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

Circuit Analysis II Lecture 8

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:
Repeated Roots -Go over today
Complex roots -Go over today somewhat
Transfer function - Introduce.
Quiz -Hand back.

Starting with Repeated roots.
1. Make sure it's a proper fraction, if it's not long division is required.
2. Write out the equation in k/x form
I.E. k₁/(s+1)
3. The equations with nothing raised to a power can be written as a k/equation, and solved like normal.
4. The equations with the power can have the first power written as k/equation...
I.E. k₂/(s+4)³
5. Any additional roots are to have the derivative taken then solve
I. E. d/ds(k₃/(s+4)²)
Then d²/ds(k₄/(s+4))


So lets try AP 12.6
We get the equation in the form (4s² + 7s + 1)/s(s+1)
Written as k/s + k₂/(s+1)² + d/ds(k₃/(s+1))

So we can easily solve for k₁ and k₂
k₁ = (4s² + 7s + 1)/(s+1)² evaluated at s = 0
thus k₁ = 1/1 = 1

Similarly for k₂ = (4s² + 7s + 1)/(s) evaluated at -1
thus k₂ = -2/-1 = 2

Now REMEMBER taking the derivative of k₃ requires you to remember that you are solving for k₃, but the equation you take the derivative of is (4s² + 7s + 1)/(s)
after taking the derivative you should end up with 4- 1/s² evaluated at -1
k₃ = 4 - 1 = 3

So we solve it and get the equation:
F(s) = 1/s + 2/(s+1)² + 3/(s+1)
Converting back to the time domain is as simple as looking at the tables and converting directly back.

Next we look at complex roots
AP 12.5
Things to remember:
1. k₂ is the -root
2. k₃ is the +root
3. k₃ = k₂^*
4. Quadratic equation:
-b +- (b² - 4ac)^1/2/2a (wikipedia)
5. (s - s_-)(s - s₊)
6. Solving the equation for k₂ where s = -5+j12
7. Solving the equation for k₃ where s = -5-j12
8. e^ajt - e^-ajt/2j = sin (at)

Moving onto the equation we solve for the roots since the other parts are quite easy.
F(s) = 10(s² + 119)/(s+5)(s+5-j12)(s+5+j12)

So solving for k₂ we get (after much pomp and ceremony and of course algebra) we get j25/6
and of course since we know that k₃ = k₂^* we can safely say that
k₃ = -j25/6

Now we convert back to time domain.
which we'll get:
f(t) = (10e^-5t + je^-(5-j12)t - j25/6e^-(5+j12)t)u(t)

Now leaving it in that form isnt good because we still have complex numbers in there, re-arranging the equation we get:
e^-5tj25/6(e^j12t - e^-j12t/2j)*2j

a little more massaging we get (remember #8 above):
f(t) = (10e^-5t - 25/3e^-5tj25/6(sin 12t)) u(t)

Next we discussed the transfer function!

Basically we look at it this way:
f_i(t) -> h(t) -> f_o(t)

1. f_i(t) represents the input, or X(s) in the s domain
2. h(t) represents the processing of the input (voltage or current), or H(s) in the s domain
3. f_o(t) represents the output, or Y(s) in the s domain

This process now only works in the S domain.

Now what we can gather from this is that if the input changes we can determine the output by finding the answer to the equation
X(s) * H(s) = Y(s)

So what this means is that if we have H(s) solved we can change either the input or the output and determine the other.

Finally we went over the quiz and were recommended some problems to work on.

1. V₀(s) = 25(s+2)/s(s² + 2s + 10)
*practice problem for complex roots.

2. I₀(s) = 10⁴/(s+80)(s+30)² A
V₀(s) = 5*10⁴s/(s+80)(s+30)² V
*practice problem for repeated roots.

3. v(t) = (10e^-1000t - 10e^-4000t)u(t)

Additionally recommended to practice 12.41

Continue on to Lecture 9 6/30/09

Circuit Analysis II Lecture 7

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.

In class we did 2 practice problems.
Problem # 1 was a circuit with a voltage source, resistor and capacitor in series, and an inductor in parallel with the capacitor.

