Re: Inverse Fourier Transform by Residue Theorem and Contour Integration
- From: "rancid moth" <rancidmoth@xxxxxxxxx>
- Date: Thu, 7 May 2009 12:25:17 +1000
"neonstarfish" <oliver.bonner@xxxxxxxxx> wrote in message
news:12725494.60028.1241552002836.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxxxxx
Q) Use the Residue theorem and contour integration to show that the
inverse Fourier transform of the function
F(w) = (1/Sqrt[2*pi]) * (1+jw) / (1+w^2)
is given by the function
f(x)= e^x for x<0; 1/2 for x=0; 0 for otherwise
-----------------------
My current strategy is to integrate about a closed contour around the pole
at w=-j, which would then equal 2*pi*j times the residue at that point
(e^x). This would then equal the integral along the real axis between -R
and R, plus the integral along the open path given by the semi circle
under the real axis which encircles the pole at -j. I would then
rearrange to solve for the desired integral.
My question is, am I going about this the right way? I've attempted to
follow the method described above, but it leads to what seems to be a dead
end (I get a horrific integral with exponentials of exponentials)...
Any help would be much appreciated!
As Wade pointed out, use the fact that exp(ixw) = exp(ixrcos(t) - xrsin(t))
and note that x*r*sin(t) is positive for 0<t<pi. so if x>0
exp(-xrsin(t)) ->0 for 0<t<pi. this is why you close in the upper half
plane. but that only works when x>0. for x<0 you observe the sign change
and so need to close in the lower half plane in order to make
exp(-x*r*sin(t))-->0 (because x<0, r>0, sin(t) needs therefore to be <0).
looks like you already have the residues worked out, so now you can see that
for x>0 f(x)=0, and x<0, f(x) = exp(x)
....so for x=0...
hint: 1/2(f(x+0) + f(x-0)) = 1/(2*pi) lim (L->oo) int(-L,L) exp(-i*x*u)du
int(-oo,oo) f(t)*exp(i*u*t) dt
....which, if you are doing fourier transforms, you already know.
.
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