496 lines
18 KiB
Markdown
496 lines
18 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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### Convolution
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Convolution is a weird thingy that does weird things.
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$$(f*g)(t)=\int^t_0f(\tau)g(t-\tau)d\tau$$
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It is commutative ($f*g=g*f$) and is useful in transforms:
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$$\mathcal L\{f*g\}=\mathcal L\{f\}\mathcal L\{g\}$$
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!!! example
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To solve $4y''+y=g(t),y(0)=3, y'(0)=-7$:
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\begin{align*}
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4\mathcal L\{y''\}+\mathcal L\{y\}&=\mathcal L\{g(t)\} \\
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4(s^2\mathcal L\{y\}-s\cdot y(0) - y'(0))+\mathcal L\{y\} &=\mathcal L\{g(t)\} \\
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\mathcal L\{y\}(4s^2+1)-12s+28&=\mathcal L\{g(t)\} \\
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\mathcal L\{y\}&=\frac{\mathcal L\{g(t)\}}{4s^2+1} + \frac{12s}{4s^2+1} - \frac{28}{4s^2+1} \\
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y&=\mathcal L^{-1}\left\{\frac{1}{4s^2+1}\mathcal L\{g(t)\}\right\} + \mathcal L^{-1}\left\{3\frac{s}{s^2+\frac 1 4}\right\}-\mathcal L^{-1}\left\{7\frac{1}{s^2+\frac 1 4}\right\} \\
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&= \mathcal L^{-1}\left\{\frac 1 2\mathcal L\left\{\sin\left(\tfrac 1 2 t\right)\right\}\mathcal L\{g(t)\} \right\}+3\cos\left(\tfrac 1 2 t\right)-14\sin\left(\tfrac 1 2t\right) \\
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&=\frac 1 2\left(\sin\left(\tfrac 1 2 t\right)*g(t)\right)+3\cos\left(\tfrac 1 2 t\right)-14\sin\left(\tfrac 1 2t\right) \\
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&=\frac 1 2\int^t_0\sin(\tfrac 1 2\tau)g(t-\tau)d\tau + 3\cos(\tfrac 1 2 t)-14\sin(\tfrac 1 2 t)
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\end{align*}
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### Impulse
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The **impulse for duration $\epsilon$** is defined by the **dirac delta function**:
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$$
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\delta_\epsilon(t)=\begin{cases}
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\frac 1\epsilon & \text{if }0\leq t\leq\epsilon \\
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0 & \text{else}
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\end{cases}
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$$
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As $\epsilon\to 0, \delta_\epsilon(t)\to\infty$. Thus:
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$$
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\delta(t-a)=\begin{cases}
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\infty & \text{if }t=a \\
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0 & \text{else}
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\end{cases} \\
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\boxed{\int^\infty_0\delta(t-a)dt=1}
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$$
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If a function is continuous, multiplying it by the impulse function is equivalent to turning it on at that particular point. For $a\geq 0$:
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$$\boxed{\int^\infty_0\delta(t-a)dt=g(a)}$$
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Thus we also have:
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$$\mathcal L\{\delta (t-a)\}=e^{-as}\implies\mathcal L^{-1}\{1\}=\delta(t)$$
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## Heat flow
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The temperature of a tube from $x=0$ to $x=L$ can be represented by the following DE:
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$$\text{temp}=u(x,t)=\boxed{u_t=a^2u_{xx}},0<x<L,y>0$$
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Two boundary conditions are requred to solve the problem for all $t>0$ — that at $t=0$ and at $x=0,x=L$.
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- $u(x,0)=f(x),0\leq x\leq L$
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- e.g., $u(0,t)=u(L,t)=0,t>0$
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Thus the general solution is:
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$$
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\boxed{u(x,t)=\sum^\infty_{n=1}a_ne^{-\left(\frac{n\pi a}{L}\right)^2t}\sin(\frac{n\pi x}{L})} \\
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f(x)=\sum^\infty_{n=1}a_n\sin(\frac{n\pi x}{L})
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$$
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### Periodicity
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The **period** of a function is an increment that always returns the same value: $f(x+T)=f(x)$, and its **fundamental period** of a function is the smallest possible period.
