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Laplace transforms:Series RLC circuit - Class Wiki

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:

(thumbnail)
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

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