hi, I produced this square wave signal, but as you can see the square peaks have no upper limit. how can i solve?
x=0.03*square(f*t,duty)
in reality I seem to see that the graph is the other way around

 Accepted Answer

Jon
Jon on 30 Jun 2022
You probably are not correctly specifiying the duty cycle. If you want the square wave to be "on" half the time you should set duty = 50, also note that the first argument to square is in radians, so each cycle is 2*pi radians. You have to scale your input accordingly.

More Answers (1)

Doesn't seem to have any issue.
t = linspace(0, 3*pi, 3001)';
f = 2;
duty = 50;
x = 0.03*square(f*t, duty);
plot(t/pi, x, t/pi, 0.03*sin(2*t))
grid on
ylim([-0.05 0.05])

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R2022a

Asked:

on 30 Jun 2022

Answered:

Jon
on 30 Jun 2022

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