4
= 300 × 700 × 0.8 × 10 -6 = 0.168 H
∴ Q = \(\frac{ωL_1}{R_1}\).
15. Given a real valued function y with period T. Its trigonometric Fourier series expansion contains no term of frequency ω = 2π \
satisfies the equation ____________
a) y = y = -y
b) y = y = y
c) y = y = -y
d) y = y = y
Answer: d
Explanation: For an even symmetry, y = y
Thus no sine component will exist because b n =0 and by half wave symmetry condition odd harmonics will exist.
Now, y = y
Combining the two conditions, we get, y = y = y
.
This set of Signals & Systems written test Questions & Answers focuses on “Fourier Series Analysis using Circuits”.
1. Given x = [2 + e -3t ] u . The final value of x is ___________
a) 2
b) 3
c) e -3t
d) 0
Answer: a
Explanation: Given x = [2 + e -3t ] u
Applying final value theorem, we get, lim t→∞ x= lim s→0 sX
= lim s→0 \(s[\frac{1}{s} + \frac{1}{s+3}]\)
= lim s→0 \(\frac{2s+3}{s+3}\)
= 2.
2. The Fourier series of the given signal is _______________
signals-systems-written-test-questions-answers-q2
a) -4/π sin x
b) 4/π sin x
c) 4/π cos x
d) -4/π cos x
Answer: b
Explanation: We know that the Fourier series is given by a 0 + \(∑_{n=1}^∞\) (a n cos nx + b n sin nx)
This can be also written as a 0 + a 1 cos x + b 1 sin x + a 2 cos 2x + b 2 sin 2x ……
Here, c 0 = a 0 = 0
So, c n = b n = \
.sinωt \,dt\)
∴ c 1 = b 1 = \(2∫_0^t sinωt \,dt\)
= \(– \frac{2}{ω} [cosωt]_0^t\)
= \(-\frac{2}{π}\)[-1-1] = 4/π
And a 1 = \
.cosωt \,dt\)
= \(\frac{2}{ω}[sinωt]_0^t\)
= 0
So, the Fourier series can be written as 0 + 0.cos x + 4/π sin x
= 4/π sin x.
3. For the circuit given below, the effective inductance of the circuit across the terminal AB, is ___________
signals-systems-written-test-questions-answers-q3
a) 9 H
b) 21 H
c) 11 H
d) 6 H
Answer: c
Explanation: LEFF across AB = L 1 + L 2 + L 3 – 2M 12 – 2M 13 + 2M 23
=
=
=
=
= 11.
4. For the circuit given below, the inductance measured across the terminals 1 and 2 was 15 H with open terminals 3 and 4. It was 30 H when terminals 3 and 4 were short-circuited. Both the inductors are having inductances 2H. The coefficient of coupling is ______________
signals-systems-written-test-questions-answers-q4
a) 1
b) 0.707
c) 0.5
d) Inderminate due to insufficient data
Answer: d
Explanation: When 2 coils are connected in series, then effective inductance,
L EFF = L 1 + L 2 ± 2M
For this case, L EFF = L 1 + L 2 – 2M
However, L 1 + L 2 + 2M or L 1 + L 2 – 2M cannot be determined.
Hence, data is insufficient to calculate the value of coupling coefficient .
5. In the circuit given below, the resonant frequency is ________________
signals-systems-written-test-questions-answers-q5
a) \
\
\
\(\frac{1}{π\sqrt{2}}\) Hz
Answer: b
Explanation: L EFF = L 1 + L 2 + 2M
= 2 + 2 + 2 × 1
Or, L EFF = 6 H
At resonance, ωL = \(\frac{1}{ωC}\)
Or, ω = \(\frac{1}{\sqrt{L_{EFF} C}} = \frac{1}{\sqrt{12}}\)
Or, I = \(\frac{1}{4π\sqrt{3}}\) Hz.
6. Q meter operator is the principle of __________
a) Series resonance
b) Current resonance
c) Self-inductance
d) Eddy currents
Answer: a
Explanation: We know that, Q = \(\frac{ωL}{R}\)
From the above relation, we can say that it works on series resonance.
7. The probability cumulative distribution function must be monotone as well as ___________
a) Increasing
b) Decreasing
c) Non-increasing
d) Non-decreasing
Answer: d
Explanation: The probability cumulative distribution function increases to 1 monotonically and there after remains constant. Therefore we can say that the probability cumulative distribution function must be monotone as well as non-decreasing.
8. The impedance seen by the source in the circuit is given by __________
signals-systems-written-test-questions-answers-q8
a) Ω
b) Ω
c) Ω
d) Ω
Answer: c
Explanation: Z 1 = 10∠30° × \
^2\)
Z 1 = Ω
Total impedance= +
= Ω.
9. In the circuit given below, the capacitor is initially having a charge of 10 C. 1 second after the switch is closed, the current in the circuit is ________
signals-systems-written-test-questions-answers-q9
a) 14.7 A
b) 18.5 A
c) 40.0 A
d) 50.0 A
Answer: a
Explanation: Using KVL, 100 = \(R\frac{dq}{dt} + \frac{q}{C}\)
100 C = RC\(\frac{dq}{dt}\) + q
Or, \(\int_{q_o}^q \frac{dq}{100C-q} = \frac{1}{RC} \int_0^t \,dt\)
100C – q = (100C – q o )e -t/RC
I = \(\frac{dq}{dt} = \frac{100C – q_o}{RC} e^{-1/1}\)
∴ e -t/RC = 40e -1 = 14.7 A.
10. Convolution is used to find ____________
a) Impulse of an LTI system
b) Frequency response of an LTI system
c) Time response of an LTI system
d) Phase response of an LTI system
Answer: c
Explanation: We know that, the time response of an LTI system is given by
Y = \
h\)
This can also be written as y = x * h , which is the convolution of two systems. Therefore convolution is used to find the time response of an LTI system.
11. The Fourier transform of a rectangular function is ___________
a) Another rectangular pulse
b) Triangular pulse
c) Sinc function
d) Impulse
Answer: c
Explanation: Substituting the square pulse function f in the below equation as shown,
F = \
e^{jωt} \,dt\)
This gives us the Sinc function.
12. The property of Fourier Transform which states that the compression in time domain is equivalent to expansion in the frequency domain is ___________
a) Duality
b) Scaling
c) Time scaling
d) Frequency shifting
Answer: b
Explanation: We know that the definition of Fourier Transform states that Fourier Transform is a function derived from a given function and representing it by a series of sinusoidal functions. The property of Fourier Transform which states that the compression in the time domain is equivalent to expansion in the frequency domain is Scaling can be found by putting the value of pulse function in the definition of Fourier Transform.
13. A coil of inductance 10 H, resistance 40 Ω is connected as shown in the figure. After the switch S has been in connection with point 1 for a very long time, it is moved to point 2 at t=0. For the value of R obtained, the time taken for 95 % of the stored energy dissipated is close to?
signals-systems-written-test-questions-answers-q13
a) 0.10 s
b) 0.15 s
c) 0.50 s
d) 1.0 s
Answer: c
Explanation: For source free circuit,
I = I o \
= 0.05 = 2 × \(e^{-\frac{60}{10} \,t}\)
Or, t = 0.61 ≈ 0.5 s.
14. The function which has its Fourier transform, Laplace transform, and Z transform unity is __________
a) Gaussian
b) Impulse
c) Sinc
d) Pulse
Answer: b
Explanation: The definition of Fourier transform is \
e^{2πjxf} \,dx\)
The definition of Laplace transform is \
e^{-st} \,dt\)
The definition of Z-transform is \
= δ in the definitions of Fourier Laplace and Z-transform, we get the transforms in each case as 1. Hence an impulse function has unity Fourier Laplace and Z-transform.
15. N = 2, 0≤t<4;
t 2 , t≥4;
The Laplace transform of M is ___________
a) \
\)
b) \
\)
c) \
\)
d) \
\)
Answer: d
Explanation: N = 2 + u (t 2 – 2)
L {2 + u (t 2 -2)} = \(\frac{2}{s}\) + L {u (t 2 -2)}
= \(\frac{2}{s} + e^{-4s}\) L { 2 – 2}
= \(\frac{2}{s} + e^{-4s}\) L {t 2 + 8t + 14}
= \
\).
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Fourier Transforms”.
1. Which of the following is the Analysis equation of Fourier Transform?
a) \ = \int_{-∞}^∞ fe^{jωt} \,dt\)
b) \ = \int_0^∞ fe^{-jωt} \,dt\)
c) \ = \int_0^∞ fe^{jωt} \,dt\)
d) \ = \int_{-∞}^∞ fe^{-jωt} \,dt\)
Answer: d
Explanation: For converting time domain to frequency domain, we use analysis equation. The Analysis equation of Fourier Transform is \ = \int_{-∞}^∞ fe^{-jωt} \,dt\).
2. Choose the correct synthesis equation.
a) \ = \frac{1}{2π} \int_{-∞}^∞ F e^{-jωt} \,dω\)
b) \ = \frac{1}{2π} \int_{-∞}^∞ F e^{jωt} \,dω\)
c) \ = \frac{1}{2π} \int_0^∞ F e^{-jωt} \,dω\)
d) \ = \frac{1}{2π} \int_0^∞ F e^{jωt} \,dω\)
Answer: b
Explanation: Synthesis equation converts from frequency domain to time domain. The synthesis equation of fourier transform is \ = \frac{1}{2π} \int_{-∞}^∞ F e^{jωt} \,dω\).
3. Find the fourier transform of an exponential signal f = e -at u, a>0.
a) \
\
\
\(\frac{1}{-a-jω}\)
Answer: a
Explanation: Given f= e -at u
We know that \=
\)
Fourier transform,
\ = \int_{-∞}^∞ fe^{-jωt} \,dt = \int_{-∞}^∞ e^{-at} ue^{-jωt} \,dt = \int_0^∞ e^{-t} \,dt\)
F = \(\frac{1}{a+jω}\), a>0.
4. Find the fourier transform of the function f = e -a|t| , a>0.
a) \
\
\
\(\frac{a}{a^2+ω^2}\)
Answer: b
Explanation: The given two-sided exponential function f = e -a|t| , a>0 can be expressed as
\=
\)
The Fourier transform is
\ = \int_{-∞}^∞ fe^{-jωt} \,dt = \int_{-∞}^0 fe^{-jωt} \,dt + \int_0^∞ fe^{-jωt} \,dt\)
\ = \frac{1}{a+jω} + \frac{1}{a-jω} = \frac{2a}{a^2+ω^2}\).
5. Gate function is defined as ______________
a) \=
\)
b) \=
\)
c) \=
\)
d) \=
\)
Answer: a
Explanation: A gate function is a rectangular function defined as
\ = rect
=
\)
Where τ is pulse width.
6. Find the fourier transform of the gate function.
a) \
\)
b) \
\)
c) \
\)
d) \
\)
Answer: c
Explanation: Gate function is defined as
\=
\)
The fourier transform is \ = \int_{-∞}^∞ fe^{-jωt} \,dt = \int_{-τ/2}^{τ/2} e^{-jωt} \,dt = \frac{2}{ω} sin
\).
7. Choose the wrong option.
a) G = rect
b) G = u
– u
c) G = τ sa
d) G = τ sinc
Answer: d
Explanation: Fourier transform of gate function, G = \
\)
Multiplying and dividing by τ we get
\ = τ \frac{sin
}{\frac{wτ}{2}} = τ \frac{sin
}{\frac{2πfτ}{2}}= τ \frac{sin}{πτf} = τ sinc\).
8. Bandwidth of the gate function is __________
a) τ Hz
b) \
2τ Hz
d) \(\frac{2}{τ}\) Hz
Answer: b
Explanation: The practical bandwidth of the gate function corresponds to the first zero crossing in the spectrum. Therefore, the bandwidth of the pulse or gate function is \(\frac{2π}{τ}\) or \(\frac{1}{τ}\) Hz.
9. Which of the following is not a fourier transform pair?
a) \ \leftrightarrow πδ + \frac{1}{jω}\)
b) \ \leftrightarrow \frac{2}{jω}\)
c) \
\)
d) \\leftrightarrow sa
\)
Answer: d
Explanation: \\leftrightarrow sa
\) is not a fourier transform pair.
\\leftrightarrow τsa
\) \\leftrightarrow G τ sinc\).
10. Find the fourier transform of the unit step function.
a) πδ + \
πδ + \
πδ – \
δ + \(\frac{1}{jω}\)
Answer: b
Explanation: We know that sgn = 2u – 1.
u = \
+1] Its Fourier transform is F[u] = \
] + \
and [sgn] = \
] = πδ + \(\frac{1}{jω}\).
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Properties of Fourier Transforms”.
1. The Fourier transform of a function x is X. What will be the Fourier transform of \
\
j2πfX
c) \
jfX
Answer: b
Explanation: We know that x = \
e^{jωt} \,dω\)
\
= \frac{1}{2π} \int_{-∞}^∞ X \frac{d}{dt} e^{jωt} \,dω = \frac{1}{2π} jω X \int_{-∞}^∞ e^{jωt} \,dω\)
= jω X = j2πfX.
2. Find the Fourier transform of \
sinc
b) sa
c) δ
d) sgn
Answer: d
Explanation: Let x = sgn
The Fourier transform of sgn is X = F[sgn] = \
= \
↔ 2πx, we have
F\
= -2πsgn
\
\
.
3. The Fourier transform of a Gaussian pulse is also a Gaussian pulse.
a) True
b) False
Answer: a
Explanation: Gaussian pulse, x = e -πt 2
Its Fourier transform is X = e -πf 2
Hence, the Fourier transform of a Gaussian pulse is also a Gaussian pulse.
4. Find the Fourier transform of f=te -at u.
a) \
\
\
\(\frac{ω}{^2} \)
Answer: b
Explanation: Using frequency differentiation property, \ \leftrightarrow j \frac{d}{dω} \,X\)
\
] = j \frac{d}{dω} F[te^{-at} \,u] = j \frac{d}{dω} \frac{1}{a+jω} = j \frac{-1}{^2} = \frac{1}{^2} \)
\
\leftrightarrow \frac{1}{^2} \).
5. Find the Fourier transform of e jω 0 t .
a) δ(ω + ω 0 )
b) 2πδ(ω + ω 0 )
c) δ(ω – ω 0 )
d) 2πδ(ω – ω 0 )
Answer: d
Explanation: We know that F[1] = 2πδ
By using the frequency shifting property, e jω 0 t x ↔ X(ω – ω 0 )
We have F[e jω 0 t ] = F[e jω 0 t ] = 2πδ(ω – ω 0 ).
6. Find the Fourier transform of u.
a) πδ + \
πδ + \
πδ – \
δ + \(\frac{1}{jω}\)
Answer: c
Explanation: We know that F[u] = πδ + \
↔ X
We have F[u] = πδ – \(\frac{1}{jω}\).
7. Find the Fourier transform of x = f + f.
a) 2Fcos2ω
b) Fcos2ω
c) 2Fsin2ω
d) Fsin2ω
Answer: a
Explanation: Using linearity property, ax + by ↔ aX + bY and
Time shifting property, x(t-t 0 ) ↔ e -jω 0 t X,
We have F[x] = F[f] e -j2ω + F[f] e j2ω = Fe -j2ω + Fe j2ω = 2Fcos2ω.
8. Find the Fourier transform of \
2πe aω u
b) 2πe aω u
c) 2πe -aω u
d) 2πe -aω u
Answer: b
Explanation: Let X = \
= \
= e -at u
As per duality property X ↔ 2πx, we have
\] = F\Big[\frac{1}{a+jt}\Big]\) = 2πx = 2πe aω u.
9. Find the Fourier transform of e -2t u.
a) \
\
\
\(e^{-2} [e^{-jω} \frac{1}{2+jω}]\)
Answer: d
Explanation: We know that e -at u ↔ \(\frac{1}{a+jw}\)
Using time shifting property, x(t-t 0 ) ↔ e -jω 0 t X we have
f[e -2t u] = \(e^{-2} [e^{-jω} \frac{1}{2+jω}]\).
