We will now cover some more complicated Laplace Transformations.
The Convolution Theorem states:
\( \mathcal {L} \{ f \ast g \} = F(s) G(s) \)
Where \(f \ast g\) is the convolution operator:
\( f \ast g = \displaystyle \int_0^t f(\tau) g(t-\tau) d\tau\)
The inverse of the theorem is simply:
\( \mathcal {L}^{-1} \{ F(s) G(s) \} = f \ast g \)
Evaluate \(\mathcal {L} \{ e^{-2t} \ast t^4 \}\).
Since there is a convolution operation, we can use the Convolution Theorem:
\( \mathcal {L} \{e^{-2t} \ast t^4 \} = \mathcal {L} \{e^{-2t} \} \mathcal {L} \{ t^4 \} = \cfrac{1}{s+2} \cfrac{5!}{s^5} \)
Evaluate \(\mathcal {L}^{-1} \left\{ \cfrac{s}{s^4-9} \right\} \).
With a bit of algebra we can rewrite as:
\(\mathcal {L}^{-1} \left \{\cfrac{s}{(s^2-3)(s^2+3)} \right \} =\mathcal {L}^{-1} \left \{ \cfrac{s}{s^2-3} \cfrac{1}{s^2+3} \right\}\)
Since this expression is the product of two terms, we can use the Convolution Theorem:
\(\mathcal {L}^{-1} \left\{ \cfrac{s}{s^4-9} \right\} = \mathcal {L}^{-1} \left\{ \cfrac{s}{s^2-3} \right\} \times \mathcal {L}^{-1} \left\{ \cfrac{1}{s^2+3} \right\}\)
\(\mathcal {L}^{-1} \left\{ \cfrac{s}{s^2-3} \right\} = \cosh {\sqrt{3}t} \)
\(\mathcal {L}^{-1} \left\{ \cfrac{1}{s^2+3} \right\} =\cfrac{1}{\sqrt{3}} \sin {\sqrt{3}t}\)
Putting it all together gives:
\(\mathcal {L}^{-1} \{ \cfrac{s}{s^4-9} \} = \cosh {\sqrt{3}t} \times \sin {\sqrt{3}t} \)
A special case is when \(g(t) = 0\) which gives:
\( f \ast 1 = \int_0^t {f(\tau) d\tau}\)
The Laplace Transformation is:
\( \mathcal{L} \{ \int_0^t {f(\tau) d\tau}\} = \cfrac{F(s)}{s} \)
The inverse Laplace Transformation is:
\( \mathcal {L}^{-1} \{ \cfrac{F(s)}{s} \} = \int_0^t {f(\tau) d\tau} \)
This allows us to solve integro-differential equations.
Solve \( f(t) = 2t -4 \int_0^t \sin (\tau) f(t-\tau) d\tau \).
First, let's rewrite the integral portion as a convolution:
\( f(t) = 2t -4 \sin (\tau) \ast f(t-\tau)\)
Next, we can take the Laplace on each side and use the convolution rule:
\( \mathcal{L} \{ f(t) \} = \mathcal{L} \{ 2t -4 \sin (\tau) \ast f(t-\tau) \} \)
\( F = \cfrac{2}{s^2} -4\cfrac{F}{s^2+1} \)
Then, we can isolate for \(F\):
\(F + 4\cfrac{F}{s^2+1} = \cfrac{2}{s^2} \)
\(F \left(1+ 4\cfrac{1}{s^2+1}\right) = \cfrac{2}{s^2} \)
\(F \left(\cfrac{s^2+5}{s^2+1}\right) = \cfrac{2}{s^2} \)
\(F = \cfrac{2(s^2+1)}{s^2(s^2+5)} \)
After, we can rewrite the right hand side using partial fractions:
\(F = \cfrac{As+B}{s^2} + \cfrac{Cs+D}{s^2+5} \)
\(F = \cfrac{(As+B)(s^2+5) + (Cs+D)(s^2)}{s^2(s^2+5)} \)
\(2s^2 + 2 = (As+B)(s^2+5) + (Cs+D)(s^2) \)
\(2s^2 + 2 = As^3 + 5As + Bs^2 + 5B + Cs^3 + Ds^2 \)
Since there is no \(s^3\) on the left hand side, \(A=0, C=0\). We are left with two equations and two unknowns:
\( 2s^2 + 2 = (B+D)s^2 + 5B \)
\( 2 = B+D \)
\( 2 = 5B \)
Solving yields \(B = \cfrac{2}{5} , D = \cfrac{8}{5}\). We can rewrite as:
\(F = \cfrac{2}{5}\cfrac{1}{s^2} + \cfrac{8}{5}\cfrac{1}{s^2+5} \)
Now take Laplace inverse on both sides:
\(\mathcal {L}^{-1} \{F\} = \mathcal {L}^{-1} \left\{\cfrac{2}{5}\cfrac{1}{s^2} + \cfrac{8}{5}\cfrac{1}{s^2+5} \right\} \)
\( f(t) = \cfrac{2}{5}t + \cfrac{8}{5 \sqrt{5}} \sin {\sqrt{5} t }\)
Consider a piecewise continuous functions with period \(T\) such as:
\(f(t) = \begin{cases} 0, & 0 \le t \lt \frac{T}{2} \\ 1, & \frac{T}{2} \lt t \le T \end{cases} \)
The Laplace Transformation is:
\( \mathcal{L} \{f\} = \cfrac{1}{1-e^{-sT}} \displaystyle \int_0^T {e^{-st} f(t)} dt\)
Evaluate \(\mathcal{L} \{f\} \) for the function above for a period of \(4\).
