We will now cover some more complicated Lapalce Transformations:
The First Translation Theorem states:
\(\mathcal {L} \{ e^{at} f(t) \} = F(s - a)\)
The inverse of the theorem is simply:
\(\mathcal {L}^{-1} \{ F(s-a) \} = e^{at} f(t)\)
Evaluate \( \mathcal {L} \{ e^{2t} \sin (\pi t) \} \).
We realize that the expression is in the format of the First Translation Theorem. We know that:
\( \mathcal {L} \{ \sin (\pi t) \} = \cfrac{\pi}{s^2 + {\pi}^2} \).
The solution will be \(F (s-a)\) where \(a = 2\):
\( \mathcal {L} \{e^{2t} \sin (\pi t) \} = \cfrac{\pi}{(s-2)^2 + {\pi}^2} \).
Evaluate \(\mathcal {L}^{-1} \left \{\cfrac{1}{(s+5)^3} \right\}\).
The expression is in the form \(\cfrac{1}{s^3}\) hinting the inverse is in the form \(t^2\):
\( \mathcal {L}^{-1} \left \{ \cfrac{1}{s^3} \right \} = \cfrac{2!}{2!} \mathcal {L}^{-1} \left \{ \cfrac{1}{s^3} \right \} = \cfrac{1}{2!} \mathcal {L}^{-1} \left \{ \cfrac{2!}{s^3} \right \} = \cfrac{t^2}{2}\)
Since the denominator was \(s+5\), we need to add \(e^{-5t}\):
\( \mathcal {L}^{-1} \left \{ \cfrac{1}{(s+5)^3} \right\} = \cfrac{1}{2} e^{-5t}t^2 \)
The Second Translation Theorem states:
\( \mathcal {L} \{ f(t-a) \mathcal {U} (t-a) \} = e^{-as}F(s) \)
The inverse of the theorem is simply:
\( \mathcal {L}^{-1} \{ e^{-as}F(s) \} = \mathcal {L} \{ f(t-a) \mathcal {U} (t-a) \} \)
Where \(\mathcal {U} (t-a) \) is the heaviside function or unit step function which is an "on/off" function:
\(\mathcal{U} (t-a) = \begin{cases} 0, & \text{if } 0 \le t \lt a \\ 1, & \text{if } t \ge a \end{cases}\)
Evaluate \(\mathcal {L} \{f(t)\}\) if \(f(t) = \begin{cases} \sinh {(t)}, & \text{if } 0 \le t \lt \frac{\pi}{8} \\ \sinh {(t)} + \cos {(t - \frac{\pi}{8})}, & \text{if } t \ge \frac{\pi}{8} \end{cases} \).
We can use the unit step function to rewrite \(f(t)\):
\(f(t) = \sinh {(t)} + \mathcal{U}(t - \frac{\pi}{8}) (\cos {(t - \frac{\pi}{8})} ) \)
Now we can use the Second Translation Theorem given:
\( \mathcal{L} \{ \cos {t} \} = \cfrac{s}{s^2 + 1}\)
\( \mathcal{L} \{ f(t) \} = \cfrac{1}{s^2-1} + e^{-\frac{\pi}{8}s} \cfrac{s}{s^2 + 1} \)
Evaluate \(\mathcal {L}^{-1} \{e^{-5s} \cfrac{s}{s^2 + 10}\}\).
We realize that the expression is in the form of the Second Translation Theorem. First, evaluate:
\( \mathcal {L}^{-1} \{ \cfrac{s}{s^2 + 10} \} = \cos {\sqrt{10}t } \)
Now we apply the Second Translation Theorem:
\( \mathcal {L}^{-1} \{ e^{-5s}\cfrac{s}{s^2 + 10} \} = \mathcal{U} (t-5) \cos {(\sqrt{10}(t-5)) } \)
The Derivative Theorem state:
\(\mathcal {L} \{ t^n f(t) \} = (-1)^n \cfrac{d^n}{ds^n} F(s)\)
You are less likely to use the inverse as the derivative of \(F(s)\) may be simplified making it challenging to identify that this theorm can be used.
Evaluate \(\mathcal{L} \{ t^2e^{3t} \}\).
Since there is a \(t^2\) multiplied to another function, we can use the Derivative Theorem:
\( \mathcal{L} \{ t^2e^{3t} \} = (-1)^2 \cfrac{d^2}{ds^2} F(s)\)
\( F(s) = \mathcal{L} \{ e^{3t} \} = \cfrac{1}{s-3}\)
\( \cfrac{d^2}{ds^2} \left ( \cfrac{1}{s-3} \right ) = \cfrac{d}{ds} \left ( \cfrac{-1}{(s-3)^2} \right )= \cfrac{2}{(s-3)^2}\)
\( \mathcal{L} \{ t^2e^{3t} \} = (-1)^2 \cfrac{d^2}{ds^2} F(s) = \cfrac{2}{(s-3)^2}\)
Note we could have also used the First Translation Theorem with \(f(t) = t^2\).