Signals & Systems Questions and Answers – Fourier Series Coefficients – 2

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This set of Signals & Systems Objective Questions & Answers focuses on “Fourier Series Coefficients – 2”.

1. The Fourier series coefficient for the signal 10δ(t) is ___________
a) 1
b) Cos (\(\frac{π}{2}\) k)
c) sin (\(\frac{π}{2}\) k)
d) 2
View Answer

Answer: d
Explanation: \(X[k] = \frac{1}{T} \displaystyle\int_{-\frac{T}{2}}^{\frac{T}{2}} Aδ(t)e^{-jkωt} \,dt\)
= \(\frac{A}{2}\)
Here, A=10, T=5
∴ X[k] = 2.

2. The Fourier series coefficient for the periodic rectangular pulses of height 2A is ____________
a) \(\frac{2A}{jkπ} \,sin\,⁡ \frac{π}{2} k\)
b) \(\frac{2A}{jkπ} \,cos⁡\, \frac{π}{2} k\)
c) \(\frac{2A}{kπ} \,sin⁡\, \frac{π}{2} k\)
d) \(\frac{2A}{kπ} \,cos⁡\, \frac{π}{2} k\)
View Answer

Answer: c
Explanation: \(X[k] = \frac{1}{T} \displaystyle\int_{-\frac{T}{2}}^{\frac{T}{2}} x(t)e^{-jkωt} \,dt\)
\(= \frac{1}{T} \displaystyle\int_{-\frac{T}{4}}^{\frac{T}{4}} Ae^{-jkωt} \,dt\)
\(= \frac{2A}{T}[\frac{e^{-jkωt}}{-jkω}]\) (from –\(\frac{T}{4}\) to \(\frac{T}{4}\))
\(=\frac{2A}{kπ} \,sin⁡\, \frac{π}{2} k\).

3. The Fourier series coefficient for the periodic signal x(t) = sin2t is _____________
a) –\(\frac{1}{4}\) δ[k-1] + \(\frac{1}{2}\) δ[k] – \(\frac{1}{4}\) δ[k+1]
b) –\(\frac{1}{4}\) δ[k-2] + \(\frac{1}{2}\) δ[k] – \(\frac{1}{4}\) δ[k+2]
c) –\(\frac{1}{2}\) δ[k-1] + δ[k] – \(\frac{1}{2}\) δ[k+1]
d) –\(\frac{1}{2}\) δ[k-2] + δ[k] – \(\frac{1}{2}\) δ[k+2]
View Answer

Answer: a
Explanation: sin2t = \((\frac{e^{jt} – e^{-jt}}{2j})^2\)
= –\(\frac{1}{4}\) (e2jt – 2 + e-2jt)
The fundamental period of sin2t is π and ω = \(\frac{2π}{π}\) = 2
∴ X[k] = –\(\frac{1}{4}\) δ[k-1] + \(\frac{1}{2}\) δ[k] – \(\frac{1}{4}\) δ[k+1].
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4. The Fourier series coefficient of time domain signal x (t) is X[k] = jδ[k-1] – jδ[k+1] + δ[k+3] + δ[k-3], the fundamental frequency of the signal is ω=2π. The signal is ___________
a) 2(cos 3πt – sin πt)
b) -2(cos 3πt – sin πt)
c) 2(cos 6πt – sin 2πt)
d) -2(cos 6πt – sin 2πt)
View Answer

Answer: c
Explanation: \(x (t) = ∑_{k=-∞}^∞ X[k]e^{j2πkt}\)
= jej2πt – je-j2πt + ej6πt + e-j6πt
= 2(cos 6πt – sin 2πt).

5. The Fourier series coefficient of time domain signal x (t) is X[k] = \((-\frac{1}{3})^{|k|}\). The fundamental frequency of signal is ω=1. The signal is _____________
a) \(\frac{4}{5 + 3 sin⁡}\)
b) \(\frac{5}{4 + 3 sin⁡t}\)
c) \(\frac{5}{4 + 3 cos⁡t}\)
d) \(\frac{4}{5 + 3 sin⁡t}\)
View Answer

Answer: d
Explanation: \(x (t) = ∑_{k=-∞}^∞ X[k]e^{jkt}\)
Or, x (t) = \(∑_{k=-∞}^{-1} (-\frac{1}{3})^{-k} e^{jk} + ∑_{k=0}^∞ (-\frac{1}{3})^k e^{jkt}\)
= \(\frac{\frac{-1}{3} e^{-jt}}{1+\frac{1}{3} e{-jt}} + \frac{1}{1 + \frac{1}{3} e^{jt}}\)
= \(\frac{4}{5 + 3 sin⁡t}\).