\( \mathcal{L} \{f\} = \cfrac{1}{1-e^{-sT}} \displaystyle \int_0^T {e^{-st} f(t)} dt\)
\( \mathcal{L} \{f\} = \cfrac{1}{1-e^{-4s}} \displaystyle \int_0^4 {e^{-st} f(t)} dt\)
Since the function is piecewise we split up the integral:
\( \mathcal{L} \{f\} = \cfrac{1}{1-e^{-4s}} \left ( \displaystyle \int_0^2 {e^{-st} f(t)} dt + \displaystyle \int_2^4 {e^{-st} f(t)} dt \right ) \)
\( \mathcal{L} \{f\} = \cfrac{1}{1-e^{-4s}} \displaystyle \int_2^4 {e^{-st}} dt \)
\( \mathcal{L} \{f\} = \cfrac{1}{1-e^{-4s}} \cfrac{-1}{s} (e^{-4s} - e^{-2s} ) \)
\( \mathcal{L} \{f\} = \cfrac{e^{-2s} - e^{-4s}}{s(1-e^{-4s})} \)
The Dirac function is a special unit impluse where the "on" time is squeezed to only a moment in time:
\(\delta (t-t_0) = \lim\limits_{a \to \infty} \delta_a (t-t_0) \)
Where the unit impluse is defined as:
\(\delta_a (t-t_0) = \begin{cases} 0, & 0 \le t \lt t_0 - a \\ \frac{1}{2a}, & t_0 - a \le t \lt t_0 + a \\ 0, & t_0 + a \ge t \end{cases} \)
That means the Dirac Function is infinitiy when \(t = t_0\). It is found in applications with temporary external forces or voltage, for example.
The Laplace Transformation is:
\( \mathcal{L} \{\delta (t-t_0) \} = e^{-st_0}\)
Solve \(y'' + 2y' +2y = \delta(t-\pi)\) subject to \(y(0) = 1, y'(0) = 0\).
First, we can take the Laplace on both sides:
\( \mathcal{L} \{ y'' + 2y' +2y \} = \mathcal{L} \{ \delta(t-\pi) \}\)
\( (s^2Y - sy(0) - y'(0)) + 2(sY - y(0)) + 2Y = e^{-\pi s } \)
\( s^2Y - s + 2sY - 2 + 2Y = e^{-\pi s } \)
\( Y(s^2 + 2s + 2) = e^{-\pi s } + s + 2 \)
\( Y = \cfrac{ e^{-\pi s } + s + 2} {s^2 + 2s + 2} \)
Next, we can complete the square in the denominator and split up the terms in the numerator:
\(Y = \cfrac{ e^{-\pi s } + s + 2} {(s+1)^2 + 1 } \)
\(Y = \cfrac{ e^{-\pi s }} {(s+1)^2 + 1 } + \cfrac{ s+1 } {(s+1)^2 + 1 } + \cfrac{ 1} {(s+1)^2 + 1 }\)
We split up the \(s+2\) to \(s+1\) and \(1\) since the denominator has a translation factor \((s+1)^2\).
Then, we can take the Laplace inverse of each term. The first term uses both the First and Second Translation Theorems:
\( \mathcal{L}^{-1} \{ \cfrac{ 1} {(s +1)^2 + 1 } \}= e^{-t} \sin(t) \)
\( \mathcal{L}^{-1} \{ \cfrac{ e^{-\pi s }} {(s+1)^2 + 1 } \}= \mathcal{U} (t-\pi) e^{-(t-\pi)} \sin(t - \pi) \)
The second and third terms use the First Translation Theorem:
\( \mathcal{L}^{-1} \{ \cfrac{ s+1} {(s +1)^2 + 1 } \}= e^{-t} \cos(t) \)
\( \mathcal{L}^{-1} \{ \cfrac{ 1} {(s +1)^2 + 1 } \}= e^{-t} \sin(t) \)
Putting it all together yields:
\( y = \mathcal{U} (t-\pi) e^{-(t-\pi)} \sin(t - \pi) + e^{-t} \cos(t) + e^{-t} \sin(t)\)