fft normalization and parseval
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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
采纳的回答
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
2 个评论
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
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
更多回答（2 个）
George Papazafeiropoulos
2014-5-25
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.
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