Coroutines

Today we talked a little about coroutines, at first glance based on the slides it seemed like they were saying that the coroutines would start at the beginning and end when they return, this was a little confusing as they seemed to be identical to procedures.

Turns out we were misunderstanding the picture.

From Wikipedia:

In computer science, coroutines are program components that generalize subroutines to allow multiple entry points for suspending and resuming execution at certain locations. Coroutines are well-suited for implementing more familiar program components such as cooperative tasks, iterators, infinite lists and pipes.

So what is a Coroutine? Well think of it this way, suppose we have a few programs running in threads, and we have an operating system. Do you think the operating system calls each of these threads to do some work, and when it has to pause and stop to go do something it just starts what it was doing all over again? We wouldn't ever get anything accomplished in computing if that was the case. This is where coroutines come into play, since we can return to the last context that was running we can just keep on running. For example, suppose we have a thread that runs a REALLY LONG for loop, the OS does a context switch and pushes the current status onto the stack. Another thread does some work then returns (or perhaps another thread needs servicing). Upon return all the data that was in use from the function is popped from the stack and back in use again. 


Now the book describes the process in a little more detail, but basically what's happening is that the next address is being pushed to the stack, then the operating system goes off and does some stuff, and pops the address off back into the program counter, and the function keeps running as if nothing happened (to it, the change was transparent).


I hope that helped ya out, let me know if you have any questions.

Circuit Analysis II Lecture 13

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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

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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}

ECE 340 Test Review

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So I have been looking at the explanations for the first problem, and right away I believe looking at the problem and analyzing it will go a long way in this case.

1.

a) Looking at the first part of problem 1, we see that it says "What is the probablility that both engines fail during the marine mission". 

Let's break the first question up into parts, Reading Both implies that we are looking for an intersection. From the introduction to the question we know that given one engine failing (the first engine) the second one fails with a probability of .1, Thus the probability of A given B is going to be .1, and we know that B is .0001, now we can say that P(ANB) = P(A|B) * P(B). Which equates to .0001 * .1

b) Breaking down "suppose the second engine fails during the mission what is the probability that the first also failed".

We can extract that we are looking for the case where P(B|A) we are looking for B given A in this case, we also know that the probability of that is = to P(ANB) / P(A) we are looking for the case where there is an intersection over the probability of just the second. This can equate to (.0001 * .1)/(.0003)

c) Looking at "What is the probability that at least ONE of the engines fail" we can extract we are looking for the UNION of the two, Because of that we know P(AUB) = P(A) + P(B) - P(ANB) we know each of the elements here, so the solution is .0003 + .0001 - (.0001 * .1)

d) Breaking the sentence "What is the probability that the first engine fails, and the second engine doesn't fail during the mission" we can extract that given A we are looking for the probability of A sans B, so that would be P(A) - P(ANB) this equates to .0003 - (.0001*.1).

2a. A group of 30 people shake hands with everyone else once,

Looking closely at this we are looking that person 30 shakes hands with 29 people, person 29 shakes hands with 28 people(can't shake 30's hand again), and so forth, so the total number of shakes that occur would be the summation of 30 .. 1 , since we can't really make an easy estimate on this one, then subtract the appropriate number of hand shakes this comes out to n(n-1)/2 or 435. Realistically if you think about it the last person can't shake hands with anyone else so I think the final result should be 434

2b.

Since the elements of the "even" set contains 6 elements {3152, 3512, 5312, 5132, 1352, 1532} and the entire sample space contains 4 varieties of that (one for each number) then the total elements are 24, this will give the solution 6/24, which is 1/4. I didn't bother to calculate the number of elements in the sample space for various reasons. BUT I did have the solution correct in that I said it was 6/(cardinality) of the sample space, which is in fact the correct answer.

3.

Since we are trying to figure out what the probability of two students having the same birthday, we can say that the solution would be (365*364*363...335) divided by all the likely scenarios of 365, which is 365^30.

so the solution is 1- (365! /335! ) / 365^30

4. 

a. { (1,1), (1,2)...(1,6).....(6,6)}

b. Since we are looking at all the EVENTS (I read outcomes on the test....durr) we can have 2^36 different variations (since there are 36 outcomes)

c.  the event that the max of the two numbers is 2 would be {(1,2), (2,1), (2,2)} the probability of this is E^#/Ohmega^# or 3/36

d. The event containing the minimum of the two numbers being 2 is {(2,2),(2,3), (2,4), (2,5),(2,6)...(3,2)} this is 4*2 + 1, 9 outcomes,the P of this is 9/36

e. IFF P(E1NE2) = P(E1) * P(E2) then it's independent. since the intersection is (2,2) and the probability of (2,2) is 1/36, P(E1) = 3/36 and P(E2) = 9/36 so 1/36 != (3/36 * 9/36) thus they are independent.

