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