Signals & Systems Questions and Answers – Trigonometric Fourier Series

This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on “Trigonometric Fourier Series”.

1. The Fourier transform of sin(2πt) e-t u (t) is ____________
a) \(\frac{1}{2j} \left(\frac{1}{1+j(ω-2π)} + \frac{1}{1+j(ω+2π)}\right)\)
b) \(\frac{1}{2j} \left(\frac{1}{1+j(ω-2π)} – \frac{1}{1+j(ω+2π)}\right)\)
c) \(\frac{1}{2j} \left(\frac{1}{1+j(ω+2π)} – \frac{1}{1+j(ω-2π)}\right)\)
d) \(\frac{1}{j} \left(\frac{1}{1+j(ω+2π)} – \frac{1}{1+j(ω-2π)}\right)\)
View Answer

Answer: b
Explanation: The Fourier Transform of e-tu (t) = \(\frac{1}{1+jω}\)
∴ Fourier transform of e-3|t| = \(\frac{6}{9+ω^2}\)
So, x (t-1) <–> X {j (ω-2π)}
∴ X (jω) = \(\frac{1}{2j} \left(\frac{1}{1+j(ω-2π)} – \frac{1}{1+j(ω+2π)}\right)\).

2. 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).

3. The unit impulse response of the system c (t) = -4e-t + 6e-2t for t>0. The step response of the same system for t ≥ 0 is _______________
a) -3e-2t – 4e-t + 1
b) -3e-2t + 4e-t – 1
c) -3e-2t – 4e-t – 1
d) 3e-2t + 4e-t – 1
View Answer

Answer: b
Explanation: Impulse response = -4e-t + 6e-2t
And step response = \(\int_{-∞}^t c(t)\,dt\)
= \(\int_0^t -4e^{-t} \,dt + \int_0^t 6e^{-2t} \,dt\)
= 4e-t – 3e-2t -1.
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4. Given a real valued function x (t) with period T. Its trigonometric Fourier series expansion contains no term of frequency ω = 2π \(\frac{(2k)}{T}\); where, k = 1, 2….. Also no terms are present. Then, x(t) satisfies the equation ____________
a) x (t) = x (t+T) = -x (t + \(\frac{T}{2}\))
b) x (t) = x (t+T) = x (t + \(\frac{T}{2}\))
c) x (t) = x (t-T) = -x (t – \(\frac{T}{2}\))
d) x (t) = x (t-T) = x (t – \(\frac{T}{2}\))
View Answer

Answer: d
Explanation: For an even symmetry, x (t) = x (t-T)
Thus no sine component will exist because bn=0 and by half wave symmetry condition odd harmonics will exist.
Now, x (t) = x (t-\(\frac{T}{2}\))
Combining the two conditions, we get, x (t) = x (t-T) = x (t-\(\frac{T}{2}\)).

5. Given a periodic rectangular waveform of frequency 1 kHz, symmetrical about t=0 and having a pulse width of 500 µs and amplitude 5 V. The Fourier series is ___________
a) 2.5 + 3.18 cos (2π × 103 t) – 1.06 cos (6π × 103 t)
b) 2.5 – 3.18 cos (2π × 103 t) – 1.06 cos (6π × 103 t)
c) 2.5 + 3.18 cos (2π × 103 t) + 1.06 cos (6π × 103 t)
d) 2.5 – 3.18 cos (2π × 103 t) + 1.06 cos (6π × 103 t)
View Answer

Answer: a
Explanation: Period, T = \(\frac{1}{f} = \frac{1}{1×10^3} = 10^{-3}s\)
Ratio, \(\frac{τ}{T} = \frac{500 × 10^{-6}}{10^{-3}} \) = 0.5
Now, \(f (t) = \frac{Aτ}{T} + 2 \frac{Aτ}{T} \frac{sin⁡ \frac{πτ}{T}}{\frac{πτ}{T}} cos⁡[\frac{2πt}{τ}] + \frac{2Aτ}{T} \frac{sin⁡ \frac{2πτ}{T}}{\frac{2πτ}{T}} cos⁡ \frac{4πt}{T}\)
= 5(0.5) + 2(5) (0.5)\(\frac{sin⁡0.5π}{0.5π}\) cos⁡(2π × 103 t) + 2(5) (0.5) \(\frac{sin⁡π}{π}\) cos⁡(4π × 103 t)
= 2.5 + 3.18 cos (2π × 103 t) – 1.06 cos (6π × 103 t).
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6. The lengths of two discrete time sequence x1[n] and x2[n] are 5 and 7 respectively. The maximum length of a sequence x1[n] * x2[n] is ____________
a) 5
b) 6
c) 7
d) 11
View Answer

Answer: d
Explanation: Maximum length of x1[n] * x2[n] is L (say).
Now, we know that, L = L1 + L2 -1 where L1 is length of x1[n] and L2 is length of x2[n] So, the maximum length of a sequence x1[n] * x2[n] = 5 + 7 – 1
= 11.