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!!! example
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The fundamental period of $\sin x$ is $2\pi$, but any $2\pi K,k\in\mathbb N$ is a period.
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The fundamental periods of $\sin \omega x$ and $\cos\omega x$ are both $\frac{2\pi}{\omega}$.
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If functions $f$ and $g$ have a period $T$, then both $af+bg$ and $fg$ also must have period $T$.
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#### Manipulating polarity
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- even: $\int^L_{-L}f(x)dx=2\int^L_0f(x)dx$
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- odd: $\int^L_{-L}f(x)dx=0$
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- even × even = even
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- odd × odd = even
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- even × odd = odd
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## Orthogonality
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$$\int^L_{-L}\cos(\frac{m\pi x}{L})\sin(\frac{n\pi x}{L})dx=0$$
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$$
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\int^L_{-L}\cos(\frac{m\pi x}{L})(\frac{n\pi x}{L})dx=\begin{cases}
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2L & \text{if }m=n=0 \\
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L & \text{if }m=n\neq 0 \\
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0 & \text{if }m\neq n
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\end{cases}
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$$
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$$
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\int^L_{-L}\sin(\frac{m\pi x}{L}\sin(\frac{n\pi x}{L})dx=\begin{cases}
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L & \text{if }m=n \\
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0 & \text{if }m\neq n
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\end{cases}
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$$
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Functions are **orthogonal** on an interval when the integral of their product is zero, and a set of functions is **mutually orthogonal** if all functions in the set are orthogonal to each other.
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If a Fourier series converges to $f(x)$:
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$$f(x)=\frac{a_0}{2} + \sum^\infty_{n=1}\left(a_n\cos(\frac{n\pi x}{L})+b_n\sin(\frac{n\pi x}{L})\right)$$
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The **Euler-Fourier** formulae must apply:
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$$
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\boxed{a_n=\frac 1 L\int^L_{-L}f(x)\cos(\frac{n\pi x}{L})dx} \\
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\\
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\boxed{b_n=\frac 1 L\int^L_{-L}f(x)\sin(\frac{n\pi x}{L})dx}
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$$
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!!! example
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The Fourier series for the square wave function: $f(x)=\begin{cases}-1 & -\pi < x < 0 \\ 1 & 0 < x < \pi\end{cases}$
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The period is clearly $2\pi\implies L=\pi$. $f(x)$ is also odd, by inspection.
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\begin{align*}
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a_n&=\frac 1\pi\int^\pi_{-\pi}\underbrace{f(x)\cos(\frac{n\pi x}{\pi})}_\text{odd × even = odd}dx=0=a_0 \\
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b_n&=\frac 1 \pi\int^\pi_{-\pi}f(x)\sin(\frac{n\pi x}{\pi})dx \\
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\tag{even}&=\frac 2\pi\int^\pi_0f(x)\sin(nx)dx \\
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\tag{$f(x)>1$ when $x>0$}&=\frac 2\pi\int^\pi_0\sin(nx)dx \\
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&=\frac 2\pi\left[\frac{-\cos nx}{n}\right]^\pi_0 \\
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&=\begin{cases}
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\frac{4}{\pi n} & \text{if $n$ is odd} \\
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0 & \text{else}
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\end{cases}
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\therefore f(x)&=\sum^\infty_{n=1}\frac 2\pi\left(\frac{1-(-1)^n}{n}\sin(nx)\right) \\
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\tag{only odd $n$s are non-zero}&=\frac4\pi\sum^\infty_{n=1}\frac{1}{2n-1}\sin[(2n-1)x]
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\end{align*}
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Thus the Fourier series is $$.
