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:

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

Substituting numbers, we get

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

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

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

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

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

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

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

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:

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

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

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

$H(s)={\dfrac {s^{2}}{(s^{2}+400)(0.01s^{2}+s+10000)}}={\dfrac {100s^{2}}{(s^{2}+400)(s^{2}+100s+1000000)}}$

The break points are:

$0\upuparrows$(40db/decade up) $20\downdownarrows$(40db/decade down) $1000\downdownarrows$(40db/decade down)

Looking at the top part of the Bode plot, we see that the graph is indeed going up from 0 to 20, and then is flat from 20 to 1000 since the break point at 20 counteracted the break point at 0. Finally, the graph is going down after 1000, and at about 40db/decade.

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Written by Nathan Reeves ~ Checked by