4.4 KiB
#ex #SoLE
Lets start modelling some electric circuits again:
t=0 and is then switched off (battery is bypassed) at t=1
Applying KVL:
0.2I_{1}'+0.1I_{3}'+2I_{1}=g(t) \qquad \text{where } g(t)=\begin{cases}6, & 0\leq t\leq 1 \\0, & 1<t\end{cases}
-I_{2}+0.1I_{3}'=0
I_{1}=I_{2}+I_{3}
Voila, a system of three linear equations.
#end of lec 21
#start of lec 22
eq 1) 0.2I_{1}'+0.1I_{3}'+2I_{1}=g(t) \qquad \text{where } g(t)=\begin{cases}6, & 0\leq t\leq 1 \\0, & 1<t\end{cases}
eq 2) -I_{2}+0.1I_{3}'=0
eq 3) I_{1}=I_{2}+I_{3}
Two differential equations, one algebraic equation. Our goal is to solve for I_{1}\ I_{2} and I_{3}
express g in terms of unit step function:
g(t)=6-6u(t-1)
combine equations 2 and 3:
-I_{1}+I_{3}+0.1I_{3}'=0
now we have just two equations:
eq 1) 0.2I_{1}'+0.1I_{3}'+2I_{1}=6-6u(t-1)
eq 2) -I_{1}+I_{3}+0.1I_{3}'=0
where I_1(0)=I_{2}(0)=I_{3}(0)=0 (the battery is just connected at t=0)
multiply both equations by 10:
2I_{1}'+1I_{3}'+20I_{1}=60(1-u(t-1))
-10I_{1}+10I_{3}+I_{3}'=0
hit it with the LT!
let:
J_{1}=\mathcal{L}\{I_{1}(t)\}(s)
J_3=\mathcal{L}\{I_{3}\}
then:
2sJ_{1}-\cancel{ 2I_{1}(0) }+sJ_{3}-\cancel{ I_{3}(0) }+20J_{1}=\mathcal{L}\{60(1-u(t-1))\}
2(s+10)J_{1}+sJ_{3}=60\left( \frac{1}{s}-\frac{e^{-s}}{s} \right)
=2(s+10)J_{1}+sJ_{3}=60 \frac{1-e^{-s}}{s}
Now for the second eq, hitting it with the LT yields:
-10J_{1}+(s+10)J_{3}=0
Let's isolate J_{1}:
\begin{matrix} 2(s+10)J_{1}+sJ_{3}=60 \frac{1-e^{-s}}{s}&\qquad\quad\times(s+10) \\ -\quad-10J_{1}+(s+10)J_{3}=0&\times s \\￣￣￣￣￣￣￣￣￣￣￣￣\end{matrix}
=\quad2(s+10)(s+10)J_{1}+\cancel{ sJ_{3}(s+10) }+10sJ_{1}-\cancel{ s(s+10)J_{3} }=(s+10)60 \frac{1-e^{-s}}{s}
=2(s^2+20s+100+5s)J_{1}=(s+10)60 \frac{1-e^{-s}}{s}
(s+5)(s+20)J_{1}=(s+10)30 \frac{1-e^{-s}}{s}
J_{1}(s)=30 \frac{s+10}{s(s+5)(s+20)}(1-e^{-s})
use partial fractions:
30 \frac{s+10}{s(s+5)(s+20)}=\frac{A}{s}+\frac{B}{s+5}+\frac{C}{s+20}
=\frac{{A(s+5)(s+20)+B(s^2+20s)+C(s^2+5s)}}{s(s+5)(s+20)}
Solve the system of equations:
A+B+C=0
25A+20B+5C=30
100A=300 \implies A=3
\implies B=-2 \qquad C=-1
J_{1}(s)=\left( \frac{3}{s}-\frac{2}{s+5}- \frac{1}{s+20} \right)(1-e^{-s})
