## Tuesday, June 30, 2009

### Circuit Analysis II Lecture 8

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) Agenda:
Repeated Roots -Go over today
Complex roots -Go over today somewhat
Transfer function - Introduce.
Quiz -Hand back.

Starting with Repeated roots.
1. Make sure it's a proper fraction, if it's not long division is required.
2. Write out the equation in k/x form
I.E. k1/(s+1)
3. The equations with nothing raised to a power can be written as a k/equation, and solved like normal.
4. The equations with the power can have the first power written as k/equation...
I.E. k2/(s+4)3
5. Any additional roots are to have the derivative taken then solve
I. E. d/ds(k3/(s+4)2)
Then d2/ds(k4/(s+4))

So lets try AP 12.6
We get the equation in the form (4s2 + 7s + 1)/s(s+1)
Written as k/s + k2/(s+1)2 + d/ds(k3/(s+1))

So we can easily solve for k1 and k2
k1 = (4s2 + 7s + 1)/(s+1)2 evaluated at s = 0
thus k1 = 1/1 = 1

Similarly for k2 = (4s2 + 7s + 1)/(s) evaluated at -1
thus k2 = -2/-1 = 2

Now REMEMBER taking the derivative of k3 requires you to remember that you are solving for k3, but the equation you take the derivative of is (4s2 + 7s + 1)/(s)
after taking the derivative you should end up with 4- 1/s2 evaluated at -1
k3 = 4 - 1 = 3

So we solve it and get the equation:
F(s) = 1/s + 2/(s+1)2 + 3/(s+1)
Converting back to the time domain is as simple as looking at the tables and converting directly back.

Next we look at complex roots
AP 12.5
Things to remember:
1. k2 is the -root
2. k3 is the +root
3. k3 = k2*
-b +- (b2 - 4ac)1/2/2a (wikipedia)
5. (s - s-)(s - s+)
6. Solving the equation for k2 where s = -5+j12
7. Solving the equation for k3 where s = -5-j12
8. eajt - e-ajt/2j = sin (at)

Moving onto the equation we solve for the roots since the other parts are quite easy.
F(s) = 10(s2 + 119)/(s+5)(s+5-j12)(s+5+j12)

So solving for k2 we get (after much pomp and ceremony and of course algebra) we get j25/6
and of course since we know that k3 = k2* we can safely say that
k3 = -j25/6

Now we convert back to time domain.
which we'll get:
f(t) = (10e-5t + je-(5-j12)t - j25/6e-(5+j12)t)u(t)

Now leaving it in that form isnt good because we still have complex numbers in there, re-arranging the equation we get:
e-5tj25/6(ej12t - e-j12t/2j)*2j

a little more massaging we get (remember #8 above):
f(t) = (10e-5t - 25/3e-5tj25/6(sin 12t)) u(t)

Next we discussed the transfer function!

Basically we look at it this way:
fi(t) -> h(t) -> fo(t)
1. fi(t) represents the input, or X(s) in the s domain
2. h(t) represents the processing of the input (voltage or current), or H(s) in the s domain
3. fo(t) represents the output, or Y(s) in the s domain
This process now only works in the S domain.

Now what we can gather from this is that if the input changes we can determine the output by finding the answer to the equation
X(s) * H(s) = Y(s)

So what this means is that if we have H(s) solved we can change either the input or the output and determine the other.

Finally we went over the quiz and were recommended some problems to work on.

1. V0(s) = 25(s+2)/s(s2 + 2s + 10)
*practice problem for complex roots.

2. I0(s) = 104/(s+80)(s+30)2 A
V0(s) = 5*104s/(s+80)(s+30)2 V
*practice problem for repeated roots.

3. v(t) = (10e-1000t - 10e-4000t)u(t)