Why dosent my fourier output simulate the input?

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Jamie  Chambers
Jamie Chambers am 18 Feb. 2017
Kommentiert: Jamie Chambers am 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

Akzeptierte Antwort

Frank Macias-Escriva
Frank Macias-Escriva am 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

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