This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on “Concept of Convolution”.

1. The resulting signal when a continuous time periodic signal x(t) 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

View Answer

Explanation: The solution lies with the definition of convolution. Given a periodic signal x (t) 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) \(\frac{1}{s^{49}} \)

b) \(\frac{1}{s^{50}} \)

c) 1

d) s^{49}

View Answer

Explanation: n times = u (t) * u (t) * …… * u (t)

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(t) = e^{-at}u(t) is ________________

a) \(\frac{e^{-at}}{a^2} \)

b) \(\frac{e^{-at}}{2a} \)

c) \(\frac{e^{-aλ}}{a^2} \)

d) \(\frac{e^{-aλ}}{2a} \)

View Answer

Explanation: R (λ) = \(\int_{-∞}^∞ x(t) x(t±λ) \,dt\)

= \(\int_{-∞}^∞ e^{-at} u(t) e^{-a(t-λ)} u(t-λ) \,dt\)

= \(\int_λ^∞ e^{-2at} e^{aλ} \,dt\)

= \(\frac{e^{aλ}}{-2a}\)[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

View Answer

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

View Answer

Explanation: From any Impulse train waveform, we can infer that it is a kind of signal having infinite discontinuity.

6. Given a signal f (t) = 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

View Answer

Explanation: \(\int_{-∞}^∞ f(t) δ(t-t_0) = f (t_0)\)

Here, t

_{0}= 2

So, \(\int_{-∞}^∞ f(t) δ(t-2)\) = f (2)

Hence, f (2) = 3(2)

^{2}+ 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

View Answer

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

View Answer

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(t) has energy E, the energy of the signal f(100t) is equal to ____________

a) E

b) 100E

c) E/100

d) 400E

View Answer

Explanation: We know that, E = \(\int_{-∞}^∞ f(t)^2 \,dt\)

Let, E

_{s}= \(\int_{-∞}^∞ f(t)^2 \,dt\)

Let 100t = p

Or, dt = dp/100

= \(\int_{-∞}^∞ f(t)^2 \,dp/100\)

So, E

_{s}= E/100.

10. Two sequences x_{1} (n) and x_{2} (n) are related by x_{2} (n) = x_{1} (- n). In the z-domain, their region of convergences are _______________

a) The same

b) Reciprocal of each other

c) Negative of each other

d) Complementary

View Answer

Explanation: x

_{1}(n) has z-transform X

_{1}(z)

The ROC = R

_{x}(say)

Again, x

_{2}(n) = x

_{1}(-n) has z-transform X

_{1}(1/z)

The ROC = 1/R

_{x}

Hence they are reciprocals.

11. If the Laplace transform of f (t) = \(\frac{w}{s^2+w^2}\). The value of lim_{t→∞} f(t) is ____________

a) Cannot be determined

b) Zero

c) Unity

d) Infinity

View Answer

Explanation: We know that,

By final value theorem, lim

_{t→∞} f(t) = lim

_{s→0}s F (s)

= 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

View Answer

Explanation: R

_{xx1}= \(\frac{1}{T} \int_{-T/2}^{T/2} x(t)x(t+T)dt\)

Which when plotted is a triangular pulse of duration 2T.

13. The power in the signal (t) = 8cos (20πt – \(\frac{π}{2}\)) + 4sin (15πt) is equal to ______________

a) 40

b) 42

c) 41

d) 82

View Answer

Explanation: Power of Signal = lim

_{T→ ∞}\(\frac{1}{T} \int_{-T/2}^{T/2} |f(t)|^2 \,dt\)

Signal power P is mean of the signal amplitude squared value of f (t).

Rms value of signal = \(\sqrt{P}\)

Now, (t) = 8cos (20πt – \(\frac{π}{2}\)) + 4sin (15πt)

= 8 sin (20πt) + 4 sin (15πt)

= \(\frac{8^2}{2} + \frac{4^2}{2}\)

= 32 + 8 = 40.

14. The Fourier transform (FT) of a function x (t) is X (f). The FT of \(\frac{dx(t)}{dt}\) will be ___________

a) \(\frac{dX(f)}{df}\)

b) 2πjf X(f)

c) X(f) jf

d) \(\frac{X(f)}{jf}\)

View Answer

Explanation: \(x (t) \rightarrow \frac{1}{2π} \int_{-∞}^∞ X(f) e^{j2πt} \,dt\)

Now, differentiating both sides,

We get, \(\frac{dx(t)}{dt} = j2π \frac{1}{2π} \int_{-∞}^∞ X(f) e^{j2πt} \,dt\)

= j2πf X(f).

15. Given the signal

X (t) = cos t, if t<0

X (t) = 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

View Answer

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 (t) is discontinuous and therefore Non-periodic.

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