ifft scale problem
in [Matlab]
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From: Frank on 9 Mar 2010 07:48 "Matt J " <mattjacREMOVE(a)THISieee.spam> wrote in message <hn3sik$5d1$1(a)fred.mathworks.com>... > "Frank " <francesco.manni01(a)fastwebnet.it> wrote in message <hn3qim$ste$1(a)fred.mathworks.com>... > > > PROBLEM: > > What I get from ifft is my time function (full of zeros at the end, that I would like to remove before plotting in the time domain) > ====================== > > Not sure why that's a difficult problem. Just throw away the zeros and the corresponding points on your axes. > > > > and also the ifft is not properly scaled, as I get 10 times the amplitude of my initial time signal. > > > > close all > > > > dx = 0.01; > > xmax = 4; > > x = 0:dx:xmax; > > y = sin(2*pi*8*x); %+ sin(2*pi*20*x); > > > > N = 2^12; > > ty = fft(y,N)/length(x); > ====== > > should be > > ty = fft(y,N)*dx; > > > > > > fs = 1/dx; > > f = (0:(N-1)/2)/N*fs; > > > > aty = real(ifft(ty,N)*N); > ================== > > Should be > > aty = abs(ifft(ty,N))/dx; Thank you for your help guys! Let me just point out that I don't see why I should take the abs of the ifft instead of the real part... I was taking the real part of the ifft just to get rid of some small complex contribution that may come from calculation roundings or by some frequency filtering I might introduce. It doesn't make much sense to take the abs of ifft if I expect to have a sine wave from ifft for example, since I will "transform" all the negative part and make it positive in the time domain. Maybe I missing something else? Kind regards Frank
From: Wayne King on 9 Mar 2010 08:52 "Frank " <francesco.manni01(a)fastwebnet.it> wrote in message <hn5g24$4m4$1(a)fred.mathworks.com>... > "Matt J " <mattjacREMOVE(a)THISieee.spam> wrote in message <hn3sik$5d1$1(a)fred.mathworks.com>... > > "Frank " <francesco.manni01(a)fastwebnet.it> wrote in message <hn3qim$ste$1(a)fred.mathworks.com>... > > > > > PROBLEM: > > > What I get from ifft is my time function (full of zeros at the end, that I would like to remove before plotting in the time domain) > > ====================== > > > > Not sure why that's a difficult problem. Just throw away the zeros and the corresponding points on your axes. > > > > > > > > and also the ifft is not properly scaled, as I get 10 times the amplitude of my initial time signal. > > > > > > close all > > > > > > dx = 0.01; > > > xmax = 4; > > > x = 0:dx:xmax; > > > y = sin(2*pi*8*x); %+ sin(2*pi*20*x); > > > > > > N = 2^12; > > > ty = fft(y,N)/length(x); > > ====== > > > > should be > > > > ty = fft(y,N)*dx; > > > > > > > > > > fs = 1/dx; > > > f = (0:(N-1)/2)/N*fs; > > > > > > aty = real(ifft(ty,N)*N); > > ================== > > > > Should be > > > > aty = abs(ifft(ty,N))/dx; > > Thank you for your help guys! Let me just point out that I don't see why I should take the abs of the ifft instead of the real part... I was taking the real part of the ifft just to get rid of some small complex contribution that may come from calculation roundings or by some frequency filtering I might introduce. It doesn't make much sense to take the abs of ifft if I expect to have a sine wave from ifft for example, since I will "transform" all the negative part and make it positive in the time domain. Maybe I missing something else? > > Kind regards > > Frank Hi Frank, if you are worried about getting some small nonzero imaginary parts due to rounding error even though your Fourier transform is conjugate symmetric, then I recommend using the 'symmetric' flag in the ifft() as I have done in a previous example. Hope that helps, Wayne
From: Matt J on 9 Mar 2010 12:18 "Frank " <francesco.manni01(a)fastwebnet.it> wrote in message <hn5g24$4m4$1(a)fred.mathworks.com>... > Thank you for your help guys! Let me just point out that I don't see why I should take the abs of the ifft instead of the real part... ======================== Forget it. You should use real() like before or the 'symmetric' option that Wayne was talking about. > I have indeed another associated question to ask you. When I had a dc component, let's say I take y = sin(...) + 5 and let the code run, I get an ifft that has a dc component different from 5... Why? =================== Is there any reason to think it's more than precision error? Are we talking about y(0)=5.00000000000000000000001
From: Frank on 10 Mar 2010 03:55 "Matt J " <mattjacREMOVE(a)THISieee.spam> wrote in message <hn5vsa$jgg$1(a)fred.mathworks.com>... > "Frank " <francesco.manni01(a)fastwebnet.it> wrote in message <hn5g24$4m4$1(a)fred.mathworks.com>... > > > Thank you for your help guys! Let me just point out that I don't see why I should take the abs of the ifft instead of the real part... > ======================== > > Forget it. You should use real() like before or the 'symmetric' option that Wayne was talking about. > > > > I have indeed another associated question to ask you. When I had a dc component, let's say I take y = sin(...) + 5 and let the code run, I get an ifft that has a dc component different from 5... Why? > =================== > > Is there any reason to think it's more than precision error? Are we talking about > > y(0)=5.00000000000000000000001 Thank you all once more. Indeed I have improved and get somehow a bit more awareness of what these matlab functions are doing and now I feel I am a bit closer to what I need ;) Anyway, the problem with the dc component was in some way associated with the improper scaling I was doing. Instead of 5 I was getting 2.3 or something like that, completely out of sense! Now it works fine... About the real() issue, I will need to put the 'symmetric' flag into ifft because I am creating some custom filter (simple stuff, with erf and other shapes) to apply to my signals and so, after the filtering, which i apply only on the positive part of the spectrum of the abs(fft(vector)), I want to be sure that it comes back to the time domain properly. Hope I have been clear enough!
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