Signals & Systems Questions and Answers – Characterization and Nature of Systems

This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on “Characterization and Nature of Systems”.

1. Given a Fourier transform pair x(t) ↔ X(jω) = \(\frac{2 sin⁡ω}{ω}\), where, x(t) = 1 for |t|<1 and 0, otherwise. Then the Fourier transform of y(t) having the shape of a triangular waveform from t=-2 to t=2 and maximum peak value=2 is ___________
a) \(\frac{4 sin^2 ω}{ω^2}\)
b) \(\frac{2 sin^2 ω}{πω^2}\)
c) \(\frac{8π sin^2 ω}{ω^2}\)
d) \(\frac{8 sin^2 ω}{ω^2}\)
View Answer

Answer: a
Explanation: We know that, y (t) = x (t)*x (t)
Or, Y (jω) = (X (jω))2 = \(\frac{4 sin^2 ω}{ω^2}\).

2. For the signal given below, the region of convergence is ____________
Find the region of convergence from the given diagram
a) ω1 < ω < ω2 in s-plane
b) Entire s-plane
c) Imaginary axis
d) Entire s-plane except imaginary axis
View Answer

Answer: b
Explanation: We know that, if x (t) is of finite duration and is absolutely integrable, then the region of convergence is the entire s-plane. Here since x (t) 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 = \(\frac{1}{3}\) F, the value of the current I(t) is ___________
a) \(\frac{3}{2} e^{3t} – \frac{1}{2} e^{-t}\)
b) \(\frac{3}{2} e^{3t} – \frac{1}{2} e^t\)
c) \(\frac{3}{2} e^{-3t} – \frac{1}{2} e^{-t}\)
d) \(\frac{3}{2} e^{-3t} – \frac{1}{2} e^t\)
View Answer

Answer: c
Explanation: Z(s) = R + Ls + \(\frac{1}{Cs}\)
= 4 + s + \(\frac{3}{s} = \frac{(s+3)(s+1)}{s}\)
Now, v (t) = δ(t)
∴ V(s) = 1
∴ I(s) = \(\frac{V(s)}{Z(s)} = \frac{1}{Z(s)}\)
∴ I(s) = \(\frac{s}{(s+3)(s+1)}\)
Taking inverse, we get, I (t) = \(\frac{3}{2} e^{-3t} – \frac{1}{2} e^{-t}\).
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4. The transfer function of an LTI system may be expressed as H (z) = \(\frac{K.(z-z_1)….(z-z_m)}{(z-P_1)….(z-P_2)}\)
We consider the following statements:
i) Poles of H(z) are called natural modes
ii) Poles of H (z) 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
View Answer

Answer: c
Explanation: We know that the poles of H (z) are called natural modes or natural frequencies.

5. The system y(t) = x3(t) is ___________
a) Stable with input x(t)
b) Stable with output y(t)
c) Stable with both input x(t) as well as output y(t)
d) Not stable
View Answer

Answer: c
Explanation: Given that y (t) = x3(t)
Let x (t) = u (t), that is bounded input
∴ y(t) = u3(t) = u(t), that is bounded output
The bounded input produces bounded output hence system is stable with both input x (t) as well as output y (t).
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6. Sinusoidal signal x (t) = 4cos (200t + \(\frac{π}{6}\)) is passed through a square law device defined by the input output relation y (t) = x2(t). The DC component in the signal is _____________
a) 3.46
b) 4
c) 2.83
d) 8
View Answer

Answer: d
Explanation: y (t) = 42 cos2 (200t + \(\frac{π}{6}\))
= \(4^2 \frac{(1+cos⁡2(200t + \frac{π}{6}))}{2}\)
= 8 + 8 cos (400t + \(\frac{π}{3}\))
Hence, DC component = 8
AC component = 8 cos (400t + \(\frac{π}{3}\)).