6. The Fourier series coefficient of the signal y(t) = x(t-t0) + x(t+t0) is _____________
a) 2 cos (\(\frac{2π}{t}\) kt0) X[k]
b) 2 sin (\(\frac{2π}{t}\) kt0) X[k]
c) 2 cos (\(\frac{2π}{t}\) kt0)
d) 2 sin (\(\frac{2π}{t}\) kt0)
View Answer

Answer: a
Explanation: x (t-t0) is periodic with period T. the Fourier series coefficient of x (t-t0) is X1[k] = \(\frac{1}{T}\) ∫ x (t-t0)e-jkωt dt
= e-jkωt0 X[k]
Similarly, the Fourier series coefficient of x (t+t0) is X2[k] = ejkωt0 X[k]
The Fourier series coefficient of x (t-t0) + x (t+t0) is
Y[k] = X1[k] + X2[k]
= e-jkωt0 X[k] + ejkωt0 X[k]
= 2 cos (\(\frac{2π}{t}\) kt0) X[k].

7. The Fourier series coefficient of the signal y(t) = Even{x(t)} is ___________
a) \(\frac{X[k]+X[-k]}{2}\)
b) \(\frac{X[k]-X[-k]}{2}\)
c) \(\frac{X[k]+X^* [-k]}{2}\)
d) \(\frac{X[k]-X^* [-k]}{2}\)
View Answer

Answer: a
Explanation: even {x (t)} = \(\frac{x(t)+x(-t)}{2}\)
The Fourier series coefficient transform of x (t) is
X1[k] = \(\frac{1}{T}\) ∫ x (-t)e-jkωt dt
= \(\frac{1}{T}\) ∫ x (α)ejkωα
= X [-k] ∴ The Fourier coefficient of Even{x(t)} = Y[k] = \(\frac{X[k]+X[-k]}{2}\).
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8. The Fourier series coefficient of the signal y(t) = Re{x(t)} is ____________
a) \(\frac{X[k]+X[-k]}{2}\)
b) \(\frac{X[k]-X[-k]}{2}\)
c) \(\frac{X[k]+X^* [-k]}{2}\)
d) \(\frac{X[k]-X^* [-k]}{2}\)
View Answer

Answer: c
Explanation: Re{x (t)} = \(\frac{x(t)+x^* (-t)}{2}\)
The Fourier coefficient of x* (t) is
X1[k] = \(\frac{1}{T}\) ∫ x* (t)e-jkωt dt = \(X_1^*\) [-k]
Or, \(X_1^*\) [k] = \(\frac{1}{T}\) ∫ x(t)ejkωt dt = X [-k]
So, X1[k] = \(X_1^*\) [-k]
∴ Y[k] = \(\frac{X[k]+X^* [-k]}{2}\).

9. The Fourier series coefficient of the signal y(t) = \(\frac{d^2 x(t)}{dt^2}\) is _____________
a) \((\frac{2πk}{T})^2 X[k]\)
b) –\((\frac{2πk}{T})^2 X[k]\)
c) j\((\frac{2πk}{T})^2 X[k]\)
d) -j\((\frac{2πk}{T})^2 X[k]\)
View Answer

Answer: b
Explanation: \(x (t) = ∑_{k=-∞}^∞ X[k]e^{j \frac{2π}{T} kt}\)
Now, \(\frac{dx(t)}{dt} = -j (\frac{2π}{T})k ∑_{k=-∞}^∞ X[k]e^{j \frac{2π}{T} kt}\)
And, \(\frac{d^2 x(t)}{dt^2} = -(\frac{2π}{T})^2 k^2 ∑_{k=-∞}^∞ X[k]e^{j \frac{2π}{T} kt}\)
∴ Y[k] = – \((\frac{2πk}{T})^2 X[k]\).
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10. The Fourier series coefficient of the signal y(t) = x(4t-1) is ______________
a) \(\frac{8π}{T} X[k]\)
b) \(\frac{4π}{T}X[k]\)
c) \(e^{-jk \frac{8π}{T}} X[k]\)
d) \(e^{jk \frac{8π}{T}} X[k]\)
View Answer

Answer: c
Explanation: The period of x (4t) is a fourth of the period of x (t). The Fourier series coefficient of x (4t) is still X[k]. Hence, the coefficient of x (4t-1) is \(e^{-jk \frac{8π}{T}} X[k]\).

