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)

Getting right into it we start on Page 486 with the equation:

F(s) = 100(s+3)/(s+6)(s

^{2}+ 6s + 25)

1. Get the roots:

-3 +- j4

k

_{1}can be solved easily and will come out to -12

k

_{2}can be solved via algebra and you'll get 6-j8

k

_{3}can be determined by taking the complex conjugate of k

_{2},which will be 6+j8.

Now we'll want to convert to polar form

|k

_{2}|e

^{jtheta}= 10

e

^{-j53.13degrees}

and then |k

_{3}|e

^{jtheta}= 10 e

^{+j53.13degrees}

After a bit of math we come upon the equation 10e

^{-3t}(e

^{j(4t-53.13degrees}+ e

^{-j(4t-53.13degrees}

Now recall that

(e

^{ic}+ e

^{-jc})/2 = cos c

What that means is we can finish this off by multiplying by two and just taking the contents of the () in the equation and make it cos(4t - 53.13

^{degrees})

We learn that if we have an equation of the form:

|K|e

^{-jtheta}/(s+alpha -jbeta) + |K|e

^{jtheta}/(s+alpha +jbeta)

We can directly convert that to the equation:

2|k|e

^{-alpha*t}cos(beta*t + theta)

Found in table 12.3 page 493

Looking at the equation 10(s+2)/(s

^{2}+ 2s + 10)

you can use the repeated roots method above, one more thing to look at though is can you convert it into a form that can be solved by the tables?

Well if you complete the square you'll get the equation:

10(s+2)/(s+1)

^{2}+ 3

^{2}

That looks ALOT like the s domain equation for e

^{-at}cos(wt)

Well we can make it look even more like it by simple manipulation and we'll get the equation:

10((s+1)/(s+1)

^{2}+ 3

^{2}+ 1/3 * 3/(s+1)

^{2}+ 3

^{2})

This can then be easily be moved to the time domain where we get:

10e

^{-t}(cos 3t + 1/3 sin3t)

Now recalling that cos (a+b) is equal to cos(a)cos(b) - sin(a)sin(b),

we can assume that since we have the 3t we can call that a or b (doesn't matter)

Now separately we look at the co-efficients for the cos and sin and call them A and B where A (and A = cos a) is the co-efficient of cos and B (and B = sin b) is the co-efficient of sin and say that:

(A

^{2}+ B

^{2})

^{1/2}(square root) cos(the remaining variable - tan

^{-1}|B/A|)

What that means is that we can take the above equation and apply to our time domain equation and get:

10.5e

^{-t}cos(3t - 18.43

^{degrees})

Which happens to be the exact same equation that they got in class and it's alot quicker.

Next we worked on Problem # 2 from the quiz.

We solved it out for the repeated roots, and got the answer's

k

_{1}= 4

k

_{2}= 200

k

_{3}= -4

*Interesting note, not ALWAYS but there are times where k

_{3}= -k

_{1}.

Transfer Function

Recall that H(s) = Y(s)/X(s)

With this equation there are always 4 possibilities

1. X(s) can be either a voltage or a current source

2. Y(s) can be either a voltage or a current source

If X(s) AND Y(s) are both voltage or current then they are dimensionless.

if one or the other is voltage or current H(s) is basically either z (impedence) or 1/z

AP 13.10a

We want to do the "unit step" of the function, well since we solved the equation earlier, we need only multiply the equation by u(t) or in the S domain 1/s

AP 13.11

We are looking for the unit impulse response, what that means is that since the unit impulse response in the s domain is 1 then the output is the h(t)

so we would convert v

_{0}(t) = 10000e

^{-70t}cos(240t + theta)

to the S domain to get H(s) (or the transfer function)

to find the step response we would need to know that tan(theta) = 7/24,

recalling the right triangle we can solve for cos and sin, and solve this equation as well for H(s).

What does this all mean?

Warning I don't know everything nor do I claim to, the below is not something that was taught to us and frankly its only my observations, correct or not that is all they are!I'm not 100% but from what little I've been able to gather is that this is the "form" of the Ohms law for functions in the S domain, it's not 100% accurate, but in general if you have some equation and you know the 2 of the items H(s), X(s), or Y(s) then you can solve for the third one.

Since Y(s) = H(s) * X(s) that would most likely equate to the V = IR in the normal equation.

Since H(s) = Y(s)/X(s) that would be like R = I/V which sounds about right.

Finally X(s) = Y(s)/H(s) which equates to I = V/R.

So obviously if we had some equation where X(s) is a Voltage, we would want to remember that if we swapped current and voltage in these equations, we would want to invert H(s).

Thats about all I can figure out from this at this point, I only know what I can extract from this information and it appears to me that this is essentially a more generalized version of the Ohms law, it seems to make sense to me!

Practice Problems:

12.41 a & b

12.43 a

AP 13.1-5

13.8-11

Problem 13.49

Examples 13.1,2

Continue on to Lecture 10 7/1/2009

## No comments:

## Post a Comment