Can't find inverse laplace

[Cantafford]

So basically I have an exercise from a book which doesn't explain anything good at all it jumps steps, lacks explanaitions, etc.

Basically the step which I'm stuck at is this:

w(t) = Ilaplace{H(s)*(1/s)} where H(s) is 1/[Ti*(s+1)].
I need to find w(t) I can't find any form in my Laplace transform table that matches what I have there. Thanks for reading.

 

[HallsofIvy]

Ti is a constant? So the Laplace transform is \(\displaystyle \dfrac{1}{Ti s(s+ 1)}\)?

The standard procedure is to use "partial fractions". There exist constants, A and B, such that \(\displaystyle \frac{1}{Ti s(s+ 1)}= \frac{A}{s}+ \frac{B}{s+ 1}\). To find A and B, multiply both sides by s(s+ 1):
\(\displaystyle \frac{1}{Ti}= A(s+ 1)+ Bs\). Since this is to be true for all s, taking two values for s gives two equations to solve for s. In particular, if s= 0, \(\displaystyle \frac{1}{Ti}= A(1)+ B(0)= A\) so \(\displaystyle A= \frac{1}{Ti}\) and if s= -1, \(\displaystyle \frac{1}{Ti}= A(0)+ B(-1)= -B\) so \(\displaystyle B= -\frac{1}{Ti}\).

Now, what does your table give for the inverse Laplace transforms of \(\displaystyle \frac{1}{Tis}\) and \(\displaystyle \frac{-1}{Ti(s+1)}\)?

 

[Cantafford]

Hello, thank you for answering,


I have done a little mistake when writting down the problem which is:
H(s) = 1 / (Ti*s + 1) instead of 1/[Ti*(s+1)]. Yes Ti is a constant.

Anyhow I have followed your instructions and solved this way:
w(t) = ILaplace [ (1 / (Ti*s + 1)) * (1/s) ]
f(t) = (1 / (Ti*s + 1)) * (1/s) = A/s + B / (Ti*s + 1), multiplied each with (Ti*s + 1) and s then from:
s(ATi + B) + A = 1 got the values A = 1 and B = -Ti therefore f(t) = 1/s - Ti/(Tis+1)
from which w(t) = ILaplace[f(t)] = 1 - e^(-t/Ti).
It looks correct to me. The book has no answer it it though so please tell me if it's fine. Thank you again!

 

[Cantafford]

Thank you. I guess this can be locked.