Why is the Sum for f(t,x) Not Normalized in Matlab FFT?

[nickthequick]

Hi,

Let's say I have an input signal

$$f(t,0)= \sum_n A_n cos(w_nt)$$

And we know that

$$f(t,x)= \sum_n A_n cos(k_nx-w_nt)$$

and we also have the relationship between k and w. We can find the coefficients A_n by taking a Fourier Transform of the relationship for f(0,t).

Here's my question. How is the sum for f(t,x) normalized? If this is unclear, below is an example:

I've been trying test cases to figure out what the relationship is by prescribing f(0,t) to be some easy function i.e.

$$f(t,0)= cos(\omega_1t) + 2cos(\omega_2t)$$

Theoretically, we know that

$$\bar{f}(t,x)= cos(k_1x-\omega_1t) + 2cos(k_2x -\omega_2t)$$

But if I did not know precisely what f(t,0) is and went through this numerically, I'd have

$$f^*(t,x)= \sum_n A_n cos(k_nx-\omega_nt)$$

where

$$A_n=\frac{1}{length(f(0,t))}fft(f(t,0))$$

My problem is $$f^* \neq \bar{f}$$. My question is, why is this the case? I think it comes down to normalizations but I'm not completely sure.

Any help would be appreciated,

Nick

PS
If you're interested, here's my code
close all;
g=10;
A= [1 .5];
tt=0:.5:2^5-.5;
omega=[1 .25];
kay=omega.^2/g;
signal = zeros(length(tt),1);
for i=1:length(tt);
signal(i,1) = sum(A.*cos(tt(i)*omega));
end
Fs= 50;
xx=1:.5:100;
yy=signal;
[YY,ff]=positiveFFT(yy,Fs); % this sets up our independent variable frequency as we want it
kk=(2*pi*ff).^2/g;

ETAETA=zeros(length(tt),length(xx));
for i =1:length(tt)
for j=1:length(xx)
ETAETA(j,i)=real(sum(YY'.*cos(kk*xx(j)-2*pi*ff*tt(i))));
end
end
figure (1)
imagesc(xx,tt,real(ETAETA));
set(gca,'YDir','normal');
caxis([-0.01 0.01])
colormap(copper);
colorbar;
xlabel('distance from wave maker'); ylabel('time');
shading('interp')
h = colorbar;
set(get(h,'YLabel'), 'String', 'SSH');
title('theoretical first order solution');
% %
% ETA3= zeros(length(xx),length(tt));
% for i=1:length(tt)
% for j=1:length(xx)
% ETA3(j,i)= sum(A.*cos(kay*xx(j)-omega*tt(i)));
% end
% end

% figure(2)
% imagesc(xx,tt,real(ETA3)');
% set(gca,'YDir','normal');
% caxis([-0.01 0.01])
% colormap(copper);
% colorbar;
% xlabel('distance from wave maker'); ylabel('time');
% shading('interp')
% h = colorbar;
% set(get(h,'YLabel'), 'String', 'SSH');
% title('theoretical first order solution');


where
positiveFFT.m is
function [X,freq]=positiveFFT(x,Fs)
N=length(x); %get the number of points
k=0:N-1; %create a vector from 0 to N-1
T=N/Fs; %get the frequency interval
freq=k/T; %create the frequency range
X=fft(x)/N; % normalize the data

%only want the first half of the FFT, since it is redundant
cutOff = ceil(N/2);

%take only the first half of the spectrum
X = X(1:cutOff);
freq = freq(1:cutOff);

 

[mmwave]

Hi Nick,

First of all, great work on exploring this problem and writing out your code for others to see. From what I can gather, your question is related to the normalization of the Fourier transform in your code and how it relates to the theoretical solution. Let me try to break it down for you.

The normalization factor in the Fourier transform is important because it helps us to compare the amplitudes of different frequency components in a signal. In your code, you are using the function positiveFFT() to calculate the Fourier transform, and it seems like you are normalizing it by dividing by the number of points in the signal (N). This is a common approach, but it may not always give you the results you expect.

In the case of your test signal f(t,0) = cos(ω1t) + 2cos(ω2t), the Fourier transform should give you two peaks at ω1 and ω2 with amplitudes of 1 and 2 respectively. However, because you are dividing by N in your code, the amplitudes of these peaks will be scaled down. This is why you are not getting the expected result of f*(t,x) = cos(k1x - ω1t) + 2cos(k2x - ω2t).

To fix this, you can try normalizing the Fourier transform by the total energy of the signal instead of N. This can be calculated by taking the sum of the squared amplitudes of the signal before taking the Fourier transform. You can then divide the Fourier transform by this energy value to get the correct amplitudes.

I hope this helps to clarify the issue you are facing. Keep up the good work and don't hesitate to reach out if you have any further questions.