Circuit Analysis II Lecture 10

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)

1. Complex Numbers
2. Repeated Roots

The test was moved to Monday (a little late notice I know)

Looking at this equation (13.59 from page 520):

I_l = 6s/(s+5000)²

using Partial Fraction Decomposition we were able to obtain k₁
k₁ = 6s/(s+5000)² evaluating s at -5000
we treat this equation pretty much like this:
k₁ = 6s/((s+5000)²)*1
so when we evaluate we would use the equation:
k₁ = 6s/1
so subbing s=-5000 we get:
k₁ = -30,000

Then we take the derivative of k₂
k₂ = d/ds 6s/1 = 6

this gives us
-30000/(s+5000)² + 6/(s+5000)

converting back to time domain we get:
(-30000te^-5000t + 6e^-5000)u(t) A


Next up is problem 13.49

We are trying to find the v_o/v_i combo of the circuits.

a) First up we manage to get the equation:
v_o = v_i/((1.25*10⁶/s) + 25000) * 1.25*10⁶/s

a little algebra we get:
v_o = 1.25*10⁶v_i/s((1.25*10⁶/s) + 25000)

a little more we get:
v_o = 1.25*10⁶v_i/(1.25*10⁶ + 25000s)

pulling out the 25000 we get:
v_o = 50v_i/(s+50)

finally we get:

v_o/v_i = 50/(s+50)

b) Combining things we get the equation:
v_o = v_i/((1.25*10⁶/s) + 25000) * 25000

A tad more algebra we get:
v_o = sv_i/(s+50)

and finally we get:
v_o/v_i = s/(s+50)

e) This one requires using an additional technique the teacher used node voltage:
(v_{o -} v_i / 25k) + (v_o/100k) + (v_o/2.5*10⁶/s) = 0

A little algebra later (multiplying by 2.5*10⁶) we get:
100(v_o - v_i) + 25v_o + sv_o = 0

Re-arranging and whatnot we get:
(125+s)v_o = 100v_i

Finally we end up with:
v_o/v_i =100/(s+125)

Last we worked with ap 13.11a&b

using the equation from 13.11a that's our transfer function:
H(s) = 9600s/s² + 140s + 62500

For part b we are supposed to look at the unit step (which is 1/s):
so subbing that in we get the equation:
v_o/(1/s) = 9600s/s²+140s+62500

moving things around a bit we get:
v_o = 9600s/s(s²+140s+62500)

cancelling things we get:
v_o = 9600/s²+140s+62500

and finally breaking it up by completing the square:
v_o = 9600/(s+70)²-4.9*10³+62500

combining the rest and separating the top to look like something from the table we get:
v_o = 240 * 40/(s+70)²⁺240²

And lastly converting to the time domain we get:
40e^-70tsin(240t)

Continue on to Lecture 14 7/9/2009