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Help recreating square wave from equation

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Mackenzie Weeks
Mackenzie Weeks 2021-5-24
评论： Mackenzie Weeks 2021-5-24
Hello,
I need help recreating a square wave from the following equation:
Thanks!

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采纳的回答

Sulaymon Eshkabilov
编辑：Sulaymon Eshkabilov 2021-5-24
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.
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Sulaymon Eshkabilov
Sulaymon Eshkabilov 2021-5-24
编辑：Sulaymon Eshkabilov 2021-5-24
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.

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更多回答（1 个）

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.
  1 个评论
Mackenzie Weeks
Mackenzie Weeks 2021-5-24
编辑：Mackenzie Weeks 2021-5-24
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)

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