How to the FFT analyze a wave in which the components are changing

1. Intuitively the DFT and FT are similar. As the other poster pointed out, the main difference is that sampling in the time domain is equivalent to repetition in the Frequency domain, so the DFT will have repeated spectra at multiples of the sample frequency.

2. Two problems here.

   i) abs(real(y)) is incorrectly being used. You need the absolute value (or magnitude) of both the imaginary and real components ... abs(y).

   ii) You're seeing the effect of spectral leakage because the DFT assumes a repetitive signal; i.e. the maths "assumes" the signal you applied is repeated infinitely. Your signal will have discontinuities in this case.

Try the following code:

t = [-10*pi:pi/10:9.9*pi]
x = sin(t)
y = fft(x)
z = abs(y)

You should see this:

This is a windowing artifact.

The linked code pads out a 10,000 sample signal with zeroes so that the length is a power of two.

%% Author :- Embedded Laboratory

%%This Project shows how to apply FFT on a signal and its physical 
% significance.

fSampling = 10000;          %Sampling Frequency
tSampling = 1/fSampling;    %Sampling Time
L = 10000;                  %Length of Signal
t = (0:L-1)*tSampling;      %Time Vector
F = 100;                    %Frequency of Signal

%% Signal Without Noise
xsig = sin(2*pi*F*t);
...

%%Frequency Transform of above Signal
subplot(2,1,2)
NFFT = 2^nextpow2(L);
Xsig = fft(xsig,NFFT)/L;
...

Note that in the above code, the FFT is taken with the FFT size NFFT which is the next power of 2 bigger than the signal length (in this case, 16,384.) From the Mathworks fft() documentation:

Y = fft(X,n) returns the n-point DFT. fft(X) is equivalent to fft(X, n) where n is the size of X in the first nonsingleton dimension. If the length of X is less than n, X is padded with trailing zeros to length n. If the length of X is greater than n, the sequence X is truncated. When X is a matrix, the length of the columns are adjusted in the same manner.

This means that you are not actually taking a FFT of a 'pure sine wave' - you are taking the FFT of a sine wave with a flat signal after it.

This is equivalent to taking the FFT of a sine wave multiplied with a square window function. The FFT spectrum is then the convolution of the sine wave frequency spectrum (an impulse function) with the square wave frequency spectrum (sinc(f).)

If you change L = 16,384 so that there is no zero-padding of the signal, you will observe a perfect FFT.

Further search keywords: "Spectral Leakage", "Window Function", "Hamming Window".

Edit: I cleaned up some material I wrote on this topic back in university, which goes into substantially more detail. I've posted that on my blog.