What should be FFT of a constant function?
17 views (last 30 days)
Show older comments
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 Comments
Image Analyst
on 11 Dec 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.
Accepted Answer
Walter Roberson
on 11 Dec 2012
Round-off error in the calculations. Look at the magnitudes: everything is down near 10E-15
3 Comments
Muthu Annamalai
on 11 Dec 2012
You have the details right, Image Analyst; while the rest of the comments are only partly true.
Image Analyst
on 11 Dec 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
More Answers (1)
Azzi Abdelmalek
on 11 Dec 2012
Edited: Azzi Abdelmalek
on 11 Dec 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)))
See Also
Categories
Find more on Multirate Signal Processing in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!