forked from Sasserisop/MATH201
100 lines
5.6 KiB
Markdown
100 lines
5.6 KiB
Markdown
# Convolution
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#convolution
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A convolution is an operation on two functions, we take two functions, convolute them and get a new function.
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Definition of convolution between $f$ and $g$:
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$$(f*g)(t):=\int _{0} ^t f(t-v)g(v)\, dv$$
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property 1) $f*g=g*f$
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proof:
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$f*g=\int _{0} ^t f(t-v)g(v)\, dv$
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u substituiton: $t-v=u \quad du=-dv$
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$f*g=-\int _{t} ^0 f(u)g(t-u) \, du$
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$f*g=\int _{0} ^t g(t-u)f(u)\, du=g*f \quad \Box$
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property 2) $(f+g)*h=f*h+g*h$
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property 3) $(f*g)*h=f*(g*h)$
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property 4) $f*0=0$
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property 5) $\mathcal{L}\{f*g\}=F(s)G(s)$
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#end of lec 19 #start of lec 20
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lets try proving property 5:
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recall property 5: $\mathcal{L}\{f*g\}=F(s)G(s)$
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$\mathcal{L}\{f*g\}=\int _{0} ^\infty \left[ e^{-st}\int_{0} ^t f(t-v)g(v) \, dv \right]\, dt$
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$\mathcal{L}\{f*g\}=\int _{0} ^\infty \left[ e^{-st}\int_{0} ^\infty u(t-v)f(t-v)g(v) \, dv \right]\, dt$
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two nested integrals!
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Using math 209, if both integrals exist, we can exchange the two integrals:
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$=\int _{0}^\infty [ g(v)\underbrace{ \int _{0}^\infty e^{-st}f(t-v)u(t-v)\, dt }_{ \mathcal{L}\{f(t-v)u(t-v)\}=e^{-vs}F(s) } ]\, dv$
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$=F(s)\int _{0} ^\infty e^{-vs}g(v)\, dv=F(s)G(s) \quad \Box$
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This is a very useful fact. We will see how it helps us solve differential equations.
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#ex #convolution #inv_LT
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Computing the following:
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$$\mathcal{L}^{-1}\left\{ \frac{1}{s^2+1}\frac{1}{s^2+1} \right\}$$
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We know the inverse LT of $\frac{1}{s^2+1}$ is $\sin(t)$
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Then, using property 5: $\mathcal{L}\{f*g\}=F(s)G(s)$
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$f*g=\mathcal{L}^{-1}\{F(s)G(s)\}=(\sin*\sin)(t)$
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$$\mathcal{L}^{-1}\{\frac{1}{s^2+1} \frac{1}{s^2+1}\}=(\sin*\sin)(t)$$
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This is pretty good but we can simplify it more.
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Using the definition of convolution:
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$\mathcal{L}^{-1}\left\{ \frac{1}{s^2+1} \frac{1}{s^2+1} \right\}=\int _{0}^t \sin(t-v)\sin(v)\, dv$
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use trig identity: $\sin \alpha \sin \beta=\frac{1}{2}(\cos(\alpha-\beta)-\cos(\beta+\alpha))$
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$=\frac{1}{2}\int _{0} ^t (\cos(t-2v)-\cos(t))\, dv$
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$=\frac{1}{2}\left( -\frac{1}{2}\sin(t-2v)|^t_{0}-t\cos t \right)=\frac{1}{2}\left( \frac{1}{2}\sin(t)+\frac{1}{2}\sin(t)-t\cos t \right)$
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$$\mathcal{L}^{-1}\{\frac{1}{s^2+1} \frac{1}{s^2+1}\}=\frac{1}{2}(\sin t-t\cos t)$$
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Note that if you tried using #partial_fractions you would fail:
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$\frac{1}{(s^2+1)(s^2+1)}=\frac{A+Bs}{s^2+1}+\frac{{C+Ds}}{(s^2+1)^2}$ <- <i>sad trombone plays</i>
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But sometimes, both convolution and partial fractions can work on the same problem. And sometimes partial fractions is easier or faster than using convolution. Keep that in mind for your exams. #remember
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One last thing, this problem could have also been solved faster and without convolution using property 30 from the [big LT table](drawings/bigLTtable.png).
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#ex #convolution #IVP #integral-differential
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solve for $y(t)$:
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$$y'+y-\int _{0} ^t y(v)\sin(t-v) \, dv =-\sin t,\qquad y(0)=1$$
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This is called an integral-differential equation.
