MATLAB: The amplitude of fft computing is a little bit different from a known value

amplitudefftfrequencymagnitudeMATLAB

I was trying test this:

n=5000;
ts=5;
t=[0:ts/n:ts];
x1=sin(2*pi*20*t);
x2=2*sin(2*pi*60*t);
x3=20*sin(2*pi*200*t);
x4=15*sin(2*pi*350*t);
x=x1+x2+x3+x4;
N=length(x);
k=[0:N-1];
T=N/Fs;
freq=k/T;
cutoff=ceil(N/2);
X=fftn(x)/(N/2);
X=abs(X);
stem(freq(1:cutoff),X(1:cutoff))

The magnitudes got different results to 1 (x1); 2 (x2); 20 (x3) and 15 (x4). And more…if "n" has more or less points for computing the fft, the frequencies (20, 60, 200, 350) also change. Why? Can someone help me? Thanks a lot.

Best Answer

You have to realize that the frequency spacing between DFT elements depends on the length of the DFT and the sampling frequency. If your frequencies do not fall directly on a DFT bin (the spacing is Fs/N where Fs is the sampling frequency and N the length of the DFT) then the energy in a particular component will be mapped to a slightly different frequency.

Here is your example the way you think it should look:

n=5000; ts=5; t=[0:ts/n:ts-ts/n];
x1=sin(2*pi*20*t); x2=2*sin(2*pi*60*t); x3=20*sin(2*pi*200*t);    
x4=15*sin(2*pi*350*t); x=x1+x2+x3+x4;
xdft = fft(x);
xdft = xdft(1:length(x)/2+1);
xdft(2:end-1) = 2*xdft(2:end-1);
xdft = xdft./length(x);
df = 1000/length(x);
freq = 0:df:500;
plot(freq,abs(xdft));
xlabel('Hz'); ylabel('Magnitude');
grid on;

You may want to review this:

amplitude estimation and zero padding

Related Solutions

MATLAB: Amplitude Error in amplitude Spectrum of FFT

I suspect that your frequency of interest is simply not falling on a DFT bin directly. Why do you think you need to use a power of two for the DFT?

For example, if I have data sampled at 1 kHz and I have 1000 points, then frequency increment in DFT bins is 1 Hz (the bins are 1 Hz apart) so a frequency of 100 Hz for example will fall directly on a bin.

   Fs = 1000;
   t = 0:1/Fs:1-1/Fs;
   x = 1.5*cos(2*pi*100*t);
   xdft = fft(x);
   freq = 0:Fs/length(x):Fs/2;
   xdft = xdft(1:length(x)/2+1)./length(x);
   xdft(2:end) = 2*xdft(2:end);
   plot(freq,abs(xdft))

See that the magnitude is estimated exactly. Now watch what happens when I pad to 1024:

   xdft = fft(x,1024);
   freq = 0:Fs/length(xdft):Fs/2;
   xdft = xdft(1:length(xdft)/2+1)./length(x);
   xdft(2:end) = 2*xdft(2:end);
   plot(freq,abs(xdft))

by padding to 1024, I have now made the frequency of interest fall between DFT bins and that makes my amplitude estimate inaccurate.

See https://www.mathworks.com/help/signal/ug/amplitude-estimation-and-zero-padding.html

in the Signal Processing Toolbox documentation

MATLAB: How to Normalize a fft to plot in frequency domain

I am not certain what you intend by ‘normalise’. You need to scale it by dividing the fft result by the length of the time-domain signal:

z = fftshift(fft(x1000)/length(x1000));

This ‘normalises’ the result, correcting for the total energy in the time-domain signal. (You can use the numel function instead of length for a vector. Read about both to understand where each is appropriate.)

Note that you are using the two-sided Fourier transform, so the signal intensity will be equally divided between the negative frequencies and positive frequencies. In a one-sided Fourier transform, correct for this by multiplying the fft output by 2 to reproduce the amplitude of the original signals.

Best Answer

Related Solutions

MATLAB: Amplitude Error in amplitude Spectrum of FFT

MATLAB: How to Normalize a fft to plot in frequency domain

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