## Thursday, June 11, 2009

### Circuit Analysis II Lecture 3

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) The links on the right, will contain all the links to this course's lecture notes.

Starting with the class with doing AP 12.2c
Looking for the laplace of :
t cos(wt)

This looks alot like
tf(t)
which is in the table and can then be converted fairly easily, so doing this you would do:
-dF(s)/ds

Substituting our equation in we get
-d(s/s2 + w
2)/ds

Now we would have to take the derivative and then we'll get the same answer as in the book.

Next we worked on problem 12.17a
the integral of e-axdx

This looks strikingly similar to
F(s)/s
so substituting it in for it we get
(1/s+a)/s

and finally
1/s(s+a)
In this case x and t are interchangeable since they are describing the same thing (a variable), which is why we could use the form from the table.
Followed by Problem 12.23a

Figure out the Laplace of:
f(t) = d/dt (e-atsin(wt))
looking at that equation we can say that everything in the parantheses would be the equivalent to f(t)

Using that we can use:
df(t)/dt = sF(s) - f(0)

Then we can convert it to
sw/((s+a2) + w2)

Next we attempted
F(s) = 40/(s2 + 1.2s +1)

To do this problem we want to do completing the square, so we would end up with :
40/((s+.6)2 + .82)
we can get w in the top to match the form
w/(s+a)2 + w2

so we just do 40/.8 which gives us 50 so we can put the equation in the correct form looking like this:
50 * (.8/((s+.6)2 + w2))

now converting back into the time domain
50e-.6tsin(.8t)

Continuing
A circuit with a voltage source (160v), a resistor(4.8), an inductor (4H) and, a capacitor(.25F) all in series, to be solved for.

*Remembering that V = IR
I(s) = (160/s)/(4.8+4s + 4/s)

moving s down to the denominator and pulling a 4 out we get the equation as such:
I(s) = 40/(s2+1.2s+1)

Since we have already done this equation just a few moments ago we can consider ourselves done.
Next we look at the equation
F(s) = 10(s+6)/(s+1)(s+3)

Using Partial Fraction Decomposition we are able to get
K1 = 25
K2 = -15

From there we sub them back in and get
25/(s+1) - 15/(s+3)

Using the Table to convert from Laplace to time domain we get:
(25e-t - 15e-3t)u(t)
We add u(t) on there to force this equation to only be valid in the Positive time domain, this works because u(t) = 0 for all time less then 0 and u(t) = 1 for all time greater then 0.
Problem 13.9 on page 550

Breakdown of how we'll solve this circuit:
1. Get the Initial values for the circuit using t<0,>
2. Calculate the rest of the circuit using the initial values, again draw the circuit.
3. We are looking for the voltage across the capacitor so we can use the voltage division.
4. Use Partial fraction decomposition to finish the circuit off and solve the formula
Your t<0>0 circuit consists of the inductor (.8H) another resistor (2K), and a capacitor (1.25uF) all in series

to find V we use the equation
(8*105*12mV/s)/2*103+.8s+(8*105/s)
That eventually ends up in the form:
12000/(s2 + 2500s + 106)

We continue and use partial fraction decomposition and leave 12000 in and we get 8 and -8 for our k's
following that through we can use the equation
-((1/8)/(s+500) - (1/8)/(s+2000))
which again looks alot like a Laplace form answer
k/s+a
convert it back and you're golden!

Practice and In-class problems and guidelines for Test #1:
1. Section 12.1,2,4,5
2. AP 12.2a,c
3. Problems 12.13a-d,12.14b,12.17,18,23,35
4. Section 13.1,2,3 (p512-3)
5. AP 13.4a,b
6. Problem 13.9
Assigned reading for the following week is:
1. AP 12.3,4,5
2. Review Chapters 3,4,7,9
3. Review Appendix B (Complex Numbers)

Continue on to Lecture 4 6/11/09