Matlab中信号的频谱

Question

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?

Answer 1

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.