## Tuesday, June 9, 2009

### Circuit Analysis II Lecture 1

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:

Next, we
discussed the concept of voltage in ECE 203 (Circuit Analysis I)
v(t) = v0 where t>= 0
later in that semester in Chapter 9 we considered where v(t) = Acos(wt + phi)

Now, in ECE 213, we are to consider "all" cases of voltage for example
v0u(t) or ku(t)

when u(t) = 1 t>0
when u(t) = 0 t<0

This is a stepwise function

Important Note: u(t) = 1

In the previous section of this course we learned about moving from the time domain to the phasor domain.

The direction of this course will be learning to move to the s domain.

One Aspect of the S domain is the use of Laplace Transforms to be able to move more difficult equations to a easier to solve format, then the ability to move that back using an Inverse Laplace Transform.

Important Note: s = jw (j omega that is)

When moving from the time domain to the s domain we will be referring to our equations as such:
Time Domain f(t)
S Domain F(s)
Therefore whenever we are looking at f(t) we are looking at the function in the time domain and with F(s) we are looking in the S domain.

Looking at tables 475 and 480 there is a list of Laplace Transforms a quick google search nets this site which looks to have a good chunk of the Laplace Transforms

Something to remember is that f1(t) * f2(t) != F1(s) * F2(s)
(That is they are NOT equal)

We proceeded to take a look at problem 12.13
the first 2 (a,b) are pretty simple because we look at the table and there is a direct correlation.

Part C uses an equation of the form sin(ab + c) which remembering our trig identites such that sin(ab)cos(c) + sin(c)cos(ab) once we have that information we are able to break the problem up pretty quickly.

Next we looked at the Laplace of the equation df(t)/dt = sF(s) - f(0)

What that equation means is that we are including an "initial condition" which is f(0).

Next we looked at the VI relationship in this new section and compared them to the time domain.

For a resistor we use the relationship v=ir in the time domain
in the s domain we would use V(s) = RI(s)

For an Inductor we use v(t) = Ldi/dt in time domain
in s domain we use V(s) = L(sI(s) - i(0))
If you note that last equation we used the equation mentioned above that follows the form df(t)/dt = sF(s) - f(0)

For a Capacitor it's the same process
i(t) = c(dv/dt) in time domain
I(s) = C(sV(s) - v(0)) in s domain

It's interesting to note that the formula for impedance is ZL = jwL
if you remove the initial condition of the equation for the inductor it's the equation sLI(s) since s = jw then it's jWLI(s) which then is essentially V = ZI

by re-arranging the equation of an inductor you have I(s) = V/sL + I0/s

Homework/Review material:
Review Chapters 3, 7, 9
Problems 12.17,18,19

All links to future and past posts regarding this topic will be found on the right column, so you can jump from lecture to lecture!