7. Let a signal a1 sin (ω1 t + φ1) be applied to an astable LTI system. The corresponding output is a2 F sin (ω2 t + φ2). Then which of the following is correct?
a) F is not necessarily a sine or cosine function but must be periodic with ω12
b) F must be a sine or cosine function with a1 = a2
c) F must be a sine function with ω12 and φ1 = φ2
d) F must be a sine or cosine function with ω12
View Answer

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 a1 = a2.
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8. The system y(t) = x2(t) is ______________
a) Invertible with input x(t) and output y(t)
b) Invertible with input x(t)
c) Invertible with output y(t)
d) Not invertible
View Answer

Answer: d
Explanation: Given y (t) = x2(t)
Let x (t) = u (t), then y (t) = u2 (t) = u (t)
Let, x (t) = -u (t), then y (t) = {-u (t)}2 = u (t)
Thus different inputs leads to same output hence system is non-invertible.

9. The system y(t) = 15ex(t) is ___________
a) Stable with input x(t)
b) Stable with output y(t)
c) Stable with both input x(t) as well as output y(t)
d) Not stable
View Answer

Answer: c
Explanation: Given that y (t) = 15 ex(t)
Let x (t) = u (t), that is bounded input
∴ y(t) = eu(t), that is bounded output
The bounded input produces bounded output hence system is stable with both input x (t) as well as output y (t).
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10. The system y(t) = x(2t) is ______________
a) Invertible with input x(t) and output y(t)
b) Invertible with input x(t)
c) Invertible with output y(t)
d) Not invertible
View Answer

Answer: a
Explanation: Given y (t) = x (2t)
Let x (t) = u (t), then y (t) = u (2t) = u (t)
Let, x (t) = -u (t), then y (t) = – u (2t) = – u (t)
Since different inputs leads to different outputs hence system is invertible with both input x (t) and output y (t).

11. The system y(t) = \(\frac{d x^2 (t)}{dt}\) is ___________
a) Stable with input x(t)
b) Stable with output y(t)
c) Stable with both input x(t) as well as output y(t)
d) Not stable
View Answer

Answer: a
Explanation: Given that y (t) = \(\frac{d x^2 (t)}{dt}\)
Let x (t) = u (t), that is bounded input
∴ y(t) = 2 \(\frac{d x(t)}{dt}\), That is not bounded output
The bounded input does not produce bounded output hence system is stable with input x (t) only.

12. The system y(t) = \(\frac{d x(t)}{dt}\) is ______________
a) Invertible with input x(t) and output y(t)
b) Invertible with input x(t)
c) Invertible with output y(t)
d) Not invertible
View Answer

Answer: d
Explanation: Given y (t) = \(\frac{d x(t)}{dt}\) Let x (t) = 1, then y (t) = 0
Let, x (t) = 2, then y (t) = 0
Thus different inputs leads to same output hence system is non-invertible.

13. The system defined by h[n] = 2n u[n-2] is _______________
a) Stable and causal
b) Causal but not stable
c) Stable but not causal
d) Unstable and non-causal
View Answer

Answer: b
Explanation: Given that, h[n] = 2n 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 (\(\frac{1}{2} – \frac{1}{2} e^{-2t}\)), the impulse response of the system is _______________
a) (\(\frac{1}{2} – \frac{1}{2} e^{-2t}\))
b) e-2t
c) (1-e-2t)
d) Constant
View Answer

Answer: b
Explanation: We know that, impulse response is given by \(\frac{d [step \,response]}{dt}\)
Step response = (\(\frac{1}{2} – \frac{1}{2} e^{-2t}\))
∴ Impulse response = \(\frac{d (\frac{1}{2} – \frac{1}{2} e^{-2t})}{dt}\)
= e-2t.

15. The system y[n] = \(∑_{k=-∞}^n x[k]\) is an example of ____________
a) Invertible system
b) Memoryless system
c) Non-invertible system
d) Averaging system
View Answer

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.

Sanfoundry Global Education & Learning Series – Signals & Systems.

To practice all areas of Signals & Systems, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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