3.8 KiB
Convolution
A convolution is an operation of function, we take two functions, convolute them and get a new function. Definition of convolution between f and g:
(f*g)(t):=\int _{0} ^t f(t-v)g(v)\, dvproperty 1) f*g=g*f
proof:
f*g=\int _{0} ^t f(t-v)g(v)\, \underset{ t-v=u }{ dv }=-\int _{t} ^0 f(u)g(t-u) \, du
=\int _{0} ^t g(t-u)f(u)\, du=g*f \quad \Box
property 2) (f+g)*h=f*h+g*h
property 3) (f*g)*h=f*(g*h)
property 4) f*0=0
property 5) \mathcal{L}\{f*g\}=F(s)G(s)
#end of lec 19 #start of lec 20
lets try proving property 5:
recall property 5: \mathcal{L}\{f*g\}=F(s)G(s)
\mathcal{L}\{f*g\}=\int _{0} ^t \left( e^{-st}\int_{0} ^t f(t-v)g(v) \, dv \right)\, dt
\mathcal{L}\{f*g\}=\int _{0} ^t \left( e^{-st}\int_{0} ^\infty u(t-v)f(t-v)g(v) \, dv \right)\, dt
two nested integrals!
using math 209, if both integrals exist, we can exchange the two integrals:
=\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
=F(s)\int _{0} ^\infty e^{-rs}g(v)\, dv=F(s)G(s) \quad \Box
This is a very useful fact. We will see how it helps us solve differential equations.
ex:
\mathcal{L}^{-1}\left\{ \frac{1}{s^2+1}\frac{1}{s^2+1} \right\}we know the inverse of 1/s^2+1 is sin(t):
then:
=(\sin*\sin)(t)
=\int _{0}^t \sin(t-v)\sin(v)\, dv
use identity: \sin \alpha \sin \beta=\frac{1}{2}(\cos(\alpha-\beta)-\cos(\beta-\alpha) DOUBLE CHECK!
=\frac{1}{2}\int _{0} ^t (\cos(t-2v)-\cos(t))\, dv
=\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)
=\frac{1}{2}(\sin t-t\cos t)#ex solve the problem:
y'+y-\int _{0} ^t y(v)\sin(t-v) \, dv =-\sin t,\qquad y(0)=1this is called an integral-differential equation.
we can convert it to a differential equation by taking the derivative of both sides (wrt to dt.):
y''+y'-y\sin(t-v)=-\cos t
ew thats a gross second order linear equation. lets solve using laplace.
sY-1+Y-\mathcal{L}\{(y*\sin)(t)\}=-\frac{1}{s^2+1}
\left( s+1-\frac{1}{s^2+1}\right)Y(s)=1-\frac{1}{s^2+1}=\frac{s^2}{s^2+1}
\frac{(s^2+1)(s+1)-1}{s^2+1}Y(s)=\frac{s^2}{s^2+1}
\frac{s^3+s^2+s+\cancel{ 1 }-\cancel{ 1 }}{s^2+1}Y(s)=\frac{s^2}{s^2+1}
Y(s)=\frac{s}{s^2+s+1}
y(t)=\mathcal{L}^{-1}\left\{ \frac{s}{s^2+s+1} \right\}=\mathcal{L}^{-1}\{\frac{s}{\left( s+\frac{1}{2} \right)^2+\left( \frac{\sqrt{ 3 }}{2} \right)^2}\}
=\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\}
=e^{-t/2}\cos\left( \frac{\sqrt{ 2 }}{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\}
y(t)=e^{-t/2}\left( \cos \frac{\sqrt{ 3 }}{2}t-\frac{1}{\sqrt{ 3 }} \sin \frac{\sqrt{ 3 }}{2}t\right)this is a good algorithmic method now for solving differential equations in software, for example solving circuits.
Transfer function
imagine we have the equation:
ay''+by'+cy=g(t), \qquad y(0)=y_{0},\ y'(0)=y_{1}ay''+by'+cy=g(t)
y(0)=y'(0)=0
gives a solution y_{*}
2)
ay''+by'+cy=0
y(0)=y_{0},\ y'(0)=y_{1}
gives a soltuion y_{**}=c_{1}y_{1}+c_{2}y_{2}
then by principle of super position:
y=y_{*}+y_{**}
solving 1) gives us:
as^2Y+bsY+cY=G(s)
Y(s)=\frac{1}{as^2+bs+c}G(s) the limit approaches 0 for large s so its a legitimate Laplace transform
let Y(s)=H(s)G(s)
where H(s)=\frac{1}{as^2+bs+c} and called the transfer function
we put in g(t) and we get out Y(s). So it "transfers".
H(s)=\frac{Y(s)}{G(s)}
\mathcal{L}^{-1}\{H\}=h(t) called the impulse response function. We will see why its called that later.
y_{*}(t)=(h*g)(t)
y(t)=(h*g)(t)+c_{1}y_{1}+c_{2}y_{2}
he's finished 8 minutes early, lets go! #end of lec 20