This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on “Inverse Fourier Transform”.
1. Find the inverse Fourier transform of X(ω) = e-2ω u(ω).
a) \(\frac{1}{2π(2+jt)}\)
b) \(\frac{1}{2π(2-jt)}\)
c) \(\frac{1}{2(2+jt)}\)
d) \(\frac{1}{π(2+jt)}\)
View Answer
Explanation: We know that x(t) = \(\frac{1}{2π} \int_{-∞}^∞ X(ω) e^{jωt} \,dω\)
x(t) = \(\frac{1}{2π} \int_{-∞}^∞ e^{-2ω} \,u(ω) e^{jωt} \,dω = \frac{1}{2π} \int_{-∞}^∞ e^{-2ω} e^{jωt} \, dω = \frac{1}{2π(2-jt)}\).
2. Find the inverse Fourier transform of X(ω) = \(\frac{1+3(jω)}{(3+jω)^2}\).
a) 3e-3t u(t) + 8e-3t u(t)
b) 3te-3t u(t) – 8e-8t u(t)
c) 3e-3t u(t) + 8te8t u(t)
d) 3e-3t u(t) – 8te-3t u(t)
View Answer
Explanation: Given X(ω) = \(\frac{1+3(jω)}{(3+jω)^2} = \frac{A}{3+jω} + \frac{B}{(3+jω)^2} = \frac{3}{3+jω} – \frac{8}{(3+jω)^2}\)
Applying inverse Fourier transform, we get
x(t) = 3e-3t u(t) – 8te-3t u(t).
3. Find the inverse Fourier transform of δ(ω).
a) \(\frac{1}{2π}\)
b) 2π
c) \(\frac{1}{π}\)
d) π
View Answer
Explanation: We know that x(t) = \(\frac{1}{2π} \int_{-∞}^∞ X(ω) e^{jωt} \,dω\)
= \(\frac{1}{2π} \int_{-∞}^∞ δ(ω) e^{jωt} \,dω = \frac{1}{2π}\).
4. Find the inverse Fourier transform of u(ω).
a) \(\frac{1}{2} δ(t) + \frac{j}{2πt}\)
b) \(\frac{1}{2} δ(t) – \frac{j}{2πt}\)
c) δ(t) + \(\frac{j}{2πt}\)
d) δ(t) – \(\frac{j}{2πt}\)
View Answer
Explanation: We know that u(ω) = \(\frac{1}{2}\)[1+sgn(ω)].
Applying linearity property,
u(ω) = -1 \([\frac{1}{2}] + F^{-1} [\frac{1}{2} sgn(ω)]\)
u(ω) = \(\frac{1}{2} δ(t) + \frac{j}{2πt}\).
5. Find the inverse Fourier transform of ej2t.
a) 2πδ(ω-2)
b) πδ(ω-2)
c) πδ(ω+2)
d) 2πδ(ω+2)
View Answer
Explanation: We know that ejω0 t ↔ 2πδ(ω-ω0)
∴ ej2t ↔ 2πδ(ω-2).
6. Find the inverse Fourier transform of jω.
a) δ(t)
b) \(\frac{d}{dt}\) δ(t)
c) \(\frac{1}{δ(t)}\)
d) ∫δ(t)
View Answer
Explanation: Time differentiation property, \(\frac{d}{dt}\) x(t) ↔ jωX(ω) and we know that δ(t) ↔ 1
∴ \(\frac{d}{dt}\) δ(t) ↔ jω.
7. Find the inverse Fourier transform of \(X(ω) = \frac{6+4(jω)}{(jω)^2 + 6(jω) + 8}\).
a) e-2t u(t) – 5e-4t u(t)
b) e-2t u(t) + 5e-4t u(t)
c) -e-2t u(t) – 5e-4t u(t)
d) -e-2t u(t) + 5e-4t u(t)
View Answer
Explanation: \(X(ω) = \frac{6+4(jω)}{(jω)^2+6(jω)+8} = \frac{A}{jω+2} + \frac{B}{jω+4} = -\frac{1}{jω+2} + \frac{5}{jω+4}\)
Applying inverse Fourier transform, we get
x(t) = -e-2t u(t) + 5e-4t u(t).
8. Find the convolution of the signals x1 (t) = e-2t u(t) and x2 (t) = e-3t u(t).
a) e-2t u(t) – e-3t u(t)
b) e-2t u(t) + e-3t u(t)
c) e2t u(t) – e3t u(t)
d) e2t u(t) – e-3t u(t)
View Answer
Explanation: Convolution property, x1 (t)*x2 (t) ↔ X1 (ω) X2 (ω)
∴ x1 (t)*x2 (t) = F-1 [X1 (ω) X2 (ω)]
Given x1 (t) = e-2t u(t)
∴ X1 (ω) = \(\frac{1}{jω+2}\)
Given x2 (t) = e-3t u(t)
∴ X1 (ω) = \(\frac{1}{jω+3}\)
x1 (t)*x2 (t) = F-1 [X1 (ω) X2 (ω)] = F-1 \([\frac{1}{jω+2} \frac{1}{jω+3}] = F^{-1} [\frac{1}{jω+2} – \frac{1}{jω+3}] \)
∴ x1 (t)*x2 (t) = e-2t u(t)-e-3t u(t).
9. Find the inverse Fourier transform of f(t)=1.
a) u(t)
b) δ(t)
c) e-t
d) \(\frac{1}{jω}\)
View Answer
Explanation: We know that the Fourier transform of f(t) = 1 is F(ω) = 2πδ(ω).
Replacing ω with t
F(t) = 2πδ(t)
As per duality property F(t) ↔ 2πf(-ω), we have
2πδ(t) ↔ 2π(1)
δ(t) ↔ 1
Hence, the inverse Fourier transform of 1 is δ(t).
10. Find the inverse Fourier transform of sgn(ω).
a) \(\frac{1}{πt}\)
b) \(\frac{j}{πt}\)
c) \(\frac{j}{t}\)
d) \(\frac{1}{t}\)
View Answer
Explanation: Given the function F(ω)=sgn(ω). The Fourier transform of a Signum function is sgn(ω) = \(\frac{2}{jω}\).
Applying the duality property F(t) ↔ 2πf(-ω), we get
F(\(\frac{2}{jt}\)) = 2πsgn(-ω).
As sgn(ω) is an odd function, sgn(-ω)=-sgn(ω).
Hence, \(\frac{2}{jt}\) ↔ -sgn(ω)
Or \(\frac{2}{πt}\) ↔ sgn(ω)
Therefore, the inverse Fourier transform of sgn(ω) is \(\frac{j}{πt}\).
Sanfoundry Global Education & Learning Series – Signals & Systems.
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