7. A waveform is given by v(t) = 10 sin2π 100 t. The magnitude of the second harmonic in its Fourier series representation is ____________
a) 0 V
b) 20 V
c) 100 V
d) 200 V
View Answer

Answer: a
Explanation: Given that v (t) = 10 sin2π 100 t
It holds the half-wave symmetry i.e. x (t) = -x (t ± \(\frac{T}{2}\))
And for half wave symmetry, a0 = 0 and ak = 0 (k is even)
So, second harmonic component a2 = 0.
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8. If the FT of x(t) is \(\frac{2}{ω}\) sin⁡(πω), then the FT of ej5t x(t) is ____________
a) \(\frac{2}{ω-5}\) sin⁡(πω)
b) \(\frac{2}{ω+5}\) sin⁡{π(ω-5)}
c) \(\frac{2}{ω+5}\) sin⁡(πω)
d) \(\frac{2}{ω-5}\) sin⁡{π(ω-5)}
View Answer

Answer: d
Explanation: Given that the FT of x (t) is \(\frac{2}{ω}\) sin⁡(πω)
Applying time shifting and scaling property, we get the FT of ej5t x (t) as \(\frac{2}{ω-5}\) sin⁡{π(ω-5)}.

9. A signal x (t) has its FT as X (f). The inverse FT of X(3f +2) is _____________
a) \(\frac{1}{2} x(\frac{t}{2})e^{j3πt}\)
b) \(\frac{1}{3} x(\frac{t}{3})e^{-j4πt/3}\)
c) 3x(3t)e-j4πt
d) x(3t + 2)
View Answer

Answer: b
Explanation: Applying the time shifting and scaling property, we get,
X [3(f+2/3)] = \(\frac{1}{3} x(\frac{t}{3})e^{-j4πt/3}\).
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10. The impulse response of a continuous time system is given by h(t) = δ(t-1) + δ(t-3). The value of the step response at t=2 is _______________
a) 0
b) 1
c) 2
d) 3
View Answer

Answer: b
Explanation: For step response, the impulse response can be integrated.
So, y (t) = u (t-1) + u (t-3)
Hence, y (2) = u (1) + u (-1)
= 1 + 0 = 1.

11. The first two components of trigonometric Fourier series of the given signal is ____________
Find the trigonometric Fourier series from the given diagram
a) 0, 4/π
b) 4/π, 0
c) 0, -4/π
d) -4/π, 1
View Answer

Answer: a
Explanation: Signal is odd and ω = \(\frac{2π}{t}\) = π
Here, c0 = a0 = 0
So, cn = bn = \(2\int_0^t f(t).sin⁡ωt \,dt\)
∴ c1 = b1 = \(2\int_0^t sin⁡ωt \,dt\)
= \(\frac{2}{ω}[cos⁡ωt]_0^t\)
= \(-\frac{2}{π}\)[-1-1] = 4/π
So, the first two components are 0, 4/π.

12. The convolution of two continuous time signals x(t) = e-t u(t) and h(t) = e-2t u(t) is ________________
a) (et – e2t) u (t)
b) (e-2t – e-t) u (t)
c) (e-t – e-2t) u (t)
d) (e-t + e-2t) u (t)
View Answer

Answer: c
Explanation: Given x (t) = e-t u (t) and h (t) = e-2t u (t).
Now, y (t) = \(∫_{-∞}^∞ x(τ)h(t-τ)dτ\)
= e-2t (et -1); t≥0
Or, y (t) = e-t – e-2t; t≥0
Since, y (t) = 0 for t<0
Therefore, y (t) = (e-t – e-2t) u (t).

13. A linear phase channel with phase delay Tp and group delay Tg must have ____________
a) Tp = Tg = constant
b) Tp ∝ f and Tg ∝ f
c) Tp = constant and Tg ∝ f
d) Tp ∝ f and Tg = constant
View Answer

Answer: a
Explanation: θ(ω) = -ωt0
Tp = \(-\frac{θ(ω)}{ω}\) = t0
Tg =\(-\frac{d θ(ω)}{dω}\) = t0
∴ Tp = Tg = t0 = constant.

14. The z-transform of cos(\(\frac{π}{3}\) n) u[n] is __________
a) \(\frac{z}{2} \frac{(2z-1)}{(z^2-z+1)}\), 0<|z|<1
b) \(\frac{z}{2} \frac{(2z-1)}{(z^2-z+1)}\), |z|>1
c) \(\frac{z}{2} \frac{(1-2z)}{(z^2-z+1)}\), 0<|z|<1
d) \(\frac{z}{2} \frac{(1-2z)}{(z^2-z+1)}\), |z|>1
View Answer

Answer: b
Explanation: Performing z-transform on anu[n], we get \(\frac{z}{z-a}\)
∴ x[n] = \(\frac{1}{2} e^{j(\frac{π}{3})n} \,u[n] + \frac{1}{2} e^{-j(\frac{π}{3})n} \,u [n]\)
Hence, X (z) = 0.5 \(\left(\frac{1}{1-e^{j(\frac{π}{3})} z^{-1}} + \frac{1}{1-e^{-j(\frac{π}{3})} z^{-1}}\right)\)
Hence, X (z) = \(\frac{z}{2} \frac{(2z-1)}{(z^2-z+1)}\), |z|>1.

15. The Laplace transform of the function cosh2(t) is ____________
a) \(\frac{s^2+2}{s(s^2+4)}\)
b) \(\frac{s^2-2}{s(s^2-4)}\)
c) \(\frac{s^2-2}{s(s^2+4)}\)
d) \(\frac{s^2+2}{s(s^2-4)}\)
View Answer

Answer: b
Explanation: L ((\(\frac{1}{2}\)(et – e-t))2)
= L \(\left(\frac{e^2t}{4} + \frac{1}{2} + \frac{e^{-2t}}{4}\right)\)
= \(\frac{1}{4} \frac{1}{s-2} + \frac{1}{2} \frac{1}{s} + \frac{1}{4} \frac{1}{s+2}\)
= \(\frac{s^2-2}{s(s^2-4)}\).

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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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