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Find the inverse Laplace transform?

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2 Answers

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)

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!