Laplace transforms: Simple Electrical Network

From Class Wiki

Jump to: navigation, search

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 I2(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 I1 where we solve the function in terms of I1 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

s2 + 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)}
Personal tools