Signals & Systems Questions and Answers – Periodic and Non-Periodic Signals

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This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on “Periodic and Non-Periodic Signals”.

1. Given the signal
X (t) = 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
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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2. The fundamental period of the signal X (t) = 10 cos2(10 πt) is __________
a) 0.2
b) 0.1
c) 0.5
d) No fundamental period exists
View Answer

Answer: b
Explanation: X (t) = 10 cos2 (10 πt)
Since, cos 2t = 2cos2 t – 1
Or, cos2 t = \(\frac{1+cos⁡2t}{2}\)
∴ X (t) = 5 + 5 cos 20πt
Now, Y (t) = cos 20πt
Fundamental period of the signal is = \(\frac{2π}{20π} = \frac{1}{10}\) = 0.1.

3. The even component of the signal X (t) = ejt is _________________
a) Sin t
b) Cos t
c) Sinh t
d) Cosh t
View Answer

Answer: b
Explanation: Let Xe (t) represents the even component of X (t)
Now, Xe (t) = \(\frac{1}{2}\)[X (t) + X (-t)]
= \(\frac{1}{2}\)[ejt + e-jt]
= cos t.
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4. The odd component of the signal X (t) = ejt is _______________
a) Sin t
b) Cos t
c) Sinh t
d) Cosh t
View Answer

Answer: a
Explanation: Let Xo (t) represents the odd component of X (t)
Now, Xo (t) = \(\frac{1}{2}\)[X (t) – X (-t)]
= \(\frac{1}{2}\)[ejt + e-jt]
= sin t.

5. The period of the signal X (t) = 24 + 50 cos 60πt is _______________
a) \(\frac{1}{30}\) s
b) 60 π s
c) \(\frac{1}{60π}\) s
d) Non-periodic
View Answer

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.
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6. The period of the signal X (t) = 10 sin 5t – 4 cos 9t is _______________
a) \(\frac{24π}{35}\)
b) \(\frac{4π}{35}\)
c) 2π
d) Non-periodic
View Answer

Answer: c
Explanation: Period of cos t = 2π
Period of cos at = \(\frac{2π}{a}\)
Here, a = 9
So, period of cos 9t = \(\frac{2π}{9}\)
Again, Period of sin t = 2π
Period of sin at = \(\frac{2π}{a}\)
Here, a = 5
So, period of sin 5t = \(\frac{2π}{5}\)
∴ Period of X (t) = LCM [Period of X1 (t), Period of X2 (t)]
∴ Period of X (t) = LCM (\(\frac{2π}{5}, \frac{2π}{9}\)) = 2π.

7. The period of the signal X (t) = 5t – 2 cos 6000 πt is ________________
a) 0.96 ms
b) 1.4 ms
c) 0.4 ms
d) Non-periodic
View Answer

Answer: d
Explanation: Period of cos t = 2π
Period of cos at = \(\frac{2π}{a}\)
Here, a = 6000π
So, period of cos 6000πt = \(\frac{2π}{6000π}\)
= \(\frac{1}{3000}\)
Again, Period of t = indefinite
∴ Period of X (t) = LCM [Period of X1 (t), Period of X2 (t)]
∴ Period of X (t) = LCM (\(\frac{1}{3000}\), ∞) = Indefinite.
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8. The period of the signal X (t) = 4 sin 6t + 3 sin \(\sqrt{3}\)t is ________________
a) \(\frac{2π}{3}\) s
b) \(\frac{2π}{\sqrt{3}}\) s
c) 2π s
d) Non-periodic
View Answer

Answer: d
Explanation: Period of sin t = 2π
Period of sin at = \(\frac{2π}{a}\)
Here, a = 6
So, period of sin 6t = \(\frac{2π}{6}\)
Again, a = \(\sqrt{3}\)
So, period of sin \(\sqrt{3}\)t = \(\frac{2π}{\sqrt{3}}\)
∴ Period of X (t) = LCM [Period of X1 (t), Period of X2 (t)]
∴ Period of X (t) = LCM (\(\frac{π}{3}, \frac{2π}{\sqrt{3}}\)) = Indefinite.

