## Wednesday, June 10, 2009

### Circuit Analysis II Lecture 2

Disclaimer: This series of posts is to serve as notes for myself as well as any others interested in the subject of Circuit Analysis II. This is the course ECE 213 at UNM.
The course will follow the book:
Electric Circuits (8th Edition) If you haven't read Lecture 1 it can be found here:Lecture 1

Also if you notice the links on the right, there is a list of links that will contain all the links to this course and you can move around freely, this will become nicer as there are more Lectures posted.

The main topic of this class period was converting actual circuits to the s domain, then using algebra to re-arrange the equations to a format of a function that the table has for you.

Key points:
• Time domain is represented by small letters
• S domain is represented by Capital Letters
• di/dt = sI(s) - i(0)
• I0 = i(0)
• ZL = sL and since s = jw then ZL = jwL
• I0 indicates Initial charge
• If your target is Voltage convert the circuit to use Current
• If your target is Current convert the circuit to use Voltage
First problem was to convert a inductor in series with a resistor

Our target was Current so we converted it using Table 13.1 to a series circuit.
Inductor = sL
Source = LI0
Resistor = R

Now we are able to treat this as a circuit using Ohms law (Remember V/IR)
So, I = (LI0)/(R + sL)

Now it's time to move back to the time domain,
Since this circuit looks most like k/(s+a)
we can massage it into a form similar
pulling L from the bottom we get the equation:
I = L(I
o)/L(s + R/L)

The L's cancel so you are left with:
I0/(s+(R/L))

Which if you say:
K = I0
a = R/L
you have the form k/s+a

Simply Converting back you have
I0e-(R/L)t

Next we tried the same process with a circuit containing a Capacitor
Again the target is Voltage so we use Current.

so using the formula v = ir
and
V = CV0Zeq

Zeq = R1R2 / R1 + R2

Inserting the "newly determined" values for the capacitor and Resistor.
We have ((1/cs)*R)/(R + (1/cs)) which is equal to:
R/(Rcs + 1)

Putting that back in our equation gives us
V = CV0 * R/(Rcs + 1)

extracting C and R from the equation gives us:
CRV0/CR(s+1/RC)

That is then transformed to
V0 / s+ (1/RC)

Now we do the same as with the Inductor circuit we try to find a similar Laplace Transform we can invert!
Again it looks like k/s+a
k = V0
a = 1/RC

V0e-t/RC

The Final Problem we did was try to use the RL circuit from 7.3 which is a RL circuit with a source:

We are going to say that L has no initial charge thus I0 = 0

Converting to the S domain we get
Inductor = sL
Resistor = R
Source = Vs/s

Now we look at the equation we can get from that:
I = (Vs/s)/(R+sL)

This can be rewritten as:
Vs/s(sL + R)

Using Partial Fraction Decomposition we would use an equation of the form:
A/s + B/s+a
and then solve for A and B

Doing this we get
A = 1/2
B = -1/2

Subbing those back in we get the equation:
Vs/L * ((1/(R/L))/s - (1/(R/L)/s+(R/L))

Simplifying and extracting L we get:
Vs/R * (1 - e-Rt/L)u(t)

Note: The u(t) added to the end will keep our function in the positive time domain which is what we want to happen.

Recommended Study/Homework
Use the S Domain for:
Ap 7.1, 7.3

Continue to Review Chapter 3, 7, 9.

Good reference videos on Differential Equations from MIT's OpenCourseWare Site: MIT OpenCourseWare Laplace Transforms (direct link to the video...very helpful on explaining how it works)
Continue on to Lecture 3 6/10/09