This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on “Fourier Analysis”.
1. The CTFT of a continuous time signal x(t) = e-A|t|, A>0 is _________
a) \(\frac{2A}{ω^2} \)
b) \(\frac{A}{A^2+ω^2} \)
c) \(\frac{2A}{A^2+ω^2} \)
d) \(\frac{A}{ω^2} \)
View Answer
Explanation: CTFT {x (t)} = X (jω) = \(\int_{-∞}^∞ x(t) e^{-jωt} \,dt\)
= \(\int_{-∞}^∞ e^{-A|t|} e^{-jωt} \,dt\)
= \(\int_{-∞}^0 e^{-A(-t)} e^{-jωt} \,dt + \int_0^∞ e^{-At} e^{-jωt} \,dt\)
= \(\int_{-∞}^0 e^{At} e^{-jωt} \,dt + \int_0^∞ e^{-At} e^{-jωt} \,dt\)
= \(\int_{-∞}^0 e^{(A-jω)t} \,dt + ∫_0^∞ e^{-(A+jω)t} \,dt\)
= \([\frac{1}{A-jω} e^{(A-jω)t}]_{-∞}^0 + [\frac{1}{-(A+jω)} e^{-(A+jω)t}]_0^∞\)
= \([\frac{1}{A-jω} + \frac{1}{A+jω} = \frac{2A}{A^2+ω^2}]\)
∴ x(jω) = \(\frac{2A}{A^2+ω^2} \).
2. A signal x(t) has the Fourier transform X(jω) having the following facts:
F-1{(1+jω) X(jω)} = Ae-2t u(t) and \(\int_{-∞}^∞ |X(jω)|^2 \,dω = 2π\)
The signal x (t) is ___________
a) \(\sqrt{3}\) (e-t – e-2t)u(t)
b) \(\sqrt{12}\) (e-t – e-2t)u(t)
c) \(\sqrt{3}\) (e-2t – e-t)u(t)
d) \(\sqrt{12}\) (e-2t – e-t)u(t)
View Answer
Explanation: F-1{(1+jω) X(jω)} = Ae-2t u(t)
Or, (1+jω) X (jω) = \(\frac{A}{2+jω} \)
Or, X (jω) = \(\frac{A}{(1+jω)(2+jω)} \)
= \(A (\frac{1}{1+jω} – (\frac{1}{2+jω}) \)
Or, x (t) = Ae-tu (t) – Ae-2tu (t)
Given that, \(∫_{-∞}^∞ |X(jω)|^2 \,dω\) = 2π
Or, \(∫_{-∞}^∞ |x(t)|^2 \,dt\) = 1
∴ \(∫_{-∞}^∞ [A^2 e^{-2t} – 2A^2 e^{-3t} + A^2 e^{-4t}]\) u(t)dt = 1
Or, \(∫_{-∞}^∞ [A^2 e^{-2t} – 2A^2 e^{-3t} + A^2 e^{-4t}]\) dt = 1
Or, \(\frac{A^2}{12}\)=1
Or, A = \(\sqrt{12}\)
We chose \(\sqrt{12}\) and not –\(\sqrt{12}\) since x (t) is non-negative.
3. The system characterized by the differential equation \(\frac{d^2 y(t)}{t^2} – \frac{dy}{dt} – 2y(t) = x(t)\) is _____________
a) Linear and stable
b) Linear and unstable
c) Nonlinear and unstable
d) Nonlinear and stable
View Answer
Explanation: \(\frac{d^2 y(t)}{t^2} – \frac{dy}{dt} – 2y(t) = x(t)\)
Now, x (t) –> h (t) so, the system is linear.
Again, taking Laplace Transform with zero initial conditions, we get, s2 Y(s) – s Y(s) – 2Y(s) = X(s)
Or, H(s) = \(\frac{Y(s)}{X(s)} = \frac{1}{s^2-s-2} = \frac{1}{(s-2)(s+1)} \)
Since pole is at s = +2, the system is unstable.