9. The period of the signal Z (t) = 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
View Answer

Answer: b
Explanation: Period of cos t = 2π
Period of cos at = \(\frac{2π}{a}\)
Here, a = 4
So, period of cos 4t = \(\frac{2π}{4}\)
= \(\frac{π}{2}\)
Again, Period of sin t = 2π
Period of sin at = \(\frac{2π}{a}\)
Here, a = 3
So, period of sin 3t = \(\frac{2π}{3}\)
∴ Period of X (t) = LCM [Period of X1 (t), Period of X2 (t)]
∴ Period of X (t) = LCM (\(\frac{2π}{5}, \frac{2π}{4}\)) = definite
Hence Z (t) is periodic with a definite period.
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10. The signal X (t) = e-4t u (t) is _______________
a) Power signal with P = \(\frac{1}{4}\)
b) Power signal with P = 0
c) Energy signal with E = \(\frac{1}{4}\)
d) Energy signal with E = 0
View Answer

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 (t)| < ∞, E = \(\int_{-∞}^∞ |x(t)|^2 \,dt\)
= \(\int_∞^∞ e^{-4t} u(t) \,dt \)
= \(\in_∞^∞ e^{-4t} \,dt = \frac{1}{4}\)
So, this is not a power signal but an energy signal.
\(P_∞ = lim_{T→∞} \frac{1}{2T} \int_{-T}^T |x(t)|^2 \,dt = ∞.\)

11. The signal X (t) = \(e^{j(2t + \frac{π}{6})}\) is ________________
a) Power signal with P = 1
b) Power signal with P = 2
c) Energy signal with E = 2
d) Energy signal with E = 1
View Answer

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 (t)| = 1, E = \(\int_{-∞}^∞ |x(t)|^2 \,dt = ∞\)
So, this is a power signal not an energy signal.
\(P_∞ = lim_{T→∞} \frac{1}{2T} \int_{-T}^T |x(t)|^2 \,dt = 1.\).

12. Signal X (t) is as shown in the figure below.
signals-systems-questions-answers-periodic-non-periodic-signals-q12
The total energy of X (t) is _______________
a) 0
b) 13
c) \(\frac{13}{3}\)
d) \(\frac{26}{3}\)
View Answer

Answer: d
Explanation: E = 2\(\int_0^5 x^2 (t) \,dt\)
= 2 \(\int_0^4 1^1 \,dt + 2\int_4^5 (5 – t^2) \,dt\)
= 8 + \(\frac{2}{3} = \frac{26}{3}\).

13. A discrete time signal is as given below
\(X [n] = cos \frac{πn}{9} + sin (\frac{πn}{7} + \frac{1}{2})\)
The period of the signal X [n] is ______________
a) 126
b) 32
c) 252
d) Non-periodic
View Answer

Answer: a
Explanation: Given that, N1 = 18, N2 = 14
We know that period of X [n] (say N) = LCM (N1, N2)
∴ Period of X [n] = LCM (18, 14) = 126.

14. A discrete time signal is as given below
\(X [n] = cos (\frac{n}{8}) cos (\frac{πn}{8})\)
The period of the signal X [n] is _____________
a) 16 π
b) 16(π+1)
c) 8
d) Non-periodic
View Answer

Answer: d
Explanation: We know that for X [n] = X1 [n] × X2 [n] to be periodic, both X1 [n] and X2 [n] should be periodic with finite periods.
Here X2 [n] = cos (\(\frac{πn}{8}\)), is periodic with fundamental period as \(\frac{8}{n}\)
But X1 [n] = cos (\(\frac{n}{8}\)) is non periodic.
∴ X [n] is a non-periodic signal.

15. A discrete time signal is as given below
\(X [n] = cos (\frac{πn}{2}) – sin (\frac{πn}{8}) + 3 cos (\frac{πn}{4} + \frac{π}{3})\)
The period of the signal X [n] is _____________
a) 16
b) 4
c) 2
d) Non-periodic
View Answer

Answer: a
Explanation: Given that, N1 = 4, N2 = 16, N3 = 8
We know that period of X [n] (say N) = LCM (N1, N2, N3)
∴ Period of X [n] = LCM (4, 16, 8) = 16.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter