This set of Fourier Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Fourier Transform and Convolution”.
1. In Fourier transform \(f(p) = \int_{-∞}^∞ e^{(ipx)} F(x)dx, e^{(ipx)} \) is said to be Kernel function.
a) True
b) False
View Answer
Explanation: In any transform, apart from function given, the other function is said to be Kernel function. So, here in Fourier transform, e(ipx) is said to be the Kernel function.
2. Fourier Transform of \(e^{-|x|} \, is \) \( \frac{2}{1+p^2} \). Then what is the fourier transform of \( e^{-2|x|} \)?
a) \(\frac{4}{(4+p^2)} \)
b) \(\frac{2}{(4+p^2)} \)
c) \(\frac{2}{(2+p^2)} \)
d) \(\frac{4}{(2+p^2)} \)
View Answer
Explanation: \(e^{-2|x|}= e^{-|2x|}= F(2x) \)
\(F\{F(2x)\} = \frac{1}{2} f(\frac{p}{2}) \)
\( = \frac{1}{2} \frac{2}{(1+\frac{p^2}{4})} \)
\( = \frac{4}{(4+p^2 )}.\)
3. What is the fourier sine transform of e-ax?
a) \(\frac{4}{(4+p^2)} \)
b) \(4 \frac{a}{(4a^2+p^2 )} \)
c) \(\frac{p}{(a^2+p^2)} \)
d) \(2 \frac{p}{(a^2+p^2)} \)
View Answer
Explanation: Fourier sine transform of \(F(x) = \int_0^∞ e^{-ax} sin(px)dx \)
\( = \frac{e^{-ax}}{(a^2+p^2 )} (-asin(px)-pcos(px)) \) from 0 to ∞
\( = \frac{p}{(a^2+p^2)} \).
4. Find the fourier sine transform of \( \frac{x}{(a^2+x^2)}. \)
a) \(2πe^{-ap} \)
b) \(\frac{π}{2} e^{-ap} \)
c) \(\frac{2}{π} e^{-ap} \)
d) \(πe^{-ap} \)
View Answer
Explanation: Fourier transform of \( e^{-ax} \, is \, \frac{p}{(a^2+p^2)} \)
Substitute x=m and p=x.
\(\frac{π}{2} e^{-am}= \int_0^∞ \frac{x}{x^2+a^2} sin(mx)dx \)
Therefore, fourier sine transform of \(\frac{x}{(a^2+x^2)} \, is \, \frac{π}{2} e^{-ap}.\)
5. Find the fourier transform of F(x) = 1, |x|<a0, otherwise.
a) \(2sin \frac{(ap)}{p} \)
b) \(2asin \frac{(ap)}{p} \)
c) \(4sin \frac{(ap)}{p} \)
d) \(4asin \frac{(ap)}{p} \)
View Answer
Explanation: \(f(p) = \int_{-a}^a e^{ipx} dx \)
\( = \frac{e^{ipx}}{ip} \) from -a to a
\( = \frac{e^{iap}-e^{-iap}}{ip} \)
\( = 2sin \frac{(ap)}{p} \).
6. In Finite Fourier Cosine Transform, if the upper limit l = π, then its inverse is given by ________
a) \(F(x) = \frac{2}{π} ∑_{p=1}^∞ fc (p)cos(px)+ \frac{1}{π} fc(0) \)
b) \(F(x) = \frac{2}{π} ∑_{p=1}^∞ fc (p)cos(px) \)
c) \(F(x) = \frac{2}{π} ∑_{p=1}^∞ fc (p)cos(\frac{px}{π}) \)
d) \(F(x) = \frac{2}{π} ∑_{p=0}^∞ fc (p)cos(px)+ \frac{1}{π} fc(0) \)
View Answer
Explanation: Now since we have fourier cosine transform, we have to use the constant \(\frac{2}{π}\). And since while writing as sum of series it also has a term if p=0. Hence, \(F(x) = \frac{2}{π} ∑_{p=1}^∞ fc (p)cos(px)+ \frac{1}{π} fc(0) \)
7. Find the Fourier Cosine Transform of F(x) = 2x for 0<x<4.
a) \(fc(p) = \frac{32}{(p^2 π^2)} (cos(pπ)-1)p \) not equal to 0 and if equal to 0 \( fc(p) = 16 \)
b) \(fc(p) = \frac{32}{(p^2 π^2)} (cos(pπ)-1)p \) not equal to 0 and if equal to 0 \( fc(p) = 32 \)
c) \(fc(p) = \frac{64}{(pπ^2)} (cos(pπ)-1)p \) not equal to 0 and if equal to 0 \( fc(p) = 16 \)
d) \(fc(p) = \frac{32}{(pπ^2)} (cos(pπ)-1)p \) not equal to 0 and if equal to 0 \( fc(p) = 64 \)
View Answer
Explanation: \(fc(p) = \int_0^4 2 xcos(\frac{pπx}{4})dx \)
\( = 2\bigg[\frac{4xsin(\frac{pπx}{4})}{pπ} + \frac{16cos(\frac{pπx}{4})}{p^2 π^2}\bigg] \) from 0 to 4
\( = \frac{32}{(p^2 π^2)} (cos(pπ)-1) \)
When \(p=0, fc(p) = \int_0^4 2 xdx = 16. \)
8. If Fourier transform of \( e^{-|x|} = \frac{2}{1+p^2} \), then find the fourier transform of \(t^2 e^{-|x|}. \)
a) \(\frac{4}{1+p^2} \)
b) \(\frac{-2}{1+p^2} \)
c) \(\frac{2}{1+p^2} \)
d) \(\frac{-4}{1+p^2} \)
View Answer
Explanation: \(F\{e^{-|x|}\} = \frac{2}{1+p^2} \)
\(F\{t^2 e^{-|x|}\} = (-i)^2 \frac{2}{1+p^2} = \frac{-2}{1+p^2} \).
