# Laplace transforms:Series RLC circuit

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## Laplace Transform Example: Series RLC Circuit

### Problem

Given a series RLC circuit with $R=10 Ohms$, $L=0.1 H$, and $C=10^{-5} F$, having power source $v(t)=10cos(20t)$, find an expression for $i(t)$ if $i(0)=0 A$ and $v_c(0)=0 V$.

### Solution

We begin with the general formula for voltage drops around the circuit:

$v(t)=Ri+L\dfrac{di}{dt}+\dfrac{1}{C}\int{i dt}$

Substituting numbers, we get

$10cos(20t)=10i+0.1\dfrac{di}{dt}+10^{5}\int{i dt}$

$\Rightarrow cos(20t)=i+0.01\dfrac{di}{dt}+10000\int{idt}$

Now, we take the Laplace Transform and get

$\dfrac{s}{s^2+20^2}=I+0.01[sI-i(0)]+10000\dfrac{I}{s}$

Using the fact that $i(0)=0A$, we get

$\dfrac{s}{s^2+400}=I+0.01sI+10000\dfrac{I}{s}$

$\Rightarrow \dfrac{s^2}{s^2+400}=sI+0.01s^2I+10000I$

$\Rightarrow \dfrac{s^2}{s^2+400}=(0.01s^2+s+10000)I$

$\Rightarrow I(s)=\dfrac{s^2}{(s^2+400)(0.01s^2+s+10000)}$

Using partial fraction decomposition, we find that

$I(s)=\dfrac{1.0004-4.003*10^{-8}s}{0.01s^2+s+10000}+\dfrac{4.003*10^{-6}s-0.04002}{s^2+400}$

$\Rightarrow I(s)=\dfrac{100.04-4.003*10^{-6}s}{s^2+100s+1000000}+\dfrac{4.003*10^{-6}s-0.04002}{s^2+400}$

$\Rightarrow I(s)=\dfrac{100.04-4.003*10^{-6}s}{(s+50)^2+997500}+\dfrac{4.003*10^{-6}s-0.04002}{s^2+400}$

$\Rightarrow I(s)=\dfrac{100.038}{(s+50)^2+(50\sqrt{399})^2}-\dfrac{4.003*10^{-6}s+.002}{(s+50)^2+(50\sqrt{399})^2}+\dfrac{4.003*10^{-6}s}{s^2+20^2}-\dfrac{0.04002}{s^2+20^2}$

$\Rightarrow I(s)=\dfrac{10.038}{50\sqrt{399}}\dfrac{50\sqrt{399}}{(s+50)^2+(50\sqrt{399})^2}-4.003*10^{-6}\dfrac{s+50}{(s+50)^2+(50\sqrt{399})^2}+4.003*10^{-6}\dfrac{s}{s^2+20^2}-\dfrac{0.04002}{20}\dfrac{20}{s^2+20^2}$

Finally, we take the inverse Laplace transform to obtain

$i(t)=0.01e^{-50t}sin(998.8t)-(4.003*10^{-6})e^{-50t}cos(998.8t)+(4.003*10^{-6})cos(20t)-0.002sin(20t) \,$

which is our answer.

### Initial/Final Value Theorems

We now want to use the Initial and Final Value Theorems on this problem.

The Initial Value Theorem states that

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

$\Rightarrow \lim_{s \to \infty}\dfrac{s^3}{(s^2+400)(0.01s^2+s+10000)}=i(0)$

$\Rightarrow i(0)=0$

In addition, when we actually evaluate $i(0)$ from our equation for $i(t)$, we find it to be 0 as well. So, things check out there.

The Final Value Theorem states that

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

$\Rightarrow \lim_{s \to 0}\dfrac{s^3}{(s^2+400)(0.01s^2+s+10000)}=i(\infty)$

$\Rightarrow i(\infty)=0$

This time, when we actually evaluate i(∞) from the equation for $i(t)$, we find it to be undefined. So here, the Final Value Theorem tells us something that is not necessarily true (in fact, because we have oscillating functions, we know that i(∞) will not be zero).

### Bode Plot

To get a Bode plot, we use the transfer function:

$H(s)=\dfrac{1}{0.01s^2+s+10000}$

We then use a program such as Octave or MATLAB to obtain the Bode plot, which looks like this:

Bode Plot

#### Break Points

We can also use break points to approximate and/or validate the Bode plot.

The break points of our function are determined by the transfer function

$H(s)=\dfrac{1}{0.01s^2+s+10000}=\dfrac{100}{(s^2+100s+1000000)}$

The break points are:

$1000 \downdownarrows$(40db/decade down)

Looking at the top part of the Bode plot, we see that the graph is indeed going down at roughly 40db/decade at 1000.

### Convolution

We now want to show how convolution can achieve the same result as our Laplace Transform methods.

Convolution means that

$i(t)=v(t)*h(t)=\int_{0}^{t}{v(\tau)h(t-\tau)d\tau}$

where $h(t)$ is the inverse Laplace transform of the transfer function.

Here,

$h(t)=0.1e^{-2500t}sin(998.7t) \,$

Thus,

$i(t)=\int_{0}^{t}{10cos(20\tau)(.01e^{-2500(t-\tau)}sin(998.7(t-\tau))d\tau}$

$\Rightarrow i(t)=-.000014e^{-2500t}cos(998.7t)-.000034e^{-2500t}sin(998.7t)+.000014cos(20t)+.0000002sin(20t)$

This doesn't look exactly like the answer we got above, but we expect this since convolution doesn't take initial conditions into account.

### State Equations

To begin the demonstration of a new method (state space equations), we want to translate the system into a set of state equations:

$\begin{bmatrix} i \\ \dfrac{di}{dt} \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ 1000000 & 10 \end{bmatrix} \begin{bmatrix} \int{idt} \\ i \end{bmatrix} + \begin{bmatrix} 0 \\ 100cos(20t) \end{bmatrix}$

$i=\begin{bmatrix} 0 & 1 \end{bmatrix} \begin{bmatrix} \int{idt} \\ i \end{bmatrix} +0$

Next, we solve the system using the matrix exponential method.

MATLAB tells us that $e^{At}$ is

$\begin{bmatrix} \dfrac{40001 e^{(5-5 \sqrt{40001}) t}+\sqrt{40001} e^{(5-5 \sqrt{40001}) t}+40001 e^{(5+5 \sqrt{40001}) t}-\sqrt{40001} e^{(5+5 \sqrt{40001}) t}}{80002} & -\dfrac{e^{(5-5 \sqrt{40001}) t}-e^{(5+5 \sqrt{40001}) t}}{10 \sqrt{40001}} \\ -\dfrac{100000 (e^{(5-5 \sqrt{40001}) t}-e^{(5+5 \sqrt{40001}) t})}{\sqrt{40001}} & \dfrac{40001 e^{(5-5 \sqrt{40001}) t}-\sqrt{40001} e^{(5-5 \sqrt{40001}) t}+40001 e^{(5+5 \sqrt{40001}) t}+\sqrt{40001} e^{(5+5 \sqrt{40001}) t}}{80002} \end{bmatrix}$

The solution, then, is

$x(t)=e^{At}x(0) + \int_{0}^{t} e^{A(t-\tau)}Bu(\tau) \, d\tau$

Since x(0)=0,

$x(t)=\int_{0}^{t} e^{A(t-\tau)}Bu(\tau) \, d\tau$

This gives the same solution as we got above.

Written by Nathan Reeves ~ Checked by