Laplace transforms: Simple Electrical Network

Problem Statement

Using the formulas

$E(t)=L\frac{d{i}_R}{dt}+R{i}_C$
$RC\frac{d{i}_R}{dt}+i_R-i_C=0$

Solve the system when V0 = 50 V, L = 4 h, R = 20 Ω, C = 10^-4 f, and the currents are initially zero.

Solution

Solve the system when V0 = 50 V, L = 4 h, R = 20 Ω, C = 10^-4 f, and the currents are initially zero.

$4\frac{d{i}_1}{dt}+20i_2=50$

$20(10^{-4})\frac{d{i}_2}{dt}+i_2-i_1=0$

Applying the Laplace transform to each equation gives

$4(s{ℒ}\left\{i_1\right\}-i_1(0))+20{ℒ}\left\{i_2\right\}=50$

$\Rightarrow 4s{I}_1(s)+20I_2(s)=\frac{50}{s}$

$0.005(s{ℒ}{i}_2-i_2(0))+{ℒ}\left\{i_2\right\}-{ℒ}\left\{i_1\right\}=0$

$\Rightarrow -500I_1(s)+[s+500]I_2(s)=0$

Solving for $I_2(s)$

$I_2(s)=\frac{6250}{s(s^2+500s+2500)}$

We find the partial decomposition

Let $I_2(s)=\frac{6250}{s(s^2+500s+2500)}=\frac{A}{s}+\frac{Bs+C}{s^2+500s+2500}$

$\Rightarrow 6250=A(s^2+500s+2500)+(Bs+C)s$

$\Rightarrow 62500=A{s}^2+500As+2500A+B{s}^2+Cs$

Comparing the coefficients we get

$A=\frac{5}{2},B=-5,C=-1250$

Thus $I_2(s)=\frac{5}{2s}-\frac{5s+1250}{s^2+500s+2500}$

Now we do the same for $I_1$ where we solve the function in terms of $I_1$ and decomposing the partial fraction resulting in

$I_1(s)=\frac{25s+12500}{s(s^2+500s+2500)}=\frac{5}{s}-\frac{5s+2475}{s^2+500s+2500}$

In order to make it nicer on us we need to complete the square as follows

$s^2+500s+2500=0$

$\Rightarrow s^2+500s=-2500$

$\Rightarrow s^2+500s+{\left(\frac{500}{2}\right)}^2=-2500+{\left(\frac{500}{2}\right)}^2$

$\Rightarrow s^2+500s+62500=6000$

$\Rightarrow (s+250{)}^2-(100\sqrt{6}{)}^2=0$

Thus

$I_2(s)=\frac{5}{2s}-5\frac{s+250}{(s+250{)}^2-(100\sqrt{6}{)}^2}-\frac{5\sqrt{6}}{12}\frac{100\sqrt{6}}{(s+250{)}^2-(100\sqrt{6}{)}^2}$

Taking the Inverse Laplace transform gives

${{ℒ}}^{-1}\{I_2(s)\}=i_2(t)=\frac{5}{2}-5e^{-250t}cosh100\sqrt{6}t-\frac{5\sqrt{6}}{12}5e^{-250t}sinh100\sqrt{6}t$

Initial Value Theorem

$\underset{s\to \mathrm{\infty }}{\lim }sI(s)=f(0^{+})$

$\underset{s\to \mathrm{\infty }}{\lim }s\frac{25s+12500}{s(s^2+500s+2500)}=i(0)$

$\Rightarrow i(0)=0$

$\underset{s\to \mathrm{\infty }}{\lim }s\frac{6250}{s(s^2+500s+2500)}=i(0)$

$\Rightarrow i(0)=0$

Final Value Theorem

$\underset{s\to 0}{\lim }sI(s)=f(\mathrm{\infty })$

$\underset{s\to \mathrm{\infty }}{\lim }s\frac{25s+12500}{s(s^2+500s+2500)}=i(\mathrm{\infty })$

$\Rightarrow i(\mathrm{\infty })=0$

$\underset{s\to 0}{\lim }s\frac{6250}{s(s^2+500s+2500)}=i(\mathrm{\infty })$

$\Rightarrow i(\mathrm{\infty })=0$

Bode Plots

The following are bode plots for the transfer functions

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$H(s{)}_1=\frac{25s+12500}{s(s^2+500s+2500)}$

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$H(s{)}_2=\frac{6250}{s(s^2+500s+2500)}$