Laplace Transform - Step Functions?

Question

Find the inverse Laplace Transform:

F(s) = (s-2)e^(-s) / (s^2 - 4s + 3)

My answer was (1/2)(u1(t))(e^(t-1) + e^(3t -3)), but the answer in the book is (u1(t))e^(2t-2)cosh(t-1)

I don't really get how the book got that. If you want to see my work below, then that is fine. I just don't understand what I did wrong.

Note: The inverse transform of 1 / (s-a) = e^(at)
The inverse transform of s^2 / (s^2 - a^2) = cosh at
The inverse transform of e^(-cs)F(s) = uc(t)f(t-c)

My work:

F(s) = (s-2)e^(-s) / (s^2 - 4s + 3)
(s^2 - 4s + 3) = (s-3)(s-1) = s^2 - 3s - s + 3 (this is a check)

(s-2) / (s-3)(s-1) = A / (s-3) + B / (s-1)
s -2 = As - A + Bs - 3B

A + B = 1
-A - 3B = -2 so A + 3B = 2
A = 1 - B
1 - B + 3B = 2
2B = 1
B = 1/2 and so A = 1/2

so now:

(1/2)e^(-s) / ((1/(s-3)) + (1/(s-1))

This Laplace Transform should be:

f(t) = (1/2)(u1(t))(e^(t-1) + e^(3t -3))

Is there someplace I went wrong?

Answer 1

First, you wrote that the inverse transform of s^2 / (s^2 - a^2) = cosh a... which is wrong. The numerator should be s, not s^2. The inverse transform of s / (s^2 - a^2) = cosh a. I don't want you to memorize something incorrectly if you'll be tested on it.

(Also, you got an answer that seems to be perfectly fine, but it is good to know that you can get an answer that looks different by doing, say, completion of the square, which might throw out the need to do partial fractions as the only option.)

F(s) = (s-2)e^(-s) / (s^2 - 4s + 3)

To get what the book has, it seems that you should do completing of the square; but completing the square is not always the way to do it, and partial fractions is sometimes needed. Since the denominator has s^2 + 4s, completing the square requires that you divide the 4 that serves as the coefficient of s by two and square it to get 4. You can then complete the square by adding 4 to the denominator and also adding its additive inverse, -4, to the denominator so that it's like you are adding zero. This makes the denominator be s^2 - 4s + 4 - 1. This is (s-2)^2 - 1.

So you have (s-2)e^(-s) / (s-2)^2 - 1. Seeing that the denominator AND numerator both have s-2 in them is a good thing because you can think of that -2 as coming from an e^2t, which is part of f(t), but it will have to be modified due to the presence of e^(-s), which means a shift will be involved.

Having e^2t as part of f(t) means that (s-2)e^(-s) / (s-2)^2 - 1 can be seen as [e^(-s)]s / s^2 - 1, which should be reminiscent of cosh t.

Here is where it comes together: You can figure out that cosh t and e^2t will be part of f(t). But since e^(-s) is part of the F(s) expression, you can know that u1(t) is part of f(t), too. So since u1(t) is part of f(t), which means that you have a shift that has to be placed on the other t's in f(t) (note that u1(t) means that the shift is t - 1), then the t in e^2t has to be replaced by t - 1 ( which means you get e^(2t-2) ), and the t in cosh t has to be replaced by t -1 as well, which means that cosh(t-1).

So u1(t))e^(2t-2)cosh(t-1) is what you get.

Answer 2

Recall that the laplace transform of a step function u_c(t)*g(t-c) = e^(-cs)*G(s)



In this inverse laplace transform, G(s) = (s-2)/(s²-4s+3) = 1/2*1/(s-1) + 1/2*1/(s-3)



(s-2)/(s-1)(s-3) = A/(s-1) + B/(s-3)



(s-3)A + (s-1)B = (s-2)



B=1/2



A=1/2



g(t) = L^(-1){G(s)} = 1/2*e^(t)+ 1/2*e^(3t)



g(t-c) = f(t-1) = 1/2*e^(t-1) + 1/2*e^(3t-3)



L^(-1){F(s)} = u1(t)*1/2*[e^(t-1)+e^(3t-3)]



You're right



The book's value is somewhat similar to yours.

Answer 3

you know there is a relation between cosh and exp functions.



Try it first, and I'll come back later if I have time.



OK, it seems to me that both answers are correct.