generate square wave using piecewise function and plot it

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!

回答(1 个)

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 个评论

Hi, thank you! But for my square wave, each period is 1ms and the duty cycle is 5%.
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.
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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