5.9 KiB
ECE 205: Advanced Calculus 1
Laplace transform
The Laplace transform is a wonderful operation to convert a function of \(t\) into a function of \(s\). Where \(s\) is an unknown variable independent of \(t\):
\[ \mathcal L\{f(t)\}=F(s)=\int^\infty_0e^{-st}f(t)dt, s > 0 \]
??? example To solve for \(\mathcal L\{\sin(at)\}\):
\begin{align*}
\mathcal L\{f(t)\}&=\int^\infty_0e^{-st}\sin(at)dt \\
\\
\text{IBP: let $u=\sin(at)$, $dv=e^{-st}dt$:} \\
&=\lim_{B\to\infty} \underbrace{\biggr[
\cancel{-\frac 1 se^{-st}\sin(at)}}_\text{0 when $s=0$ or $s=\infty$}+\frac a s\int e^{-st}\cos(at)dt
\biggr]^B_0 \\
&=\frac a s\lim_{B\to\infty}\left[\int e^{-st}\cos(at)dt \right]^B_0 \\
\text{IBP: let $u=\cos(at)$, $dv=e^{-st}dt$:} \\
&=\frac a s \lim_{B\to\infty}\left[
-\frac 1 s e^{-st}\cos(at)-\frac a s\underbrace{\int e^{-st}\sin(at)dt}_{\mathcal L\{\sin(at)\}}
\right]^B_0 \\
&=\frac{a}{s^2}-\frac{a^2}{s^2}\mathcal L\{\sin(at)\} \\
\mathcal L\{\sin(at)\}\left(1+\frac{a^2}{s^2}\right)&=\frac{a}{s^2} \\
\mathcal L\{\sin(at)\}&=\frac{a}{a^2+s^2}, s > 0
\end{align*}A piecewise continuous function on \([a,b]\) is continuous on \([a,b]\) except for a possible finite number of finite jump discontinuities.
- This means that any jump discontinuities must have a finite limit on both sides.
- A piecewise continuous function on \([0,\infty)\) must be piecewise continuous \(\forall B>0, [0,B]\).
The exponential order of a function is \(a\) if there exist constants \(K, M\) such that: \[|f(t)|\leq Ke^{at}\text{ when } t\geq M\]
!!! example - \(f(t)=7e^t\sin t\) has an exponential order of 1. - \(f(t)=e^{t^2}\) does not have an exponential order.
Linearity
A piecewise continuous function \(f\) on \([0,\infty)\) of an exponential order \(a\) has a defined Laplace transform for \(s>a\).
Laplace transforms are linear. If there exist LTs for \(f_1, f_2\) for \(s>a_1, a_2\), respectively, for \(s=\text{max}(a_1, a_2)\):
\[\mathcal L\{c_1f_1 + c_2f_2\} = c_1\mathcal L\{f_1\} + c_2\mathcal L\{f_2\}\]
??? example We find the Laplace transform for the following.
$$
f(t)=\begin{cases}
1 & 0\leq t < 1 \\
e^{-t} & t\geq 1
\end{cases}
$$
Clearly $f(t)$ is piecewise ocontinuous on $[0,\infty)$ and has an exponential order of -1 when $t\geq 1$ and 0 when $0\leq t<1$. Thus $\mathcal L\{f(t)\}$ is defined for $s>0$.
\begin{align*}
\mathcal L\{f(t)\}&=\int^1_0 e^{-st}dt + \int^\infty_1e^{-st}e^{-t}dt \\
\tag{$s\neq 0$}&=\left[-\frac 1 s e^{-st}\right]^1_0 + \int^\infty_1e^{t(-s-1)}dt \\
&=-\frac 1 se^{-s}+\frac 1 s + \lim_{B\to\infty}\left[ \frac{1}{-s-1}e^{t(-s-1)} \right]^B_1 \\
\tag{$s\neq 0,s>-1$}&=\frac{-e^{-s}+1}{s} -\frac{e^{-s-1}}{-s-1}
\end{align*}
We solve for the special case $s=0$:
\begin{align*}
\mathcal L\{f(t)\}&=\int^1_0 e^{0}dt + \int^\infty_1e^{-st}e^{-t}dt \\
&=1 -\frac{e^{-s-1}}{-s-1} \\
\end{align*}
$$
\mathcal L\{f(t)\}=
\begin{cases}
\frac{-e^{-s}+1}{s}-\frac{e^{-s-1}}{-s-1} & s\neq 0, s>-1 \\
1-\frac{e^{-s-1}}{-s-1} &s=0
\end{cases}
$$If there exists a transform for \(s>a\), the original function multiplied by \(e^{-bt}\) exists for \(s>a+b\).