The values are as follows:
Source = 30v
Resistor = 800 ohms
Capacitor = 6.25 uF
Inductor = 2.5H

1. We convert to s-domain
2. We find a node voltage equation
3. We solve for the node
(there happens to be only one node so we only need to solve for the one node).

Then it's Solved!

Next we worked on problem 13.42

1. Convert to s-domain
2. Get node voltage equations
3. Make v₂ in both equations the same
(simply multiply both equations by the coefficient)
4. Add the equations and get one equation.
5. Multiply thorough all the parts, and then combine like items,

We end with the correct answer (can be found in the back of the book)

Don't forget we have a quiz tommorrow (June 25th, 2009)


Continue on to Lecture 8 6/29/09

Circuit Analysis II Lecture 6

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.

First we worked on problem 13.8c
it turns out that there is only one node so the equation is:
v₁ - 20/s / 2 + v₁ / 1.25s + v₁ / 20 /s - 5/s = 0

Now we can simply solve for v₁

Next we went over Thevinin starting on page 520, figure 13.17 - 19

First we migrate to the s-domain, so we end up with
480/s for the source
2*10^-3s
Since we are looking at the circuit and the 60ohm resistor is not connected so it has no current running through it, so we can get rid of that particular resistor.

We end up using voltage division because we want to know the voltage at the inductor in this particular example.

v_th = (480/s / 20 + .002s * .002s)
so pulling things out and canceling things out we get
v_th = 480 / s+10⁴

Next we want to convert it to look at it from the a and b terminals, so what we'll want to do is get rid of the source, remember:

voltage = short
current = open

doing that we can get z_th
z_th = (20s / s + 10000) + 60

Cleaning it up we get:
z_th = (60s + 6*10⁵ + 20s/s + 4)

and finally we can get:
80(s+7500)/(s+10⁴)

Now we add a 5uF capacitor to the a and b terminals, so looking for the current we get:
I_c = (480/s+10⁴) / (80(s+7500)/(s+10⁴))

Basically what we're looking at is
I_c = v_th / z_th + z_c

after solving it we get 65/(s+5000)²

We would be able to do partial fraction decomposition to continue but that is outside the scope of this class period, so we didn't continue.

Next we worked on problem 13.20 on page 552
1. We want to move to the s-domain
2. Next we find R_eq
note: we use voltage division in this case.
3) Using Partial fraction decomposition we want to get the k's
4) Convert back to t-domain

Next we worked on Problem 12.42a page 504
F(s) = (10s² 85s +95) / (s² 6s + 5)

First reaction might be to try to reduce it, but it turns out this is not a proper fraction and therefore we will need to do long division to make it a proper fraction.

Once we do that we get
10 + (25s + 45)/(s² + 6s + 5)
reducing it we get:
10 + (25s + 45)/(s+1)(s+5)

using partial fraction decomposition for the second part we get,
20 and 5 for the k's

since 10 is a constant we use (delta)(t) so our equation ends up as:
10(delta)(t) + 5e^-t + 20e^-5t

Finally we tried one last one and that was:
problem 12.21a
20e^-5(t-2)u(t-2)
which using the laplace transform from the table f(t-a)u(t-a)
we get
e^-2s-20/(s+5)

Continue on to Lecture 7 6/24/09

Circuit Analysis II Lecture 5

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.

First,
AP 12.3 page 485

F(s) = (6s² + 26s +26)/(s+1)(s+2)(s+3)

using Partial fraction decomposition we get
K₁ = 3
K₂ = 2
K₃ = 1

Of course converting back to the time domain we get
(3e^-t + 2e^-2t + e^-3t)u(t)

Next we do Problem 13.5 from page 519
Our plan is to move to the s-domain, analyze the circuit and then move back to the time domain (or t-domain)

The circuit in the s-domain has the values as such:
5/s for the current
15/s for the voltage
1/s for the capacitor
s for the inductor

Additionally to note is that we have no initial charge so we don't need to "convert" the inductor to a inductor current/voltage source combination, as well as the capacitor.

