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

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

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) $$\frac{2A}{kπ} \,cos⁡\, \frac{π}{2} k$$

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]

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].

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)

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) $$\frac{4}{5 + 3 sin⁡t}$$

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

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) $$\frac{X[k]-X^* [-k]}{2}$$

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] = $$\frac{X[k]+X[-k]}{2}$$.

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) $$\frac{X[k]-X^* [-k]}{2}$$

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$$(\frac{2πk}{T})^2 X[k]$$

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] = – $$(\frac{2πk}{T})^2 X[k]$$.

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) $$e^{jk \frac{8π}{T}} X[k]$$

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

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])

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) $$\frac{5}{s} – \frac{e^{-s}}{s^2} + \frac{4e^{-s}}{s}$$

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) $$\frac{2}{s} + e^{-4s} (\frac{2}{s^3} + \frac{8}{s^2} + \frac{14}{s})$$

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) $$\frac{3e^{-7s}}{s^3}$$

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

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