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:

$H(s)=\frac{s^2}{(s^2+400)(0.01s^2+s+10000)}=\frac{s^2}{0.01s^4+s^3+10004s^2+400s+4000000}$

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

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Written by Nathan Reeves ~ Checked by