invert the LT: (use \mathcal{L}^{-1}\{e^{-as}F(s)\}=f(t-a)u(t-a))
I_{1}=3-2e^{-5t}-e^{-20t}-u(t-1)(3-2e^{-5(t-1)}-e^{-20(t-1)})
Nice! Now we solve for I_{3}. Let's use: -10J_{1}+(s+10)J_{3}=0
J_{3}=\frac{10}{s+10}J_{1}
J_{3}=300 \frac{1}{s(s+5)(s+20)}(1-e^{ -s })
partial fraction it so we can eventually take the inverse LT:
skip some steps:
J_{3}=\left( \frac{3}{s}-\frac{4}{s+5}+\frac{1}{s+20} \right)(1-e^{ -s })
I_{3}=3-4e^{ -5t }+e^{ -20t }-(3-4e^{ -5(t-1) }+e^{ -20(t-1) })u(t-1)
Finally, I_{2} is simply:
I_{2}=I_{1}-I_{3}
Our final answer is:
\begin{matrix}I_{1}=3-2e^{-5t}-e^{-20t}-u(t-1)(3-2e^{-5(t-1)}-e^{-20(t-1)}) \\I_{3}=3-4e^{ -5t }+e^{ -20t }-(3-4e^{ -5(t-1) }+e^{ -20(t-1) })u(t-1) \\I_{2}=I_{1}-I_{3}\end{matrix}Every 2nd order linear equation can be written as a system of 2 1st order linear equations:
ay''+by'+cy=f
y'=z
\begin{cases}y'-z=0 \\az'+bz+cy=f\end{cases}
last example of the chapter:
#ex #SoLE #IVP
x'+y=0, \qquad x(0)=0
x+y'=1-u(t-2) \qquad y(0)=0
This is a review problem in chapter 7 of the textbook.
Hit equation 1 and 2 with the LT:
sX+Y=0
X+sY=\frac{{1-e^{ -2s }}}{s}
Let's isolate X
multiply equation 1 by s
and subtract eq 1 from eq 2:
-(s^2-1)X=\frac{{1-e^{ -2s }}}{2}
X(s)=e^{-2s} \frac{1}{s(s-1)(s+1)}-\frac{1}{s(s-1)(s+1)}
use partial fractions:
\frac{A}{s}+\frac{B}{s-1}+\frac{C}{s+1}=\frac{1}{s(s-1)(s+1)}
A(s^2-1)+B(s^2+s)+C(s^2-s)=1
A+B+C=0
B-C=0
-A=1\implies A=-1
\implies B=\frac{1}{2} \qquad C=\frac{1}{2}
X(s)=\left( -\frac{1}{s}+\frac{1}{2} \frac{1}{s-1}+\frac{1}{2} \frac{1}{s+1} \right)(e^{ -2s }-1)
inverse laplace:
x(t)=1-\frac{1}{2}e^t-\frac{1}{2}e^{-t}-u(t-2)\left( 1-\frac{1}{2}e^{t-2} -\frac{1}{2}e^{-(t-2)}\right)Now solve for y(t) too:
Y=-sX=-s (\frac{-1}{s(s-1)(s+1)}+ \frac{e^{ -2s }}{s(s-1)(s+1)})
=\frac{1}{(s-1)(s+1)}- \frac{e^{-2s}}{(s-1)(s+1)}
partial 'frac it:
\frac{1}{(s-1)(s+1)}=\frac{A}{s-1}+\frac{B}{s+1}
A(s+1)+B(s-1)=1
A+B=0
A-B=1
\implies A=\frac{1}{2} \quad B=-\frac{1}{2}
=\frac{1}{2}(1-e^{-2s})\left( \frac{1}{s-1}-\frac{1}{s+1} \right)
\implies y(t)=\frac{1}{2}(e^t-e^{-t})-\frac{1}{2}u(t-2)(e^{t-2}-e^{-(t-2)})we are done #end of lec 22