how to scale fft?

Question

Rabih Sokhen 2022-2-10

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1647010-how-to-scale-fft

评论： Rabih Sokhen 2022-3-3

采纳的回答： Matt J

在 MATLAB Online 中打开

hy guys.

i have a question concerning the scaling in fft .

i have seen alot of code on this forum, but some of the code scale the fft with length of the time domain while other's with increment.

code:

clear all; clc
dt=0.1; 
t=-5:dt:5; 
z=numel(t);
df = 1/dt;                         
f = linspace(-df/2, df/2, z);
h_t=rectpuls(t,1/f0);
h_f=fftshift(fft(h_t,n));

% how to scale h_f? should i divide h_f with z or dt and why?

% thank you in advance

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Answer 1

Matt J 2022-2-10

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https://ww2.mathworks.cn/matlabcentral/answers/1647010-how-to-scale-fft#answer_892950

编辑：Matt J 2022-2-10

The DFT is a tool for computing several other theoretical transforms. It can be used to compute a Discrete Fourier Series, which is when you would scale by the length. It can also be used to approximate a continuous Fourier transform, which is when you scale by the time increment.

There is also some disagreement across different professions (esp., engineering versus physics) on scaling factors in the definitions of these transforms. Therefore, the safest way to determine the scaling that you need is to just compare the formula in Matlab's FFT defintiion to the formula for the thing you want to compute.

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Paul 2022-2-10

编辑：Paul 2022-2-12

在 MATLAB Online 中打开

To illustrate ...

If the desire is to use fft() to compute the DFT to approximate the Continuous Time Fourier Transform (CTFT) of a finite duration signal, then zero padding should be considered because the DFT inherenntly assumes the underlying signal is the periodic extension of the seuqence, which can lead to misleading results.

Consider an example with a rectangular pulse and its CTFT

syms t
f(t) = rectangularPulse(0,1,t);
syms w
ctft(w) = fourier(f(t),t,w);  % CTFT of f(t)

Samples of f(t) at 20 Hz

dt = 0.05;
tvec = 0:dt:1;
fvals = ones(1,numel(tvec));

Compute the Discrete Time Fourier Transform (DTFT) of fvals.

[dtft,wdtft] = freqz(fvals,1,1024);

Plot the CTFT and the DTFT. We see that dt*DTFT is a reasonable approximation to the CTFT up to the Nyquist frequency (pi/dt)

figure

fplot(abs(ctft),[0 65]);

hold on

plot(wdtft/dt,dt*abs(dtft))

xlim([-1 pi/dt])

legend('CTFT','DTFT')

Now add to the plot the DFT of fvals

figure

fplot(abs(ctft),[0 65]);

hold on

plot(wdtft/dt,dt*abs(dtft))

dft = fft(fvals);

wdft = (0:numel(dft)-1)/numel(dft)*2*pi/dt;

stem(wdft(1:11),dt*abs(dft(1:11)),'MarkerFaceColor','auto')

xlim([-1 pi/dt])

legend('CTFT','DTFT','DFT no pad')

The DFT is nonzero only at w = 0 because the periodic extension of fvals is a constant sequence over all time. Note that the DFT are samples of the of the DTFT as they must be.

Now take the DFT with zero padding to pick up the ups and downs of the DTFT.

figure

fplot(abs(ctft),[0 65]);

hold on

plot(wdtft/dt,dt*abs(dtft))

dft = fft(fvals,64);

wdft = (0:numel(dft)-1)/numel(dft)*2*pi/dt;

stem(wdft(1:31),dt*abs(dft(1:31)),'MarkerFaceColor','auto')

xlim([-1 pi/dt])

legend('CTFT','DTFT','DFT w/pad')

In summary for a finite duration signal, the DFT of time domain samples of the signal are frequency domain samples of the DTFT, and dt*DTFT is an approximation to the CTFT up to the Nyquist frequency.

Edit: I just realized that making the samples of the rectangular pulse unity at the leading and trailing edges is is why the DTFT is larger at DC and has nulls offset in frequency compared to the CTFT. I guess one has to decide how what it means to sample at the discotinuities. I think reasonable options are to use a sample equal to 1 at the leading edge and zero at the trailing edge, i.e., the sampling starts at t = 0+, or make the samples = 1/2 at both edges. Here's the final plot using the second approach

figure

fplot(abs(ctft),[0 65]);

hold on

fvals = [0.5 ones(1,numel(tvec)-2) 0.5];

[dtft,wdtft] = freqz(fvals,1,1024);

plot(wdtft/dt,dt*abs(dtft));

dft = fft(fvals,64);

wdft = (0:numel(dft)-1)/numel(dft)*2*pi/dt;

stem(wdft(1:31),dt*abs(dft(1:31)),'MarkerFaceColor','auto')

xlim([-1 pi/dt])

legend('CTFT','DTFT','DFT w/pad')

Matt J 2022-2-11

No, you must always multiply by dt. The issue of choosing how much zero-padding you need to do is separate.

Rabih Sokhen 2022-3-3

thank you Matt, sry for the late response .

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how to scale fft?

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