How to plot Gauss sums ?

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Mohamed Ahmed
Mohamed Ahmed on 24 Jan 2022
Commented: Paul on 25 Jan 2022
I'm trying to plot the Gauss sums according to the equation shown in the image s(t), but I keep receiving errors.
Can you please show me what am I doing wrong ?
clear all
close all
clc
%%
Fs = 1000; % Sampling frequency
T = 1/Fs; % Sampling period
L = 1024; % Length of signal
t = 2*(0:L-1)*T; % Time vector
x = 0;
k = 0;
s = 0;
p = primes(L);
% s(t) = cumsum((k/p)(1:length(p)-1)).*exp(1i*k*t);
for k=1:p-1
s(t) = s(t) + (k/p).*exp(1i*k*t);
end
figure
subplot(2,2,1)
plot(t,s)
title('signal')

Accepted Answer

Paul
Paul on 24 Jan 2022
Edited: Paul on 24 Jan 2022
There are at least two problems with the code. Assuming you want to compute numerical values of s, the code can't reference s in a functional form, like s(t). For numerical objects, the "t" in s(t) is an index, which should be numerical integers or a logical. So for one value of p, you need to compute one value of an array s for each value of t, and then for each value of t, you need that inner loop over k. If you want more than one value of p, you'll need to make s a 2D array (one dimension for t and the other for p). Here is modification to the code for a single value of p
clear all
close all
clc
%%
Fs = 1000; % Sampling frequency
T = 1/Fs; % Sampling period
L = 1024; % Length of signal
t = 2*(0:L-1)*T; % Time vector
x = 0;
k = 0;
s = 0;
p = primes(L);
p = 13; % odd prime
% s(t) = cumsum((k/p)(1:length(p)-1)).*exp(1i*k*t);
s = 0*t; % initialzize t
for jj = 1:numel(t)
for k=1:p-1
s(jj) = s(jj) + (k/p).*exp(1i*k*t(jj));
end
end
It's likely that at least the inner loop could be vectorized. However, this code has another problem. As stated in the text of the Question, the term (k/p) does not mean "k divided by p." Rather it is the Legendre symbol L(k,p). So you'll need a function that computes L(k,p).
Check out
doc jacobiSymbol
which would be perfect to use if p is an odd prime. But the problem statement isn't limited to only odd primes. However the wikipedia page for Legendre symbol seems to imply that it is only defined for odd primes. OTOH, the Mathworld page doesn't quite say that p is odd prime is assumed in the defintion (it might just be poorly worded).
  6 Comments
Paul
Paul on 25 Jan 2022
s(t) is a sum of complex exponentials. The Fourier transform of a complex exponential is Dirac delta, so the spectrum of s(t) should be an train of p-1 delta functions, shouldn't it?

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