# Frequency spectrum of signal - Matlab

Here is the code I use to plot a function in frequency domain in Matlab:

    dt = 1/10000; % sampling rate
et = 0.1; % end of the interval
t = 0:dt:et; % sampling range
y = 2+sin(2.*pi.*50.*t)+18.*sin(2.*pi.*90.*t)+6.*sin(2.*pi.*180.*t); %     sample the signal
subplot(2,1,1); % first of two plots
plot(t,y); grid on % plot with grid
xlabel('Time (s)'); % time expressed in seconds
ylabel('Amplitude'); % amplitude as function of time
Y = fft(y); % compute Fourier transform
n = size(y,2)/2; % 2nd half are complex conjugates
amp_spec = abs(Y)/n; % absolute value and normalize
subplot(2,1,2); % second of two plots
freq = (0:100)/(2*n*dt); % abscissa viewing window
stem(freq,amp_spec(1:101)); grid on % plot amplitude spectrum
xlabel('Frequency (Hz)'); % 1 Herz = number of cycles/second
ylabel('Amplitude'); % amplitude as function of frequency


The problem is, when I zoom in my graph I don't see peaks exactly on 50Hz, 90Hz and 180Hz.

What did I do wrong in my code?

• I think you mean specifically DFT X[k] of x[n]. Because otherwise DTFT $X(e^{j\omega})$ would compute whatever frequency there is in the discrete signal in exactness, due to its continuous frequency variable. – Fat32 Apr 8 '15 at 1:01