Laplace transforms:Series RLC circuit: Difference between revisions

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


<math>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}</math>
<math>H(s)=\dfrac{1}{0.01s^2+s+10000}</math>


We then use a program such as Octave or MATLAB to obtain the Bode plot, which looks like this:
We then use a program such as Octave or MATLAB to obtain the Bode plot, which looks like this:
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The break points of our function are determined by the transfer function
The break points of our function are determined by the transfer function


<math>H(s)=\dfrac{s^2}{(s^2+400)(0.01s^2+s+10000)}=\dfrac{100s^2}{(s^2+400)(s^2+100s+1000000)}</math>
<math>H(s)=\dfrac{1}{0.01s^2+s+10000}=\dfrac{100}{(s^2+100s+1000000)}</math>


The break points are:
The break points are:


<math>0 \upuparrows</math>(40db/decade up)
<math>20 \downdownarrows</math>(40db/decade down)
<math>1000 \downdownarrows</math>(40db/decade down)
<math>1000 \downdownarrows</math>(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.
Looking at the top part of the Bode plot, we see that the graph is indeed going down at roughly 40db/decade at 1000.


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Revision as of 22:24, 1 November 2009

Laplace Transform Example: Series RLC Circuit

Problem

Given a series RLC circuit with , , and , having power source , find an expression for if and .

Solution

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

Substituting numbers, we get

Now, we take the Laplace Transform and get

Using the fact that , we get

Using partial fraction decomposition, we find that

Finally, we take the inverse Laplace transform to obtain

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

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

The Final Value Theorem states that

This time, when we actually evaluate i(∞) from the equation for , 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:

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

The break points are:

(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.


Written by Nathan Reeves ~ Checked by