10. Find the Fourier transform of sinc.
a) G π
b) G 2π
c) \
d) G π
Answer: b
Explanation: Using duality property, X ↔ 2πx
We get sinc ↔ G 2π .
11. If the Fourier transform of g is G, then match the following and choose the right answer.
(i) The Fourier transform of g(t-2) is (A) G(ω)e^-j2ω
(ii) The Fourier transform of g(t/2) is (B) G(2ω)
(C) 2G(2ω)
(D) G(ω-2)
a) -B, -A
b) -A, -C
c) -D, -C
d) -C, -A
Answer: b
Explanation: Using time shifting property, x(t – t 0 ) ↔ e -jω 0 t X
g ↔ e -j2ω G
Time scaling property, x ↔ \
\)
g ↔ 2G.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Inverse Fourier Transform”.
1. Find the inverse Fourier transform of X = e -2ω u.
a) \
\
\
\(\frac{1}{π}\)
Answer: b
Explanation: We know that x = \
e^{jωt} \,dω\)
x = \
e^{jωt} \,dω = \frac{1}{2π} \int_{-∞}^∞ e^{-2ω} e^{jωt} \, dω = \frac{1}{2π}\).
2. Find the inverse Fourier transform of X = \
3e -3t u + 8e -3t u
b) 3te -3t u – 8e -8t u
c) 3e -3t u + 8te 8t u
d) 3e -3t u – 8te -3t u
Answer: d
Explanation: Given X = \
= 3e -3t u – 8te -3t u.
3. Find the inverse Fourier transform of δ.
a) \
2π
c) \
π
Answer: d
Explanation: We know that x = \
e^{jωt} \,dω\)
= \
e^{jωt} \,dω = \frac{1}{2π}\).
4. Find the inverse Fourier transform of u.
a) \
+ \frac{j}{2πt}\)
b) \
– \frac{j}{2πt}\)
c) δ + \
δ – \(\frac{j}{2πt}\)
Answer: a
Explanation: We know that u = \
].
Applying linearity property,
u = -1 \
]\)
u = \
+ \frac{j}{2πt}\).
5. Find the inverse Fourier transform of e j2t .
a) 2πδ
b) πδ
c) πδ
d) 2πδ
Answer: a
Explanation: We know that e jω 0 t ↔ 2πδ(ω-ω 0 )
∴ e j2t ↔ 2πδ.
6. Find the inverse Fourier transform of jω.
a) δ
b) \
c) \
∫δ
Answer: b
Explanation: Time differentiation property, \
↔ jωX and we know that δ ↔ 1
∴ \
↔ jω.
7. Find the inverse Fourier transform of \ = \frac{6+4}{^2 + 6 + 8}\).
a) e -2t u – 5e -4t u
b) e -2t u + 5e -4t u
c) -e -2t u – 5e -4t u
d) -e -2t u + 5e -4t u
Answer: d
Explanation: \ = \frac{6+4}{^2+6+8} = \frac{A}{jω+2} + \frac{B}{jω+4} = -\frac{1}{jω+2} + \frac{5}{jω+4}\)
Applying inverse Fourier transform, we get
x = -e -2t u + 5e -4t u.
8. Find the convolution of the signals x 1 = e -2t u and x 2 = e -3t u.
a) e -2t u – e -3t u
b) e -2t u + e -3t u
c) e 2t u – e 3t u
d) e 2t u – e -3t u
Answer: a
Explanation: Convolution property, x 1 *x 2 ↔ X 1 X 2
∴ x 1 *x 2 = F -1 [X 1 X 2 ]
Given x 1 = e -2t u
∴ X 1 = \(\frac{1}{jω+2}\)
Given x 2 = e -3t u
∴ X 1 = \(\frac{1}{jω+3}\)
x 1 *x 2 = F -1 [X 1 X 2 ] = F -1 \([\frac{1}{jω+2} \frac{1}{jω+3}] = F^{-1} [\frac{1}{jω+2} – \frac{1}{jω+3}] \)
∴ x 1 *x 2 = e -2t u-e -3t u.
9. Find the inverse Fourier transform of f=1.
a) u
b) δ
c) e -t
d) \(\frac{1}{jω}\)
Answer: b
Explanation: We know that the Fourier transform of f = 1 is F = 2πδ.
Replacing ω with t
F = 2πδ
As per duality property F ↔ 2πf, we have
2πδ ↔ 2π
δ ↔ 1
Hence, the inverse Fourier transform of 1 is δ.
10. Find the inverse Fourier transform of sgn.
a) \
\
\
\(\frac{1}{t}\)
Answer: b
Explanation: Given the function F=sgn. The Fourier transform of a Signum function is sgn = \
↔ 2πf, we get
F
= 2πsgn.
As sgn is an odd function, sgn=-sgn.
Hence, \
Or \
Therefore, the inverse Fourier transform of sgn is \(\frac{j}{πt}\).
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Discrete Fourier Transform”.
1. Given that S 1 and S 2 are two discrete time systems. Consider the following statements:
i) If S 1 and S 2 are linear, then S is linear
ii) If S 1 and S 2 are non-linear, then S is non-linear
iii) If S 1 and S 2 are causal, then S is causal
iv) If S 1 and S 2 are time invariant, then S is time invariant
The true statements from the above are ____________
a) i, ii, iii
b) ii, iii, iv
c) I, iii, iv
d) I, ii, iii, iv
Answer: c
Explanation: Only statement ii is false.
For example, S 1 : y[n] = x[n] +b and S 2 : y[n] = x[n]-b, where, b≠0.
S{x[n]} = S2 {S 1 {x[n]}} = S 2 {x[n] +b} = x[n]
Hence, S is linear.
2. For two discrete time systems, consider the following statements:
i) If S 1 and S 2 are linear and time invariant, then interchanging their order does not change the system.
ii) If S 1 and S 2 are linear and time variant, then interchanging their order does not change the system
The correct statement from the above is __________
a) Both i & ii
b) Only i
c) Only ii
d) Neither i, nor ii
Answer: b
Explanation: S 1 : y[n] = n x[n]
And S 2 : y[n] = n x [n+1]
If x[n] = δ[n], then S 2 {S1 {δ[n]}} = S 2 [0] = 0
S 1 {S 2 {δ[n]}} = S 1 {δ[n+1]} = – δ[n+1]≠0.
3. The following input-output pairs have been observed during the operation of a time invariant system
i) x 1 [n] = {1, 0, 2} y 1 [n] = {0, 1,2}
ii) x 2 [n] = {0,0, 3} y 2 [n] = {0,1,0,2}
iii) x 3 [n] = {0,0,0,1} y 3 [n] = {1,2,1}
The conclusion regarding the linearity of the system is _____________
a) Linear
b) Non-linear
c) One more observation is required
d) Conclusion cannot be drawn from observation
Answer: b
Explanation: System is not linear. This is evident from the observation of the pairs, x 3 [n] – y 3 [n] and x 2 [n] and y 2 [n]. If the system were linear y 2 [n] would be of the form y 2 [n] = {3, 6, 3}.
4. S 1 and S 2 are two DT systems which are connected together to form a new system. Consider the following statements:
i) If S 1 and S 2 are non-causal, then S is non-causal
ii) If S 1 and/or S 2 are unstable, then S is unstable
The correct statement from the above is ____________
a) Both i and ii
b) Only i
c) Only ii
d) Neither i nor ii
Answer: d
Explanation: S 1 : y[n] = x [n+1] ……
S 2 : y[n] = x [n-2] ………
S: y[n] = x [n-1] which is causal ……….
S 1 : y[n] = e x[n] stable, S 2 : y[n] = ln ………
But S: y[n] = x[n] ……….
5. Given a signal x[n] = δ[n] + 0.9 δ [n − 6]. The Discrete Time Fourier Transform for 8 points is __________
a) 1 – 0.9 \
1 + 0.9 \
1 + 0.9 \
1 – 0.9 \(e^{j \frac{2π}{8} k6}\)
Answer: b
Explanation: Given N = 8.
Now, x[k] = \
e^{-j \frac{2π}{8} kn}\)
= 1 + 0.9 \(e^{-j \frac{2π}{8} kn}\)
Here, n = 6, from given question.
Hence, x[k] = 1 + 0.9 \(e^{-j \frac{2π}{8} k6}\).
6. The Z transform of δ is ___________
a) z -n
b) z -m
c) \
\(\frac{1}{z-m}\)
Answer: b
Explanation: δ is a delayed impulse function which is delayed by m units. We know that the Z-transform of a delayed function f is times the Z-transform of the function f . So, the Z transform of δ is z -m .
7. A 10 V is connected across a load whose V-I characteristics is given by 7I = V 2 + 2V. The internal resistance of the battery is of magnitude 1Ω. The current delivered by the battery is ____________
a) 6 A
b) 5 A
c) 7 A
d) 8 A
Answer: b
Explanation: 7I = V 2 + 2V ………………….
Now, V = 10 – 1 × I
Putting the value of V in eqt , we get,
&I = 2 + 2 ………………….
Or, I = 100 + I2 – 20I + 20 – 2I
Or, I2 – 29I + 120 = 0
∴ \
∴ I = 5 A.
8. The period of the signal x = 10 sin 12 π t + 4 cos18 π t is ____________
a) \
\
\
\(\frac{1}{3}\)
Answer: d
Explanation: There are two waveforms of frequencies 6 and 9, respectively. Hence the combined frequency is the highest common factor between 6 and 9 which is 3. Therefore the period is \(\frac{1}{3}\).
9. Given a series RLC circuit with V = 5V, R = 200 kΩ, C = 10µF. Sampling frequency of the circuit is 10 Hz. The expression and the ROC of the z-transform of the sampled signal are ____________
a) \(\frac{5z}{z-e^{-5′}}\), |z|<e -5
b) \(\frac{5z}{z-e^{-0.05′}}\), |z|<e -0.05
c) \(\frac{5z}{z-e^{-0.05′}}\), |z|>e -0.05
d) \(\frac{5z}{z-e^{-5′}}\), |z|>e -5
Answer: c
Explanation: I = \
= V e -t/RC = 5 e -t/RC
= 5 \(e{\frac{-t}{200 × 10 × 10^{-6} × 10^3}}\) = 5 e -t/2
Given that, the Sampling frequency of the circuit = 10 Hz
Hence, x = 5e -n/2 X 10 = 5e -0.05n
Now, X = \
^n\)
= 5. \(\frac{1}{1-e^{-0.05} Z^{-1′}}\), ROC |z|>e -0.05
= \(\frac{5z}{z-e^{-0.05′}}\), |z|>e -0.05 .
10. Given a series RLC circuit with V = 5V, R = 200 kΩ, C = 10µF. Sampling frequency of the circuit is 10 Hz. The samples x , where n=0,1,2,…., is ___________
a) 5(1-e -0.05n )
b) 5e -0.05n
c) 5(1-e -5n )
d) 5e -5n
Answer: b
Explanation: The charging current in circuit I = I (0 + ) e -t/RC
Since the capacitor acts as short circuit, I (0 + ) = \
= \(\frac{V}{R}\) e -t/RC
Voltage across resistor = R I
= V e -t/RC = 5 e -t/RC
= 5 \(e{\frac{-t}{200 × 10 × 10^{-6} × 10^3}}\) = 5 e -t/2
Given that, the Sampling frequency of the circuit is 10 Hz
∴ x = 5e -n/2 X 10 = 5e -0.05n .
11. For the circuit given below, if the frequency of the source is 50 Hz, then a value of to which results in a transient free response is _________________
signals-systems-questions-answers-discrete-fourier-transform-q11
a) 0
b) 1.78 ms
c) 7.23 ms
d) 9.21 ms
Answer: b
Explanation: T = \(\frac{L}{R} \)
Or, T = \(\frac{0.01}{5}\) = 0.002 s = 2 ms
For the ideal case, transient response will die out with time constant.
Practically, T will be less than 2 ms.
12. If G represents the Fourier Transform of a signal g which is real and odd symmetric in time, then G is ____________
a) Complex
b) Imaginary
c) Real
d) Real and non- negative
Answer: b
Explanation: Fourier transform of g is G
Given that, g is real, odd and symmetric with respect to time.
∴G* = – G; G purely imaginary.
13. If R 1 is the region of convergence of x and R 2 is the region of convergence of y, then the region of convergence of x convoluted y is ___________
a) R 1 + R 2
b) R 1 – R 2
c) R 1 ∩ R 2
d) R 1 ∪ R 2
Answer: c
Explanation: The z-transform of x = X . Let the region of convergence be R 1
The z-transform of y = y . Let the region of convergence be R 2
The z-transform of x * y is X .Y [from property]
So, the region of convergence is R 1 ∩ R 2 .
14. The system under consideration is an RC low-pass filter with R = 1 kΩ and C = 1 µF. Let H denotes the frequency response of the RC, low-pass filter. Let f1 be the highest frequency, such that 0≤|f|≤f 1 , \(\frac{|H|}{H}\)≥0.95 Then f 1 is ___________
a) 327.8
b) 163.9
c) 52.2
d) 104.4
Answer: c
Explanation: H = \
= \
| = \
= 1
Given that \(\frac{|H|}{H}\)≥0.95
Or, 1 + 4π 2 f 1 2 R 2 C 2 ≤ 1.108
Simplifying, f 1 ≤ \(\frac{0.329}{2πRC}\)
∴f 1 ≤ 52.2 Hz.
15. The response of the LTI system for \
= \frac{dx}{dt}\). Given that y(0 – ) = 2, \
= 0, x = u is __________
a) 2e -t cos t u
b) 0.5 e -t sin t u
c) 2e -t cos t u + 0.5 e -t sin t u
d) 0.5 e -t cos t u + 2e -t sin t u
Answer: c
Explanation: s 2 Y – 2s + 2sY – 2 + 5Y = 1
∴ (s 2 +2s+5) Y = 3+2s
Or, Y = \
= 2e -t cos t u + 0.5 e -t sin t u.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Common Fourier Transforms”.
1. The Fourier series of an odd periodic function, contains __________
a) Only odd harmonics
b) Only even harmonics
c) Only cosine terms
d) Only sine terms
Answer: d
Explanation: We know that, for a periodic function, if the dc term i.e. a0 = 0, then it is an odd function. Also, we know that an odd function consists of sine terms only since sine is odd. Hence the Fourier series of an odd periodic function contains only sine terms.
2. The trigonometric Fourier series of a periodic time function can have only ___________
a) Only cosine terms
b) Only sine terms
c) Both cosine and sine terms
d) Dc and cosine terms
Answer: d
Explanation: The Fourier series of a periodic function is given by,
X = \(∑_{n=0}^∞ a_n \,cosnωt + ∑_{n=1}^∞ b_n \,sinnωt\)
Thus the series has cosine terms of all harmonics i.e., n = 0,1,2,…..
The 0th harmonic which is the DC term = a 0 .
So, the trigonometric Fourier series of a periodic time function can have only Dc and cosine terms.
3. The Fourier transform of the signal δ + δ is ____________
a) \
\
2 cos ω
d) 2 sin ω
Answer: c
Explanation: Fourier transform of x (t+t 0 ) = e jωtX
Hence, X = \
+ δ]^1 e^{-jωt} \,dt\)
= e jω + e -jω
= 2 cos ω.
4. The trigonometric Fourier series of an even function of time does not have ___________
a) The dc term
b) The cosine terms
c) The sine terms
d) The odd harmonic terms
Answer: c
Explanation: For periodic even function, the trigonometric Fourier series does not contain the sine terms since sine terms are in odd functions. The function only has dc term and cosine terms of all harmonics. So, the sine terms are absent in the trigonometric Fourier series of an even function.
5. The Fourier transform of the signal \
\) is _____________
a) \
\
\
\(\frac{1}{1-ae^{-jω}}\)
Answer: d
Explanation: X = \
e^{-jωt} \,dt\) = \
^m \)
= \(\frac{1}{1-ae^{-jω}}\).
