Question 5.6.1: Determine L^−1[F(s)] for each of the following functions:

(a) Because of the presence of e^{−2s} in F(s), we will use the second shifting property. But first, we find the partial fraction decomposition

\frac {1}{s(s +1)^{2}}= \frac {1}{s} − \frac {1}{s +1}− \frac {1}{(s +1)^{2}}

and note that

L^{−1} [\frac {1}{s(s +1)^{2}}]= L^{−1} [\frac {1}{s} − \frac {1}{s +1}− \frac {1}{(s +1)^{2}}]

Now, in order to compute the inverse transform of the given function, we use the second shifting property to address the presence of e^{−2s} in each term and thus find that

L^{−1} [\frac {e^{−2s}}{s(s +1)^{2}}]= u(t −2)[1−e^{−(t−2)} −(t −2)e^{−(t−2)}]

(b) Partial fractions shows that

F(s) = \frac {2}{s^{4} +4s^{2}}= \frac {1}{2} (\frac {1}{s^{2}}−\frac {1}{s^{2} +4})

Using the inverses of familiar transforms of f (t ) = t and f (t ) = sin2t, we see

L^{−1}[F(s)] = \frac {1}{2}(t − \frac {1}{2}sin2t)

(c) Given the function

F(s) = \frac {4se^{−2πs}}{(s^{2} +2s +5)(s^{2} +9)}

we see that the presence of e^{−2πs} implies the inverse of the second shifting property will be used. As is now custom, we first use partial fractions to break the rational part of F(s) into a sum of simpler expressions. Doing so and completing the square to re-express s^{2} +2s +5,

\frac {4s}{(s^{2} +2s +5)(s^{2} +9)}=\frac {1}{13} (−\frac {4s−18}{s^{2}+9}+ \frac {4s−10}{s^{2}+2s+5})

=\frac {1}{13}( −\frac {4s}{s^{2}+9}+ \frac {18}{s^{2}+9}+\frac {4(s+1)}{(s+1)^{2}+4}− \frac {14}{(s+1)^{2}+4})

Letting G(s) = 4s/(s^{2} +2s +5)(s^{2} +9), it now follows from familiar rules with inverse transforms and the first shifting property that

L^{−1}[G(s)]=− \frac {4}{13}cos3t + \frac {18}{39}sin3t + \frac {4}{13}e^{−t} cos2t − \frac {7}{13}e^{−t} sin2t

Finally, since [ F(s)] = e^{−2πs}G(s), the second shifting property implies

L^{−1} [F(s)] = u(t −2π)(− \frac {4}{13}cos 3(t −2π)+ \frac {6}{13}sin 3(t −2π))+u(t −2π)(\frac {4}{13}e^{−t} cos 2(t −2π)− \frac {7}{13}e^{−(t−2π)} sin 2(t −2π)

The 2π shift in each of the sine and cosine functions can be removed; for instance, cos 3(t −2π) = cos3t . Doing so throughout shows that

L^{−1}[F(s)] = u(t −2π)   (− \frac {4}{13}cos3t + \frac {6}{13}sin3t + \frac {4}{13}e^{−(t−2π)} cos2t − \frac {7}{13}e^{−(t−2π)} sin2t)