4. An LTI system with impulse response h1 [n] = -2(\(\frac{1}{4})^n\) u[n] is connected in parallel with another causal LTI system with impulse response h2 [n]. The resulting interconnection has frequency response H (ejω) = \(\frac{-12+5e^{-jω}}{12+7e^{-jω}+e^{-j2ω}}\). Then h2[n] is ___________
a) (\(\frac{1}{3}\))n u[n-1]
b) (\(\frac{1}{3}\))n u[n]
c) (\(\frac{1}{4}\))n u[n]
d) (\(\frac{1}{4}\))n u[n-1]
View Answer
Explanation: H (ejω) = H1 (ejω) + H2 (ejω)
Or, \(\frac{-12+5e^{-jω}}{12+7e^{-jω}+e^{-j2ω}} = \frac{1}{1-\frac{1}{3} e^{-jω}} + \frac{(-2)}{1-\frac{1}{4} e^{-jω}}\)
∴ H2(ejω) = \(\frac{1}{1-\frac{1}{3} e^{-jω}}\)
So, h2 [n] = (\(\frac{1}{3}\))n u[n].
5. A pulse of unit amplitude and width a, is applied to a series RL circuit having R = 1 Ω, L = 1H. The current I(t) at t = ∞ is __________
a) 0
b) Infinite
c) 2 A
d) 1 A
View Answer
Explanation: Since, the circuit has a resistance; the current will eventually decrease and ultimately die down.
Now, I (t) = \((\frac{1}{R})[u (t) (1-e^{\frac{-Rt}{L}}) – u (t-a) (1-e^{\frac{-R(t-a)}{L}})]\)
But R = 1Ω and L = 1H
∴ I(t) = [u(t)(1-e-t) – u(t-a)(1-e-(t-a))]
This also leads to I (∞) = 0.
6. The rms value of a rectangular wave of period T, having value +V for a duration, T1(<T) and –V for the duration, T-T1 = T2 is __________
a) V
b) \(\sqrt{V}\)
c) \(\frac{\sqrt{V}}{2}\)
d) 0
View Answer
Explanation: \(\sqrt{V^2}\)n: Mean square value = \(\frac{1}{T}[∫_0^{T_1} V^2 \,dt + ∫_1^T V^2 \,dt] \)
Where, T1 + T2 = T
= \(\frac{1}{T}\)[V2 T1 + V2 T – V2 T1] = V2
∴ RMS = V.
7. X (ejω) = \(\frac{(b-a) e^{jω}}{e^{-j2ω}-(a+b) e^{jω} + ab)}\), |b|<1<|a|
The value of x[n] is __________
a) bn u [n] + an u [n-1]
b) bn u [n] – an u [-n-1]
c) bn u [n] + an u [-n-1]
d) bn u [n] – an u [n+1]
View Answer
Explanation: X (ejω) = \(\frac{(b-a) e^{jω}}{e^{-j2ω}-(a+b) e^{jω} + ab)}\)
= \(\frac{(b-a) e^{jω}}{1-(a+b) e^{-jω}} + ab \,e^{-j2ω})\)
= \(\frac{1}{1-be^{-jω}} + \frac{(-1)}{1-ae^{-jω}}\)
∴ x [n] = bn u [n] + an u [-n-1].
8. Frequency and time period are ____________
a) Proportional to each other
b) Inverse of each other
c) Same
d) equal
View Answer
Explanation: Frequency is the number of occurrences of a repeating event per unit time.
Time period, denoted by T, is the duration of one cycle and is equal to the reciprocal of the frequency. Mathematically we can write T = 1/f.
9. The Fourier series for the function f (x) = sin2x is ______________
a) 0.5 + 0.5 sin 2x
b) 0.5 – 0.5 sin 2x
c) 0.5 + 0.5 cos 2x
d) 0.5 – 0.5 cos 2x
View Answer
Explanation: f(x) = sin2x
Now, f(x) = sin2x = \(\frac{1-cos2x}{2}\)
= 0.5 – 0.5 cos 2x.
10. The continuous time system described by the equation y(t) = x(t2) comes under the category of ____________
a) Causal, linear and time varying
b) Causal, non-linear and time varying
c) Non-causal, non-linear and time invariant
d) Non-causal, linear and time variant
View Answer
Explanation: Let y (t) = x (t2). We can infer that y (t) depends on x (t2) i.e. on future values of input if t>1. Hence, the system is non-casual.