So using Nodal Analysis we get:
Node 1:
-5/s + v₁/1/s + v₁ - v₂/s = 0

Node 2:
v₂ - v₁/s + v₂/3 + v₂ - 15/s / 15 = 0

We were asked to solve for v₁ and v₂

we were given that it ends up as :
s(s+3) / s(s+.5)(s+2)

Using Partial Fraction Decomposition:
We were asked to solve for all the k's.

Once we finish, convert it to t-domain.

We were then asked to try it using mesh current.

Hw: Try 13.42, and Ap 13.8c using node voltage

Review Thevinin, Norton Equivalent for next class.

We then went over the test, as well as did all the problems.
I have problem 7 solved, and 8 solved, from class, the rest of the problems can be given if you just either leave a comment or email me.

Continue on to Lecture 6 6/23/09

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:
I_dcu(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:
I_dc/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) = (I_dc/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

Circuit Analysis II Lecture 3

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.

Starting with the class with doing AP 12.2c
Looking for the laplace of :
t cos(wt)

This looks alot like
tf(t)
which is in the table and can then be converted fairly easily, so doing this you would do:
-dF(s)/ds

Substituting our equation in we get
-d(s/s² + w²)/ds

Now we would have to take the derivative and then we'll get the same answer as in the book.

Next we worked on problem 12.17a
the integral of e^-axdx

This looks strikingly similar to
F(s)/s
so substituting it in for it we get
(1/s+a)/s

and finally
1/s(s+a)

In this case x and t are interchangeable since they are describing the same thing (a variable), which is why we could use the form from the table.

Followed by Problem 12.23a

Figure out the Laplace of:
f(t) = d/dt (e^-atsin(wt))
looking at that equation we can say that everything in the parantheses would be the equivalent to f(t)

Using that we can use:
df(t)/dt = sF(s) - f(0)

Then we can convert it to
sw/((s+a²) + w²)

Next we attempted
F(s) = 40/(s² + 1.2s +1)

To do this problem we want to do completing the square, so we would end up with :
40/((s+.6)² + .8²)
we can get w in the top to match the form
w/(s+a)² + w²

so we just do 40/.8 which gives us 50 so we can put the equation in the correct form looking like this:
50 * (.8/((s+.6)² + w²))

now converting back into the time domain
50e^-.6tsin(.8t)

Continuing
A circuit with a voltage source (160v), a resistor(4.8), an inductor (4H) and, a capacitor(.25F) all in series, to be solved for.

*Remembering that V = IR
I(s) = (160/s)/(4.8+4s + 4/s)

moving s down to the denominator and pulling a 4 out we get the equation as such:
I(s) = 40/(s²+1.2s+1)

Since we have already done this equation just a few moments ago we can consider ourselves done.

Next we look at the equation
F(s) = 10(s+6)/(s+1)(s+3)

Using Partial Fraction Decomposition we are able to get
K₁ = 25
K₂ = -15

From there we sub them back in and get
25/(s+1) - 15/(s+3)

Using the Table to convert from Laplace to time domain we get:
(25e^-t - 15e^-3t)u(t)

We add u(t) on there to force this equation to only be valid in the Positive time domain, this works because u(t) = 0 for all time less then 0 and u(t) = 1 for all time greater then 0.

Problem 13.9 on page 550

Breakdown of how we'll solve this circuit:

1. Get the Initial values for the circuit using t<0,>
2. Calculate the rest of the circuit using the initial values, again draw the circuit.
3. We are looking for the voltage across the capacitor so we can use the voltage division.
4. Use Partial fraction decomposition to finish the circuit off and solve the formula

Your t<0>0 circuit consists of the inductor (.8H) another resistor (2K), and a capacitor (1.25uF) all in series

to find V we use the equation
(8*10⁵*12mV/s)/2*10³+.8s+(8*10⁵/s)
That eventually ends up in the form:
12000/(s² + 2500s + 10⁶)

We continue and use partial fraction decomposition and leave 12000 in and we get 8 and -8 for our k's
following that through we can use the equation
-((1/8)/(s+500) - (1/8)/(s+2000))
which again looks alot like a Laplace form answer
k/s+a
convert it back and you're golden!

