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?
The problem is that in order to get perfect spectra (peaks on 50,90,180 and 0 otherwise), your interval should be a multiplier of all the frequencies.
Explanation: consider y=sin(2.*pi.*t). Use your code to plot it with:
1) et = 1-dt;
2) et = 1;
In the first case, if you zoom in, you will see that peak is perfectly on 1 Hz. In the second case it is not (also very close).
Why? Because you are working with the finite number of points, and, consequently, finite number of frequencies. If 1 Hz is among this set of frequencies (1st case), you will get perfect peak at 1 Hz at zeros at all others frequencies.
In case 2 there is no 1 Hz frequency in your set, so you will get peak on the nearest frequencies (and also it will have finite width).
Ultimately, in your original code, there are no 50, 90, 180 Hz frequencies in your full set of frequencies.