## Friday, November 5, 2010

### Circuit Analysis II Lecture 16

****THE BELOW IS INCOMPLETE, but I'm releasing it in case anyone can get some value out of what was written.****
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) <Concepts:
1. High/Low Pass Filters can be RC or RL circuits depending on what element you are using for vo

a. For RC the Low Pass filter is on the Capacitor.
b. For RL the Low Pass filter is on the Resistor.
c. a/s+b is low pass
s/s+b is high pass

2. Band Pass/Reject Filters are RLC circuits
a. Ks/(as2 + bs + c)

3. The cutoff frequency is A/(2)1/2

4. Q = (L/CR2)1/2

5. wo = 1/(LC)2

To recap:
If there is a quadratic equation on the bottom, it's likely it's a bandpass.
If it's not and there is an s on the top then it's a highpass filter and if not then it's a lowpass.
Don't forget if given in HZ convert to rads/sec via equation w = 2piF

>Examples:
1. Ap 14.1 page 577
2pi * 8000 = 1/10,000c
c = 1/(80000*2*pi)
c = 1.98*10-9
c = 1.98nF

2. Given a 1-10Khz band, C=1uF what are R and L
1/LC = wc1 * wc2
LC = 1/wc1 * wc2
L = 1/C*wc1 * wc2
Solving we get 2.53mH
From there we use the equation R/L = wc2 - wc1
Remembering 2piF = w,
R = 2pi*L*(9000)
R = 143.06 ohms

3. 14.20 page 602
Given c = 20nF, Q = 5, wc = 20kHz
(20000*2*pi)2 * 20*10-9 = 1/L
L = 3.17mH
5 = (3.17*10-3/20*10-9*R2)2
25 * 20*10 = L/R2
R = (1.25/394784176044)1/2
R = (3.166*10-12)1/2