Hi Matt,
I made one minor change at the beginning of your code, replacing the first four lines with
M = 100;
m = linspace(0,1,M)-1/2;
f = 2;
x = sin(2*pi*f*m);
This does the same thing, but putting the 2 pi into the sine argument is more consistent with common practice and with the rest of the code. And it's useful later.
Since you have zerofilled by a factor of 100, frequency f = 2 appears at point 200 in the FFT array. Plus one more point since Matlab is one-based and f=0 is at point 1. So, point 201 is the goal.
The problem is that the oscillations are for sine, but you are looking at the abs amplitudes of the positive frequency components and picking the largest one of those. Writing the continuous-m integral expression for simplicity [ and using f0 in place of f ] , the fourier amplitude for frequency fn is
an = Integral (1/(2*pi)) * ...
[ exp(2*pi*i*f0*m) - exp(-*pi*i*f0*m) ] * exp(-2*pi*i*fn*m) dm
The first term does have max abs value at fn = f0, just like you expect. However, the second term also makes a contribution, and you are finding max abs of the sum of the two terms. That peaks out at the wrong spot, point 195.
Replacing the signal with
x = exp(2*pi*i*f*m)
i.e. positive frequency only, gives the correct answer. Except not quite. Now it peaks out at point 203. The problem is that the linspace array does not represent the periodic wave correctly, because it includes both end points. But if you go to
M = 100;
m = ((0:M-1)/M)-1/2;
f = 2;
x = exp(2*pi*i*f*m);
which includes only the first end point, then it all works out. The max is at point 201. The max of the chirp transform is at point 2001, which sounds right. The waves in the second plot all align perfectly.
