- #1

- 46

- 0

F(s) = (s-1)/(s+1)^3

Find:

f(t)

Solution:

Using the equation that when F(s) = n!/(s-a)^(n=1), L^(-1){F(s)} = t^n*e^(at)

So far I find that f(t) = e^(-t)*(-t^2+__)

The book says that f(t) = e^(-t)*(t-t^2)

How did they get the t?

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- Thread starter kahless2005
- Start date

- #1

- 46

- 0

F(s) = (s-1)/(s+1)^3

Find:

f(t)

Solution:

Using the equation that when F(s) = n!/(s-a)^(n=1), L^(-1){F(s)} = t^n*e^(at)

So far I find that f(t) = e^(-t)*(-t^2+__)

The book says that f(t) = e^(-t)*(t-t^2)

How did they get the t?

- #2

- 10

- 0

[tex]

F(s) \equiv \frac{s-1}{(s+1)^3} \equiv \frac{1}{(s+1)^2} - \frac{2}{(s+1)^3}

[/tex]

and apply the inverse laplace transform

when F(s) = n!/(s-a)^(n=1), L^(-1){F(s)} = t^n*e^(at)

for each term.

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