fft normalization and parseval

Question

yasser 2014-5-25

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/131053-fft-normalization-and-parseval

回答： yasser 2014-5-26

    h=complex(randi([-1 1],128,1),randi([-1 1],128,1));
    h=h/std(h); % no h is normalized to unit pow thus var(h)=1
    t=128*ifft(h);
    k=var(t);
    f=fft(t)/128
    g=var(f);

my problem is that k(power in time ) not = g(power in freq) also i have done normalizatoin so that parseval therom applies

any help please

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Answer 1

Matt J 2014-5-25

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https://ww2.mathworks.cn/matlabcentral/answers/131053-fft-normalization-and-parseval#answer_138189

编辑：Matt J 2014-5-25

在 MATLAB Online 中打开

also i have done normalizatoin so that parseval therom applies

No, you haven't, I'm afraid. For a length-N fft, the proper normalization is

   Y=fft(X)/sqrt(N);
   XX=sqrt(N)*ifft(Y);

For example,

>> X=rand(1,10);
>> Y=fft(X)/sqrt(10); XX= sqrt(10)*ifft(Y); 
>> norm(X), norm(Y), norm(XX)
ans =
      2.3117
ans =
      2.3117
ans =
      2.3117

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Matt J 2014-5-25

yasser Commented

@Matt J thanks but please provide a ref for your normalization factor as i saw lots of codes doing as i did without the sqrt issue

Matt J 2014-5-25

编辑：Matt J 2014-5-25

As you can see from the formula in the FFT documentation

the formula for the transform that MATLAB uses is non-orthogonal by a factor of sqrt(N). Normalizing by N and 1/N is what is needed when using FFTs to compute Fourier Series coefficients, see the formulas here

http://en.wikipedia.org/wiki/Fourier_series#Approximation_and_convergence_of_Fourier_series

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Answer 2

George Papazafeiropoulos 2014-5-25

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/131053-fft-normalization-and-parseval#answer_138183

在 MATLAB Online 中打开

In the following code:

var1=128;
h=complex(randi([-1 1],var1,1),randi([-1 1],var1,1));
h=h/std(h); % no h is normalized to unit pow thus var(h)=1
t=var1*ifft(h);
k=var(t)
f=fft(t)/var1;
g=var(f)

you have specified var1=128. Try to increase var from 128 to larger values. The two results will eventually converge.