6. For a periodic signal = 30 sin 1000 + 10 cos 3000 + 6 sin, the fundamental frequency in rad/s is __________
a) 1000
b) 3000
c) 5000
d) 15000
Answer: a
Explanation: First term has w 1 = 1000 rad/s
Second term has w 2 = 3000 rad/s
Third term has w 3 = 500 rad/s
Now, w 1 is the fundamental frequency, w 2 is the third harmonic and w 3 is the 5th harmonic. So, fundamental frequency = 1000 rad/s.
7. The Fourier transform of the signal sin e -t u is ____________
a) \Missing or unrecognized delimiter for \right\)
b) \Missing or unrecognized delimiter for \right\)
c) \Missing or unrecognized delimiter for \right\)
d) \Missing or unrecognized delimiter for \right\)
Answer: b
Explanation: The Fourier Transform of e -t u = \(\frac{1}{1+jω}\)
∴ Fourier transform of e -3|t| = \
↔ X {j }
∴ X = \Missing or unrecognized delimiter for \right\).
8. Which of the following cannot be the Fourier series expansion of a periodic signal?
a) x 1 = 2 cos + 3 cos 3
b) x 2 = 2 cos + 7 cos
c) x 3 = cos + 0.5
d) x 4 = 2 cos 1.5 + sin 3.5t
Answer: b
Explanation: x 1 = 2 cost + 3 cost is periodic signal with fundamental frequency w 0 = 1.
x 2 = 2 cos πt + 7 cos t The frequency of first term w 1 = π frequency of 2nd term is w 2 = 1. Since, \
is not periodic.
x 3 = cos t + 0.5 is a periodic function with w 0 = 1
x 4 = 2 cost + sint first term has frequency w 1 = 1.5π and 2nd term has frequency w 2 = 3.5π. Since, \(\frac{ω_1}{ω_2} = \frac{3}{7}\), which is rational.
Since x 2 is not periodic, so it cannot be expanded in Fourier series.
9. The Fourier transform of the signal sgn is ____________
a) \
\
\
\(\frac{1}{jω} + 1\)
Answer: c
Explanation: sgn = 1, 0≤t<∞ and
-1, -∞<t<0
X = \
e^{-jωt} \,dt + \int_0^∞ e^{-jωt} \,dt\)
= \(\frac{e^{-jωt}}{jω} + \frac{e^{-jωt}}{-jω}\)
= \(\frac{2}{jω}\).
10. The Fourier transform of the signal u is ____________
a) π δ
b) \
π δ + \
–\(\frac{1}{jω}\)
Answer: c
Explanation: Fourier transform of 1 = 2π δ
And Fourier transform of \
Hence, Fourier transform of sgn = \
= \
= w
Fourier transform of w = π δ + \(\frac{1}{jω}\).
11. The Fourier transform of the signal \Missing open brace for subscript \
\
\
\(\frac{π}{a} e^{a|ω|}\)
Answer: b
Explanation: Fourier transform of e -a|t| = \(\frac{2a}{a^2+ω^2}\)
Now, e -a|t| = e -at , t>0 and eat, t<0
Using Duality property, \(\frac{2a}{a^2+w^2}\) ↔ 2πe -a|ω|
Or, \(\frac{1}{a^2+w^2} \leftrightarrow \frac{π}{a} e^{-a|ω|}\).
12. The Fourier transform of the signal e -|t| in [-2,2] is _____________
a) \
\
\
\(\frac{2 + 2e^{-2} cos2ω + 2ωe^{-2} \,sin2ω}{1-ω^2}\)
Answer: c
Explanation: X = \(\int_{-2}^0 e^t e^{-jωt} \,dt + \int_0^2 e^{-t} e^{-jωt} \,dt\)
= \(\frac{1-e^{-^2}}{1-jω} + \frac{1-e^{-^2}}{1+jω}\)
= \(\frac{2 – 2e^{-2} cos2ω + 2ωe^{-2} \,sin2ω}{1+ω^2}\).
13. The Fourier transform of the signal u – 2u + u is ____________
a) \
\
\
\(\frac{2 sinω+2}{jω}\)
Answer: b
Explanation: X = \(\int_{-1}^0 e^{-jωt} \,dt – \int_0^1 e^{-jωt} \,dt\)
= \(\frac{2 cosω-2}{jω}\).
14. The Inverse Fourier transform of the signal 2πδ + πδ + πδ is ______________
a) 2π
b) π
c) 1 + cos 4πt
d) 2 π
Answer: c
Explanation: x = \
+ πδ + πδ] e jωt dω
= 1 + \(\frac{1}{2} e^{j4πt} + \frac{1}{2} e^{-j4πt}\)
= 1 + cos 4πt
15. Given x = 1, |t|<T1; 0, T1<|t|<\
is ___________
a) \
\
\
\(\frac{T_0}{T_1}\)
Answer: b
Explanation: Fourier series the function can be written as, x = a 0 + \(∑_{n=1}^∞ a_n \,cosnωt + b_n \,sinnωt\)
The DC component is given by, a 0 = \
\,dt\)
Or, a 0 = \
\,dt\)
= \(\frac{1}{T_0}\) [0 + 2T 1 + 0]
= \(\frac{2T_1}{T_0}\).
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Discrete-Time Fourier Transform”.
1. Given a discrete time signal x[k] defined by x[k] = 1, for -2≤k≤2 and 0, for |k|>2. Then, y[k] = x[3k-2] is ______________
a) y[k] = 1, for k = 0, 1 and 0 otherwise
b) y[k] = 1, for k = 1 and -1 for k=-1
c) y[k] = 1, for k = 0, 1 and -1 otherwise
d) y[k] = 1, for k = 0, 1 and 0 otherwise
Answer: a
Explanation: y[k] = x [3k-2]
Now, y [0] = x [-2] = 1
Or, y [1] = x [1] = 1
Or, y [2] = x [4] = 0
∴y[k] = 1, for k = 0, 1 and 0 otherwise.
2. The time system which operates with a continuous time signal and produces a continuous time output signal is _________
a) CTF system
b) DTF System
c) Time invariant System
d) Time variant System
Answer: a
Explanation: DTF System operates with a discrete signal, on the other hand time invariant system is a system whose output does not depend explicitly on time. For continuous time system, the inputs as well as output both are CT signals.
3. A discrete time signal is given as X [n] = cos \
. The period of the signal X [n] is ______________
a) 126
b) 32
c) 252
d) Non-periodic
Answer: a
Explanation: Given that, N 1 = 18, N 2 = 14
We know that period of X [n] = LCM (N 1 , N 2 )
∴ Period of X [n] = LCM = 126.
4. What is the steady state value of The DT signal F , if it is known that F = \
\
Cannot be determined
c) 0
d) \(\frac{1}{8}\)
Answer: c
Explanation: The steady state value of the DT signal F exists since all poles of the given Laplace transform have negative real part.
∴F = lim s→0 s F
= lim s→0 \(\frac{s}{^2 }\)
= 0.
5. F and G are the one-sided z-transforms of discrete time functions f and g, the z-transform of ∑fg is given by _____________
a) ∑fgz -n
b) ∑fgz n
c) ∑fg z -n
d) ∑fgz -n
Answer: a
Explanation: Given that F and G are the one-sided z-transforms.
Also, f and g are discrete time functions, which means that property of Linearity, time shifting and time scaling will be similar to that of continuous Fourier transform. Since, for a continuous Fourier transform, the value of ∑fg is given by∑fgz -n .
∴ z-transform of ∑fg is given by∑fgz -n .
6. A discrete time signal is as given below
X [n] = cos
cos
The period of the signal X [n] is _____________
a) 16 π
b) 16
c) 8
d) Non-periodic
Answer: d
Explanation: We know that for X [n] = X 1 [n] × X 2 [n] to be periodic, both X 1 [n] and X 2 [n] should be periodic with finite periods.
Here X 2 [n] = cos
, is periodic with fundamental period as 8/n
But X 1 [n] = cos
is non periodic.
∴ X [n] is a non-periodic signal.
7. A Discrete signal is said to be even or symmetric if X is equal to __________
a) X
b) 0
c) –X
d) –X
Answer: a
Explanation: We know that any signal be it discrete or continuous is said to be even or symmetric when that signal f = f . Here given signal is X . It is a discrete time signal. So, the signal will be even symmetric if X = X .
8. The system described by the difference equation y – 2y + y = X – X has y = 0 and n<0. If x = δ, then y will be?
a) 2
b) 1
c) 0
d) -1
Answer: c
Explanation: Given equation = y – 2y + y = X – X has y = 0
For n = 0, y 2y + y = x – x
∴ y = x – x
Or, y = 0 for n<0
For n=1, y = -2y + y = x – x
Or, y = x – x + 2x – 2x
Or, y = x +x – 2x
For n=2, y = x – x + 2y – y
Or, y = x – x + 2x + 2x – 4x – x + x
∴y = d + d + d – 3d .
9. A discrete time signal is as given below
X [n] = cos
– sin
+ 3 cos
\) The period of the signal X [n] is _____________
a) 16
b) 4
c) 2
d) Non-periodic
Answer: a
Explanation: Given that, N 1 = 4, N 2 = 16, N 3 = 8
We know that period of X [n] = LCM (N 1 , N 2 , N 3 )
∴ Period of X [n] = LCM = 16.
10. The Nyquist frequency for the signal x = 3 cos 50πt + 10 sin 300πt – cos 100t is ___________
a) 50 Hz
b) 100 Hz
c) 200 Hz
d) 300 Hz
Answer: d
Explanation: We know that Nyquist frequency is twice the maximum frequency, i.e. f s = 2 f m .
The maximum frequency present in the signal is ω m = 300 π or f m = 150 Hz. Therefore the Nyquist frequency f s = 2 f m = 300 Hz.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Sampling”.
1. Find the Nyquist rate and Nyquist interval of sin.
a) 2 Hz, \
\
\
2 Hz, 2 sec
Answer: a
Explanation: We know that sin ω 0 t ↔ jπ[δ(ω+ω 0 ) – δ(ω-ω 0 )]
sin 2πt ↔ jπ[δ-δ]
Here ω m = 2π
But ω m = 2πf m
∴ f m = 1 Hz
Nyquist rate, F s = 2f m = 2 Hz
Nyquist interval, T = \(\frac{1}{2f_m} = \frac{1}{2} \)sec.
2. Find the Nyquist rate and Nyquist interval of sinc[t].
a) 1 Hz, 1 sec
b) 2 Hz, 2 sec
c) \
2 Hz, \(\frac{1}{2}\)sec
Answer: a
Explanation: We know that sinc[t] ↔ G 2π
Here ω m = 2π
2πf m = π
∴ 2f m = 1
Nyquist rate, F s = 2f m = 1 Hz
Nyquist interval, T = \(\frac{1}{2f_m}\) = 1 sec.
3. Find the Nyquist rate and Nyquist interval of Asinc[t].
a) 2 Hz, 2 sec
b) 1 Hz, 1 sec
c) \
1 Hz, \(\frac{1}{2}\) sec
Answer: b
Explanation: Nyquist rate and Nyquist interval are independent of Amplitude . But time scaling will change the rate.
We know that sinc[t] ↔ G 2π
Here ω m = 2π
2πf m = π
∴ 2f m = 1
Nyquist rate, F s = 2f m = 1 Hz
Nyquist interval, T = \(\frac{1}{2f_m}\) = 1 sec.
∴F s = 1 Hz, T = 1 sec.
4. Find the Nyquist rate and Nyquist interval of sinc[200t].
a) 200 Hz, \
200 Hz, 200 sec
c) \
100 Hz, 100 sec
Answer: a
Explanation: Here ω m =200π
2πf m =200π
2f m =200 Hz
Nyquist rate, F s = 2f m = 200 Hz
Nyquist interval, T = \(\frac{1}{2f_m} = \frac{1}{200}\) sec.
5. Which of the following is the process of ‘aliasing’?
a) Peaks overlapping
b) Phase overlapping
c) Amplitude overlapping
d) Spectral overlapping
Answer: d
Explanation: Aliasing is defined as the phenomenon in which a high frequency component in the frequency spectrum of the signal takes the identity of a lower frequency component in the spectrum of the sampled signal.
Aliasing can occur if either of the following condition exists:
• The signal is not band-limited to a finite range.
• The sampling rate is too low.
6. Find the Nyquist rate and Nyquist interval for the signal f=\
500 Hz, 2 sec
b) 500 Hz, 2 msec
c) 2 Hz, 500 sec
d) 2 Hz, 500 msec
Answer: b
Explanation: Given f = \(\frac{sin500πt}{πt}\)
Frequency, ω m = 500π
2πf m = 500π
2f m = 500 Hz
Nyquist rate, F s = 2f m = 500 Hz
Nyquist interval, T = \(\frac{1}{2f_m} = \frac{1}{500}\) = 2 msec.
7. Find the Nyquist rate and Nyquist interval for the signal f = \
1000 Hz, 1 msec
b) 1 Hz, 1000 sec
c) 1000 Hz, 1 sec
d) 1000 Hz, 1000 sec
Answer: a
Explanation: Given f = \(\Big[\frac{sin500πt}{πt}\Big]^2 = \frac{1-cos1000πt}{^2}\)
Frequency, ω m = 1000π
2πf m = 1000π
2f m = 1000 Hz
Nyquist rate, F s = 2f m = 1000 Hz
Nyquist interval, T = \(\frac{1}{2f_m} = \frac{1}{1000}\) = 1 msec.
8. Find the Nyquist rate and Nyquist interval for the signal f = 1 + sinc300πt.
a) 300 Hz, 3 msec
b) 300 Hz, 3.3 msec
c) 30 Hz, 3 msec
d) 3 Hz, 3 msec
Answer: b
Explanation: Given f = 1 + sinc300πt
Frequency, ω m = 300π
2πf m = 300π
2f m = 300 Hz
Nyquist rate, F s = 2f m = 300 Hz
Nyquist interval, T = \(\frac{1}{2f_m} = \frac{1}{300}\) = 3.3 msec.
9. Find the Nyquist rate and Nyquist interval for the signal f = rect.
a) ∞ Hz, 0 sec
b) 0 Hz, ∞ sec
c) ∞ Hz, ∞ Hz
d) 0 Hz, 0 sec
Answer: a
Explanation: Given f = rect, which is a rectangular pulse signal having pulse width of 1/200 seconds. Since the signal is a finite duration signal, it is not band-limited. The signal spectrum consists of infinite frequencies.
Hence, Nyquist rate is infinity and Nyquist interval is zero.
10. The sampling frequency of a signal is F s = 2000 samples per second. Find its Nyquist interval.
a) 0.5 sec
b) 5 msec
c) 5 sec
d) 0.5 msec
Answer: b
Explanation: Given F s = 2000 samples per second
Nyquist interval, T = \(\frac{1}{F_s} = \frac{1}{2000}\) = 0.5 msec.
11. Determine the Nyquist rate of the signal x = 1 + cos 2000πt + sin 4000πt.
a) 2000 Hz
b) 4000 Hz
c) 1 Hz
d) 6000 Hz
Answer: b
Explanation: Given x = 1 + cos 2000πt + sin 4000πt
Highest frequency component in 1 is zero
Highest frequency component in cos2000πt is ω m1 = 2000π
Highest frequency component in sin4000πt is ω m2 = 4000π
So the maximum frequency component in x is ω m = 4000π [highest of 0, 2000π, 4000π]
∴ 2πf m = 4000π
2f m = 4000
Nyquist rate, F s = 2f m = 4000 Hz.
12. Find the Nyquist rate and Nyquist interval for the signal f = -10 sin 40πt cos 300πt.
a) 40 Hz, 40 sec
b) 340 Hz, 340 sec
c) 300 Hz, 300 sec
d) 340 Hz, \(\frac{1}{340}\) sec
Answer: d
Explanation: sin 40πt has highest frequency ω m1 = 40π
cos300πt has highest frequency ω m2 = 300π
As we know, multiplication in time domain is equivalent to convolution in frequency domain, the convoluted spectra will have highest frequency component ω m = ω m1 + ω m2 = 40π + 300π
ω m = 340π
2πf m = 340π
2f m = 340
Nyquist rate, F s = 2f m = 340 Hz
Nyquist interval, T = \(\frac{1}{F_s} =\frac{1}{340}\) sec.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “The Laplace Transform”.
