Help recreating square wave from equation

Hello,

I need help recreating a square wave from the following equation:

Thanks!

Answer 1

[Sulaymon Eshkabilov]

Hi,

IHere is the correct code:

t = ..
T = ..
n=1:2:15;  % By increasing n = 1:2:25, you will get much better approximation.
U1 = sum((4/pi)*(1./n(:)).*sin(2*pi*n(:).*t/T));
figure
plot(t, U1, 'r')

Good luck.

[Mackenzie Weeks]

Thanks for the response!

That gives me this plot:

and i think it's supposed to look something like this ( https://au.mathworks.com/help/signal/ref/square.html):

[Sulaymon Eshkabilov]

Run this or you can increase more n end value:

n=1:2:75;  % By increasing, you will get much better approximation

You can run and plot n=1 and n=1:2:75.

Note that the Gibbs phenomenon (ripples on both ends of the rectangle wave) will be present no matter how big the number of series. Study Gibbs phenomenon: https://en.wikipedia.org/wiki/Gibbs_phenomenon

Rectangle wave shown in your given mathworks source is obtained with square() builtin function that is different from this Fourier series approximation.

Good luck.

[Mackenzie Weeks]

Oh ok I see! Thank you!!

Answer 2

[Sulaymon Eshkabilov]

Hi,

It is a quite straightforward exercise. You can create this signal using a colon operator (:), or [for .. end] or [while .. end] loop. Colon operator is the most efficient one.

Good luck.

[Mackenzie Weeks]

Hi

I'm confused by your answer.

This is currently what I have so far, which creates this plot. But im pretty sure the square wave should be closer to the wave itself? I'm supposed to plot where n = 1 3 5 7 and 15.

clc;clear all;

%u(t)= (4/pi) * (1/n)sin * (2*pi*n*
t=0:0.01:1; 
T = 1; 
n = 5;
x=(4/pi)*(1/n)*sin(2*pi*n*(t/T));
sq= square(x);

figure(1); 
plot(t,x)
hold on;
plot(t,sq)