# Laplace transforms: Simple Electrical Network

## Contents

### Problem Statement

Using the formulas

$E(t)=L\dfrac{di_R}{dt}+Ri_C$
$RC\dfrac{di_R}{dt}+i_R-i_C=0$


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.

$4\frac{di_1}{dt}+20i_2=50$

$20(10^{-4})\frac{di_2}{dt}+i_2-i_1=0$

Applying the Laplace transform to each equation gives

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

$\Rightarrow4sI_1(s)+20I_2(s)=\frac{50}{s}$

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

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

Solving for $I_2(s)$

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

We find the partial decomposition

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

$\Rightarrow6250=A(s^2+500s+2500)+(Bs+C)s$

$\Rightarrow62500=As^2+500As+2500A+Bs^2+Cs$

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

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

## Initial Value Theorem

$\lim_{s \to \infty}sI(s)=f(0^+)$

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

$\Rightarrow i(0)=0$

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

$\Rightarrow i(0)=0$

## Final Value Theorem

$\lim_{s \to 0}sI(s)=f(\infty)$

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

$\Rightarrow i(\infty)=0$

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

$\Rightarrow i(\infty)=0$

## Bode Plots

The following are bode plots for the transfer functions

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

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