What should be FFT of a constant function?
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As per theory,FFT of a constant fn. is a DC value. But when I take an array 'A' of ones of size 172, i.e. A[1 1 1 1 1.....172 times]
FFT(A)
gives DC as well as AC cofts. Why?
2 Commenti
Image Analyst
il 11 Dic 2012
The FFT breaks up the signal into a weighted sum of sinusoidal signals. Any point in an FFT not at the center is the weight (coefficient) of one of the sinusoidal signals that goes into making up the final signal. The center frequency is flat (no sinusoid) and is often called the DC component.
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Walter Roberson
il 11 Dic 2012
Round-off error in the calculations. Look at the magnitudes: everything is down near 10E-15
3 Commenti
Muthu Annamalai
il 11 Dic 2012
You have the details right, Image Analyst; while the rest of the comments are only partly true.
Image Analyst
il 11 Dic 2012
If you're interested in seeing the sinc effect, I posted some nice demo code here: http://www.mathworks.com/matlabcentral/answers/56139#comment_116309
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Azzi Abdelmalek
il 11 Dic 2012
Modificato: Azzi Abdelmalek
il 11 Dic 2012
If you mean by AC, sinusoidal signal, when you calculate its FFT, it's important to define the interval. To give a sens to your FFT, you have to calculate it in one period. For the sinusoidal signal, the period is 2*pi, then theoretically, the result will be one value at k=1 and not at k=0 like in a constant signal.
t=0:.1:2*pi-0.1;
g=fft(sin(t));
stem(abs(g))
For a constant the result will be a constant for k=0; and 0 elsewhere
h=ones(1,length(t))
figure;
stem(abs(fft(h)))
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