IFFT operation on the vector
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Hello,
I want to make an ifft operation on some vector. according to help of ifft, it do not require any multipcation constant. does this statement true all the time?
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Wayne King
il 21 Feb 2013
Modificato: Wayne King
il 21 Feb 2013
As long as you do not zero pad, you should get back the original vector without using any scaling factor.
x = randn(10,1);
xdft = fft(x);
y = ifft(xdft);
max(abs(x-y))
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Più risposte (1)
Michael Adelman
il 21 Feb 2013
1 Commento
Wayne King
il 21 Feb 2013
Nothing, it will just add zeros at the end.
x = randn(10,1);
xdft = fft(x,16);
ifft(xdft)
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