This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on “Properties of Fourier Transforms”.
1. The Fourier transform of a function x(t) is X(ω). What will be the Fourier transform of \(\frac{dX(t)}{dt}\)?
a) \(\frac{X(f)}{jf}\)
b) j2πfX(f)
c) \(\frac{dX(f)}{dt}\)
d) jfX(f)
View Answer
Explanation: We know that x(t) = \(\frac{1}{2π} \int_{-∞}^∞ X(ω) e^{jωt} \,dω\)
\( \frac{d}{dt} \,x(t) = \frac{1}{2π} \int_{-∞}^∞ X(ω) \frac{d}{dt} e^{jωt} \,dω = \frac{1}{2π} jω X(ω) \int_{-∞}^∞ e^{jωt} \,dω\)
= jω X(ω) = j2πfX(f).
2. Find the Fourier transform of \(\frac{j}{πt}\).
a) sinc(ω)
b) sa(ω)
c) δ(ω)
d) sgn(ω)
View Answer
Explanation: Let x(t) = sgn(t)
The Fourier transform of sgn(t) is X(ω) = F[sgn(t)] = \(\frac{2}{jω}\)
Replacing ω with t
–> X(t) = \(\frac{2}{jt}\)
As per duality property X(t) ↔ 2πx(-ω), we have
F\(\Big[\frac{2}{jt}\Big]\) = 2πsgn(-ω) = -2πsgn(ω)
\(\frac{2}{jt}\) ↔ -2πsgn(ω)
\(\frac{2}{πt}\) ↔ sgn(ω).
3. The Fourier transform of a Gaussian pulse is also a Gaussian pulse.
a) True
b) False
View Answer
Explanation: Gaussian pulse, x(t) = e-πt2
Its Fourier transform is X(f) = e-πf2
Hence, the Fourier transform of a Gaussian pulse is also a Gaussian pulse.
4. Find the Fourier transform of f(t)=te-at u(t).
a) \(\frac{1}{(a-jω)^2} \)
b) \(\frac{1}{(a+jω)^2} \)
c) \(\frac{a}{(a-jω)^2} \)
d) \(\frac{ω}{(a-jω)^2} \)
View Answer
Explanation: Using frequency differentiation property, \(tx(t) \leftrightarrow j \frac{d}{dω} \,X(ω)\)
\(F[te^{-at} u(t)] = j \frac{d}{dω} F[te^{-at} \,u(t)] = j \frac{d}{dω} \frac{1}{a+jω} = j \frac{-1(j)}{(a+jω)^2} = \frac{1}{(a+jω)^2} \)
\(te^{-at} \,u(t) \leftrightarrow \frac{1}{(a+jω)^2} \).
5. Find the Fourier transform of ejω0t.
a) δ(ω + ω0)
b) 2πδ(ω + ω0)
c) δ(ω – ω0)
d) 2πδ(ω – ω0)
View Answer
Explanation: We know that F[1] = 2πδ(ω)
By using the frequency shifting property, ejω0t x(t) ↔ X(ω – ω0)
We have F[ejω0t] = F[ejω0t (1)] = 2πδ(ω – ω0).
6. Find the Fourier transform of u(-t).
a) πδ(ω) + \(\frac{1}{ω}\)
b) πδ(ω) + \(\frac{1}{jω}\)
c) πδ(ω) – \(\frac{1}{jω}\)
d) δ(ω) + \(\frac{1}{jω}\)
View Answer
Explanation: We know that F[u(t)] = πδ(ω) + \(\frac{1}{jω}\)
Using time reversal property, x(-t) ↔ X(-ω)
We have F[u(-t)] = πδ(ω) – \(\frac{1}{jω}\).
7. Find the Fourier transform of x(t) = f(t – 2) + f(t + 2).
a) 2F(ω)cos2ω
b) F(ω)cos2ω
c) 2F(ω)sin2ω
d) F(ω)sin2ω
View Answer
Explanation: Using linearity property, ax(t) + by(t) ↔ aX(ω) + bY(ω) and
Time shifting property, x(t-t0) ↔ e-jω0t X(ω),
We have F[x(t)] = F[f(t)] e-j2ω + F[f(t)] ej2ω = F(ω)e-j2ω + F(ω)ej2ω = 2F(ω)cos2ω.
8. Find the Fourier transform of \(\frac{1}{a+jt}\).
a) 2πeaω u(ω)
b) 2πeaω u(-ω)
c) 2πe-aω u(ω)
d) 2πe-aω u(-ω)
View Answer
Explanation: Let X(t) = \(\frac{1}{a+jt}\)
Replacing t with ω
X(ω) = \(\frac{1}{a+jw}\)
x(t )= e-at u(t)
As per duality property X(t) ↔ 2πx(-ω), we have
\(F[X(t)] = F\Big[\frac{1}{a+jt}\Big]\) = 2πx(-ω) = 2πeaω u(-ω).
9. Find the Fourier transform of e-2t u(t-1).
a) \(e^{-2} [e^{-jω} \frac{1}{2-jω}]\)
b) \(e^2 [e^{-jω} \frac{1}{2-jω}]\)
c) \(e^{-2} [e^{jω} \frac{1}{2-jω}]\)
d) \(e^{-2} [e^{-jω} \frac{1}{2+jω}]\)
View Answer
Explanation: We know that e-at u(t) ↔ \(\frac{1}{a+jw}\)
Using time shifting property, x(t-t0) ↔ e-jω0t X(ω) we have
f[e-2t u(t-1)] = \(e^{-2} [e^{-jω} \frac{1}{2+jω}]\).
10. Find the Fourier transform of sinc(t).
a) Gπ (ω)
b) G2π (ω)
c) \(G_{\frac{π}{2}}\) (ω)
d) Gπ (-ω)
View Answer
Explanation: Using duality property, X(t) ↔ 2πx(-ω)
We get sinc(t) ↔ G2π (ω).
11. If the Fourier transform of g(t) is G(ω), then match the following and choose the right answer.
(i) The Fourier transform of g(t-2) is (A) G(ω)e^-j2ω
(ii) The Fourier transform of g(t/2) is (B) G(2ω)
(C) 2G(2ω)
(D) G(ω-2)a) (i)-B, (ii)-A
b) (i)-A, (ii)-C
c) (i)-D, (ii)-C
d) (i)-C, (ii)-A
View Answer
Explanation: Using time shifting property, x(t – t0) ↔ e-jω0 t X(ω)
g(t – 2) ↔ e-j2ω G(ω)
Time scaling property, x(at) ↔ \( \frac{1}{a} X(\frac{w}{a})\)
g(t/2) ↔ 2G(2ω).
Sanfoundry Global Education & Learning Series – Signals & Systems.
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