the final result of laplace transform still contain laplace function?

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!

Answer 1

[David Goodmanson]

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.

[Ziying Huang]

Thank you so much !

It helps me a lot!