Laplace transforms:Series RLC circuit
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Laplace Transform Example: Series RLC Circuit
Given a series RLC circuit with R = 10Ohms, L = 0.1H, and C = 10 − 5F, having power source v(t) = 10cos(20t), find an expression for i(t) if i(0) = 0A and vc(0) = 0V.
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 i(0) = 0A, 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 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
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).
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:
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:
Looking at the top part of the Bode plot, we see that the graph is indeed going down at roughly 40db/decade at 1000.
We now want to show how convolution can achieve the same result as our Laplace Transform methods.
Convolution means that
where h(t) is the inverse Laplace transform of the transfer function.
This doesn't look exactly like the answer we got above, but we expect this since convolution doesn't take initial conditions into account.
To begin the demonstration of a new method (state space equations), we want to translate the system into a set of state equations:
Next, we solve the system using the matrix exponential method.
MATLAB tells us that eAt is
The solution, then, is
This gives the same solution as we got above.
Written by Nathan Reeves ~ Checked by