Peaks in FFT transformation

Hi Kaustubh, excuse me for the time of the answer.

Do you want to know about the peaks in the FFT transformation (here), right?

Well, this blog is about Scilab, and you can find what you want in any book of Digital Signal Processing.

But God loves you and I'll try to explain what the peaks mean.

Think in a pure frequency signal x(t) like a sine or cosine, it has only the own frequency (f0). The FFT transform is two impulses (X(f) = d(-f0) + d(f0)) (similar to this figure)

If you plot the FFT transform of x(t), then you obtain the peaks in -f0 and f0.

Now, think in a signal composed by two pure frequency signals, y(t) = x1(t) + x2(t), of different frequencies (f1 and f2).

The FFT transform of y(t) is four impulses, two for x1(t) and two for x2(t).

If x1(t) and x2(t) have different amplitudes, for example: x1(t) = cos(2t) and x2(t) = 3 cos(5t), then the peaks of the FFT transform of y(t) are weighted,

Following the example, the FFT transform of x1(t) is X1(f) = d(-f1) + d(f1) and the FFT transformation of x2(t) is X2(t) = 3(d(-f2) + d(f2)). So, the FFT transform of y(t) = x1(t) + x2(t) is Y(f) = X1(f) + X2(f) = (d(-f1) + d(f1)) + 3(d(-f2) + d(f2)).


Now, if you have a real signal xr(t), it has many frequencies and it FFT transform XR(f) presents many peaks, one by each frequency. The peaks' amplitude represents the amplitude of each frequency in xr(t).

My friend, I said this not a blog about Signal Processing, but I tried to help you.
If you have more question, I'll try to help you again.