diff --git a/content/Cauchy-Euler equations (lec 10-11).md b/content/Cauchy-Euler equations (lec 10-11).md index 4784e82..f5c9875 100644 --- a/content/Cauchy-Euler equations (lec 10-11).md +++ b/content/Cauchy-Euler equations (lec 10-11).md @@ -17,7 +17,7 @@ $\frac{dy}{dx}=\frac{dy}{dt}{\frac{dt}{dx}}=\frac{ dy }{ dt }{\frac{1}{x}}\Right compute 2nd derivative of y wrt to x: $\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2} \frac{dt}{dx}\cdot \frac{dt}{dx}+\frac{dy}{dt}{\frac{d^2t}{dx^2}}=\frac{1}{x^2}{\frac{d^2y}{dt^2}}-\frac{\frac{1}{x^2}dy}{dt}$ $\underset{ \text{Important} }{ x^2{\frac{d^2y}{dx^2}}=\frac{d^2y}{dt^2}-\frac{dy}{dt} }$ -plugging those derivatives in we get: +plugging those derivatives in we get: #remember $$a\frac{d^2y}{dt^2}+(b-a){\frac{dy}{dt}}+cy(t)=f(e^t)$$ ^ this is a constant coefficient second order non-homogenous equation now! We can solve it now using prior tools.  diff --git a/content/Convolution (lec 19-20).md b/content/Convolution (lec 19-20).md new file mode 100644 index 0000000..30e89f1 --- /dev/null +++ b/content/Convolution (lec 19-20).md @@ -0,0 +1,84 @@ +# 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: +$$(f*g)(t):=\int _{0} ^t f(t-v)g(v)\, dv$$ +property 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) $\mathcal{L}\{f*g\}=F(s)G(s)$ +he will see us tomorrow at 10oclock. ;) +#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: +$$y'+y-\int _{0} ^t y(v)\sin(t-v) \, dv =-\sin t,\qquad y(0)=1$$ +this 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 do it another way + +$sY-1+Y-\mathcal{L}\{(y*\sin)(t)\}=-\frac{1}{s^2+1}$ +$\left( s+1-\frac{1}{s^2+1}\right)Y(s)=1-\frac{1}{s^2+1}=\frac{s^2}{s^2+1}$ +$\frac{(s^2+1)(s+1)-1}{s^2+1}Y(s)=\frac{s^2}{s^2+1}$ +$\frac{s^3+s^2+s+\cancel{ 1 }-\cancel{ 1 }}{s^2+1}Y(s)=\frac{s^2}{s^2+1}$ +$Y(s)=\frac{s}{s^2+s+1}$ +$y(t)=\mathcal{L}^{-1}\left\{ \frac{s}{s^2+s+1} \right\}=\mathcal{L}^{-1}\{\frac{s}{\left( s+\frac{1}{2} \right)^2+\left( \frac{\sqrt{ 3 }}{2} \right)^2}\}$ +$=\mathcal{L}^{-1}\left\{ \frac{s+\frac{1}{2}-\frac{1}{2}}{\left( s+\frac{1}{2} \right)^2+\left( \frac{\sqrt{ 3 }}{2} \right)^2} \right\}$ +$=e^{-t/2}\cos\left( \frac{\sqrt{ 2 }}{2}t \right)-\frac{1}{2} \frac{2}{\sqrt{ 3 }}\mathcal{L}^{-1}\left\{ \frac{\frac{\sqrt{ 3 }}{2}}{\left( s+\frac{1}{2} \right)^2+\left( \frac{\sqrt{ 3 }}{2} \right)^2} \right\}$ +$$y(t)=e^{-t/2}\left( \cos \frac{\sqrt{ 3 }}{2}t-\frac{1}{\sqrt{ 3 }} \sin \frac{\sqrt{ 3 }}{2}t\right)$$ +this is a good algorithmic method now for solving differential equations in software, for example solving circuits. + +## Transfer function +imagine we have the equation: +$$ay''+by'+cy=g(t), \qquad y(0)=y_{0},\ y'(0)=y_{1}$$ +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 \ No newline at end of file diff --git a/content/Laplace transform (lec 14-16).md b/content/Laplace transform (lec 14-16).md index 21c65ed..a6565ba 100644 --- a/content/Laplace transform (lec 14-16).md +++ b/content/Laplace transform (lec 14-16).md @@ -17,7 +17,7 @@ compute the LT of this funny function: $f(t)=\begin{cases}1 &\text{if } 0\leq t\leq 1 \\ 2 &\text{if } 1 log, inv trig, algebraic, trig, exp set u to the first in the list above