MATLAB: To Get Correct FFT Result the length(x) must be An Integral Multiple of 100

Question

Hi, all

I'm trying to get the amplitude- frequency relationship and phase- frenquency relationship using fft. My signal is given by:

t=0:0.005:14; % Time [s] Where mod(length(t),100)~=0
% t=t=0:0.005:14; t(end)=[]; % Where mod(length(t),100)==0
Vs=10*cos(2*pi*20*t-pi/5)+13*cos(2*pi*60*t+pi/6);
Spectrum=fft(Vs)/length(Vs);
Spectrum(2:end-1)=Spectrum(2:end-1)*2;

To get the amplitude and the phase of the signal of specified frequency, I used:

Amplitude=abs(Spectrum);
Phase=angle(Spectrum);

The question is, if the length(Vs) is an integral multiple of 100 (100,2700, etc.), the programe returns a correct amplitude and phase like:

Both Amplitude and Phase value at the DC, 20Hz, 40Hz is the setting value. However, once the length(Vs) is not of an integral multiple of 100, the same program returns a totally wrong result like:

I don't know the reason, and I want a method by which I can get right Spectrum no matter what the length(Vs) is.(On the premise of guaranteeing sampling theorem).

Thanks

Answer 1

The desired plot will be obtained when N, the number of elements of t, is an integer multiple of the ratios Fs/Fn, where Fs is the sample frequency and Fn are the frequencies in Vs (in Hz). In this case we have

Fs = 200;
Fn = [20 60];

Given these vaues, selecting N = k*100 will satisy the criterion:

syms k N
N = k*sym(100);
N*sym(Fn)./sym(Fs)  % integer valued for all integer k
ans = 

Here is the "good" case in the question:

N = 2800;
N.*Fn./Fs   % integer

ans = 1×2
   280   840
tfinal = func(N,Fs,Fn)
tfinal = 13.9950

Here is another case, where N satisfies the criteria, but is not a muliple of 100

N = 2820;
N.*Fn./Fs   % integer
ans = 1×2
   282   846
tfinal = func(N,Fs,Fn)
tfinal = 14.0950

As you already showed, choosing N = 2801, which does not satisfy the criterion, results in an undesired plot

N = 2801;
N.*Fn./Fs  % not both integers
ans = 1×2
  280.1000  840.3000
tfinal = func(N,Fs,Fn)
tfinal = 14

Note this result is not wrong. It is correct based on the input parameters.

function tfinal = func(N,Fs,Fn)
t = (0:(N-1))/Fs;
Vs = 10*cos(2*pi*Fn(1)*t-pi/5)+13*cos(2*pi*Fn(2)*t+pi/6);
Spectrum = fft(Vs)/length(Vs);
Spectrum(2:end-1) = Spectrum(2:end-1)*2;
Amplitude = abs(Spectrum);
Phase = angle(Spectrum);
figure;
stem((0:N-1)/N*Fs,Amplitude)
xlim([0 Fs/2])
xlabel('Hz')
tfinal = t(end);
end