1 个投票

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

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Jon
Jon 2022-6-30

1 个投票

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.

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Sam Chak
Sam Chak 2022-6-30
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2 个投票

Ran in:
@Federico Paolucci
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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