how to plot a periodic function?

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Hey there,
I have to construct a periodic function of a signal according to the image shown in order to apply fourier series into it. I tried using piecewise command but I failed to get an appropriate output. What is the point that I'm missing? My code is shown below:
clc;
clear all;
a=3.2;
t1=5;
t2=7;
t3=12;
t4=17;
%constructing partial periodic function
syms a(x)
T=17;
i=-2;
interval=[0 -i*T];
pw=[];
while i<=diff(interval/(2*T))
a(x)=piecewise(i*T<=mod(x,17)<=t1+(i*T),3.2,t1+(i*T)<=mod(x,17)<=t2+(i*T),-1.6*(x-i*T)+11.2*4*i,t2+(i*T)<=mod(x,17)<=t3+(i*T),-0.64*(x-i*T)+4.48*4*i,t3+(i*T)<=mod(x,17)<=t4+(i*T),-3.2);
i=i+1;
pw=[pw a ];
end
pw
pw(x) = 
fplot(pw,interval)

Accepted Answer

Paul
Paul on 1 Dec 2022
Hi Zafer,
To compute the Fourier series, you really only need to define one period of the function, because the defining integrals are only taken over one period. So you can use piecewise to define one period of the function.
a=3.2;
t1=5;
t2=7;
t3=12;
t4=17;
syms f(t)
f(t) = piecewise(t < t1,a, t1<=t<t2,-a/(t2-t1)*(t-t1)+a, t2<=t<t3, -a/(t3-t2)*(t-t2), t3<=t<t4,-a, 0);
fplot(f(t),[0 t4])
If you want to plot the periodic fucntion, then use mod as the argument into f, not in the definition of f
fplot(f(mod(t,t4)),[-50 50])
  1 Comment
Zafer Duyenli
Zafer Duyenli on 2 Dec 2022
thank you, Paul! Appreciated for the answer!

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More Answers (2)

Torsten
Torsten on 1 Dec 2022
Edited: Torsten on 1 Dec 2022
a = 3.2;
t1 = 5;
t2 = 7;
t3 = 12;
t4 = 17;
fun = @(x) a*(x>=0 & x<=t1) + (-a/(t2-t1)*(x-t2)).*(x>t1 & x<=t2) + (-a/(t3-t2)*(x-t3)-a).*(x>t2 & x<=t3) + (-a)*(x>t3 & x <t4);
Fun = @(x) fun(mod(x,t4));
x = 0:0.01:17;
plot(x,Fun(x))
grid on

Walter Roberson
Walter Roberson on 1 Dec 2022
Edited: Walter Roberson on 1 Dec 2022
You do not need any loop. You generate one cycle based upon t1 t2 t3 t4 and piecewise(). Then you substitute mod(Time,t4) to the piecewise, to end up with a piecewise that is cyclic every interval of t4.
Once you have the piecewise() then you can ask to rewrite(EXPRESSION, 'heaviside') to get an expression that, in theory, can be pass through fourier() . In practice the mod() is going to cause problems.
  1 Comment
Zafer Duyenli
Zafer Duyenli on 2 Dec 2022
Thank you for the help! It really makes things easier for me!

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