What should be FFT of a constant function?

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?

[Azzi Abdelmalek]

What do you mean by AC coefts?

[Image Analyst]

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.

Answer 1

[Walter Roberson]

Round-off error in the calculations. Look at the magnitudes: everything is down near 10E-15

[Image Analyst]

Theoretically because the function (array) is also of finite length, the spectrum will be multiplied by a sinc function, though for an array 172 long the sinc will be pretty narrow - practically a delta function - so the effect of it will be miniscule. You'd have very small values multiplying more very small values so it's essentially zero. But it's good to understand the theory regardless.

[Muthu Annamalai]

You have the details right, Image Analyst; while the rest of the comments are only partly true.

[Image Analyst]

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

Answer 2

[Azzi Abdelmalek]

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)))

[Richa Sharma]

Thank You very much for your response. Actually, my inputs are digital images.I was trying to check the fft response for constant background in images. I am going for 1D FFT in matlab. Say, my image is I and I write J=FFT(I) Then,for constant background in I, first coft. in columns of J should have high coft. value, the rest should be zeros is what I was expecting.But I was getting nonzero cofts for other frequency bins also.I think it is due to Round-off error in the calculations as Roberson said.Kindly correct me if I am wrong. Regards