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the final result of laplace transform still contain laplace function?

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Ziying Huang
Ziying Huang on 12 Mar 2021
Commented: Ziying Huang on 15 Mar 2021
here is the original code
I intend to transform it to frequency domain using laplace function.
but the final answer still contain laplace function
can somebody tell me why?
really appreciated!

Answers (1)

David Goodmanson
David Goodmanson on 12 Mar 2021
Edited: David Goodmanson on 12 Mar 2021
Hello ZH
Matlab doesn't know if taoc is positive or negative. If taoc is negative, the calculation can't really be done. That's because the standard laplace transform assumes f(t) = 0 for negative t, and if taoc is negative, the heaviside function goes into negative t. Assuming w0 is positive, then
syms t vdd s w0
syms taoc positive
v = (1-cos(w0*t))*(heaviside(t)-heaviside(t-taoc));
Z = laplace(v,t,s)
Z = subs(Z,taoc,10*pi/w0)
Z =
exp(-(10*pi*s)/w0)*(s/(s^2 + w0^2) - 1/s) - s/(s^2 + w0^2) + 1/s
I left out a couple of constants for simplicity's sake. Since taoc is an exact multiple of pi, the answer is simpler than it would be otherwise.

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