how to get fourier transform of a signal - matlab

Question

In the below code I am trying to get the fourier transform of a stationary signal (x3). But at run time the plot i get is absolutely something wrong and does not show any frequencies of the signals x3.
kindly please guide me and help me to get the fourier transform correctly.
Code:
%% Time specifications:
Fs = 8000; % samples per second
dt = 1/Fs; % seconds per sample
StopTime = 1; % seconds
t = (0:dt:StopTime-dt); % seconds
x1 = (10)*cos(2*pi*3*(t));
x2 = x1 + (10)*cos(2*pi*5*(t));
x3 = x2 + (10)*cos(2*pi*10*(t));
%% here i try to Plot fourier transform of the signal x3:
NFFT = 2^nextpow2(StopTime); % Next power of 2 from length of y
Y = fft(y,NFFT)/StopTime;
f = Fs/2*linspace(0,1,NFFT/2+1);
figure;
plot(f,2*abs(Y(1:NFFT/2+1)));
%% Plot the signal versus time:
figure;
hold on;
plot(t,x1,'r');
plot(t,x2,'g');
plot(t,x3,'b');
Update_1

Answer 1

You can not see what you expected because the value of NFFT is 1 means when you write NFFT/2+1 as an index of Y it will not be an integer value so MATLAB warns you. You can calculate NFFT like this:
NFFT = 2^nextpow2(length(t))
instead of writing
NFFT = 2^nextpow2(StopTime)
Well, try this:
Fs = 8000; % samples per second
dt = 1/Fs; % seconds per sample
StopTime = 1; % seconds
t = (0 : dt : StopTime-dt); % seconds
x1 = 10 * cos(2*pi*3*t);
x2 = x1 + 10 * cos(2*pi*5*t);
x3 = x2 + 10 * cos(2*pi*10*t);
%% here i try to Plot fourier transform of the signal x3:
NFFT = 2 ^ nextpow2(length(t)); % Next power of 2 from length of y
Y = fft(x3, NFFT) / StopTime;
f = Fs / 2 * linspace(0, 1, NFFT/2 + 1);
figure;
plot(f, 2 * abs( Y( 1:length(f) ) ) ); % // Also try this: plot(f(f <= 200), 2 * abs( Y( 1:length(f(f <= 200)) ) ) )
%% Plot the signal versus time:
figure;
hold on;
plot(t, x1, 'r');
plot(t, x2, 'g');
plot(t, x3, 'b');
Plots:
EDIT:
1- Actually you don' t have to use nextpow() function. If you use it, fft() function works faster. Because, due to time efficiency, fft() works like that recursively divide the signal by 2 for each time. Then calculates discrete fourier transform for each part and gather them. This means that the FFT is most efficient when the signal vector length is a power of 2.
2- Dividing fft result by StopTime part doesn' t make any sense to me either. Dividing fft result by NFFT may be more convenient theoretically.