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### Separation of variables
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To solve IBVPs, where $X(x)$ and $T(t)$ are exclusively functions of their respective variables:
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$$u(x,t)=X(x)T(t)$$
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Substituting it into the IBVP results in a **separation constant** $-\lambda$.
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$$\boxed{\frac{T'(t)}{a^2T(t)}=\frac{X''(x)}{X(x)}=-\lambda}$$
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Possible values for the separation constant are known as **eigenvalues**, and their corresponding **eigenfunctions** contain the unknown constant $a_n$:
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$$
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\lambda_n=\left(\frac{n\pi}{L}\right)^2 \\
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X_n(x)=a_n\sin(\frac{n\pi x}{L})
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$$
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### Wave equation
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A string stretched between two secured points at $x=0$ and $x=L$ can be represented by the following IBVP:
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$$
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u_{tt}=a^2u_{xx},0<x<L,t>0 \\
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u(0,t)=u(L,t)=0,t>0 \\
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u(x,0)=f(x), 0\leq x\leq L \\
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u_t(x,0)=g(x), 0\leq x\leq L
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$$
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The following conditions must be met:
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$$
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u(x,t)=\sum^\infty_{n=1}\sin(\frac{n\pi x}{L})\left(\alpha_n\cos(\frac{n\pi a}{L}t)+\beta_n\sin(\frac{n\pi a}{L}t)\right) \\
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\boxed{f(x)=\sum^\infty_{n=1}\alpha_n\sin(\frac{n\pi x}{L}),0\leq x\leq L} \\
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\boxed{g(x)=\sum^\infty_{n=1}\frac{n\pi a}{L}\beta_n\sin(\frac{n\pi x}{L}), 0\leq x\leq L}
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$$
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### Fourier symmetry
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To find a Fourier series for functions defined only on $[0, L]$ instead of $[-L, L]$, a **periodic extension** can be used.
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A **half-range sine expansion (HRS)** is used for odd functions:
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$$
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f_o(x)=\begin{cases}
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f(x) & x\in(0, L) \\
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-f(-x) & x\in(-L, 0)
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\end{cases}
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$$
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A **half-range cosine expansion (HRC)** is used for even functions:
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$$
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f_e(x)=\begin{cases}
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f(x) & x\in(0, L) \\
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f(-x) & x\in(-L, 0)
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\end{cases}
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$$
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Thus if a Fourier series on $(0,L)$ exists, it can be expressed as either a **Fourier sine series** (via HRS) or a **Fourier cosine series** (via HRC).
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!!! example
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For $f(x)=\begin{cases}\frac\pi 2 & [0,\frac\pi 2] \\ x-\frac\pi 2 & (\frac\pi2,\pi]\end{cases}$:
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\begin{align*}
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a_n&=\frac 2 L\int^L_0f(x)\cos(\frac{n\pi x}{L})dx \\
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&=\frac 2\pi \int^{\pi/2}_0\frac\pi 2\cos(\frac{n\pi x}{\pi})dx + \frac 2 \pi\int^\pi_{\pi/2}(x-\frac\pi2)\cos(\frac{n\pi x}{\pi})dx \\
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&=\frac{2}{n^2\pi}[(-1)^n-\cos(\frac{n\pi}{2})+\frac{n\pi}{2}\sin(\frac{n\pi}{2}) \\
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\\
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a_0&=\frac2\pi\int^\pi_0f(x)\cos(0)dx \\
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&=\frac{3\pi}{4} \\
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\\
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\therefore f(x)&=\frac{3\pi}{8}+\sum^\infty_{n=1}\frac{2}{n^2\pi^2}[(-1)^n-\cos(\frac{n\pi}{2})+\frac{n\pi}{2}\sin(\frac{n\pi}{2})]\cos(nx),x\in[0,\pi]
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\end{align*}
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!!! example
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For:
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$$
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u_t=2u_{xx},0<x<\pi,t:0 \\
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u(0,t)=u(\pi,t)=0,t>0 \\
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u(x,0)=\begin{cases}
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\frac\pi 2 & [0,\frac\pi 2] \\
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x-\frac\pi 2 & (\frac\pi 2,\pi]
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\end{cases}
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$$
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We have $L=\pi,a=\sqrt 2$.