1. The necessary condition for convergence of the Laplace transform is the absolute integrability of fe -σt .
a) True
b) False
Answer: a
Explanation: The necessary condition for convergence of the Laplace transform is the absolute integrability of fe -σt .Mathematically, this can be stated as
\
e^{-σt}|\)dt<∞
Laplace transform exists only for signals which satisfy the above equation in the given region.
2. Find the Laplace transform of e -at u and its ROC.
a) \
\
\
\(\frac{1}{s+a}\), Re{s}>-a
Answer: d
Explanation: Laplace transform, L{x} = X = \
e^{-st} \,dt\)
L{e -at ) u} = \
e^{-st} \,dt = \int_0^∞ e^{-at} e^{-st} \,dt = \frac{1}{s+a}\) when >0
>0
σ>-a
ROC is Re{s}>-a.
3. Find the Laplace transform of δ.
a) 1
b) 0
c) ∞
d) 2
Answer: a
Explanation: Laplace transform, L{x} = X = \
e^{-st} \,dt\)
L{δ} = \
e^{-st} \,dt\)
[xδ = xδ]
= \
dt\)
= 1.
4. Find the Laplace transform of u and its ROC.
a) \
\
\
\(\frac{1}{1-s}\), σ≤0
Answer: b
Explanation: Laplace transform, L{x} = X = \
e^{-st} \,dt\)
L{u} = \
e^{-st} \,dt = \int_0^∞ e^{-st} \,dt = \frac{1}{s}\) when s>0 i.e,σ>0.
5. Find the ROC of x = e -2t u + e -3t u.
a) σ>2
b) σ>3
c) σ>-3
d) σ>-2
Answer: d
Explanation: Given x = e -2t u + e -3t u
Laplace transform, L{x} = X = \
e^{-st} \,dt\)
X = \(\frac{1}{s+2} + \frac{1}{s+3}\)
ROC is {σ > -2}∩{σ > -3}
Hence, the ROC is σ > -2.
6. Find the Laplace transform of cosωt u.
a) \
\
\
\(\frac{ω}{s^2-ω^2}\)
Answer: a
Explanation: Laplace transform, L{x} = X = \
e^{-st} \,dt\)
X = L{cosωt u} = \
] = \frac{1}{2} L[e^{jωt} \,u] + \frac{1}{2} L[e^{jωt} \,u]\)
X = \
+ \frac{1}{2}
= \frac{s}{s^2+ω^2}\).
7. Find the Laplace transform of e -at sinωt u.
a) \
\
\
\(\frac{ω}{^2+ω^2}\)
Answer: d
Explanation: Laplace transform, L{x} = X = \
e^{-st} \,dt\)
L{x} = X = \
] – \frac{1}{2j} L[e^{-t} \,u]\)
X = \(\frac{1}{2j} [\frac{1}{s+} – \frac{1}{s+}] = \frac{1}{2j} [\frac{2jω}{^2+ω^2}] = \frac{ω}{^2+ω^2}\)
e^ -at sinωt u \
>-a.
8. Find the Laplace transform of the signal x=e t sin2t for t≤0.
a) \
\
\
\(-\frac{2}{^2+2^2}\)
Answer: b
Explanation: Given x = e t sin2t for t≤0
∴ x = e t sin2t u
L{x} = X = \
e^{-st} \,dt = \int_{-∞}^∞ e^t \,sin2t \,u e^{-st} \,dt\)
= \Missing or unrecognized delimiter for \right = \frac{1}{2j} \int_{-∞}^0 [e^{t} – e^{t}]dt\)
= \Missing or unrecognized delimiter for \right\)
=\(-\frac{2}{^2+2^2}\).
9. Find the Laplace transform of the signal x=te -2|t| .
a) \
\
\
\(-\frac{1}{^2} – \frac{1}{^2}\)
Answer: a
Explanation: Given x=te -2|t|
L{x} = X = \
e^{-st} \,dt = \int_{-∞}^∞ te^{-2|t|} e^{st} \,dt \)
=\(\int_{-∞}^0 te^{2t} e^{-st} \,dt + \int_0^∞ te^{-2t} e^{-st} \,dt = -\frac{1}{^2} + \frac{1}{^2}\).
10. Find the Laplace transform of 3 u.
a) \
\
\
\(\frac{s}{
}\)
Answer: a
Explanation: Given x= 3 u = \
X = L{x} = \
] = \frac{1}{4}\) {L[cos6t u]+3L[cos2t u]}
= \Missing or unrecognized delimiter for \right = \frac{s
}{
}\).
11. Find the Laplace transform of [1 +sin 2t cos 2t]u.
a) \
\
\
\(\frac{s^2+2s+16}{s}\)
Answer: b
Explanation: Given x=[1 + sin 2t cos 2t]u =
u
L{x} = X = L[u + \
] = L[u] + \
] = \(\frac{1}{s} + \frac{1}{2} \frac{4}{
} = \frac{s^2+2s+16}{s
}\).
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Properties of the Laplace Transform”.
1. Find the Laplace transform of x = u + u.
a) \
\
\
\(\frac{sin2s}{s}\)
Answer: b
Explanation: Given x = u + u
We know that the Laplace transform u ↔ \(\frac{1}{s}\)
Time shifting property states that L{x(t±t 0 )} = Xe ±st 0
L{u}=\
= L{u+u} = \(\frac{e^{-2s}+e^{-2s}}{s} = \frac{cosh2s}{s}\).
2. Find the Laplace transform of the signal x = e -2t cosu.
a) \
\
\
\(\frac{s+2}{^2+^2}\)
Answer: d
Explanation: Given x = e -2t cosu
We know that L{cosωt u} = \(\frac{s}{s^2+ω^2}\)
∴L{cosu} = \(\frac{s}{s^2+^2}\)
Frequency shifting property states that L{e -at x} = X
L{e -2t cosu} = L{cosu}| s=s+2 = \(\Big[\frac{s}{s^2+^2}\Big]_{s=s+2} = \frac{s+2}{^2+^2}\).
3. Find the Laplace transform of the signal x = sin
u
.
a) \
\
\
\(\frac{2}{^2+1}\)
Answer: d
Explanation: We know that sint u ↔ \
↔ \
\)
\
u
\leftrightarrow \frac{1}{
} \frac{1}{\Big[
^2+1\Big]} \leftrightarrow \frac{2}{^2+1}\).
4. Find the Laplace transform of the signal x = \
1
b) s
c) \
s 2
Answer: b
Explanation: Given x = \
L{\(\frac{dδ}{dt}\)} = sL{δ} = s × 1 = s.
5. Find the Laplace transform of the signal x = te -αt .
a) \
\
\
\(\frac{1}{s+α}\)
Answer: b
Explanation: We know that L{e -αt } = \
n f ↔ \(\frac{d^n F}{ds^n}\)
L{te -αt } = –\(\frac{d}{ds} [\frac{1}{s+α}] = \frac{1}{^2}\).
6. Find the Laplace transform for f = \(\frac{1}{t}\) [e -2t – e -3t ]u.
a) ln\Missing or unrecognized delimiter for \right\)
b) ln\Missing or unrecognized delimiter for \right\)
c) ln\Missing or unrecognized delimiter for \right\)
d) ln\Missing or unrecognized delimiter for \right\)
Answer: b
Explanation: Given f = \(\frac{1}{t}\) [e -2t – e -3t ]u
We know that L{e -2t ) u} = \(\frac{1}{s+2}\); L{ -3t u} = \
ds \leftrightarrow \frac{f}{t}\)
L{\(\frac{1}{t}\) [e -2t – e -3 )]u} = \Missing or unrecognized delimiter for \rightds = [ln – ln]|_s^∞ = ln\leftMissing or unrecognized delimiter for \right\).
7. Find the initial value of f if F = \
0
b) -1
c) ∞
d) 1
Answer: d
Explanation: Given F = \(\frac{s}{^2+ω^2}\)
The initial value is f(0 + ) = lim s→∞ sF
= lim s→∞ s \(\frac{s}{^2+ω^2} = lim_{s→∞} \frac{1}{^2+^2} = 1\).
8. Find the final value of the function F given by \
1
b) 0
c) -1
d) ∞
Answer: a
Explanation: Given F = \
=lim s→0 sF
= lim s→0 s \(\frac{}{s
} = lim_{s→0} \frac{1}{s+1} = 1\).
9. Determine the initial value x(0 + ) for the Laplace transform X = \
-1
b) 0
c) 1
d) ∞
Answer: b
Explanation: Given X = \(\frac{4}{s^2+3s-5}\)
Initial value, x(0 + ) = lim s→∞ sX = lim s→∞ s
= lim x→0 \(\frac{4x}{1+3x-5x^2} = 0\) [let s = 1/x].
10. Find x if X is given by \
1
b) -1
c) \
–\(\frac{1}{2}\)
Answer: d
Explanation: Given X = \
= lim s→0 sX = lim s→0 \(\frac{}{s} = -\frac{2}{4} = -\frac{1}{2}\).
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Common Laplace Transforms – 1”.
1. The Laplace transform of f = (e -2t – 1) 2 is ________________
a) \
\
\
–\(\frac{2}{s+2} + \frac{1}{s}\)
Answer: c
Explanation: (e -2t – 1) 2 = e -4t – 2e -2t + 1
We know that, L {e -at } = \(\frac{1}{s+a}\)
L {1} = \(\frac{1}{s}\)
∴L {(e -2t – 1) 2 } = L {e -4t – 2e -2t + 1} = \(\frac{1}{s+4} – \frac{2}{s+2} + \frac{1}{s}\).
2. Given f = t 2 e -2x cos . The value of L {f} is __________________
a) \
\
\
\(\frac{2
}{
^3}\)
Answer: a
Explanation: Let g = cos ; h = e -2x cos = e -2x g
Then, f = t 2 h
Let G = L {g }, H = L {h }, F = L {f }
So, G = \
= \
= \
] = \frac{2
}{
^3}\).
3. The inverse Laplace transform of F = \Missing open brace for subscript 2 e -k
b) 2 e -k u
c) e -k u
d) 2 u
Answer: b
Explanation: Let G = \
= L -1 {G} = 2e -ct
Again, F = L -1 {G e -bs } = 2 e -k u .
4. The inverse Laplace transform of F = \Missing open brace for subscript -2e -2t + 2e -t
b) 2e -2t + 2e -t
c) -2e -2t – 2e -t
d) 2e -t + e -2t
Answer: a
Explanation: s 2 + 3s + 2 =
Now, F = \
F |s=-2
= \
F |s=-1
= \
= \
= L -1 {F}
= -2e -2t + 2e -t for t≥0
5. The Laplace transform of signal u is ___________
a) \
\
\
Zero
Answer: b
Explanation: X = \
e^{-st} \,dt\)
Here, the given signal is u .
Hence, u = 1 for all t>2 and =0 for all t<2.
So, limit is from 2 to ∞
∴ Laplace of signal= \(\int_2^∞ e^{-st} \,dt = \frac{e^{-2s}}{s}\).
6. The Laplace transform of the signal u is _________
a) \
\
\
\(\frac{-e^{-2s}}{s}\)
Answer: a
Explanation: X = \
e^{-st} \,dt\)
Here, the given signal is u .
Hence, u = 1 for all t>-2 and = 0 for all t<-2.
So, limit is from 0 to ∞
∴ Laplace of signal = \(\int_0^∞ e^{-st} \,dt = \frac{1}{s}\).
7. The Laplace transform of the signal e -2t u is ___________
a) \
\
\
\(\frac{-e^{-s}}{s+2}\)
Answer: a
Explanation: X = \
e^{-st} \,dt\)
Here, the given signal is u .
Hence, u = 1 for all t>-1 and = 0 for all t<-1.
So, limit is from 0 to ∞
∴ Laplace of signal = \(\int_0^∞ e^{-2t} e^{-st} \,dt = \frac{1}{s+2}\).
8. The Laplace transform of the signal e 2t u is ___________
a) \
\
\
\(\frac{e^{-2s}}{s-2}\)
Answer: c
Explanation: X = \
e^{-st} \,dt\)
= \
e^{-st} \,dt\)
= \(\int_0^2 e^{t} \,dt\)
= \(\frac{e^{2-1}}{2-s}\)
= \(\frac{1-e^{-2}}{s-2}\).
9. The Laplace transform of the signal sin 5t is _____________
a) \
\
\
\(\frac{s}{s^2+25}\)
Answer: c
Explanation: We know that, sin 5t = \
= \(\int_0^∞ \frac{
}{2j} e^{-st} \,dt \)
= \(\frac{5}{s^2+25}\).
10. The Laplace transform of the signal u – u is ______________
a) \
\
\
\(\frac{-2}{s}\)
Answer: b
Explanation: X = \
e^{-st} \,dt\)
Here, the given signal is u .
Hence, u = 1 for all t>0 and = 0 for all t<0.
Again, the given signal is u .
Hence, u = 1 for all t>2 and = 0 for all t<2.
So, limit is from 0 to 2
∴ Laplace of signal= \(\int_0^2 e^{-st} \,dt\)
= \(\frac{-e^{-2s}+1}{s}\).
11. The Laplace transform of the signal \(\frac{d}{dx}\)(te -t u) is ______________
a) \
\
\
\(\frac{e^{-s}}{^2}\)
Answer: b
Explanation: q = t e -t u
Laplace transform of q is q = \
= \
Laplace transform of x is x = \(\frac{s}{^2}\).
12. The Laplace transform of the signal t u * cos 2πt u is _____________
a) \
\
\
\(\frac{s^3}{
}\)
Answer: a
Explanation: p = t u
Laplace transform of p , p = \
= cos 2πt u
Laplace transform of q , q = \
= p * q
Laplace transform of x , x = \(\frac{1}{s
}\).
13. The Laplace transform of the signal t 3 u is _____________
a) \
\
\
\(\frac{-6}{s^4}\)
Answer: c
Explanation: p = t u
Laplace transform of p , p = \
= – t p
Laplace transform of q , q = \
= \
= -t q
Laplace transform of x , x = \(\frac{6}{s^4}\).
14. The Laplace transform of the signal u * e -2t u is ___________
a) \
\
\
\(\frac{e^{-2}}{s+2}\)
Answer: d
Explanation: p = u
Laplace transform of p , p = \
= u
Laplace transform of q , q = \
= e -2t u
Laplace transform of r , r = \
= e -2t u
Laplace transform of v , v = \
= q * v
Laplace transform of x , x = \(\frac{e^{-2}}{s+2}\).
15. The Laplace transform of the signal \Missing open brace for subscript \
\
\
\(\frac{-s}{^2+4}\)
Answer: b
Explanation: Laplace transform of e-at cost u = \
= e -3t cos2t u
Laplace transform of p , p = \
dt = \frac{1}{s} \int_{-∞}^0 \,pdt + \frac{p}{s}\)
Or, X = \(\frac{}{s[^2+4]}\).
This set of Signals & Systems Puzzles focuses on “Common Laplace Transforms – 2”.
1. The Laplace transform of the function e 4t + 5 is ___________
a) \
\
\
\(\frac{1}{s+4} – \frac{5}{s}\)
Answer: b
Explanation: L {e 4t + 5} = \(\frac{1}{s-4} + \frac{5}{s}\).
2. The Laplace transform of the function cos + 7sin is ____________
a) \
\
\
\(\frac{s+14}{s^2+4}\)
Answer: d
Explanation: L {cos + 7 sin } = \(\frac{s}{s^2+4} + \frac{7 * 2}{s^2+4}\)
= \(\frac{s+14}{s^2+4}\).
3. Given F = \(\frac{3s+5}{s^2+7}\). The value of L -1 {F} is _______________
a) \ + \frac{5}{\sqrt{7}} \,sin \)
b) \- \frac{5}{\sqrt{7}} sin \)
c) \ \)
d) \
\)
Answer: a
Explanation: Let, f = L -1 {F}
= L -1 \(\Big\{\frac{3s}{s^2+7} + \frac{5}{s^2+7}\Big\}\)
= L -1 \
+ \frac{5}{\sqrt{7}} \,sin \).
