generate square wave using piecewise function and plot it

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Hi,
I'd like to generate a square wave using piecewise function rather than square function, is there someone that can help me? m`Many thanks!

Answers (1)

Walter Roberson
Walter Roberson on 13 Jun 2021
PulseLength = 50;
CycleLength = 1000;
ReduceCycle = @(x) x - CycleLength*floor(x/CycleLength)
ReduceCycle = function_handle with value:
@(x)x-CycleLength*floor(x/CycleLength)
syms t
y(t) = piecewise( ReduceCycle(t) < PulseLength, 1, 0)
y(t) = 
fplot(y, [0 3100]); ylim([-1.1 1.1])
S(t) = sin(2*pi*t/CycleLength)
S(t) = 
fplot(S, [0 3100]); ylim([-1.1 1.1])
ys(t) = y(t) * S(t)
ys(t) = 
fplot(ys, [0 3100]); ylim([-1.1 1.1])
  3 Comments
Han Gao
Han Gao on 13 Jun 2021
Thanks for your reply and it is working. One more question is I'd like to present the FFT spectrum and how to do a fft of the synthetic signal? I know FFT is designed only to work numerically on discrete data. But I don't know how to deal with the symbolic expression. Thank y ou.
Walter Roberson
Walter Roberson on 13 Jun 2021
PulseLength = 50/1000;
CycleLength = 1000/1000;
ReduceCycle = @(x) x - CycleLength*floor(x/CycleLength)
ReduceCycle = function_handle with value:
@(x)x-CycleLength*floor(x/CycleLength)
syms t
y(t) = piecewise( ReduceCycle(t) < PulseLength, 1, 0)
y(t) = 
fplot(y, [0 3100]/1000); ylim([-1.1 1.1])
S(t) = sin(2*pi*t/CycleLength)
S(t) = 
fplot(S, [0 3100]/1000); ylim([-1.1 1.1])
ys(t) = y(t) * S(t)
ys(t) = 
fplot(ys, [0 3100]/1000); ylim([-1.1 1.1])
yf = fourier(ys)
yf = 
... not very useful ;-)
You should create a vector of specific numeric times to sample at, and ys(times) to get specific values there. double() that and you have something you can fft()
Watch out that you use a full cycle of times without the time that would be the start of a new cycle. For example for 100 Hz if you sampled for 1 second, do not use times 0:1/100:1, and instead use 0:1/100:1-1/100 . Otherwise the first and the last point would both be the start of a cycle, and that will distort your fft.

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