Can't find inverse laplace

Cantafford

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Dec 25, 2013
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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.
 
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Ti is a constant? So the Laplace transform is 1Tis(s+1)\displaystyle \dfrac{1}{Ti s(s+ 1)}?

The standard procedure is to use "partial fractions". There exist constants, A and B, such that 1Tis(s+1)=As+Bs+1\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):
1Ti=A(s+1)+Bs\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, 1Ti=A(1)+B(0)=A\displaystyle \frac{1}{Ti}= A(1)+ B(0)= A so A=1Ti\displaystyle A= \frac{1}{Ti} and if s= -1, 1Ti=A(0)+B(1)=B\displaystyle \frac{1}{Ti}= A(0)+ B(-1)= -B so B=1Ti\displaystyle B= -\frac{1}{Ti}.

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