Normalization of zero padded signals

13 views (last 30 days)
Marc Fuller
Marc Fuller on 21 Oct 2022
Edited: Matt J on 24 Oct 2022
I have a simple question regarding zero padding and normalization. Consider an impulse resonse of a 4 point moving average filter. and its fft zero padded to 1024 points..
x=[1/4 1/4 1/4 1/4]
X=fft(x,1024 )
xpowrsum=dot(x,x)
Xpowrsum=dot(abs(X),abs(X))/1024
plot(fftshift(abs(X)))
FFT of 4 PT moving average
By Parsevals theorem the two energies are equal as expected. However, the fft without scaling shows the correct frequency response with a gain of 1 at 0 Hz. So why do I always read the FFT should be scaled by the number of samples before zero padding (in this case 4) if I am interested in the magnitude response of the filter?

Answers (2)

Matt J
Matt J on 21 Oct 2022
Edited: Matt J on 21 Oct 2022
So why do I always read the FFT should be scaled by the number of samples before zero padding (in this case 4) if I am interested in the magnitude response of the filter?
The FFT is a tool with many applications, each with its own appropriate scaling.
Scaling by 1/N is done when the FFT is being used to evaluate the Discrete Fourier Series.
When it is being used to approximate the continuous Fourier transform, it is scaled by the time sampling interval 1/Fs.
To achieve Parseval's equality, the fft should be scaled by 1/sqrt(N):
x=[1/4 1/4 1/4 1/4];
X=fft(x,1024 )/sqrt(1024);
xpowrsum=norm(x).^2
xpowrsum = 0.2500
Xpowrsum=norm(X).^2
Xpowrsum = 0.2500
  6 Comments
Matt J
Matt J on 23 Oct 2022
Edited: Matt J on 23 Oct 2022
One example to motivate the 1/N factor is to consider a periodic signal like,
If the goal is to recover the coefficients of the sinusoidal terms (5 and 3), we can see in the following code that the 1/N is necessary.
N=10;
n=(0:9)';
x=5+3*exp(1j*2*pi*n/N);
c=fft(x)/N
c =
5.0000 + 0.0000i 3.0000 - 0.0000i 0.0000 + 0.0000i 0.0000 - 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i -0.0000 + 0.0000i 0.0000 + 0.0000i -0.0000 + 0.0000i 0.0000 + 0.0000i

Sign in to comment.


Marc Fuller
Marc Fuller on 23 Oct 2022
So after thinking about this, here is the reason you should not normalize a filter.
Consider y(t)=x(t) convolved with h(t)- where h(t) is some filter such as the moving average filter in this question.
If we want to be consistent, suppose we transform y(t) with normaiization to Y(f) by dividing by the length of the fft, N
Now if we transfrom x(t) convolved with h(t) We have X(f)H(f) by the convolution theorem. We only want to normalize X(f) to have the product identical with Y(f). since Y(f)/N= (X(f)H(f)/N.
  9 Comments
Matt J
Matt J on 24 Oct 2022
Edited: Matt J on 24 Oct 2022
It requires no toolboxes and works on arrays of any dimension.

Sign in to comment.

Categories

Find more on Fourier Analysis and Filtering in Help Center and File Exchange

Products


Release

R2022a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!