Exercises

(1)
Which is the inverse Laplace transform of \(\displaystyle F(s)=\frac {s-1}{s^2+2s+5}\)?
(a)
\(f(t)=\frac {1}{2}e^{-t}\left (2\cos {(2t)}-\sin {(2t)}\right )\)
(b)
\(f(t)=e^{-t}\left (\cos {(2t)}-2\sin {(2t)}\right )\)
(c)
\(f(t)=\frac {1}{2}e^{-t}\left (2\cos {(2t)}+\sin {(2t)}\right )\)
(d)
\(f(t)=e^{-t}\left (\cos {(2t)}+\sin {(2t)}\right )\)
(e)
\(f(t)=e^{-t}\left (\cos {(2t)}-\sin {(2t)}\right )\)
(f)
\(f(t)=e^{-t}\sin {(2t)}\)
(2)
Evaluate \(\displaystyle \mathcal {L}^{-1}\left (\frac {s+9}{(s^2+2s+2)(s^2-1)}\right )\) at \(t=4\) to the nearest tenth.
(3)
Evaluate \(\displaystyle \mathcal {L}^{-1}\left (\dfrac {28s^2+2}{16s^4+s^2}\right )\) at \(t=2\pi \). Round to the nearest thousandth.
(4)
Evaluate \(\displaystyle \mathcal {L}^{-1}\left (\frac {s^4-3s^3+18s^2+81}{s^6+18s^4+81s^2}\right )\) at \(t=1\) to the nearest hundredth.
(5)
Let \(a,b\) be constants, \(a>0\), and \(f(t)\) be a continuous function. Given \(\displaystyle F(s)=\mathcal {L}(f(t))=\int _0^{\infty } f(t)e^{-st}\,dt\), which is \(\mathcal {L}^{-1}(F(as-b))\) in terms of \(a\), \(b\), and the function \(f\)?

Hint: use \(u\)-substitution to determine an expression for \(F(as)\); then use the First-Shifting Property to determine an expression for
\(F(a(s-b/a))=F(as-b)\).

(a)
\(\displaystyle f\left (\frac {t}{a}\right )\)
(b)
\(\displaystyle f\left (\frac {t}{a}\right )e^{bt}\)
(c)
\(\displaystyle f\left (\frac {t}{a}\right )e^{bt/a}\)
(d)
\(\displaystyle f\left (\frac {t}{a}\right )e^{-bt}\)
(e)
\(\displaystyle f\left (\frac {t}{a}\right )e^{-bt/a}\)
(f)
\(\displaystyle \frac {1}{a}f\left (\frac {t}{a}\right )\)
(g)
\(\displaystyle \frac {1}{a}f\left (\frac {t}{a}\right )e^{bt}\)
(h)
\(\displaystyle \frac {1}{a}f\left (\frac {t}{a}\right )e^{bt/a}\)
(i)
\(\displaystyle \frac {1}{a}f\left (\frac {t}{a}\right )e^{-bt}\)
(j)
\(\displaystyle \frac {1}{a}f\left (\frac {t}{a}\right )e^{-bt/a}\)