\[\mathcal L\{f(t)\}=F(s), s>a\implies \mathcal L\{e^{-bt}f(t)\}=F(s),s>a+b\]
Inverse transform
The inverse is found by manipulating the equation until you can look it up in the Laplace Table.
The inverse transform is also linear.
Inverse of rational polynomials
If the transformed function can be expressed as a partial fraction decomposition, it is often easier to use linearity to reference the table.
\[\mathcal L^{-1}\left\{\frac{P(s)}{Q(s)}\right\}\]
- \(Q, P\) are polynomials
- \(\text{deg}(P) > \text{deg}(Q)\)
- \(Q\) is factored
??? example \[\begin{align*} \mathcal L^{-1}\left\{\frac{s^2+9s+2}{(s-1)(s^2+2s-3)}\right\} &=\mathcal L^{-1}\left\{\frac{A}{s-1}+\frac{B}{s+3} + \frac{Cs+D}{(s-1)^2}\right\} \\ &\implies A=2,B=3,C=-1 \\ &=2\mathcal L^{-1}\left\{\frac{1}{s-1}\right\} + 3\mathcal L^{-1}\left\{\frac{1}{(s-1)^2}\right\}-\mathcal L^{-1}\left\{\frac{1}{s+3}\right\} \\ &=2e^t+3te^t-e^{-3t} \end{align*}\]
Inverse of differentiable equations
If a function \(f\) is continuous on \([0,\infty)\) and its derivative \(f'\) is piecewise continuous on \([0,\infty)\), for \(s>a\):
\[ \mathcal L\{ f'\}=s\mathcal L\{f\}-f(0) \\ \mathcal L\{ f''\} = s^2\mathcal L\{f\}-s\cdot f(0)-f'(0) \]
Solving IVPs
Applying the Laplace transform to both sides of an IVP is valid to remove any traces of horrifying integration.
!!! example \[\begin{align*} y''-y'-2y=0, y(0)=1, y'(0)=0 \\ \mathcal L\{y''-y'-2y\}&=\mathcal L\{0\} \\ s^2\mathcal L\{y\}-s\cdot y(0)-y'(0) - s\mathcal L\{y\} +y(0) - 2\mathcal L\{y\}&=0 \\ \mathcal L\{y\}(s^2-s-2)-s+1&=0 \\ \mathcal L\{y\}&=\frac{s-1}{(s-2)(s+1)} \\ &= \\ \mathcal L^{-1}\{\mathcal L\{y\}\}&=\mathcal L^{-1}\left\{ \frac 1 3\cdot\frac{1}{s-2} + \frac 2 3\cdot\frac{1}{s+1} \right\} \\ y&=\frac 1 3\mathcal L^{-1}\left\{\frac{1}{s-2}\right\} + \frac 2 3\mathcal L^{-1}\left\{\frac{1}{s+1}\right\} \\ \tag{from Laplace table}&=\frac 1 3 e^{2t} + \frac 2 3 e^{-t} \end{align*}\]
Heaviside / unit step
The Heaviside and unit step functions are identical:
\[ H(t-c)=u(t-c)=u_c(t)=\begin{cases} 0 & t < c \\ 1 & t \geq c \end{cases} \]
Piecewise continuous functions can be manipulated into a single equation via the Heaviside function.
For a Heaviside transform \(\mathcal L\{u_c(t)g(t)\}\), if \(g\) is defined on \([0,\infty)\), \(c\geq 0\), and \(\mathcal L\{g(t+c)\}\) exists for some \(s>s_0\):
\[ \mathcal L\{u_c(t)g(t)\}=e^{-sc}\mathcal L\{g(t+c)\},s>s_0 \]
Likewise, under the same conditions, shifting it twice restores it back to the original.
\[ \mathcal L\{u_c(t)f(t-c)\}=e^{-sc}\mathcal L\{f\} \]