## Tuesday, July 14, 2009

### Circuit Analysis II Lecture 15

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) Ch 14
Low pass filters
High pass filters
Band pass filters

A simple circuit with a resistor and inductor in series across a voltage source.

HL(s) = VoL/V(s) = sL/r+sL

re-writing it with some manipulation we get:
s/(s+(R/L))

Something to note is that wc = R/L (wc is the cutoff frequency)

so for HR(s) = VoR/V(s) = R/R+SL = (R/L)/(s+(R/L))

So what does this mean?
Substituting in wc in the equations we'll get:
HR(s) = wc/s+wc
HL(s) = s/s+wc

Looking at the graph for HR(s) we have a graph that starts at 1 at w/wc and it slopes down acting similar to a 1/x graph.

Turns out this is a low pass filter.

wc is our cutoff, and for wc>x it's passed through

So it turns out if you look at HL(s) it's a High pass filter.

Now if you look at a RC circuit it turns out that the values flip, what that means is that the filter across the capacitor would be a low pass filter, and across the resistor is a high pass.

Problem 14.3
P582

So in order to find this we just need to know R and L, since we do we can determine that it's 5k/3.5*10-3 = 1.43Mrads/s

Finally we took a RLC circuit and determined that because of the L and the C in series we were able to determine that it was a bandpass filter because there was more or less a range of frequencies that were aloud to pass through or block.

Finally we went over the solution for 13.71, I plan to post the answers to 13.71, 9.30a and 9.31 once I've solved them so stay tuned.

HW:
AP 14.1, 14.2a,b
Ex 14.1-5, 14.8
p 14.1a, 2, 5, 10-12, 14,19, 20

Good luck

Continue on to Lecture 22 7/28/2009