## Tuesday, July 28, 2009

### Circuit Analysis II Lecture 22

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) Final Review:
Solve for power across the resistor.
You are given the equation:
ak = (vm/k(pi))sin(k(pi)/2) and bk = (vm/k(pi))(1-cos(k(pi)/2))

When looking at k = 3 (Third harmonic) we find that the a3 and b3 will come out to -vm/3(pi) and positive

So for example when vm = 1 and RL = 15ohms

We can get An by just taking the co-efficients of the cos and sin from above and solve the equation .5 * An2/RL

We can plug into the equation:
(((1/3(pi)2 + 1/3(pi)2).5)2)/30

Solving that equation we get .75mW (since we're looking for power)

Now this method only works if we're looking for the power across a resistor on our output.

Also from the last quiz we were trying to decide what to do with 21.99 well converting to s domain and whatnot we basically get A from A|H(jw)|cos(wt+phi)
so plugging in 0 (where w = 0) into H(0j) we get 4/5 so since cos(0) = 1 we get A*4/5 which ends up being around 17.something

Good luck on the final!

Start at the beginning Lecture 1 6/8/09