4. The Laplace transform of the function 10 + 5t + t 2 – 4t 3 is ___________
a) \
\
\
\(\frac{10}{s} + \frac{5}{s^2} + \frac{2}{s^3} + \frac{24}{s^4}\)
Answer: a
Explanation: L {10 + 5t + t 2 – 4t 3 } = \(\frac{10}{s} + \frac{5}{s^2} + \frac{2}{s^3} – \frac{4 3!}{s^4}\)
= \(\frac{10}{s} + \frac{5}{s^2} + \frac{2}{s^3} – \frac{24}{s^4}\).
5. The Laplace transform of the function (t 2 + 4t + 2)e 3t is ___________
a) \
\
\
\(\frac{2}{^3} – \frac{4}{^2} – \frac{2}{s-3}\)
Answer: b
Explanation: L {(t 2 + 4t + 2)e 3t } = L {t 2 e 3t + 4te 3t + 2e 3t }
= \(\frac{2}{^3} + \frac{4}{^2} + \frac{2}{s-3}\).
6. Given f = [cos ] 2 . The value of L {f} is _______________
a) \
\)
b) \
\)
c) \
\)
d) \
\)
Answer: a
Explanation: We know that by trigonometric identity, cos 2 = \
)
Also, we know that, L {cos } = \(\frac{s}{s^2+a^2}\)
And L {1} = \(\frac{1}{s}\)
So, L {cos 2 } = L \
\).
7. F = 0, 0≤t<6;
F = 3, t≥6;
The Laplace transform of F is __________
a) \
\
\
\(\frac{6e^{-6s}}{s^2}\)
Answer: c
Explanation: F = 3u
L {3u } = e -6s L {3}
= \(\frac{3e^{-6s}}{s}\).
8. G = 3, 0≤t<5;
G = 10, 5≤t<8;
G = 0; t≥8;
The Laplace transform of G is ___________
a) \
\
\
\(\frac{3}{s} – \frac{7e^{-5s}}{s} – \frac{10e^{-8s}}{s}\)
Answer: c
Explanation: G = 3 + u + u = 3 + 7u − 10u
L {3 + 7u − 10u } = \(\frac{3}{s} + \frac{7e^{-5s}}{s} – \frac{10e^{-8s}}{s}\).
9. H = 0, 0≤t<3;
H = 6sin , t≥3;
The Laplace transform of H is ___________
a) \
\
\
\(\frac{6e^{-3s}}{s^2-1}\)
Answer: c
Explanation: H = 6u sin
L {6u sin } = 6e −3s L {sin }
= \(\frac{6e^{-3s}}{s^2+1}\).
10. J = 4, 0≤t<2;
J = 4 + 5 e t-2 , t≥2;
The Laplace transform of J is ____________
a) \
\
\
\(\frac{4}{s} + \frac{5e^{-2s}}{^2}\)
Answer: d
Explanation: J = 4 + 5u et -2
L {4 + 5u et -2 } = \(\frac{4}{s}\) + 5 L {u e t-2 }
= \(\frac{4}{s}\) + 5e -2s L {t e t }
= \(\frac{4}{s} + \frac{5e^{-2s}}{^2}\).
11. U = 0, 0≤t<7;
U = 3 , t≥7;
The Laplace transform of U is ___________
a) \
\
\
\(\frac{3e^{-7s}}{s^3}\)
Answer: a
Explanation: U = u 3
L {u 3 } = e -7s L {t 3 }
= \(\frac{3!e^{-7s}}{s^4} = \frac{6e^{-7s}}{s^4}\).
12. V = 5, 0≤t<1;
V = t, t≥1;
The Laplace transform of V is ___________
a) \
\
\
\(\frac{5}{s} – \frac{e^{-s}}{s^2} + \frac{4e^{-s}}{s}\)
Answer: b
Explanation: V = 5 + u
L {5 + u } = \(\frac{5}{s}\) + L {u }
= \(\frac{5}{s}\) + e -s L {t-4}
= \
\)
= \(\frac{5}{s} + \frac{e^{-s}}{s^2} – \frac{4e^{-s}}{s}\).
13. W = 2, 0≤t<4;
W = t 2 , t≥4;
The Laplace transform of W is ___________
a) \
\)
b) \
\)
c) \
\)
d) \
\)
Answer: d
Explanation: W = 2 + u (t 2 -2)
L {2 + u (t 2 -2)} = \(\frac{2}{s}\) + L {u (t 2 -2)}
= \(\frac{2}{s}\) + e -4s L { 2 -2}
= \(\frac{2}{s}\) + e -4s L {t 2 + 8t + 14}
= \
\).
14. The Laplace transform of the function sin cos is ____________
a) \
\
\
\(\frac{s^2-16}{2}\)
Answer: a
Explanation: L
)
= \
)
= \(\frac{1}{2} \frac{4}{s^2+16}\)
= \(\frac{2}{s^2+16}\).
15. The Laplace transform of f = sin cos is ____________
a) \
\
\
\(\frac{1}{2}\)
Answer: b
Explanation: Using trigonometric identity,
We get, sin cos = \
∴ L{ sin cos } = L{\
}
We know that, L {sin at} = \(\frac{a}{s^2+a^2}\)
∴L{\
} = \(\frac{1}{2}{\frac{4}{s^2+16}}\).
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Region of Convergence”.
1. The impulse response h of a LTI system is given by h = exp u where, u denotes the unit step function. The frequency response H is ____________
a) \
\
\
\(\frac{jω}{2+jω}\)
Answer: c
Explanation: Impulse response h = e -2t u
Taking Fourier transform, H = \
e^{-jωt} \,dt\)
= \(\int_0^∞ e^{-2t} e^{-jωt} \,dt\)
= \(\int_0^∞ e^{-t} \,dt\)
= \(\frac{1}{2+jω}\).
2. The impulse response h of a LTI system is given by h = exp u where, u denotes the unit step function. The output of this system to the sinusoidal input x = 2 cos for all t, is _____________
a) 0
b) 2 -0.25 cos
c) 2 -0.5 cos
d) 2 -0.5 cos
Answer: d
Explanation: X = 2 cos2t
And ω=2 rad/s
H = \
| = \
= |H | × 2 cos [2t + *H ] = \
\)
= 2 -0.5 cos .
3. The ROC of u = \Missing open brace for subscript | z | > 1
b) | z | < 1
c) -1 < | z | < 1
d) -1 < | z |
Answer: a
Explanation: Z [u ] = \
] = ∑_{r=0}^∞\) \
^{-}\)
The series is convergent for | 1/z | < 1 i.e., for | z | > 1
Hence, ROC is | z | > 1.
4. The Radius of convergence of f = 5 n /n!, n≥0 is __________
a) -2<|z|<2
b) Z-plane
c) | z |<2
d) |z| = 2
Answer: b
Explanation: \
+ \frac{1}{2!}
^2 + \frac{1}{3!}
^3 + ……\)
The above series is convergent for all values of z.
Hence, the radius of convergence is the entire z-plane.
5. The ROC of u = 4 n , for n<0; 2 n , for n≥0 is ___________
a) 0<z<1
b) z<4
c) 2<z
d) 2<z<4
Answer: d
Explanation: \] = ∑_{-∞}^∞ u Z^{-n}\)
= \(∑_{-∞}^{-1} 4^n Z^{-n} + ∑_{n=0}^∞ 2^n Z^{-n}\)
= \(\frac{z}{4} \Big\{1 + \frac{z^1}{4} + \frac{z^2}{4} +⋯….\Big\} + \Big\{1 + \frac{2}{z} + \frac{2^2}{z} + ….\Big\}\)
= \(\frac{z}{4}. \frac{1}{1-\frac{z}{4}} + \frac{1}{1 – \frac{2}{z}}\)
= \(\frac{z}{4-z} + \frac{z}{z-2}\)
= \(\frac{2z}{}\)
Now, the above series will be convergent if | \(\frac{z}{4}\) |<1 and | \(\frac{2}{z}\) |<1.
That is 2<z<4.
6. The system under consideration is an RC low-pass filter with R = 1 kΩ and C = 1 µF. Let H denotes the frequency response of the RC, low-pass filter. Let t 1 be the group delay function of the given RC low-pass filter and f 2 = 100 Hz. Then t g (f 2 ) in ms is ___________
a) 0.717
b) 7.17
c) 71.7
d) 4.505
Answer: a
Explanation: H = \
= – tan -1
Group delay = t g = – \(\frac{d θ}{dω}\)
= – \(\frac{d tan^{-1} }{dω}\)
∴ t g = \(\frac{RC}{1+ω^2 R^2 C^2}\)
Now, t g = \(\frac{RC}{1+4π^2 f^2 R^2 C^2}\)
So, t g (f 2 ) = \(\frac{RC}{1+4π^2 f_2^2 R^2 C^2}\) = 0.717 ms.
7. Given that X is the Laplace transform of the signal cos 2t u . The time signal corresponding to X is _________________
a) u
b)
u
c) u
d)
u
Answer: a
Explanation: We know that, s X + X is having Laplace transform \
Or, y = u .
8. Given that X is the Laplace transform of the signal cos 2t u . The signal X corresponds is _______________
a) cos
u
b) \
u
c) cos6t u
d) \
Answer: b
Explanation: We know that the time signal of X
= a x
Here, a = \
= \
u .
9. The value of the radius of convergence of f = 2 n , n<0 is ____________
a) 0<| z |<1
b) -2<| z |
c) | z |<2
d) z-plane
Answer: c
Explanation: Z [f ] = \
z^{-n}\)
= \
\)
= \(\frac{1}{1-\frac{z}{2}} . \frac{z}{2} = \frac{z}{2-z}\)
The series is convergent if | \(\frac{z}{2}\) |<1, that is | z |<2
Hence, ROC is | z |<2.
10. The impulse response of a LTI system which is continuous is H = e -|t| . The system is ___________
a) Causal and stable
b) Causal but not stable
c) Stable but not causal
d) Neither causal nor stable
Answer: c
Explanation: For t<0,
H ≠ 0
Therefore the system is not causal
Again, \
| \,dt = \frac{1}{3}\) < ∞
∴ The system is stable.
11. For the system, y = u{x } which of the following holds true?
a) System is Linear, time-invariant, causal and stable
b) System is time-invariant, causal and stable
c) System is causal and stable
d) System is stable
Answer: b
Explanation: Let x 1 = v , then y 1 = u {v }
Let x 2 = k v , then y 2 = u {k v } = k y 1
Hence, non-linear
y 1 = u {v }
y 2 = u {v (t-t 0 )} = y 1 (t-t 0 )
Hence, time-invariant
Since the response at any time depends only on the excitation at time t=t 0 , and not on any further values, hence causal.
12. Given, R = 10 Ω, L = 100 mH and C = 10 μF. Selectivity is ____________
signals-systems-questions-answers-region-convergence-q12
a) 10
b) 1.2
c) 0.15
d) 0.1
Answer: d
Explanation: Selectivity = \(\frac{1}{Q}\)
Q = \(\frac{ωL}{R} = \frac{1000×100×10^{-3}}{10}\)
∴ Q = 10
So, Selectivity = 0.1.
13. For the circuit given below, if the frequency of the source is 50 Hz, then a value of to which results in a transient free response is _________________
signals-systems-questions-answers-region-convergence-q13
a) 0
b) 1.78 ms
c) 7.23 ms
d) 9.21 ms
Answer: b
Explanation: T = \(\frac{L}{R}\)
Or, T = \(\frac{0.01}{5}\) = 0.002 s = 2 ms
For the ideal case, transient response will die out with time constant.
Practically, T will be less than 2 ms.
14. A series RC circuit excited by voltage V is __________
a) A memory less system
b) A causal system
c) A dynamic system
d) Static system
Answer: c
Explanation: Dynamic systems are those systems which consist of memory. In the series RC circuit excited by voltage V, the capacitor C is an energy storing element which acts as a memory for the circuit. Therefore since system has memory it is not a memoryless system. Also, a causal system depends only on the past and present value. But since the future value of the charge is also under consideration in this type of circuit, so the system is not causal. Since charge moves about in the circuit due to the applied voltage V, hence the system is not static system. Therefore the system is a dynamic system.
15. The steady state value of F , if it is known that F = \Missing open brace for subscript \
\
1
d) Cannot be determined
Answer: a
Explanation: F = lim s→0 s F
= lim s→0 \(\frac{sb}{s}\)
= lim s→0 \(\frac{b}{}\)
= \(\frac{b}{a}\).
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Properties of ROC”.
1. Given a system function H = \Missing open brace for subscript \
Infinite
c) 0
d) 8
Answer: c
Explanation: H = \
= \
J = L = \
= \(\frac{1}{s^2+4} . \frac{2}{s+3}\)
VSS = lim s→0 sV
= 0.
2. If H = \
is ___________
a) exp\
\)
b) exp\
\)
c) exp
d) exp
Answer: c
Explanation: Let us consider x = e 2jnft
So, y = \
xdτ\)
= \
e^{j2nπf} \,dτ\)
= e j2nft H
Or, H = \
= e j2nft .
3. The z-transform of –u is ___________
a) \
\
\
\(\frac{z}{1-z^{-1}}\)
Answer: c
Explanation: z [-u ] = \
] z^{-n}\)
= – [z + z 2 + z 3 + z 4 …]
= \(\frac{z}{z-1}\)
= \(\frac{1}{1-z^{-1}}\).
4. The value of z(a k u[-k]) is _______________
a) \
\
\
\(\frac{a}{a-z}\)
Answer: b
Explanation: z (a k u [-k]) = \
^m \)
= \(\frac{a^{-1} z}{1-a^{-1} z} = \frac{z}{a-z}\).
5. If a system has N different poles, then the system can have ______________
a) N ROC’s
b) ROC’s
c) ROC’s
d) 2N ROC’s
Answer: c
Explanation: Let us consider 2 poles. For 2 poles, we will have 3 ROC conditions. Hence, if a system has N poles then the system will have ROC’s.
6. Given 2 signals k u and u . These two signals are superimposed. This superimposed signal is _______________
a) \
\
\
\(\frac{z}{z+3} + \frac{1}{z+1}\)
Answer: a
Explanation: We know that superposition means addition of these 2 signals.
So, superimposed f[k] = k u + u
Hence, z[k] = \(\frac{z}{z+3} + \frac{1}{z-1}\).
7. X 1 = 2z + 1 + z -1 and X 2 = z + 1 + 2z -1 is ________________
a) Even signal
b) Odd signal
c) In time power signal
d) In time energy signal
Answer: d
Explanation: Given X 1 = 2z + 1 + z -1 and X 2 = z+1+2z -1 .
So, x 1 = {2, 1, 1}; x 2 = {1, 1, 2}.
8. The value of z{[k-1] u} is _______________
a) \
\
\
\(\frac{z}{^2}\)
Answer: b
Explanation: z {[k-1] u } = z {k u – u }
= \(\frac{z}{^2} – \frac{z}{z-1}\)
= \(\frac{z-z}{^2}\)
= \(\frac{z-z^2+z}{^2}\)
= \(\frac{2z-z^2}{^2}\).
9. The area under Gaussian pulse \Missing open brace for subscript Unity
b) Infinity
c) Pulse
d) Zero
Answer: a
Explanation: Putting \(π^{t^2}\) = x, we get,
\(\int_{-∞}^∞ e^{{-π}^{{t}^2}} \,dt = \int_{-∞}^∞ e^{-x} 2π \sqrt{\frac{x}{π}} \,dx\)
\(= 2\sqrt{π} \int_{-∞}^∞ \sqrt{x} e^{-x} \,dx\)
= 1.
10. The system x = 7
k u-6
k u is ___________
a) Causal
b) Anti-causal
c) Non-causal
d) Cannot be determined
Answer: c
Explanation: Taking the z-transform, we get,
X = \Missing or unrecognized delimiter for \right – 6\leftMissing or unrecognized delimiter for \right\)
∴ the ROC for given condition is as derived above.
