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) = \(\frac{2 sinω}{ω}\), 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

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 ____________

a) ω_{1} < ω < ω_{2} in s-plane

b) Entire s-plane

c) Imaginary axis

d) Entire s-plane except imaginary axis

View Answer

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

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) = \(\frac{3}{2} e^{-3t} – \frac{1}{2} e^{-t}\).

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

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

5. The system y(t) = x^{3}(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

Explanation: Given that y (t) = x

^{3}(t)

Let x (t) = u (t), that is bounded input

∴ y(t) = u

^{3}(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).

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) = x^{2}(t). The DC component in the signal is _____________

a) 3.46

b) 4

c) 2.83

d) 8

View Answer

Explanation: y (t) = 4

^{2}cos

^{2}(200t + \(\frac{π}{6}\))

= \(4^2 \frac{(1+cos2(200t + \frac{π}{6}))}{2}\)

= 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 (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}

View Answer

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(t) = x^{2}(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

Explanation: Given y (t) = x

^{2}(t)

Let x (t) = u (t), then y (t) = u

^{2}(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) = 15e^{x(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

Explanation: Given that y (t) = 15 e

^{x(t)}

Let x (t) = u (t), that is bounded input

∴ y(t) = e

^{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).

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

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

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

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] = 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

View Answer

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 (\(\frac{1}{2} – \frac{1}{2} e^{-2t}\)), 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

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

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.**

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