Find the inverse Laplace transform of F(s)=(2s+2)/(s^2+2s+5).

Answer: f(t)=2e^-t*cos 2t

Find the inverse Laplace transform of F(s)=(2s+2)/(s^2+2s+5).

Answer: f(t)=2e^-t*cos 2t

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This problem can also be done without resorting to complex numbers.

Complete the square on the denominator:

s^{2} + 2s + 5 = (s+1)^{2} + 4

Since the denominator is now expressed in terms of s+1, express the numerator the same way:

2s + 2 = 2(s+1)

Now the whole fraction is in terms of s+1. A Laplacian translation theorem says we can substitute "s" for "s+1" if we compensate by multiplying the inverse Laplacian by e^{-t}:

f(t) = L^{-1}{2(s+1)/[(s+1)^{2} + 4]}

= e^{-t }L^{-1}{2s/(s^{2} + 4)}

And the rest is easy:

f(t) = 2e^{-t} L^{-1}{s/(s^{2} + 4)}

= 2e^{-t }cos(2t)

The denominator of the function has two complex roots:

s^{2} + 2s + 5 = (s+1+2i)(s+1 -2i) (1)

Thus,

F(s) = 2(s+1)/[(s + 1 +2i)(s+1 -2i)] = 2(s+1) {(-i/4) [ 2i/(s+1 -2i) - 2i/(s+1+2i)] =

2(s+1)/{(1/2)[ 1/(s+1 -2i) - 1/(s+ 1 +2i)]. (2)

Now let's do thje following transformations:

(s+1)/(s+1 -2i) = (s+1 - 2i +2i)/(s+1 - 2i) = 1 + 1/(s+1 -2i) and

(s+1)/(s+1 + 2i) = (s+1 + 2i - 2i)/(s+ 1+ 2i) = 1 - 1/(s+1 + 2i).

Now if you take the difference of the rational functions shown in (2), you will obtain

F(s) = [1/(s+1 -2i) - 1/(s+1 + 2i)]

Thus,

L^{-1} F(s) = L^{-1} [1/(s+1 - 2i)] - L^{-1} [1/(s+1 + 2i)] (3)

You already know (from inverse transformation tables and your prior calculations) that

L^{-1} (1/s-a)) = e ^{
at}

If you substitute "a" by s+1-2i one time and by s +1 + 2i in the second part of (3), then you will have

L^{-1} F(s) =e ^{(-1 +2i)t} + e^{(-1-2i)t }= 2e^{-t}^{ } [(e
^{- 2it} +e ^{2it})/2] (4)

But the expression in parenthesis in (4) is cos 2t (use Muavre's formula to make sure that sin-s functions cancel each other). Thus, your answer is

f(t) = L^{-1} F(s) = 2 e^{-t} cos (2t) (5)

What about calculkating Laplace transforms using Texas instruments, I don't have TI - 89 to discuss available options. But I can find old versions of Texas instruments, and I'll try to figure it out later.

Good luck to you!

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