∴ the bounded signal as a whole is non-causal.
11. The spectral density of white noise is ____________
a) Exponential
b) Uniform
c) Poisson
d) Gaussian
Answer: b
Explanation: The distribution of White noise is homogeneous over all frequencies. Power spectrum is the Fourier transform of the autocorrelation function. Therefore, the power spectral density of white noise is uniform.
12. In the circuit given below, the value of Z L for maximum power to be transferred is _____________
signals-systems-questions-answers-properties-roc-q12
a) R
b) R + jωL
c) R – jωL
d) jωL
Answer: c
Explanation: The value of load for maximum power transfer is given by the complex conjugate of Z AB
Z AB = R + jX L
= R + jωL
∴ Z L for maximum power transfer is given by Z L = R – jωL.
13. A current I given by I = – 8 + 6\) A is passed through three meters. The respective readings will be?
a) 8, 6 and 10
b) 8, 6 and 8
c) – 8, 10 and 10
d) -8, 2 and 2
Answer: c
Explanation: PMMC instrument reads only DC value and since it is a center zero type, so it will give – 8 values.
So, rms = \(\sqrt{8^2 +
^2}\) = 10 A
Moving iron also reads rms value, so its reading will also be 10 A.
14. The CDF for a certain random variable is given as F = {0, -∞<x≤0; kx2, 0<x≤10; 100k, 10<x<∞;} The value of k is __________
a) 100
b) 50
c) 1/50
d) 1/100
Answer: d
Explanation: From the given F,
We get, \(\frac{dF}{dx}\) = 0 + 2kx + 0
= 2kx
∴ \(\int_0^{10}\) 2 kx dx =1
Or, 100k = 1
Or, k = 1/100.
15. For a stable system which of the following is correct?
a) |z| < 1
b) |z| = 1
c) |z| > 1
d) |z| ≠ 1
Answer: a
Explanation: We know that, for the system to be stable, the ROC should include the unit circle.
|z| = 1, represents the unit circle but does not include it.
|z| > 1, represents the region outside the unit circle. In other words, it excludes the region of the unit circle.
|z| ≠ 1 does not represent a unit circle.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Inverse Laplace Transform”.
1. Find the inverse Laplace transform for \
te t u
b) te -t u
c) tu
d) e t u
Answer: b
Explanation: Given X = \
= L -1 [X] = \(L^{-1} \Big[\frac{1}{^2}\Big] = e^{-t} L^{-1} [\frac{1}{s^2}]\) = e -t tu = te -t u.
2. Find the inverse Laplace transform for \
te -t u
b) e -t sint u
c) e -t cost u
d) e -t u
Answer: b
Explanation: Given X = \
= L -1 [X] = \(L^{-1} \Big[\frac{1}{^2+1}\Big] = e^{-t} L^{-1} \Big[\frac{1}{s^2+1}\Big]\) = e -t sint u.
3. Find the inverse Laplace transform for \
te -t u
b) e -t sint u
c) e -2t u
d) e 2t u
Answer: c
Explanation: Given X = \
= L -1 [X] = \(L^{-1} \Big[\frac{s}{^2}\Big] = L^{-1} \Big[\frac{s+2}{^2} – \frac{2}{^2}\Big] = L^{-1} \Big[\frac{1}{s+2}\Big] – 2L^{-1} \Big[\frac{1}{^2}\Big] \)
= e -2t – 2e -2t L -1 \([\frac{1}{s^2}]\) = e -2t u.
4. Find the inverse Laplace transform for \
[2e -2t cost + e -2t sint]u
b) [e -2t cost + 2e -2t sint]u
c) [2e -2t cost – e -2t sint]u
d) [e -2t cost – 2e -2t sint]u
Answer: d
Explanation: Given X = \
= L -1 [X] = \(L^{-1} \Big[\frac{s}{^2+1}\Big] = L^{-1} \Big[\frac{s+2}{^2+1} – \frac{2}{^2+1}\Big] \)
\(= L^{-1} \Big[\frac{s+2}{^2+1}\Big] – 2L^{-1} \Big[\frac{1}{^2+1}\Big] = e^{-2t} L^{-1} \Big[\frac{s}{s^2+1}\Big] – 2e^{-2t} L^{-1} \Big[\frac{1}{s^2+1}\Big]\)
= [e -2t cost – 2e -2t sint]u.
5. Find the inverse Laplace transform of X = \
\
t\)
b) \
t\)
c) \
t\)
d) \
t\)
Answer: b
Explanation: Given X = \(\frac{s}{s^2 a^2+b^2} = \frac{1}{a^2} \Big[\frac{s}{s^2+^2}\Big]\)
We know that L -1 \Missing or unrecognized delimiter for \right\) = cosωt
∴x = L -1 [X] = \
t\).
6. Find the inverse Laplace transform of X = \
\
\
\
\(\frac{1}{2a}\) t cosat
Answer: b
Explanation: Given X = \
= L -1 [X] = \(L^{-1} \Big[\frac{s}{
^2}\Big] = \frac{1}{2a} \Big[-\frac{d}{ds} \Big\{\frac{a}{s^2+a^2}\Big\}\Big] \)
= \(\frac{1}{2a} tL^{-1} \Big[\frac{a}{s^2+a^2}\Big] = \frac{1}{2a}t sinat\).
7. If F 1 = \(\frac{1}{s+2}\) and F 2 = \
= F 1 F 2 .
a) [e -2t + e -3t ]u
b) [e -2t – e -3t ]u
c) [e 2t + e 3t ]u
d) [e 2t + e -3t ]u
Answer: b
Explanation: Given F 1 = \(\frac{1}{s+2}\) and F 2 = \
= F 1 F 2 = \Missing or unrecognized delimiter for \right\leftMissing or unrecognized delimiter for \right = \frac{1}{s+2} – \frac{1}{s+3}\)
Applying inverse Laplace transform, we get
f = [e -2t – e -3t ]u.
8. Find the inverse Laplace transform for X = \
cosh2t
b) \
sinh2t
d) \(\frac{1}{2}\) sinh2t
Answer: b
Explanation: Given X = \
=L -1 [X] = \Missing or unrecognized delimiter for \right = \frac{1}{2} cosh2t.\)
9. Find the inverse Laplace transform for X = \
\).
a) \
\
\
\(\frac{e^{bt} + e^{-at}}{t}\)
Answer: b
Explanation: Given X = \
\)
x = L -1 [X] = L -1 \
\Big]\)
L[x] = \
\) = ln-ln
L[tx] = –\
-ln] = \
= \
\) = e -bt – e -at
x = \(\frac{e^{-bt} – e^{-at}}{t}\).
10. Find the inverse Laplace transform for the function X = \
e -2t cos2t u – e -2t sin2t u
b) 2e -2t cos2t u – \(\frac{5}{2}\) e -2t sin2t u
c) 2e -2t cos2t u – e -2t sin2t u
d) e -2t cos2t u – \(\frac{5}{2}\) e -2t sin2t u
Answer: b
Explanation: Given function X = \
= 2e -2t cos2t u – \(\frac{5}{2}\) e -2t sin2t u.
11. Find the inverse Laplace transform for the function X = \
e -t u – u + e - u-u
b) e -t u + e - u
c) e - u – u
d) e -t u – u
Answer: a
Explanation: Given function X = \
= L -1 [X] = \(L^{-1} \Big[\frac{1+e^{-2s}}{3s^2+2s}\Big] = L^{-1} \Big[\frac{1}{3s^2+2s}\Big] + L^{-1} \Big[\frac{e^{-2s}}{3s^2+2s}\Big]\)
\(L^{-1} \Big[\frac{1}{3s^2+2s}\Big] = L^{-1} \Big\{\frac{1}{3s[s+]}\Big\} = L^{-1} \Big\{\frac{-1}{s} + \frac{1}{[s+]}\Big\}\) = e -t u – u
\Missing or unrecognized delimiter for \right_{t=t-2}\) = e - u-u
∴x = e -t u – u + e - u-u.
12. Given x=e -t u. Find the inverse Laplace transform of e -3s X.
a) \(\frac{1}{2}\) e -/2 u
b) \(\frac{1}{2}\) e -/2 u
c) \(\frac{1}{2}\) e /2 u
d) \(\frac{1}{2}\) e /2 u
Answer: b
Explanation: Given x = e -t u
X = L[x] = L[e -t u] = \
= \(\frac{1}{2s+1} = \frac{1/2}{s+}\)
L -1 [X] = \(L^{-1} [\frac{1/2}{s+}] = 1\frac{1}{2}\) e -t/2 u
L -1 [e -3s X] = L -1 [X] t=t-3 = \(\frac{1}{2}\) e -/2 u
∴L -1 [e -3s X] = \(\frac{1}{2}\) e -/2 u if x = e -t u.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Characterization and Nature of Systems”.
1. Given a Fourier transform pair x ↔ X = \
= 1 for |t|<1 and 0, otherwise. Then the Fourier transform of y having the shape of a triangular waveform from t=-2 to t=2 and maximum peak value=2 is ___________
a) \
\
\
\(\frac{8 sin^2 ω}{ω^2}\)
Answer: a
Explanation: We know that, y = x *x
Or, Y = ) 2 = \(\frac{4 sin^2 ω}{ω^2}\).
2. For the signal given below, the region of convergence is ____________
signals-systems-questions-answers-characterization-nature-systems-q2
a) ω 1 < ω < ω 2 in s-plane
b) Entire s-plane
c) Imaginary axis
d) Entire s-plane except imaginary axis
Answer: b
Explanation: We know that, if x is of finite duration and is absolutely integrable, then the region of convergence is the entire s-plane. Here since x is of finite duration and is also integrable in the given range, so the ROC is the entire s-plane.
3. For a series RLC circuit excited by an impulse voltage of magnitude 1 and having R = 4Ω, L = 1H, C = \
is ___________
a) \
\
\
\(\frac{3}{2} e^{-3t} – \frac{1}{2} e^t\)
Answer: c
Explanation: Z = R + Ls + \
= δ
∴ V = 1
∴ I = \
= \
= \(\frac{3}{2} e^{-3t} – \frac{1}{2} e^{-t}\).
4. The transfer function of an LTI system may be expressed as H = \
Poles of H are called natural modes
ii) Poles of H are called natural frequencies.
The correct option is ________________
a) i-true, ii-false
b) i-false, ii-true
c) i-true, ii-true
d) i-false, ii-false
Answer: c
Explanation: We know that the poles of H are called natural modes or natural frequencies.
5. The system y = x 3 is ___________
a) Stable with input x
b) Stable with output y
c) Stable with both input x as well as output y
d) Not stable
Answer: c
Explanation: Given that y = x 3
Let x = u , that is bounded input
∴ y = u 3 = u, that is bounded output
The bounded input produces bounded output hence system is stable with both input x as well as output y .
6. Sinusoidal signal x = 4cos
is passed through a square law device defined by the input output relation y = x 2 . The DC component in the signal is _____________
a) 3.46
b) 4
c) 2.83
d) 8
Answer: d
Explanation: y = 4 2 cos 2
= \
Hence, DC component = 8
AC component = 8 cos
.
7. Let a signal a1 sin (ω 1 t + φ 1 ) be applied to an astable LTI system. The corresponding output is a 2 F sin (ω 2 t + φ 2 ). Then which of the following is correct?
a) F is not necessarily a sine or cosine function but must be periodic with ω 1 =ω 2
b) F must be a sine or cosine function with a 1 = a 2
c) F must be a sine function with ω 1 =ω 2 and φ 1 = φ 2
d) F must be a sine or cosine function with ω 1 =ω 2
Answer: b
Explanation: We can infer that the output of an LTI system has the same function sine or cosine with the same frequency and different phase. Hence F must be a sine or cosine function with a 1 = a 2 .
8. The system y = x 2 is ______________
a) Invertible with input x and output y
b) Invertible with input x
c) Invertible with output y
d) Not invertible
Answer: d
Explanation: Given y = x 2
Let x = u , then y = u 2 = u
Let, x = -u , then y = {-u } 2 = u
Thus different inputs leads to same output hence system is non-invertible.
9. The system y = 15e x is ___________
a) Stable with input x
b) Stable with output y
c) Stable with both input x as well as output y
d) Not stable
Answer: c
Explanation: Given that y = 15 e x
Let x = u , that is bounded input
∴ y = e u , that is bounded output
The bounded input produces bounded output hence system is stable with both input x as well as output y .
10. The system y = x is ______________
a) Invertible with input x and output y
b) Invertible with input x
c) Invertible with output y
d) Not invertible
Answer: a
Explanation: Given y = x
Let x = u , then y = u = u
Let, x = -u , then y = – u = – u
Since different inputs leads to different outputs hence system is invertible with both input x and output y .
11. The system y = \Missing open brace for subscript Stable with input x
b) Stable with output y
c) Stable with both input x as well as output y
d) Not stable
Answer: a
Explanation: Given that y = \
= u , that is bounded input
∴ y = 2 \
only.
12. The system y = \Missing open brace for subscript Invertible with input x and output y
b) Invertible with input x
c) Invertible with output y
d) Not invertible
Answer: d
Explanation: Given y = \
= 1, then y = 0
Let, x = 2, then y = 0
Thus different inputs leads to same output hence system is non-invertible.
13. The system defined by h[n] = 2 n u[n-2] is _______________
a) Stable and causal
b) Causal but not stable
c) Stable but not causal
d) Unstable and non-causal
Answer: b
Explanation: Given that, h[n] = 2 n u [n-2]
For causal system, h[n] = 0 for n<0.
Hence, the given system is stable.
Since, \(∑_{n=2}^∞ 2^n = ∞\)
So, given system is not stable.
14. If the response of LTI continuous time system to unit step input is
, the impulse response of the system is _______________
a)
b) e -2t
c) (1-e -2t )
d) Constant
Answer: b
Explanation: We know that, impulse response is given by \
∴ Impulse response = \(\frac{d
}{dt}\)
= e -2t .
15. The system y[n] = \Missing open brace for subscript Invertible system
b) Memoryless system
c) Non-invertible system
d) Averaging system
Answer: a
Explanation: Given that, y[n] = \(∑_{k=-∞}^n x[k]\) We know that if a system is invertible, then an inverse system exists that, when cascaded with the original system, yields an output w[n] equal to the input x[n] to the first system.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “The Z-Transform”.
1. When do DTFT and ZT are equal?
a) When σ = 0
b) When r = 1
c) When σ = 1
d) When r = 0
Answer: b
Explanation: Discrete Time Fourier Transform, X(e -jω ) = \
e^{-jωn}\)
Z-Transform, X = \
z^{-n}\), z = r e jω
When r=1, z = e jω and hence DTFT and ZT are equal.
2. Find the Z-transform of δ.
a) z
b) z 2
c) 1
d) z 3
Answer: d
Explanation: Given x = δ
We know that δ = \
= \
z^{-n} = \sum\limits_{n=-\infty}^{\infty} δ z^{-n}\) = z 3 .
3. Find the Z-transform of a n u;a>0.
a) \
\
\
\(\frac{1}{1+az}\)
Answer: a
Explanation: Given x = a n u
We know that \=
\)
X = \
z^{-n} = \sum\limits_{n=-\infty}^{\infty} a^n u z^{-n}\)
= \
z^{-n} = \sum\limits_{n=0}^{∞}
^n =
^{-1}\)
= \(\frac{1}{1-az^{-1}} = \frac{z}{z-a}\).
4. Find the Z-transform of cosωn u.
a) \
\
\
\(\frac{z}{z^2+2z cosω+1}\)
Answer: b
Explanation: Given x = cosωn u
We know that \=
\)
Z[cosωn u] = \
\Big] = \frac{1}{2} Z[e^{jωn} u] + \frac{1}{2} Z[e^{-jωn} u]\)
\Missing or unrecognized delimiter for \right = \frac{1}{2} \Big[\frac{z
+ z
}{
}\Big]\)
\(= \frac{1}{2} \Big\{\frac{z[2z-
]}{z^2-z
+1}\Big\} = \frac{z}{z^2-2z cosω+1}\).