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\begin{align*}
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u(x,t)&=\sum^\infty_{n=1}\alpha_ne^{-left(\frac{n\pi\sqrt 2}{\pi}\right)^2t}\sin(\frac{n\pi x}{\pi})
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&=\sum^\infty_{n=1}\apha_ne^{-2n^2t}\sin(nx) \\
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\alpha_n&=\frac 2 L\int^L_0f(x)\sin(\frac{n\pi x}{L})dx \\
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&=\frac2\pi\int^{\pi/2}_0\frac\pi 2\sin(nx)dx+\frac2\pi\int^\pi_{\pi/2}(x-\frac\pi2\sin(nx)dx \\
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&=\frac 1 n[1+(-1)^{n+1}-\cos(\frac{n\pi}{2})-\frac{2}{n\pi}\sin(\frac{n\pi}{2}]
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\end{align*}
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### Convergence of Fourier series
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!!! definition
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- $f(x^+)=\lim_{h\to0^+}f(x+h)$
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- $f(x^-=\lim_{h\to0^-}f(x+h)$
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If $f$ and $f'$ are piecewise continuous on $[-L, L]$ for $x\in(-L,L)$, where $a_n$ and $b_n$ are from the Euler-Fourier formulae:
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$$\frac{a_0}{2}+\sum^\infty_{n=1}a_n\cos(\frac{n\pi x}{L})+b_n\sin(\frac{n\pi x}{L})=\boxed{\frac 1 2[f(x^+)+f(x^-)]}$$
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At $x=\pm L$, the series converges to $\frac 1 2[f(-L^+)+f(L^-)]$. This implies:
|
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- A continuous $f$ converges to $f(x)$
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- A discontinuous $f$ has the Fourier series converge to the average of the left and right limits
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- Extending $f$ to infinity using periodicity allows it to hold for all $x$
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||
|
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!!! example
|
||
The square wave function $f(x)=\begin{cases}-1 & -\pi<x<0 \\ 1 & 0<x<\pi\end{cases},f(x+2\pi)=f(x)$:
|
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|
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$f$ and $f'$ are piecewise continuous, but the function is discontinuous at $k\pi,k\in\mathbb Z$. Thus at $x=\pm\pi$, the series converges to $\frac 1 2(-1+1)=0$. At $x=0$, the series converges to $\frac 1 2(1+(-1))=0$.
|
||
|
||
If $f$ is 2L-periodic and continuous on $-\infty,\infty$, and $f'$ is piecewise continuous on $[-L,L]$, the Fourier series converges **uniformly** to $f$ on $[-L,L]$ and thus any interval.
|
||
|
||
More formally, for every $\epsilon>0$, there exists an integer $N_0$ depending on $\epsilon$ such that $|f(x)-[\frac{a_0}{2}+\sum^N_{n=1}a_n\cos(\frac{n\pi x}{L})+b_n\sin(\frac{n\pi x}{L})]|<\epsilon$ for all $N\geq N_0$ and all $x\in(-\infty,\infty)$.
|
||
|
||
More intuitively, for a high enough summation of the Fourier series, the value must lie in an **$\epsilon$-corridor** of $f(x)$ such that $f(x)$ is always between $f(x)\pm\epsilon$.
|
||
|
||
!!! example
|
||
- The Fourier series for the triangle wave function **is** uniformly convergent.
|
||
- The Fourier series for the square wave function **is not** uniformly convergent, which means that Gibbs overshoots would not fit in an arbitrarily small $\epsilon$-corridor.
|
||
|
||
The **Weierstrass M-test** states that if $|a_n(x)|\leq M_n$ for all $x\in[a,b]$ and if $\sum^\infty_{n=1}M_n$ converges, then $\sum^\infty_{n=1}a_n(x)$ converges uniformly to $f(x)$ on $[a,b]$.
