Laplace transforms: Simple Electrical Network

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

$4(s{\mathcal {L}}\left\{i_{1}\right\}-i_{1}(0))+20{\mathcal {L}}\left\{i_{2}\right\}=50$

$\Rightarrow 4sI_{1}(s)+20I_{2}(s)={\frac {50}{s}}$

$0.005(s{\mathcal {L}}{i_{2}}-i_{2}(0))+{\mathcal {L}}\left\{i_{2}\right\}-{\mathcal {L}}\left\{i_{1}\right\}=0$

$\Rightarrow -500I_{1}(s)+[s+500]I_{2}(s)=0$

Solving for $I_{2}(s)$

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

We find the partial decomposition

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

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

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

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

$H(s)_{1}={\frac {25s+12500}{s(s^{2}+500s+2500)}}$

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