9. If \(Fs\{e^{-ax}\} = \frac{p}{a^2+p^2}\), find the \(Fs\{-a \, e^{-ax}\}.\)
a) \(4 \frac{p}{a^2+p^2} \)
b) \(\frac{-p^2}{a^2+p^2} \)
c) \(4 \frac{p^2}{a^2+p^2} \)
d) \(\frac{p}{a^2+p^2} \)
View Answer
Explanation:\(-a \, e^{-ax} = \frac{d}{dx}(e^{-ax}) = F’(x) \)
\(Fs\{F’(x)\} = -pfc(p) \)
\( = \frac{-p^2}{a^2+p^2} \).
10. Find the fourier transform of \(\frac{∂^2 u}{∂x^2}\) . (u’(p,t) denotes the fourier transform of u(x,t)).
a) (ip)2 u’(p,t)
b) (-ip)2 u’(p,t)
c) (-ip)2 u(p,t)
d) (ip)2 u(p,t)
View Answer
Explanation: \( F\{\frac{∂^2 u}{∂x^2}\} = \int_{-∞}^∞ \frac{∂^2 u}{∂x^2} e^{ipx} dx \)
\( = e^{ipx} \frac{∂u}{∂x} \) from (-infinity to infinity) \(– \int_{-∞}^∞ ip \, e^{ipx} u \)
\( = (ip)^2u’(p,t) \)
11. What is the fourier transform of e-a|x| * e-b|x|?
a) \(\frac{4ab}{(a^2+p^2)(b^2+p^2)} \)
b) \(\frac{2ab}{(a^2+p^2)(b^2+p^2)} \)
c) \(\frac{4}{(a^2+p^2)(b^2+p^2)} \)
d) \(\frac{a^2 b^2}{(a^2+p^2)(b^2+p^2)} \)
View Answer
Explanation: Fourier transform of \( e^{-a|x|} = \frac{2a}{a^2+p^2} \)
Fourier transform of \(e^{-b|x|} = \frac{2b}{b^2+p^2} \)
fourier transform of \(e^{-a|x|}* e^{-b|x|}= \frac{2a}{a^2+p^2}.\frac{2b}{b^2+p^2} \)
\( = \frac{4ab}{(a^2+p^2)(b^2+p^2)} \).
12. What is the Fourier transform of eax? (a>0)
a) \(\frac{p}{a^2+p^2} \)
b) \(2 \frac{a}{a^2+p^2} \)
c) \(-2 \frac{a}{a^2+p^2} \)
d) cant’t be found
View Answer
Explanation: Fourier transform of eax, does not exist because the function does not converge. The function is divergent.
13. \(F(x) = x^{(\frac{-1}{2})} \)is self reciprocal under Fourier cosine transform.
a) True
b) False
View Answer
Explanation: \(Fc \{x^{(\frac{-1}{2})}\} = \int_0^∞ x^{(\frac{-1}{2})} cos(px)dx = constant * p^{(\frac{-1}{2})} \)
Inverse fourier transform of \( p^{(\frac{-1}{2})} = constant * x^{(\frac{-1}{2})} \)
Hence the function \(x^{(\frac{-1}{2})} \)is self reciprocal.
14. Find the fourier sine transform of of e-ax * e-ax.
a) \(\frac{p^2}{a^2+p^2} \)
b) \(\frac{p^2}{(a^2+p^2)^2} \)
c) \(4 \frac{p^2}{(a^2+p^2)^2} \)
d) \(\frac{-p^2}{(a^2+p^2 )^2} \)
View Answer
Explanation: Fourier sine transform of \(e^{-ax} = \frac{p}{a^2+p^2}\)
fourier sine transform of \(e^{-ax}* e^{-ax} = \frac{p}{a^2+p^2} . \frac{p}{a^2+p^2} \)
\( = \frac{p^2}{(a^2+p^2 )^2} \).
15. Find the fourier sine transform of F(x) = -x when x<c and (π – x) when x>c and 0≤c≤π.
a) \(\frac{π}{c} cos(pc) \)
b) \(\frac{π}{p} cos(pc) \)
c) \(\frac{π}{c} cos(pπ) \)
d) \(p \frac{π}{c} cos(pc) \)
View Answer
Explanation: \(fs(p) = – \int_0^c x sin(px)dx + \int_c^π (π-x) sin(px)dx \)
\( = \frac{π}{p} cos(pc). \)
Sanfoundry Global Education & Learning Series – Fourier Analysis.
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