Find the inverse Laplace transform?

Question

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

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

Answer 1

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

Complete the square on the denominator:

          s² + 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/(s² + 4)}

And the rest is easy:

          f(t) = 2e^-t L^-1{s/(s² + 4)}

                = 2e^-t cos(2t)

Answer 2

The denominator of the function has two complex roots:

                     s² + 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!