5. For causal sequences, the ROC is the exterior of a circle of radius r.
a) True
b) False
Answer: a
Explanation: Consider a causal sequence, x = r n u
X = \
z^{-n} = \sum\limits_{n=-∞}^{∞} r^n u z^{-n} = \sum\limits_{n=0}^{∞} r^n z^{-n} = \sum\limits_{n=0}^{∞}
^n\)
The above summation converges for |rz -1 |<1, i.e. for |z|>r
Hence, for the causal sequences, the ROC is the exterior of a circle of radius r.
6. x = a n u and x = -a n u have the same X and ROC.
a) True
b) False
Answer: b
Explanation: a n u ↔ \(\frac{1}{1-az^{-1}} = \frac{z}{z-a}\); ROC:|z|>|a|
-a n u ↔ \
= a n u and x = -a n u have the same X and differ only in ROC.
7. Find the Z-transform of y = xu.
a) z 2 X – z 2 x – zx
b) z 2 X + z 2 x – zx
c) z 2 X – z 2 x + zx
d) z 2 X + z 2 x + zx
Answer: a
Explanation: Given y = xu
Y = Z[y] = Z[xu] = \
u z^{-n} = \sum\limits_{n=0}^{∞} xz^{-n}\)
Let n + 2 = p,i.e.n = p – 2
Y = \
z^{-} = z^2 ∑_{p=2}^∞ xz^{-p} = z^2 ∑_{p=0}^∞ xz^{-p} – x – x z^{-1}\)
=z 2 X – z 2 x – zx.
8. Find the Z-transform of x = a |n| ; |a|<1.
a) \
\
\
\(\frac{1}{z-a} – \frac{1}{z-}\)
Answer: a
Explanation: a^ |n| = a^n u + a -n u = a n u + \
^n\) u
Z[a |n| ] = Z[a n u] + Z[\
^n\) u] = \(\frac{z}{z-a} – \frac{z}{z-}\).
9. Find the Z-transform of u.
a) \
\
\
\(\frac{z}{1+z}\)
Answer: a
Explanation: Given x = u
Z[x] = X = \
z^{-n} = ∑_{n=-∞}^∞ u z^{-n} = ∑_{n=-∞}^0 z^{-n}\)
=\(∑_{n=0}^∞ z^n = \frac{1}{1-z}\).
10. For a right hand sequence, the ROC is entire z-plane.
a) True
b) False
Answer: b
Explanation: If x is finite duration causal sequence ,the X converges for all values of z except at z = 0. Hence, the ROC is entire z-plane except at z = 0.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Properties of Z-Transforms – 1”.
1. The z-transform of δ[n-k]>0 is __________
a) Z k , Z>0
b) Z -k , Z>0
c) Z k , Z≠0
d) Z -k , Z≠0
Answer: d
Explanation: Performing z-transform on δ[n-k], we get,
X = \(∑_{n=0}^∞ x[n]z^{-n}\)
= Z -k , Z≠0.
2. The z-transform of δ[n+k]>0 is __________
a) Z -k , Z≠0
b) Z k , Z≠0
c) Z -k , all Z
d) Z k , all Z
Answer: d
Explanation: Performing z-transform on δ[n+k], we get,
X = \(∑_{n=0}^∞ x[n]z^{-n}\)
= Z k , all Z
3. The z-transform of u[n] is _________
a) \
\
\
\(\frac{z}{1-z^{-1}}\), |Z|>1
Answer: a
Explanation: Performing z-transform on u[n], we get,
X = \(∑_{n=0}^∞ x[n]z^{-n}\)
= \(\frac{1}{1-z^{-1}}\), |Z|>1.
4. The z-transform of \
^n\) is __________
a) \
\
\
\(\frac{z^5 – 0.25^5}{z^4 }\), all z
Answer: d
Explanation: X = \
^4 \)
= \(\frac{1-
^5}{1-
^1}\)
= \(\frac{z^5 – 0.25^5}{z^4 }\), all z.
5. The z-transform of \
^4\) u[-n] is ___________
a) \
\
\
\(\frac{1}{1-4z}\), |Z|<\(\frac{1}{4}\)
Answer: d
Explanation: X = \
^n\)
= \
^n \)
= \(\frac{1}{1-4z}\), |Z|<\(\frac{1}{4}\).
6. The z-transform of 3 n u[-n-1] is ___________
a) \
\
\
\(\frac{3}{3-z}\), |Z|<3
Answer: b
Explanation: X = \
^n \)
= \
^n \)
= \(\frac{\frac{1}{3} z}{1-\frac{1}{3}z}\), |z|<3
= \(\frac{z}{3-z}\).
7. The z-transform of \
^{[n]}\) is ____________
a) \
\
\
\(\frac{5z}{}\), –\(\frac{3}{2} \) < z< –\(\frac{2}{3}\)
Answer: b
Explanation: X = \
^n + ∑_{n=0}^∞
^n \)
= \(\frac{-1}{
} + \frac{1}{
}\)
= \(\frac{-5z}{}\), \(\frac{2}{3}\) < |z| < \(\frac{3}{2} \).
8. The z-transform of cos
u[n] is __________
a) \
\
\
\(\frac{z}{2} \frac{}{
}\), |z|>1
Answer: b
Explanation: Performing z-transform on a n u[n], we get \
=\Missing or unrecognized delimiter for \right\)
Hence, X = \(\frac{z}{2} \frac{}{
}\), |z|>1.
9. The z-transform of {3,0,0,0,0,6,1,-4} is ___________
a) 3z 5 + 6 + z -1 – 4z -2 , 0≤|z|<∞
b) 3z 5 + 6 + z -1 – 4z -2 , 0<|z|<∞
c) 3z 5 + 6 + z – 4z -2 0<|z|<∞
d) 3z 5 + 6 + z -1 – 4z -2 , 0≤|z|<∞
Answer: b
Explanation: Performing z-transform on x (n+n 0 ), we get z n 0 X
Now, x[n] = 3δ[n+5] + 6δ[n] + δ[n-1] – 4δ[n-2]
So, X = 3z 5 + 6 + z -1 – 4z -2 , 0<|z|<∞.
10. The z-transform of x[n]= {2,4,5,7,0,1} is ___________
a) 2z 2 + 4z + 5 +7z + z 3 , z≠∞
b) 2z -2 + 4z -1 + 5 + 7z + z 3 , z≠∞
c) 2z -2 + 4z -1 + 5 + 7z + z 3 , 0<|z|<∞
d) 2z 2 + 4z + 5 + 7z -1 + z 3 , 0<|z|<∞
Answer: d
Explanation: Performing z-transform on x (n+n 0 ), we get z n 0 X
Now, x[n] = 2δ[n+2] + 4δ[n+1] + 5δ[n] + 7δ[n-1] + δ[n-3]
So, X = 2z 2 + 4z + 5 + 7z -1 + z 3 , 0<|z|<∞.
11. The z-transform of x[n]= {1,0,-1,0,1,-1} is __________
a) 1+2z -2 -4 z -4 + 5z -5
b) 1-z -2 + z -4 – z -5
c) 1-2z 2 + 4z 4 – 5z 5
d) 1-z 2 + z 4 – z 5
Answer: b
Explanation: Performing z-transform on x (n-n 0 ), we get z -n 0 X
Now, x[n] = δ[n] – δ[n-2] + δ[n-4] – δ[n-5]
So, X = 1-z -2 + z -4 – z -5 , z≠0.
12. Given the z-transform pair
\Missing open brace for subscript \
\
\
\(\frac{^2}{^2-16}\)
Answer: c
Explanation: Performing z-transform on x , we get z n 0 X
Now, z-transform of y[n] = x [n-2] is given by,
Y = z -2 X
= \(\frac{1}{z^2-16}\).
13. Given the z-transform pair
\Missing open brace for subscript \
\
\
\(\frac{z^2}{z^2-64}\)
Answer: b
Explanation: y[n] = \
= X
∴ Y = \(\frac{z^2}{z^2-4}\).
14. Given the z-transform pair
\Missing open brace for subscript \
\
\
\(\frac{16z^2}{
^2}\)
Answer: c
Explanation: y[n] = x [-n]*x[n] Performing z-transform on y[n], we get, Y = X
X
∴ X
↔ x [-n].
15. Given the z-transform pair
\Missing open brace for subscript \
\
\
\(\frac{z}{
^2}\)
Answer: d
Explanation: y[n] = x[n]*x [n-3] Performing z-transform on y[n], we get, Y = X z -3 X
Or, Y = \(\frac{z}{
^2}\).
This set of Tough Signals & Systems Questions and Answers focuses on “Properties of Z-Transforms – 2”.
1.Find the Z-transform of the causal sequence x = {1,0,-2,3,5,4}.
a) 1 – 2z -2 + 3z -3 + 5z -4 + 4z -5
b) 1 – 2z 2 + 3z 3 + 5z 4 + 4z 5
c) z -1 – 2z 2 + 3z 3 + 5z 4 + 4z 5
d) z – 2z 3 + 3z 4 + 5z 5 + 4z 6
Answer: a
Explanation: Given sequence values are :
x=1, x=0, x=-2, x=3, x=5, x=4.
We know that
\ = \sum\limits_{n=-∞}^{∞} x z^{-n}\)
X = x + x z -1 + x z -2 + x z -3 + x z -4 + x z -5
X = 1 – 2z -2 + 3z -3 + 5z -4 + 4z -5 .
2. Find the Z-transform of the anticausal sequence x = {4,2,3,-1,-2,1}.
a) 4z 5 + 2z 4 + 3z 3 – z 2 – 2z + 1
b) 4z -5 + 2z -4 + 3z -3 -z -2 – 2z -1 + 1
c) -4z 5 – 2z 4 – 3z 3 + z 2 + 2z – 1
d) -4z -5 – 2z -4 – 3z -3 + z -2 + 2z -1 – 1
Answer: a
Explanation: Given sequence values are :
x=4, x=2, x=3, x=-1, x=-2, x=1
We know that
\ = \sum\limits_{n=-∞}^{∞} x z^{-n}\)
X = x z 5 + x z 4 + x z 3 + x z 2 + xz + x
X = 4z 5 + 2z 4 + 3z 3 – z 2 – 2z + 1.
3. Find the Z-transform of x = u.
a) \
\
\
\(\frac{-1}{z+1}\)
Answer: c
Explanation: Given x = u
Time reversal property of Z-transform states that
If x ↔ X, then x ↔ X
Z[u] = \Missing or unrecognized delimiter for \right_{z=} = \frac{1/z}{-1} = \frac{1}{1-z}\).
4. Find the Z-transform of x = u.
a) \
\
\
\(\frac{z^2}{2-z}\)
Answer: b
Explanation: Given x = u
Time shifting property of Z-transform states that
If x ↔ X, then x ↔ z -m X
Z[u] = Z{u[-]}=z 2 Z[u] = \(\frac{z^2}{1-z}\).
5. Find the Z-transform of x = n 2 u.
a) \
\
\
\(\frac{z}{^3}\)
Answer: b
Explanation: Given x = n 2 u
We know that X = Z[x] = Z[u] = \
↔ X, then n k x ↔ k z k \(\frac{d^k X}{dz^k}\)
Z[n 2 u] = z 2 \(\frac{d^2 X}{dz^2} = z^2 \frac{d^2}{dz^2}[\frac{z}{1-z}] = \frac{z}{^3}\).
6. Find the Z-transform of x = 2 n u.
a) \
\
\
\(\frac{4}{z}\)
Answer: d
Explanation: Given x = 2 n u
Time shifting property of Z-transform states that
If x ↔ X, then x ↔ z -m X
Z[u] = z -2 Z[u] = \
↔ X, then a n x ↔ X
Z[2 n u] = Z[u]| z= = \(\Big[\frac{1}{z}\Big]_{z=}\)
\( = \frac{1}{[-1]} = \frac{4}{z}\).
7. Find the Z-transform of x = n[a n u].
a) \
\
\
\(\frac{a}{z^2}\)
Answer: d
Explanation: Given x = n[a n u]
We know that a n u ↔ \
↔ X, then nx ↔ -z \
] = Z{n[a n u]} = -z \(\frac{dX}{dz} = -z \frac{d}{dz} \Big[\frac{z}{z-a}\Big] = \frac{a}{z^2}\).
8. Find the Z-transform of x =
n u*
n u.
a) \
\
\
\(\frac{z}{z-} – \frac{z}{z+}\)
Answer: a
Explanation: We know that a n u ↔ \(\frac{z}{z-a}\)
Let x 1 =
n u and x 2 =
n u
∴X 1 = \(\frac{z}{z-}\) and X 2 = \
= x 1 * x 2
The convolution property of Z-transform states that
x 1 * x 2 ↔ X 1 X 2
∴Z[x] = X = Z[x 1 *x 2 ] = X 1 X 2 = \(\frac{z}{z-} \frac{z}{z-}\).
9. Find x if X = \
1
b) 0
c) ∞
d) 0.6
Answer: b
Explanation: Given X = \
↔ X, then x = Lt z→1 X
x = Lt z→1 X = Lt z→1 \(\frac{Z+1}{^2}\) = 0.
10. Find x if X = \
∞
b) -1
c) 1
d) 0
Answer: d
Explanation: Given X = \
↔ X, then x = Lt z→∞ X
x = Lt z→∞ X = \(Lt_{z→∞} \frac{1}{z} \frac{[1+]}{[1+][1+]}\) = 0.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Inverse Z-Transform”.
1. Given the z-transform pair 3nn2 u[n] ↔ X . The time signal corresponding to X is ___________
a) n 2 3 n u[2n]
b) \
^n n^2 u[n] \)
c) \
^n n^2 u[n] \)
d) 6nn2u[n]
Answer: c
Explanation: Y = X ↔ y[n] = \(\frac{1}{2^n}\) x[n]
Or, y[n] = \(\frac{1}{2^n}\) n 2 3 n u[n]
Or, y[n] = \
^n n^2 u[n] \).
2. Given the z-transform pair 3 n n 2 u[n] ↔ X . The time signal corresponding to X(z -1 ) is ___________
a) n 2 3 -n u[n]
b) n 2 3 -n u[-n]
c) \
\(\frac{1}{n^2} 3^{\frac{1}{n}} u[-n]\)
Answer: b
Explanation: Y = X
↔ y[n] = X [-n]
Or, y[n] = n 2 3 -n u[-n].
3. Given the z-transform pair 3 n n 2 u[n] ↔ X . The time signal corresponding to \Missing open brace for subscript 3 3 n-1 u[n-1]
b) n 3 3 n u[n-1]
c) 3 3 n-1 u[n-1]
d) 3 3 n-1 u[n]
Answer: c
Explanation: Y = \
↔ y = – X [-n-1] Or, y = – 3 3 n-1 u[n-1].
4. Given the z-transform pair 3 n n 2 u[n] ↔ X . The time signal corresponding to \
is ___________
a) \
b)
c) \
d)
Answer: a
Explanation: Y = \
↔ y[n] = \
.
5. Given the z-transform pair 3 n n 2 u[n] ↔ X . The time signal corresponding to {X} 2 is ___________
a) {x[n]} 2
b) x[n]*x[n]
c) x[n]*x[-n]
d) x[-n]*x[-n]
Answer: b
Explanation: Y = X H
Y = X X ↔ y[n] Or, y [n] = x[n]*x[n]
6. The system described by the difference equation y – 2y + y = X – X has y = 0 and n<0. If x = δ, then y will be?
a) 2
b) 1
c) 0
d) -1
Answer: c
Explanation: Given equation = y – 2y + y = X – X has y = 0
For n = 0, y 2y + y = x – x
∴ y = x – x
Or, y = 0 for n<0
For n=1, y = -2y + y = x – x
Or, y = x – x + 2x – 2x
Or, y = x + x – 2x
For n=2, y = x – x + 2y – y
Or, y = x – x + 2x + 2x – 4x – x + x
∴y = d + d + d – 3d .