|
||
|
||
!!! example
|
||
$\sum^\infty_{n=1}\frac{1}{n^2}\cos(nx)$ converges uniformly on any finite closed interval $[a,b]$.
|
||
|
||
$|\frac{\cos(nx)}{n^2}|\leq\frac{1}{n^2}$ for all $x$, and $\sum^\infty_{n=1}\frac{1}{n^2}$ also converges. Thus the result follows from the M-test.
|
||
|
||
### Differentiating Fourier series
|
||
|
||
You can termwise differentiate the Fourier series of $f(x)$ only if:
|
||
|
||
- $f(x)$ is continuous on $(-\infty,\infty)$ and 2L-periodic
|
||
- $f'(x),f''(x)$ are both piecewise continuous on $[-L,L]$
|
||
|
||
You can termwise integrate the Fourier series of $f(x)$ only if $f(x)$ is piecewise continuous on $[-L,L]$.
|
||
|
||
Then, for any $x\in[-L,L]$:
|
||
|
||
$$\int^x_{-L}f(t)dt=\int^x_{-L}\frac{a_0}{2}dt+\sum^\infty_{n=1}\int^x_{-L}(a_n\cos(\frac{n\pi t}{L})+b_n\sin(\frac{n\pi t}{L}))dt$$
|
||
|
||
### Complex Fourier series
|
||
|
||
By employing Euler's theorem, sine and cosine can be transformed into exponential forms.
|
||
|
||
$$
|
||
\cos(\frac{n\pi x}{L})=\frac{e^{i\frac{n\pi x}{L}} + e^{-i\frac{n\pi x}{L}}}{2} \\
|
||
\sin(\frac{n\pi x}{L})=\frac{-ie^{i\frac{n\pi x}{L}} + ie^{-i\frac{n\pi x}{L}}}{2}
|
||
$$
|
||
|
||
Thus the **complex Fourier series** is given by:
|
||
|
||
$$
|
||
f(x)=\sum^\infty_{n=-\infty}c_ne^{i\frac{n\pi x}{L}} \\
|
||
c_n=\frac{1}{2L}\int^L_{-L}f(x)e^{-i\frac{n\pi x}{L}}dx = \frac 1 2(a_n-ib_n)
|
||
$$
|
||
|
||
To convert it to a real Fouier series:
|
||
|
||
- $a_0=2c_0$
|
||
- $a_n=c_n+\overline{c_n}$
|
||
- $b_n=i(c_n-\overline{c_n})$
|
||
|
||
!!! example
|
||
The complex Fourier series for the sawtooth wave function: $f(x)=x,-1<x<1,f(x+2)=f(x)$. Thus we have a period of 2 and $L=1$.
|
||
|
||
\begin{align*}
|
||
c_0&=\frac 1 2\int^1_{-1}\underbrace{xe^{0}}_\text{odd}dx \\
|
||
&=0 \\
|
||
\\
|
||
c_n&=\frac 1 2\int^1_{-1}xe^{-in\pi x}dx \\
|
||
\tag{IBP}&=\frac 1 2\left[\frac{xe^{-in\pi x}}{-in\pi}-\int\frac{1}{-in\pi}e^{-in\pi x}dx\right]^1_{-1} \\
|
||
&=\frac 1 2\left[\frac{xe^{-n\pi x}}{-in\pi}+\frac{1}{n^2\pi^2}e^{-in\pi x}\right]^1_{-1} \\
|
||
&=\frac{(-1)^ni}{n\pi} \\
|
||
\\
|
||
\therefore f(x)&=\sum^\infty_{\substack{n=-\infty \\ n\neq0}}\frac{(-1)^ni}{n\pi}e^{in\pi x}
|
||
\end{align*}
|
||
|
||
## Resources
|
||
|
||
- [Laplace Table](/resources/ece/laplace.pdf)
|