Why dosent my fourier output simulate the input?
1 view (last 30 days)
Show older comments
Jamie Chambers
on 18 Feb 2017
Commented: Jamie Chambers
on 20 Feb 2017
This is my input signal:
clc
clear all
syms L psi f(x) t
L=2*pi;
psi=pi;
f(x)=(t-psi)^2;
ezplot(f(x),[0,2*pi])
xlabel ('period 0,2*pi')
ylabel ('amplitude')
title ('input signal')
pretty(f(x))
I'm trying to plot the Fourier using this code:
clc
clear all
syms n m L t psi x
L=2*pi;
psi=pi
f(x)=(t-psi)^2
A0=int(f(x),t,0,2*pi)/L
ezplot(A0,[0,2*pi])
hold on
for m=1:10
An=int(f(x)*cos(n*t),t,0,L)*(L/1);
An=subs(An,n,m);
Bn=int(f(x)*sin(n*t),t,0,L)*(L/1);
Bn=subs(Bn,n,m);
Fo=A0+sum((An*cos(n*pi)/L)*f(x)+(Bn*sin(n*pi)/L)*f(x))
Fo=subs(Fo,n,m);
ezplot(Fo,[0,2*pi])
ylim auto
hold on
end
I've tried numerous attempts but cant seen to generate the simulated input???? what a I doing wrong? Regards J
Accepted Answer
Frank Macias-Escriva
on 19 Feb 2017
Try this code bellow:
syms t n;
L = 2*pi;
f(t) = (t-pi)^2;
figure('Name', 'Original signal');
ezplot(f(t), [0, L]);
ylim auto;
N = 100;
A0 = int(f(t), t, 0, L)/L;
A(n) = int(f(t)*cos(n*t), t, 0, L) * (2/L);
B(n) = int(f(t)*sin(n*t), t, 0, L) * (2/L);
Fo(t) = A0 + symsum(A(n)*cos(n*t) + B(n)*sin(n*t), n, 1, N);
figure('Name', 'Generated signal');
ezplot(Fo(t), [0, L]);
ylim auto;
In this code, you avoid using for loops and keep using the symbolic toolbox for the entire solution. Also, note te use of "symsum" instead of "sum". Play with N for getting different accuracies of the generated signal.
Best,
fm
More Answers (0)
See Also
Categories
Find more on Fourier Analysis and Filtering in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!