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

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.

2. The fundamental period of the signal X (t) = 10 cos^{2}(10 πt) is __________

a) 0.2

b) 0.1

c) 0.5

d) No fundamental period exists

View Answer

Explanation: X (t) = 10 cos

^{2}(10 πt)

Since, cos 2t = 2cos

^{2}t – 1

Or, cos

^{2}t = \(\frac{1+cos2t}{2}\)

∴ 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) = e^{jt} is _________________

a) Sin t

b) Cos t

c) Sinh t

d) Cosh t

View Answer

Explanation: Let X

_{e}(t) represents the even component of X (t)

Now, X

_{e}(t) = \(\frac{1}{2}\)[X (t) + X (-t)]

= \(\frac{1}{2}\)[e

^{jt}+ e

^{-jt}]

= cos t.

4. The odd component of the signal X (t) = e^{jt} is _______________

a) Sin t

b) Cos t

c) Sinh t

d) Cosh t

View Answer

Explanation: Let X

_{o}(t) represents the odd component of X (t)

Now, X

_{o}(t) = \(\frac{1}{2}\)[X (t) – X (-t)]

= \(\frac{1}{2}\)[e

^{jt}+ 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

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.

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

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 X

_{1}(t), Period of X

_{2}(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

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 X

_{1}(t), Period of X

_{2}(t)]

∴ Period of X (t) = LCM (\(\frac{1}{3000}\), ∞) = Indefinite.

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

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 X

_{1}(t), Period of X

_{2}(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

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 X

_{1}(t), Period of X

_{2}(t)]

∴ Period of X (t) = LCM (\(\frac{2π}{5}, \frac{2π}{4}\)) = definite

Hence Z (t) is periodic with a definite period.

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

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

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.

The total energy of X (t) is _______________

a) 0

b) 13

c) \(\frac{13}{3}\)

d) \(\frac{26}{3}\)

View Answer

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

Explanation: Given that, N

_{1}= 18, N

_{2}= 14

We know that period of X [n] (say N) = LCM (N

_{1}, N

_{2})

∴ 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

Explanation: We know that for X [n] = X

_{1}[n] × X

_{2}[n] to be periodic, both X

_{1}[n] and X

_{2}[n] should be periodic with finite periods.

Here X

_{2}[n] = cos (\(\frac{πn}{8}\)), is periodic with fundamental period as \(\frac{8}{n}\)

But X

_{1}[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

Explanation: Given that, N

_{1}= 4, N

_{2}= 16, N

_{3}= 8

We know that period of X [n] (say N) = LCM (N

_{1}, N

_{2}, N

_{3})

∴ Period of X [n] = LCM (4, 16, 8) = 16.

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