# 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) $$\frac{1}{j} \left(\frac{1}{1+j(ω+2π)} – \frac{1}{1+j(ω-2π)}\right)$$

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)

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

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.

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 – $$\frac{T}{2}$$)

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)

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

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

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.

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)}

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)

Explanation: Applying the time shifting and scaling property, we get,
X [3(f+2/3)] = $$\frac{1}{3} x(\frac{t}{3})e^{-j4πt/3}$$.

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

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 ____________

a) 0, 4/π
b) 4/π, 0
c) 0, -4/π
d) -4/π, 1

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)

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

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

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) $$\frac{s^2+2}{s(s^2-4)}$$

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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