Fourier Analysis Multiple Choice Questions 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
Answer: a
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
Answer: a
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
Answer: c
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
Answer: b
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
Answer: a
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
Answer: a
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
Answer: a
Explanation: (fc(p) = int_0^4 2 xcos(frac{pπx}{4})dx )
( = 2bigg[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
Answer: b
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 (Fc{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
Answer: b
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
Answer: a
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
Answer: a
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
Answer: d
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
Answer: a
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 cosine transform 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
Answer: b
Explanation = fourier cosine transform of (e^{-ax} = frac{p}{a^2+p^2})
fourier cosine 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 xc 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
Answer: b
Explanation: (fs(p) = – int_0^c x sin(px)dx + int_c^π (π-x) sin(px)dx )
( = frac{π}{p} cos(pc). )
Global Education & Learning Series – Fourier Analysis.
To practice all areas of Fourier Analysis, here is complete set of 1000+ Multiple Choice Questions and Answers.