Signals and Systems Questions and Answers – Concept of Convolution

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

Answer: c
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) s49
View Answer

Answer: a
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-atu(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

Answer: d
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

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.
∴ Pavg = 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

Answer: a
Explanation: From any Impulse train waveform, we can infer that it is a kind of signal having infinite discontinuity.
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6. Given a signal f (t) = 3t2+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

Answer: c
Explanation: \(\int_{-∞}^∞ f(t) δ(t-t_0) = f (t_0)\)
Here, t0 = 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

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
View Answer

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(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

Answer: c
Explanation: We know that, E = \(\int_{-∞}^∞ f(t)^2 \,dt\)
Let, Es = \(\int_{-∞}^∞ f(t)^2 \,dt\)
Let 100t = p
Or, dt = dp/100
= \(\int_{-∞}^∞ f(t)^2 \,dp/100\)
So, Es = E/100.

10. Two sequences x1 (n) and x2 (n) are related by x2 (n) = x1 (- 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

Answer: b
Explanation: x1(n) has z-transform X1(z)
The ROC = Rx (say)
Again, x2(n) = x1(-n) has z-transform X1(1/z)
The ROC = 1/Rx
Hence they are reciprocals.

11. If the Laplace transform of f (t) = \(\frac{w}{s^2+w^2}\). The value of limt→∞ f(t) is ____________
a) Cannot be determined
b) Zero
c) Unity
d) Infinity
View Answer

Answer: b
Explanation: We know that,
By final value theorem, limt→∞⁡ f(t) = lims→0 s F (s)
= lims→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

Answer: d
Explanation: Rxx1 = \(\frac{1}{T} \int_{-T/2}^{T/2} x(t)x(t+T)dt\)
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

Answer: a
Explanation: Power of Signal = limT→ ∞ \(\frac{1}{T} \int_{-T/2}^{T/2} |f(t)|^2 \,dt\)
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

Answer: b
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
b) Periodic but with no fundamental period
c) Non-periodic and discontinuous
d) Non-periodic but continuous
View Answer

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 (t) is discontinuous and therefore Non-periodic.

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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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