Suppose the following (I'm assuming that you have done something similar for t, s, and nu)
Fs = 8192; % sampling rate (Hz)
t = 0:1/Fs:1-1/Fs; % time period (seconds)
s = sin(2000*2*pi*t); % 2000 Hz signal
N = 8192; % N-point FFT (block size)
nu = 0:1:N-1; % number of points
We use your Discrete Fourier Transform function (is that what it is?) to get
dftY = FourierDir(t,s,nu);
and plot the results
fHz = 0:Fs/N:(N-1)*Fs/N;
figure;
plot(fHz,abs(dftY),'b');
Note that in the plot, we see the signal frequency at 2000 Hz with a height of 0.5 (not quite the amplitude of the signal).
Now if we do something similar with the MATLAB fft function as
yFft = fft(s,N);
and plot on the same figure
hold on;
plot(fHz,abs(yFft)*(2/N),'g'); % multiply by 2/N since input signal is real
we see that the FFT'd signal has a frequency of 2000 Hz and an amplitude of 1.0 (the 2/N factor gives us the amplitude of the input signal).
So all that is different between the MATLAB fft output and yours is a factor of two. I guess it all depends on what your function, FourierDir, is returning. If we want the fft data to be the same, then we could just do
hold on;
yFft = fft(s,N)*(2/N)/2; % now, yFft should be same as yDft
plot(fHz,abs(yFft),'g');