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We can convert it to a differential equation by taking the derivative of both sides (wrt to $t$.):
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$y''+y'-y\sin(t-v)=-\cos t$
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ew that's a gross second order linear equation. Laplace is the only tool we've learned so far that has a chance at defeating this problem.
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But let's solve instead by taking the LT of the original equation, since that integral can be expressed as a convolution.
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Hit it with the LT!
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$sY-1+Y-\mathcal{L}\{(y*\sin)(t)\}=-\frac{1}{s^2+1}$
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using property 5, $\mathcal{L}\{f*g\}=F(s)G(s)$:
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$sY-1+Y-Y \frac{1}{s^2+1}=-\frac{1}{s^2+1}$
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isolate $Y(s)$:
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$\left( s+1-\frac{1}{s^2+1}\right)Y(s)=1-\frac{1}{s^2+1}=\frac{s^2}{s^2+1}$
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$\frac{(s^2+1)(s+1)-1}{s^2+1}Y(s)=\frac{s^2}{s^2+1}$
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$\frac{s^3+s^2+s+\cancel{ 1 }-\cancel{ 1 }}{s^2+1}Y(s)=\frac{s^2}{s^2+1}$
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$Y(s)=\frac{s}{s^2+s+1}$
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This cant be solved with partial fractions, but if you remember from lecture 16, it can be solved by completing the square.
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$y(t)=\mathcal{L}^{-1}\left\{ \frac{s}{s^2+s+1} \right\}=\mathcal{L}^{-1}\left\{\frac{s}{\left( s+\frac{1}{2} \right)^2+\left( \frac{\sqrt{ 3 }}{2} \right)^2}\right\}$
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$=\mathcal{L}^{-1}\left\{ \frac{s+\frac{1}{2}-\frac{1}{2}}{\left( s+\frac{1}{2} \right)^2+\left( \frac{\sqrt{ 3 }}{2} \right)^2} \right\}$
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$=e^{-t/2}\cos\left( \frac{\sqrt{ 3 }}{2}t \right)-\frac{1}{2} \frac{2}{\sqrt{ 3 }}\mathcal{L}^{-1}\left\{ \frac{\frac{\sqrt{ 3 }}{2}}{\left( s+\frac{1}{2} \right)^2+\left( \frac{\sqrt{ 3 }}{2} \right)^2} \right\}$
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$$y(t)=e^{-t/2}\left( \cos \frac{\sqrt{ 3 }}{2}t-\frac{1}{\sqrt{ 3 }} \sin \frac{\sqrt{ 3 }}{2}t\right)$$
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This is a good algorithmic method now for solving differential equations in software, for example solving circuits.
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## Transfer function
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Imagine we have the equation:
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$$ay''+by'+cy=g(t), \qquad y(0)=y_{0},\ y'(0)=y_{1}$$
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1)
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$ay''+by'+cy=g(t)$
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$y(0)=y'(0)=0$
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gives a solution $y_{*}$
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2)
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$ay''+by'+cy=0$
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$y(0)=y_{0},\ y'(0)=y_{1}$
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gives a soltuion $y_{**}=c_{1}y_{1}+c_{2}y_{2}$
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then by principle of super position:
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$y=y_{*}+y_{**}$
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solving 1) gives us:
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$as^2Y+bsY+cY=G(s)$
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$Y(s)=\frac{1}{as^2+bs+c}G(s)$ the limit approaches 0 for large s so it's a legitimate Laplace transform
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let $Y(s)=H(s)G(s)$
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where $H(s)=\frac{1}{as^2+bs+c}$ and called the transfer function
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we put in $g(t)$ into this system and we get out $Y(s)$. So it "transfers".
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$H(s)=\frac{Y(s)}{G(s)}$
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$\mathcal{L}^{-1}\{H\}=h(t)$ called the impulse response function. We will see why its called that later.
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$y_{*}(t)=(h*g)(t)$ (the first * only denotes a name, it doesn't mean convolution)
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$y(t)=(h*g)(t)+c_{1}y_{1}+c_{2}y_{2}$
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What does this mean? It means the particular solution for a linear second order equation is composed of the convolution between $g(t)$, which is the force applied to the system, and the impulse response function. But I don't know what this means physically yet.
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He's finished 8 minutes early, lets go!
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#end of lec 20 |