Laplace transforms:Series RLC circuit

Laplace Transform Example: Series RLC Circuit

Problem

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

Solution

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

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

Substituting numbers, we get

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

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

Now, we take the Laplace Transform and get

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

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

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

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

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

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

Using partial fraction decomposition, we find that

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

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

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

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

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

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

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

$\Rightarrow \underset{s\to \mathrm{\infty }}{\lim }\frac{s^3}{(s^2+400)(0.01s^2+s+10000)}=i(0)$

$\Rightarrow i(0)=0$

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

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

$\Rightarrow \underset{s\to 0}{\lim }\frac{s^3}{(s^2+400)(0.01s^2+s+10000)}=i(\mathrm{\infty })$

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

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:

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

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

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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

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

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

$i(t)=v(t)*h(t)=\underset{0}{\overset{t}{\int }}v(\tau )h(t-\tau )d\tau$

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

Here,

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

Thus,

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

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

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:

$\left[\begin{array}{c}i\\ \frac{di}{dt}\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 1000000& 10\end{array}\right]\left[\begin{array}{c}\int idt\\ i\end{array}\right]+\left[\begin{array}{c}0\\ 100cos(20t)\end{array}\right]$

$i=\left[\begin{array}{cc}0& 1\end{array}\right]\left[\begin{array}{c}\int idt\\ i\end{array}\right]+0$

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

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

$\left[\begin{array}{cc}\frac{40001e^{(5-5\sqrt{40001})t}+\sqrt{40001}{e}^{(5-5\sqrt{40001})t}+40001e^{(5+5\sqrt{40001})t}-\sqrt{40001}{e}^{(5+5\sqrt{40001})t}}{80002}& -\frac{e^{(5-5\sqrt{40001})t}-e^{(5+5\sqrt{40001})t}}{10\sqrt{40001}}\\ -\frac{100000(e^{(5-5\sqrt{40001})t}-e^{(5+5\sqrt{40001})t})}{\sqrt{40001}}& \frac{40001e^{(5-5\sqrt{40001})t}-\sqrt{40001}{e}^{(5-5\sqrt{40001})t}+40001e^{(5+5\sqrt{40001})t}+\sqrt{40001}{e}^{(5+5\sqrt{40001})t}}{80002}\end{array}\right]$

The solution, then, is

$x(t)=e^{At}x(0)+\underset{0}{\overset{t}{\int }}{e}^{A(t-\tau )}Bu(\tau )d\tau$

Since x(0)=0,

$x(t)=\underset{0}{\overset{t}{\int }}{e}^{A(t-\tau )}Bu(\tau )d\tau$

This gives the same solution as we got above.

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Written by Nathan Reeves ~ Checked by