7. The value of \Missing open brace for subscript \
\
\
\(\frac{a^{n+1} + b^{n+1}}{a+b}\)
Answer: b
Explanation: We know that, \(Z^{-1}{\frac{z}{z-a}} = a^n\) and\(Z^{-1}{\frac{z}{z-b}} = b^n\)
∴\(Z^{-1}\Big\{\frac{z^2}{}\Big\} = Z^{-1}{\frac{z}{z-a} . \frac{z}{z-b}} = a^n * b^n\)
= \(∑_{m=0}^n a^m.b^{n-m}\)
= \(b^n ∑_{m=0}^n \frac{a^m}{b}\)
= \(b^n . \frac{\frac{a^{n+1}}{b}-1}{\frac{a}{b}-1}\)
= \(\frac{a^{n+1} – b^{n+1}}{a-b}\).
8. Given the z-transform pair
\Missing open brace for subscript \
\
\
\(\frac{-32z}{
^2}\)
Answer: a
Explanation: y[n] = n x[n] n x[n] ↔ Y = \(-z\frac{dX}{dz}\)
Now, \(-z\frac{dX}{dz} = \frac{32z^2}{
^2}\).
9. Given the z-transform pair
\Missing open brace for subscript \
\
\
\(\frac{^2}{^2-16}\)
Answer: b
Explanation: x ↔ z -n 0 X
Now, y[n] = x [n+1] + x [n-1] ↔ Y
Y = X
∴ Y = \(\frac{z^2 }{z^2-16}\).
10. The value of inverse Z-transform of log
is _______________
a) n /n for n = 0; 0 otherwise
b) n /n
c) 0, for n = 0; n /n, otherwise
d) 0
Answer: c
Explanation: Putting z = \
= log \Missing or unrecognized delimiter for \right\)
= – log = -y + \(\frac{1}{2}\) y 2 – \(\frac{1}{3}\) y 3 + …..
= -z -1 + \(\frac{1}{2}\) z -2 – \(\frac{1}{3}\) z -3 + …..
Thus, u n = 0, for n = 0; n /n otherwise.
11. The inverse Z-transform of z/ 2 is ______________
a) n+1
b) n-1 n
c) n-1
d) n+1 n
Answer: b
Explanation: U = \
= \
^{n-1} nz^{-n}\)
Hence, un = n-1 n.
This set of Tricky Signals & Systems Questions and Answers focuses on “Ideal LPF, HPF, BPF and BSF Characteristics”.
1. In the circuit given below, C = C 1 = C 2 . The gain of the multiple-feedback band-pass filter is ___________
tricky-signals-systems-questions-answers-q1
a) \
\
\
\(A_0 = \frac{R_1}{2R_2}\)
Answer: c
Explanation: The total output C = C INPUT + C OUTPUT that is the gain capacitor.
∴ The total Resistance is equal to the Resistance input and Resistance output.
Again, the total resistance gain = \(\frac{R_1 R_2}{R_1+ R_2}\)
Hence, the gain = \(A_0 = \frac{R_2}{2R_1}\).
2. Two network functions are given below.
H = \(\frac{1}{s^2 + \sqrt{2} s + 1}\), H 2 = \Missing open brace for subscript Low-pass
b) Band-pass
c) High-pass
d) Band reject
Answer: a
Explanation: | H 1 | 2 = \(\frac{1}{[^2 + \sqrt{2} jω + 1] [^2 + \sqrt{2} jω + 1]}\)
Similarly, | H 2 | 2 = \(\frac{1}{1+ ω^6}\)
Therefore at ω = 0, 1 and∞, we have | H | 2 = 1, \(\frac{1}{2}\) and 0 respectively.
Hence, the filter is a Low-pass filter.
3. A particular band-pass function has a network function as H = \Missing open brace for subscript \
\
\
\(\frac{\sqrt{3}}{4}\)
Answer: d
Explanation: H = \(\frac{Ks}{s^2+as+b}\)
Then, quality factor is given as \(\frac{\sqrt{b}}{a}\)
Here, b = 3, a = 4
∴ Q = \(\frac{\sqrt{3}}{4}\).
4. The filter which passes all frequencies above f c by attenuating significantly, all frequencies below f c is _______________
a) Low-pass
b) High-pass
c) Band-pass
d) Band-stop
Answer: b
Explanation: A high-pass filter is one which passes all frequencies above f c by attenuating significantly, all frequencies below f c .
5. For the circuit given below, the cut-off frequency of the filter is ________________
tricky-signals-systems-questions-answers-q5
a) 5283 kHz
b) 5283 Hz
c) 2653.1 kHz
d) 2653.1 Hz
Answer: d
Explanation: We know that, F = \(\frac{1}{2π×R×C}\)
Where, R = 1200 Ω, C 1 = 0.05 × 10 -6 F
∴ F = \(\frac{1}{2π×1200×C}\)
= \(\frac{1}{2π×60×10^{-6}}\)
= \(\frac{10^6}{2π×60}\) = 2653.1 Hz.
6. For a Band Pass Filter, the slope of the filter is given as 40dB/decade. The order of the Band Pass Filter is __________
a) 2
b) 3
c) 4
d) 6
Answer: c
Explanation: The Bode plot is a logarithmic plot which helps in fitting a large scale of values into a small scale by the application of logarithm. Plotting the slope 40dB/decade on the Bode plot, we get n = 4. Hence the order of the Band Pass Filter is 4.
7. The circuit given below represents which type of filter circuit?
tricky-signals-systems-questions-answers-q5
a) Low-pass Filter
b) High-pass Filter
c) Band-pass Filter
d) Band-stop Filter
Answer: b
Explanation: We know that, the position of Resistance and Capacitance determines whether it is Low-pass Filter or High-Pass Filter. If R is connected directly to source and the capacitor connected in parallel to it, then it is a Low-pass Filter and if the position of R and C are inter change then high pass filter is formed.
Here since R and C are in series and also R is not connected directly to the power source, hence the filter is a High-pass Filter.
8. The circuit given below represents which type of filter circuit?
tricky-signals-systems-questions-answers-q8
a) Low-pass Filter
b) High-pass Filter
c) Band-pass Filter
d) Band-stop Filter
Answer: c
Explanation: The given circuit is a first order Band Pass Filter. Also, the Roll-off of the filter depends upon the order of the filter. For a first order it is 20dB/decade, for second order it is 40dB/decade, and so on.
9. For the circuit given below, the Roll-off value of the filter is _____________
tricky-signals-systems-questions-answers-q9
a) 20 dB/decade
b) 40 dB/decade
c) 60 dB/decade
d) 80 dB/decade
Answer: d
Explanation: The given filter is a first order Band Pass Filter. Also, the Roll-off of the filter is depends upon the order of the filter. For a first order it is 20dB/decade, for second order it is 40dB/decade, and so on. Therefore, the Roll-off of the filter = 20 dB/decade. Roll of first order low pass Butterworth filter is 20dB/decade. Now here two stages of second order Low-pass Butterworth filter are cascaded.
∴ Roll-off = 20*4 = 80 dB/decade.
10. The circuit given below, represents which filter?
tricky-signals-systems-questions-answers-q9
a) Low-pass
b) High-pass
c) Band-pass
d) Band-stop
Answer: b
Explanation: From the given circuit, we can infer that Roll off of the Filter circuit is 80dB/decade. This Roll-off value is obtained as second order high pass filter followed by another 2nd order HPF results in an HPF.
Therefore the circuit represents a High-pass Filter.
11. In the circuit given below, the Roll-off of the filter is _______________
tricky-signals-systems-questions-answers-q11
a) 20 dB/decade
b) 40 dB/decade
c) 60 dB/decade
d) 80 dB/decade
Answer: a
Explanation: The given filter is a first order Band Pass Filter. Also, the Roll-off of the filter depends upon the order of the filter. For a first order it is 20dB/decade, for second order it is 40dB/decade, and so on.
Therefore, the Roll-off of the filter = 20 dB/decade.
12. In which of the filter circuits given below, will the bandwidth be equal to the critical frequency?
a) Low-pass
b) High-pass
c) Band-pass
d) Band-stop
Answer: a
Explanation: Bandwidth can be calculated by considering,
Largest positive value – Smallest Positive Value
Here, in case of the Low-pass filter only, the largest positive value will of course be the critical frequency, beyond which frequencies have to be blocked. Hence, the bandwidth in a Low-pass filter equals the critical frequency.
13. For the circuit given below, the cut-off frequency of the filter is ________________
tricky-signals-systems-questions-answers-q13
a) 3225.8 Hz
b) 7226 Hz
c) 3225.8 kHz
d) 7226 kHz
Answer: c
Explanation: We know that, F = \(\frac{1}{2π×
×
}\)
Where, R 1 = 4700 Ω, R 2 = 4700 Ω, C 1 = 0.047×10 -6 F, C 2 = 0.047 × 10 -6 F
∴ F = \(\frac{1}{2π××
}\)
= \(\frac{1}{2π×48796.81×10^{-12}}\)
= \(\frac{10^6}{2π×0.04879681}\)
= \(\frac{10^6}{0.30654156042}\) = 3225.8 kHz.
14. For providing a Roll-off greater than 20dB/decade/pole, filters with which characteristics are useful?
a) Butterworth
b) Chebyshev
c) Bessel
d) Butterworth & Bessel
Answer: b
Explanation: Roll off is a term commonly refers to the steepness of the transmission function wrt to the frequency.
For a Chebyshev filter, the Roll-off value greater than 20. This characteristic feature is useful when a rapid roll-off is required because it provides a Roll-off rate is more than 20.
On the other hand, both Butterworth and Bessel have the Roll-off rate less than or equal to 20 dB/decade/pole.
15. A Low-pass filter circuit has a cut-off frequency of 1.23 kHz. The bandwidth of the filter is ______________
a) 2.46 kHz
b) 1.23 kHz
c) 0.615 kHz
d) 3.69 kHz
Answer: b
Explanation: The bandwidth is defined as the highest cut-off frequency to the lowest cut-off frequency. Here the lowest cut-off frequency is Zero.
For a Low-pass filter, Cut-off Frequency = Bandwidth of the filter
∴ Bandwidth = 1.23 kHz.
This set of Signals & Systems Multiple Choice Questions & Answers focuses on “Concept of Convolution”.
1. The resulting signal when a continuous time periodic signal x having period T, is convolved with itself is ___________
a) Non-Periodic
b) Periodic having period 2T
c) Periodic having period T
d) Periodic having period T/2
Answer: c
Explanation: The solution lies with the definition of convolution. Given a periodic signal x having period T. When convolution of a periodic signal with period T occurs with itself, it will give the same period T.
2. Convolution of step signal 49 times that is 49 convolution operations. The Laplace transform is ______________
a) \
\
1
d) s 49
Answer: a
Explanation: n times = u * u * …… * u
Laplace transform of the above function = \(\frac{1}{s^n} \), where n is number of convolutions.
∴ Laplace transform for 49 convolutions = \(\frac{1}{s^{49}} \).
3. The auto correlation of x = e -at u is ________________
a) \
\
\
\(\frac{e^{-aλ}}{2a} \)
Answer: d
Explanation: R = \
x \,dt\)
= \
e^{-a} u \,dt\)
= \(\int_λ^∞ e^{-2at} e^{aλ} \,dt\)
= \(\frac{e^{aλ}}{-2a}\)[0-e -2aλ ]
= \(\frac{e^{-aλ}}{2a} \).
4. For any given signal, average power in its 6 harmonic components as 10 mw each and fundamental component also has 10 mV power. Then, average power in the periodic signal is _______________
a) 70
b) 60
c) 10
d) 5
Answer: b
Explanation: We know that according to Parseval’s relation, the average power is equal to the sum of the average powers in all of its harmonic components.
∴ P avg = 10 × 6 = 60.
5. One of the types of signal is an Impulse train. The type of discontinuity in an impulse train is ______________
a) Infinite
b) Zero
c) One
d) Finite
Answer: a
Explanation: From any Impulse train waveform, we can infer that it is a kind of signal having infinite discontinuity.
6. Given a signal f = 3t 2 +2t+1, which is multiplied by 2 unit delayed version of impulse and integrated over period -∞ to ∞. The resultant is ______________
a) 1
b) 6
c) 17
d) 16
Answer: c
Explanation: \
δ
= f
\)
Here, t 0 = 2
So, \
δ\) = f
Hence, f = 3 2 + 2 + 1
= 12 + 4 + 1 = 17.
7. A PT is a device which is ___________
a) Electrostatically coupled
b) Electrically coupled
c) Electromagnetically coupled
d) Conductively coupled
Answer: c
Explanation: A PT cannot be electrostatically coupled since CRO are electrostatically coupled. Also, they cannot be conductively coupled. But since they are kind of electrically coupled hence electromagnetically coupled is the only correct option.
8. The CT supplies current to the current coil of a power factor meter, energy meter and, an ammeter. These are connected as?
a) All coils in parallel
b) All coils in series
c) Series-parallel connection with two in each arm
d) Series-parallel connection with one in each arm
Answer: b
Explanation: Since the CT supplies the current to the current coil, therefore the coils are connected in series so that the current remains the same. If they were connected in parallel then the voltages would have been same but the currents would not be the same and thus efficiency would decrease.
9. If a signal f has energy E, the energy of the signal f is equal to ____________
a) E
b) 100E
c) E/100
d) 400E
Answer: c
Explanation: We know that, E = \
^2 \,dt\)
Let, E s = \
^2 \,dt\)
Let 100t = p
Or, dt = dp/100
= \
^2 \,dp/100\)
So, E s = E/100.
10. Two sequences x 1 and x 2 are related by x 2 = x 1 . In the z-domain, their region of convergences are _______________
a) The same
b) Reciprocal of each other
c) Negative of each other
d) Complementary
Answer: b
Explanation: x 1 has z-transform X 1
The ROC = R x
Again, x 2 = x 1 has z-transform X 1
The ROC = 1/R x
Hence they are reciprocals.
11. If the Laplace transform of f = \(\frac{w}{s^2+w^2}\). The value of lim t→∞ f is ____________
a) Cannot be determined
b) Zero
c) Unity
d) Infinity
Answer: b
Explanation: We know that,
By final value theorem, lim t→∞ f = lim s→0 s F
= lim s→0 \(\frac{s.w}{s^2+w^2}\)
= 0.
12. The auto-correlation function of a rectangular pulse of duration T is _____________
a) A rectangular pulse of duration T
b) A rectangular pulse of duration 2T
c) A triangular pulse of duration T
d) A triangular pulse of duration 2T
Answer: d
Explanation: R xx1 = \
xdt\)
Which when plotted is a triangular pulse of duration 2T.
13. The power in the signal = 8cos
+ 4sin is equal to ______________
a) 40
b) 42
c) 41
d) 82
Answer: a
Explanation: Power of Signal = lim T→ ∞ \
|^2 \,dt\)
Signal power P is mean of the signal amplitude squared value of f .
Rms value of signal = \ = 8cos
+ 4sin
= 8 sin + 4 sin
= \(\frac{8^2}{2} + \frac{4^2}{2}\)
= 32 + 8 = 40.
14. The Fourier transform of a function x is X . The FT of \Missing open brace for subscript \
2πjf X
c) X jf
d) \(\frac{X}{jf}\)
Answer: b
Explanation: \ \rightarrow \frac{1}{2π} \int_{-∞}^∞ X e^{j2πt} \,dt\)
Now, differentiating both sides,
We get, \
e^{j2πt} \,dt\)
= j2πf X.
15. Given the signal
X = cos t, if t<0
X = Sin t, if t≥0
The correct statement among the following is?
a) Periodic with fundamental period 2π
a) Periodic but with no fundamental period
a) Non-periodic and discontinuous
a) Non-periodic but continuous
Answer: c
Explanation: From the graphs of cos and sin, we can infer that at t=0, the function becomes discontinuous.
Since, cos 0 = 1, but sin 0 = 0
As 1 ≠ 0, so, the function X is discontinuous and therefore Non-periodic.