11. The discrete time Fourier coefficients of \(∑_{m=-∞}^∞ δ[n-4m]\) is ____________
a) –\(\frac{1}{4}\) for all k
b) \(\frac{1}{4}\) for all k
c) –\(\frac{1}{2}\) for all k
d) \(\frac{1}{2}\) for all k
View Answer

Answer: b
Explanation: N=4, ω = \(\frac{2π}{4} = \frac{π}{2}\)
\(X[k] = \frac{1}{4} ∑_{n=4}^3 x[n]e^{-j(\frac{π}{2})nk}\)
= \(\frac{1}{4}\) x[0] = \(\frac{1}{4}\) for all k.

12. The discrete time Fourier coefficient of cos2(\(\frac{π}{8}\) n) is ______________
a) \(\frac{π}{2}\)(δ(k+1] + 2δ[k] + δ[k-1])
b) \(\frac{1}{4}\)j(δ(k+1] + 2δ[k] + δ[k-1])
c) \(\frac{1}{4}\)(δ(k+1] + 2δ[k] + δ[k-1])
d) \(\frac{π}{4}\)(δ(k+1] + 2δ[k] + δ[k-1])
View Answer

Answer: c
Explanation: N=8, ω = \(\frac{2π}{8} = \frac{π}{4}\)
X[n] = cos2 (\(\frac{π}{8}\) n) = \(\frac{1}{4}(e^{j(\frac{π}{8})n} + e^{-j(\frac{π}{8})n})^2\)
\(= \frac{1}{4}(e^{j(\frac{π}{8})n} + 2 + e^{-j(\frac{π}{8}n})^2\)
Or, X[k] = \(\frac{1}{4}\)(δ(k+1] + 2δ[k] + δ[k-1]).

13. V(t) = 5,    0≤t<1;
    t,           t≥1;
The Laplace transform of V (t) is ___________
a) \(\frac{5}{s} + \frac{e^{-s}}{s^2} + \frac{4e^{-s}}{s}\)
b) \(\frac{5}{s} + \frac{e^{-s}}{s^2} – \frac{4e^{-s}}{s}\)
c) \(\frac{5}{s} – \frac{e^{-s}}{s^2} – \frac{4e^{-s}}{s}\)
d) \(\frac{5}{s} – \frac{e^{-s}}{s^2} + \frac{4e^{-s}}{s}\)
View Answer

Answer: b
Explanation: V (t) = 5 + u (t) (t-5)
L {5 + u (t) (t-5)} = \(\frac{5}{s}\) + L {u (t) (t-5)}
= \(\frac{5}{s} + e^{-s}\) L {t-4}
= \(\frac{5}{s} + e^{-s} (\frac{1}{s^2} – \frac{4}{s})\)
= \(\frac{5}{s} + \frac{e^{-s}}{s^2} – \frac{4e^{-s}}{s}\).

14. W(t) = 2,   0≤t<4;
   t2,          t≥4;
The Laplace transform of W (t) is ___________
a) \(\frac{2}{s} – e^{-4s} (\frac{2}{s^3} – \frac{8}{s^2} – \frac{14}{s})\)
b) \(\frac{2}{s} + e^{-4s} (\frac{2}{s^3} – \frac{8}{s^2} – \frac{14}{s})\)
c) \(\frac{2}{s} – e^{-4s} (\frac{2}{s^3} + \frac{8}{s^2} + \frac{14}{s})\)
d) \(\frac{2}{s} + e^{-4s} (\frac{2}{s^3} + \frac{8}{s^2} + \frac{14}{s})\)
View Answer

Answer: d
Explanation: W (t) = 2 + u (t) (t2-2)
L {2 + u (t) (t2-2)} = \(\frac{2}{s}\) + L {u (t) (t2-2)}
= \(\frac{2}{s} + e^{-4s}\) L {(t+4)2 -2}
= \(\frac{2}{s} + e^{-4s}\) L {t2 + 8t + 14}
= \(\frac{2}{s} + e^{-4s} (\frac{2}{s^3} + \frac{8}{s^2} + \frac{14}{s})\).

15. U(t) = 0,   0≤t<7;
   (t-7)3,          t≥7;
The Laplace transform of U (t) is ___________
a) \(\frac{6e^{-7s}}{s^4}\)
b) \(\frac{e^{-7s}}{s^4}\)
c) \(\frac{6e^{-7s}}{s^3}\)
d) \(\frac{3e^{-7s}}{s^3}\)
View Answer

Answer: a
Explanation: U (t) = u (t) (t-7)3
L {u (t) (t-7)3} = e-7s L {t3}
= \(\frac{3!e^{-7s}}{s^4} = \frac{6e^{-7s}}{s^4}\).

Sanfoundry Global Education & Learning Series – Signals & Systems.

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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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