# the final result of laplace transform still contain laplace function?

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Ziying Huang 2021 年 3 月 12 日
コメント済み: Ziying Huang 2021 年 3 月 15 日
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! サインインしてコメントする。

### 回答 (1 件)

David Goodmanson 2021 年 3 月 12 日

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
clear
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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Ziying Huang 2021 年 3 月 15 日
Thank you so much !
It helps me a lot!

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