Laplace Transform¶
Definition¶
\[F(s) = L\big\{f(t)\big\} = \int_{0}^{\infty}{e^{-st}f(t)\ dt}
\]
The Laplace Transforms of Basic Functions¶
| \(\qquad f(t)\qquad\) | \(\qquad F(s)\qquad\) |
|---|---|
| \(1\) | \(1/s\) |
| \(t^n\) | \(\frac{n!}{s^{n+1}}\) |
| \(\exp{(at)}\) | \(\frac{1}{s-a}\) |
| \(t^ne^{at}\) | \(\frac{n!}{(s-a)^{n+1}}\) |
| \(\sin{(kt)}\) | \(\frac{k}{s^2 + k^2}\) |
| \(\cos{(kt)}\) | \(\frac{s}{s^2+k^2}\) |
| \(\sinh{(kt)}\) | \(\frac{k}{s^2-k^2}\) |
| \(\cosh{(kt)}\) | \(\frac{s}{s^2-k^2}\) |
Constraints¶
-
constraint 1 For a function \(f(t)\), \(\exists\) constant \(c\), \(M > 0\), and \(T>0\) s.t.
\[|f(t)| \leq M\ e^{ct}, \ \forall \ t > T \]\(\rightarrow f(t)\) 不能跑得比 \(\exp {(ct)}\) 快
-
constraint 2
\(f(t)\) should be piecewise continuous on \([0, \infty)\)
\(\rightarrow\ \forall \ t \in [0, \infty)\), \(\quad f(t)\) 為 discontinuous 的點有限
note : these constraints are sufficient conditions
五大定理
Five Theorem¶
Transforms of Derivatives¶
\[
\begin{align}
L\big\{f'(t)\big\} &= \int_0^{\infty}{e^{-st}\ f'(t)\ dt} = \Bigg[e^{-st}\ f(t)\Bigg]^{\infty}_0 + s\ \int_0^{\infty}{e^{-st}\ f(t)\ dt} \\\\
&=sL\big\{f(t)\big\} - f(0)
\\\\\\
L\big\{f''(t)\big\} &= s^2\ F(s)-sf(0)-f'(0)
\\\\
L\big\{f'''(t)\big\} &= s^3F(s) - s^2f(0)-sf'(0)-f''(0)
\\\\
L\big\{f^{(n)}(t)\big\} &= s^nF(s)-s^{n-1}f(0) - s^{n-2}f'(0)-\dots -sf^{(n-2)}-f^{(n-1)}(0)
\end{align}
\]
First Translation Theorem¶
translation for \(s\)
\[
L\big\{e^{at}\ f(t)\big\}=F(s-a)
\]
proof :
\[
\begin{align}
L\big\{e^{at}\ f(t)\big\}&=\int_0^{\infty}{e^{-st}\ e^{at}\ f(t)\ dt} = \int_0^{\infty}{e^{-(s-a)t}\ f(t)\ dt}
\\\\
&= F(s-a)
\end{align}
\]
Second Translation Theorem¶
translation for \(t\)
\[
L\big\{f(t-a)\cdot u(t-a)\big\} = e^{-as}\ F(s)
\]
or
\[
L\big\{g(t)\ u(t-a)\big\}=e^{-as}\ L\big\{g(t+a)\big\}
\]
in which \(u(t)\) is an unit step function.
proof :
\[
\begin{align}
L\big\{f(t-a)\ u(t-a)\big\} &= \int_0^{\infty}{e^{-st}\ f(t-a)\ u(t-a)\ dt}\\\\
&=\int_a^{\infty}{e^{-st}\ f(t-a)\ dt}
\qquad (\text{let }t_1=t-a)\\\\
&= \int_0^{\infty}{e^{-s(t_1+a)}\ f(t_1)\ dt_1} = e^{-as}\ F(s)
\end{align}
\]
Derivatives of Transforms¶
\[
L\big\{t^n\ f(t)\big\} = (-1)^n\ \frac{d^n}{ds^n}F(s)
\]
Convolution¶
Definition :
\[
\begin{gather}
f(t)*g(t)=\int_{-\infty}^{\infty}{f(\tau)\ g(t-\tau)\ d\tau}
\end{gather}
\]
When \(f(t)=0\) for \(t<0\) and \(g(t)=0\) for \(t<0\)
We have the same definition as below.
\[
\begin{gather}
f(t)*g(t)=\int_0^t{f(\tau)\ g(t-\tau)\ d\tau}
\end{gather}
\]
Convolution Theorem¶
\[
L\big\{f(t)*g(t)\big\}=F(s)\ G(s)
\]
Integration¶
from theorem above
\[
L\bigg\{\int_0^t{f(\tau)\ d\tau}\bigg\}=L\big\{f(t)*1\big\}=\frac{F(s)}{s}
\]
Transform of a Periodic Function¶
When
\[f(t+T)=f(t)\]
then
\[
L\big\{f(t)\big\}=\frac{1}{1-e^{-sT}}\int_0^T{e^{-st}\ f(t)\ dt}
\]
\[
\begin{align}
\\\\
\frac{1}{1-e^{-sT}}
\end{align}
\]
來自無窮等比級數
Dirac Delta Function¶
\[
\delta(t-t_0)= \Bigg\{
\begin{eqnarray}
\infty \quad \text{for}\ t=t_0\\
0 \quad \text{for}\ t\neq t_0
\end{eqnarray}
\]
Integration¶
\[\int_{-\infty}^{\infty}{\delta (t-t_0)\ dt}=1\]
Sifting¶
\[
\int_p^q{f(t)\ \delta (t-t_0)\ dt}=f(t_0)
\]
\[t_0 \in [p,q]\]
Laplace Transform¶
\[
L\big\{\delta (t-t_0)\big\}=e^{-t_0s
}
\]
\[
\text{when}\ t_0 >0
\]
proof :
\[
\begin{align}
\because \ \delta(t-t_0)&=u'(t-t_0)\\\\
\therefore \ L\big\{\delta(t-t_0)\big\}&=sL\big\{u(t-t_0)\big\}-u(0)\\\\
\end{align}
\]
due to second translation theorem, we have
\[
\begin{align}
L\big\{\delta(t-t_0)\big\}=\frac{e^{-t_0s}}{s}\ s=e^{-t_0s}
\end{align}
\]