Circuit Analysis II Lecture 13

****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)

Agenda for the day:
13.6 - AP 13.12
13.7


AP 13.12

x(t) = Acos(wt+phi) is given

Looking at the equation:
v_o/i_g = ((2+s)s * 10/s)/((2+s)+10/s)

working the algebra through to simplify you get:
10*(2+s/(s² + 2s + 10))

This is our H(s), so using that H(s) we sub it into our steady state equation:


y_ss(t) = A|H(jw)|cos(wt+phi+theta)

Now we want to find H(jw) so remember that s = jw so we can just grab w and sub j and w in for s in the H(s) equation, that will look like this:
10(2+j4)/((j4)² + 2*j4 + 10)

a little manipulation we get:
20+j40/(-16 + 8j + 10)

next we combine and multiply by the conjugate:
(20+j40)/(-6 + 8j) * (-6-8j)/(-6-8j)

finally we get:
2-j4

Converting to polar we get:
4.47e^-j63.43


Finally substituting back in our equation we get:
10*20^1/2cos(4t-63.43°)

Exercise 13.72
we have a resistor of 50 ohms and two inductors whose values are 2F and 8F in series.
v_i is attached between the resistor and the 8F inductor. the v_o is located across the 8F inductor.

Converting to the S domain and using voltage division we get:
v_o = v_i/(50+2s+8s) * 8s

Solving it out for v_o/v_i we get:
.8s/(s+5)

Now we'll sub in 75/s for v_i

so we'll get:
75*.8s/s(s+5)

Finishing the algebra we get:
60/(s+5)

Now we want to use the convolution integral on this one:

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 v_o

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. w_o = 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 = w_c1 * w_c2
LC = 1/w_c1 * w_c2
L = 1/C*w_c1 * w_c2
Solving we get 2.53mH
From there we use the equation R/L = w_c2 - w_c1
Remembering 2piF = w,
R = 2pi*L*(9000)
R = 143.06 ohms

3. 14.20 page 602
Given c = 20nF, Q = 5, w_c = 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}

Circuit Analysis II Lecture 15

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)

Ch 14
Low pass filters
High pass filters
Band pass filters

A simple circuit with a resistor and inductor in series across a voltage source.

H_L(s) = V_oL/V(s) = sL/r+sL

re-writing it with some manipulation we get:
s/(s+(R/L))

Something to note is that w_c = R/L (w_c is the cutoff frequency)

so for H_R(s) = V_oR/V(s) = R/R+SL = (R/L)/(s+(R/L))

So what does this mean?
Substituting in w_c in the equations we'll get:
H_R(s) = w_c/s+w_c
H_L(s) = s/s+w_c

Looking at the graph for H_R(s) we have a graph that starts at 1 at w/w_c and it slopes down acting similar to a 1/x graph.

Turns out this is a low pass filter.

w_c is our cutoff, and for w_c>x it's passed through

So it turns out if you look at H_L(s) it's a High pass filter.

Now if you look at a RC circuit it turns out that the values flip, what that means is that the filter across the capacitor would be a low pass filter, and across the resistor is a high pass.

Problem 14.3
P582

So in order to find this we just need to know R and L, since we do we can determine that it's 5k/3.5*10^-3 = 1.43Mrads/s

Finally we took a RLC circuit and determined that because of the L and the C in series we were able to determine that it was a bandpass filter because there was more or less a range of frequencies that were aloud to pass through or block.

Finally we went over the solution for 13.71, I plan to post the answers to 13.71, 9.30a and 9.31 once I've solved them so stay tuned.

HW:
AP 14.1, 14.2a,b
Ex 14.1-5, 14.8
p 14.1a, 2, 5, 10-12, 14,19, 20

Good luck

Continue on to Lecture 22 7/28/2009

Circuit Analysis II Lecture 14

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)

Requested to make sure we ABSOLUTELY DID (or at least tried):
Problems 13.7, 9.30a, 9.31

We also got our test back.

That's all!

Continue on to Lecture 15 7/13/2009

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

Circuit Analysis II Lecture 9

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)(s² + 6s + 25)

1. Get the roots:
-3 +- j4

k₁ can be solved easily and will come out to -12
k₂ can be solved via algebra and you'll get 6-j8
k₃ can be determined by taking the complex conjugate of k₂,which will be 6+j8.

Now we'll want to convert to polar form

|k₂|e^jtheta = 10
e^{-j53.13degrees}
and then |k₃|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*tcos(beta*t + theta)
Found in table 12.3 page 493

Looking at the equation 10(s+2)/(s² + 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 + 3²

That looks ALOT like the s domain equation for e^-atcos(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 + 3² + 1/3 * 3/(s+1)² + 3²)

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:
(A² + B²)^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^-tcos(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
k₁ = 4
k₂ = 200
k₃ = -4

*Interesting note, not ALWAYS but there are times where k₃ = -k₁.

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 v₀(t) = 10000e^-70tcos(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