Inverse Fourier transforms
in [Matlab]
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From: Mat Hunt on 23 Jun 2010 06:58 Hi Greg, Still not really getting you. I have a very formal training in terms of Fourier analysis, it was a pure maths course I did in my third year at university. I think of a Fourier transform as a map essentially which is defined by: F(k)=int_{-infinity}^{+infinity}f(x)e^{-ikx}dx The inverse transform being defined as: f(x)=1/(2*pi)*\int_{-infinity}^{+infinity}F(k)e^{ikx}dk In my work I have calculated what F(k), I now need to find what f(x) is. You said that I need to do fftshift(ifft(F(k)) to get f, now I tried to read your post to figure out what I need to do but I couldn't quite get it. Can you explain again as if I were a simpleton :o) The thing with the residue calculus was just a way of testing if I got the correct answer or not. Mat
From: Greg Heath on 23 Jun 2010 13:38 On Jun 23, 6:58 am, "Mat Hunt" <hunt_...(a)hotmail.com> wrote: > Hi Greg, > > Still not really getting you. I have a very formal training in terms of Fourier analysis, it was a pure maths course I did in my third year at university. > > I think of a Fourier transform as a map essentially which is defined by: > > F(k)=int_{-infinity}^{+infinity}f(x)e^{-ikx}dx > > The inverse transform being defined as: > > f(x)=1/(2*pi)*\int_{-infinity}^{+infinity}F(k)e^{ikx}dk > > In my work I have calculated what F(k), I now need to find what f(x) is. > > You said that I need to do fftshift(ifft(F(k)) to get f, now I tried to read your post to figure out what I need to do but I couldn't quite get it. Can you explain again as if I were a simpleton :o) You said it. I didn't. ;>) I repeat: The MATLAB fft and ifft functions assume the nonnegative intervals x = dx*(0:N-1) and k = dk*(0:N-1) corresponding to the Fourier pair f(x) and F(k). If f and F are considered periodic, fftshift yields the translation to the "zero centered" bipolar intervals with indexing xb = dx * [-ceil((N-1)/2) : floor((N-1)/2)] kb = dk * [-ceil((N-1)/2) : floor((N-1)/2)] corresponding to the Fourier pair fb(xb) and Fb(kb). ifftshift is the inverse of fftshift. When N is even, fftshift and ifftshift yield the same result. However, when N is odd, x = ifftshift(xb) X = fft(x) Xb = fftshift(X) and to return X = ifftshift(Xb) x = ifft(X) xb = fftshift(x) The trick is to remember that fftshift will "center" the waveform about the coordinate zero. If in doubt type fftshift([-2:2]) into the command line. Hope this helps. Greg
From: Greg Heath on 23 Jun 2010 13:53 On Jun 23, 1:38 pm, Greg Heath <he...(a)alumni.brown.edu> wrote: > On Jun 23, 6:58 am, "Mat Hunt" <hunt_...(a)hotmail.com> wrote: > > > Hi Greg, > > > Still not really getting you. I have a very formal training in terms of Fourier analysis, it was a pure maths course I did in my third year at university. > > > I think of a Fourier transform as a map essentially which is defined by: > > > F(k)=int_{-infinity}^{+infinity}f(x)e^{-ikx}dx > > > The inverse transform being defined as: > > > f(x)=1/(2*pi)*\int_{-infinity}^{+infinity}F(k)e^{ikx}dk > > > In my work I have calculated what F(k), I now need to find what f(x) is.. > > > You said that I need to do fftshift(ifft(F(k)) to get f, now I tried to read your post to figure out what I need to do but I couldn't quite get it. Can you explain again as if I were a simpleton :o) > > You said it. I didn't. ;>) > > I repeat: > > The MATLAB fft and ifft functions assume the nonnegative intervals x = > dx*(0:N-1) and k = dk*(0:N-1) corresponding > to the Fourier pair f(x) and F(k). > > If f and F are considered periodic, fftshift yields the translation to > the "zero centered" bipolar intervals with > indexing > > xb = dx * [-ceil((N-1)/2) : floor((N-1)/2)] > kb = dk * [-ceil((N-1)/2) : floor((N-1)/2)] > > corresponding to the Fourier pair fb(xb) and Fb(kb). > > ifftshift is the inverse of fftshift. > > When N is even, fftshift and ifftshift yield > the same result. > > However, when N is odd, > > x = ifftshift(xb) > X = fft(x) > Xb = fftshift(X) > > and to return > > X = ifftshift(Xb) > x = ifft(X) > xb = fftshift(x) WHOOPS! Screwed up the notation. Correction; f = ifftshift(fb) F = fft(f) Fb = fftshift(F) and to return F = ifftshift(Fb) f = ifft(F) fb = fftshift(f) > The trick is to remember that fftshift will > "center" the waveform about the coordinate zero. > > If in doubt type fftshift([-2:2]) into the command > line. > Hope this helps. Greg
From: Mat Hunt on 24 Jun 2010 07:25 Hi Greg, I had a play with it and to be honest, I can't get the thing to work. I don't understand the need for something like the fftshift function. I expected (naively) that I would give the point at which I would want the function evaluated at. Looking at the definition of the Fourier transform, they have the ability to do this, yet they choose not to for some reason. I looked at what fftshift did to -2:2 and I have to say that I really didn't understand why it was doing that. I looked at the maximum and minimum of the vector I got after doing ifft and it didn't match up to how I got doing it another method. I think I am going to keep with my old method of computing the inverse Fourier transform as it gives sensible. I wanted to compare my method with matlab's inbuilt method but after all this messing around, it still doesn't work. I shall mention this to the developers as something that they might like to comsider. Thanks for your time. Mat
From: Mat Hunt on 24 Jun 2010 18:13
So can anyone give me a clear and simple methodology to go from a Fourier transformed function F(k), back to the original function at a point, x say? So this would be a single number? I would have thought the the people at matlab would have thought this one through. Mat |