Again, α x1(t) → y1 (t) = α x1(t2) and β x2(t) –> y2 (t) = β x2(t2)
Therefore α x1(t) + β x2(t) –> y (t) = α x1(t2) + β x2(t2) = y1 (t) + y2 (t), which implies that the system is linear.
Again, x (t) = u (t) – u (t-z) –> y (t) and X1(t) = x (t – 1) –> y1 (t).
So, we get, y1 (t) ≠ y (t –1), which implies that the system is time varying.
11. The running integrator, given by y(t) = \(∫_{-∞}^∞ x(t) \,dt\) has ____________
a) No finite singularities in it’s double sided Laplace transform Y(s)
b) Produces an abounded output for every causal bounded input
c) Produces a bounded output for every anti-causal bounded input
d) Has no finite zeroes in it’s double sided Laplace transform Y (s)
View Answer
Explanation: The running integrator \(∫_{-∞}^t x(t)\,dt = 0\) for every causal system. As causal systems have no memory and the initial value is zero, the output is followed by input. So, y (t) will always be bounded if this function is a causal bounded system.
12. A signal x (t) is given by
x(t) = 1, -T/4<t≤3T/4
= -1, 3T/4<t≤7T/4
= -x (t+T)
Which among the following gives the fundamental Fourier terms of x (t)?
a) \(\frac{4}{π} cos(\frac{πt}{T} + \frac{π}{4})\)
b) \(\frac{4}{π} cos(\frac{πt}{T} – \frac{π}{4})\)
c) \(\frac{4}{π} sin(\frac{πt}{T} – \frac{π}{4})\)
d) \(\frac{4}{π} sin(\frac{πt}{T} + \frac{π}{4})\)
View Answer
Explanation: Given signal,
x(t) = 1, -T/4<t≤3T/4
= -1, 3T/4<t≤7T/4
= -x (t+T)
Now by property of symmetry of Fourier transform of x (f), we get the fundamental Fourier term as, \(\frac{4}{π} sin(\frac{πt}{T} – \frac{π}{4})\).
13. The type of systems which are characterized by input and the output capable of taking any value in a particular set of values are called as __________
a) Analog
b) Discrete
c) Digital
d) Continuous
View Answer
Explanation: We know that continuous systems have a restriction on the basis of the upper bound and lower bound. However within this set, the input and output can assume any value. Hence, there are infinite values attainable in this system.
14. In Maxwell’s capacitance bridge for calculating unknown inductance, the various values at balance are, R1 = 300 Ω, R2 = 700 Ω, R3 = 1500 Ω, C4 = 0.8 μF. Calculate R1, L1 and Q factor, if the frequency is 1100 Hz.
a) 240 Ω, 0.12 H, 3.14
b) 140 Ω, 0.168 H, 8.29
c) 140 Ω, 0.12 H, 5.92
d) 240 Ω, 0.36 H, 8.29
View Answer
Explanation: From Maxwell’s capacitance, we have
R1 = \(\frac{R_2 R_3}{R_4} = \frac{300 × 700}{1500}\) = 140 Ω
L1 = R2 R3 C4
= 300 × 700 × 0.8 × 10-6 = 0.168 H
∴ Q = \(\frac{ωL_1}{R_1}\).
15. Given a real valued function y (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, y(t) satisfies the equation ____________
a) y (t) = y (t+T) = -y (t+\(\frac{T}{2}\))
b) y (t) = y (t+T) = y (t+\(\frac{T}{2}\))
c) y (t) = y (t-T) = -y (t-\(\frac{T}{2}\))
d) y (t) = y (t-T) = y (t-\(\frac{T}{2}\))
View Answer
Explanation: For an even symmetry, y (t) = y (t-T)
Thus no sine component will exist because bn=0 and by half wave symmetry condition odd harmonics will exist.
Now, y (t) = y (t-\(\frac{T}{2}\))
Combining the two conditions, we get, y (t) = y (t-T) = y (t-\(\frac{T}{2}\)).
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