Practice and In-class problems and guidelines for Test #1:

1. Section 12.1,2,4,5
2. AP 12.2a,c
3. Problems 12.13a-d,12.14b,12.17,18,23,35
4. Section 13.1,2,3 (p512-3)
5. AP 13.4a,b
6. Problem 13.9

Assigned reading for the following week is:

1. AP 12.3,4,5
2. Review Chapters 3,4,7,9
3. Review Appendix B (Complex Numbers)
   

Continue on to Lecture 4 6/11/09

Circuit Analysis II Lecture 2

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)

If you haven't read Lecture 1 it can be found here:Lecture 1

Also if you notice the links on the right, there is a list of links that will contain all the links to this course and you can move around freely, this will become nicer as there are more Lectures posted.

The main topic of this class period was converting actual circuits to the s domain, then using algebra to re-arrange the equations to a format of a function that the table has for you.

Key points:

• Time domain is represented by small letters
• S domain is represented by Capital Letters
• di/dt = sI(s) - i(0)
• I0 = i(0)
• ZL = sL and since s = jw then ZL = jwL
• I0 indicates Initial charge
• If your target is Voltage convert the circuit to use Current
• If your target is Current convert the circuit to use Voltage
  

First problem was to convert a inductor in series with a resistor


Our target was Current so we converted it using Table 13.1 to a series circuit.
Inductor = sL
Source = LI0
Resistor = R

Now we are able to treat this as a circuit using Ohms law (Remember V/IR)
So, I = (LI₀)/(R + sL)

Now it's time to move back to the time domain,
Since this circuit looks most like k/(s+a) we can massage it into a form similar
pulling L from the bottom we get the equation:
I = L(I_o)/L(s + R/L)

The L's cancel so you are left with:
I0/(s+(R/L))

Which if you say:
K = I0
a = R/L
you have the form k/s+a

Simply Converting back you have
I₀e^-(R/L)t

Next we tried the same process with a circuit containing a Capacitor
Again the target is Voltage so we use Current.

so using the formula v = ir
and
V = CV₀Z_eq

Z_eq = R₁R₂ / R₁ + R₂

Inserting the "newly determined" values for the capacitor and Resistor.
We have ((1/cs)*R)/(R + (1/cs)) which is equal to:
R/(Rcs + 1)

Putting that back in our equation gives us
V = CV₀ * R/(Rcs + 1)

extracting C and R from the equation gives us:
CRV₀/CR(s+1/RC)

That is then transformed to
V₀ / s+ (1/RC)

Now we do the same as with the Inductor circuit we try to find a similar Laplace Transform we can invert!
Again it looks like k/s+a
k = V₀
a = 1/RC

So we get the answer
V₀e^-t/RC

The Final Problem we did was try to use the RL circuit from 7.3 which is a RL circuit with a source:

We are going to say that L has no initial charge thus I₀ = 0

Converting to the S domain we get
Inductor = sL
Resistor = R
Source = V_s/s

Now we look at the equation we can get from that:
I = (V_s/s)/(R+sL)

This can be rewritten as:
V_s/s(sL + R)

Using Partial Fraction Decomposition we would use an equation of the form:
A/s + B/s+a
and then solve for A and B

Doing this we get
A = 1/2
B = -1/2

Subbing those back in we get the equation:
V_s/L * ((1/(R/L))/s - (1/(R/L)/s+(R/L))

Simplifying and extracting L we get:
V_s/R * (1 - e^-Rt/L)u(t)

Note: The u(t) added to the end will keep our function in the positive time domain which is what we want to happen.

Recommended Study/Homework
Use the S Domain for:
Ap 7.1, 7.3

Continue to Review Chapter 3, 7, 9.

Good reference videos on Differential Equations from MIT's OpenCourseWare Site:
MIT OpenCourseWare Diff EQ
MIT OpenCourseWare Laplace Transforms (direct link to the video...very helpful on explaining how it works)
Continue on to Lecture 3 6/10/09

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:




First we received the Syllabus.

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!

Associated Links:
http://www.vibrationdata.com/Laplace.htm
Circuits Book

Continue on to Lecture 2 6/09/09
If you have any Comments, Suggestions or Questions please leave them in the comments and I will do my best to address them.