## Friday, June 12, 2009

### Circuit Analysis II Lecture 4

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) The links on the right, will contain all the links to this course's lecture notes.

Beginning with
F(s) = -20(s+15000)/s2 + 15000s + 50*106

By factoring it out we can find the equation
-20(s+15000)/(s+10000)(s+5000)

Our game plan will include fraction decomposition, then we just convert back to the time domain using the Tables in the book.

Decomposition: -20(s+15000)/(s+5000) (where s = -10000) k1= 20

and,
-20(s+15000)/(s+10000) (where s = -5000) k2 = -40

This brings us to:
20/(s+10000)-40/(s+5000)

Because this looks like e-ax we can convert it as such:
f(t) = (20e-10000t - 40e-5000t)u(t)

Again we need to remember to use u(t) to keep it in the positive domain which is the only place Laplace Transforms work.

Next we work on the equation:
9600s/s2 + 140s + 62500

We can find a perfect square and get
9600s/(s+70)2 + 2402

if we take and replace s with (s+70-70) this then looks like something we can work with
so we can break it up looking at it in the respect of what does our "k" have to be to make the number 240 (which is our w), it happens to be that it has to be 2800

so we end up with:
9600e-70tcos(240t) - 2800e-70tsin240t

Problem 12.18c the book actually has the wrong answer in it so,
remember that the second derivative of t2 is actually 2 and since we can get (2!/s3) * s3 will give us 2 then the previous information will cause it to be 2-2 which is 0

Problem 12.35, this problem requires us to know the equation that we get from section 12.28, and that is based on section 12.6
So Section 6 primarily talks about RLC circuits which you have a basic equation for:
Idcu(t) = v(t)/R + 1/L(integral)v(x)dx + Cdv(t)/dt

Here are some things we can note from this equation:
1. We can transform each part.
2. We can then solve in the S domain
3. This will come out as a rational expression and can be solved.
4. We can convert back to time domain

Some equations to note:
V(s)/R = v(t)/R
V(s)/sL = (integral)v(x)dx
C[sV(s) - v(0)] = Cdv(t)/dt

So the equivalent circuit in the S domain will now look like this:
Idc/s = V(s)/R + V(s)/sL + C[sV(s) - v(0)]

Re-arranging to make it set to V we get the equation:
V(s) = (Idc/C)/s2 + s/RC + 1/LC

Now the expression is in an algebraic form and can be manipulated!
This solves Problem 12.28 and allows us to simply plug in the values from 12.35.

120000/s2 + 10000s + 16000000

This can be factored down to
120,000/(s+8000)(s+2000)

Using partial fraction decomposition we can get the values 20 and -20
which can be easily be converted using the Laplace tables.

We can have table 13.1, 12.2,12.1 copied for the test

Some things to make sure you know what to do for the test are:
RC, RLC, RL circuits with and without source as well as with or without an initial charge

Practice problems:
12.23b

f(t) = (integral)e-axcos(wx)dx
This follows nearly identical to an operational transform from the book:
Time Domain S domain
(integral)f(x)dx = F(s)/s

and the f(x) part looks like
Time Domain S domain
e-atcos(wt) = s+a/(s+a)2 + w2

Problem 12.21a
f(t) = -20e-5(t-2)u(t-2)

This looks alot like the transform f(t-a)u(t-a) which = e-asF(s)
so using that information we can gather that
a = 2 for the e-as
and for F(s) we would use f(t) so replacing t-2 with t we get -5t for the exponent,
so we have for F(s) a=5

Converting to the S domain we get the equation:
-20e-2s/s+5

Finally we tried
(8t-8)u(t-1)

we simply pull the 8 out to get
8(t-1)u(t-1)
Using the same equation we did for the last problem. we'll get
8e-s/s2

Continue on to Lecture 5 6/22/09