f. Given E2, we can expect that the probability of getting E1 is the intersection of the two / cardinality of E2 so the solution is (1/36)/(3/36) = 1/3

5.

a. Decoding this we are looking for a zero to be received, this means we are looking for the intersection of the Event that a zero was received with the event that a 1 was transmitted and received as a 0 and the event that 0 was transmitted and received as a 0.  This gives P(0RN1T) + P(0RN0T) = P(0R), we also know that P(0R|1T)*P(1T) = P(0RN1T), and the same goes for the other one, so we now can say that P(0R) = P(0R|1T)*P(1T) + P(0R|0T)*P(0T) which plugging in the numbers we get P(0R) = .6*.001 + .4*(1-.01) = .3966

b. Decoding what is the probability that a 1 transmitted given a 1 was received.
We are looking for the event where a 1 was transmitted given a 1 was received this gives P(1T|1R) this can be equated to P(1TN1R)P(1T)/P(1R)which is saying that we are looking at the probability of the intersection occuring times the probability a 1 was transmitted, divided by the fact that a 1 was received. Since we know the probability that a 0 was transmitted given a 1 was received, we can say that the "opposite" would be 1-P(0T|1R) also since we solved above the event that a 0 was received we should be able to plug that in too which nets us: 1-P(0T|1R)*P(1T) /  1-P(0R) these come out to (.99 * .6 ) /.3966  is approxmiately 1.5.

c. Since .6 of the bits will have .001 chance of error, and .4 of the bits have .01 chance of an error, we are looking for the intersection of the error since a bit transmitted as a 1 can't be "converted" to a 1, we can say the sets are disjoint, and simply add the probabilities together, so .4*.01 + .6*.001

6.
Given that P(A|B) is independent we can say P(A|B) = P(ANB)/P(B) since they're given as independent, we can say that P(ANB) = P(A) * P(B) replacing that in the equation we can say that P(A|B) = P(A) Since the P(B)'s cancel out. the same will hold for P(B|A) which will equal P(B)

7.

a. Trying to decode this question we're looking for the case where a 1 was sent, this means that the majority must have ruled in our favor, so we would have to do N choose K for 4,5,6,7 and multiply appropriately (p^k)*(1-p^n-k) this would give us the solution

b.
The probability of a number being a heads (using the above sample space) we can assume that the probabilities will be the summation of the N choose K for 1,3,5,7, solving for that, I can only assume that for all cases where the # is even, is 1-P(odd).

Insert Graph here:

8. 

i. As defined in your lectures a random variable is a mapping of the elements to the real number set. thus the answer is b)

ii. As defined in the book/lectures p(F1uF2) is = P(F1) + P(F2) ONLY if the intersection is a null set, thus the solution is b)

iii. I immediately recognized the equation to determine independence (c), BUT I didn't think there would be more then one, and didn't really look, obviously P(F1UF2) = P(F1) + P(F2) - P(F1NF2) I didn't look closely but you can "swap" P(F1NF2) with P(F1UF2) (using simple math), I'm blind and didn't see this on the test, so the actual solution is f, C and E

iv. I spoke with you in this class, since I didn't know if F=G I couldn't determine if the probability of F was equal to that of G, from the sounds of things we SHOULD have assumed F = G which means the probability is P(G) <= P(F), this gives the solution b, BUT assuming that F = G, then isn't the P(F) = P(G) and it's not "less then" since they're literally equal to each other?

Page Replacement Algorithms

There are many page replacement algorithms, this week we were asked to implement the fifo and lru page replacement algorithms. But, first a little primer.

FIFO Page Replacement
The fifo page replacement algorithm is essentially a queue setup that pushes the first items into the list out first when the list is full, so for example, if you have the numbers 1,2,3 and they are put into the array in that order, when the 4th number comes along your 1 gets pushed out, and you are left with 2,3,4. That simple!

The FIFO page replacement algorithm was pretty simple to implement, all I had to do was create an array that was the size of my frames, then I simply setup a "current pointer" to that array head initially, and a pointer to the tail and head of the array. When we found a frame we didn't do anything. When we didn't find a frame, we simply assigned the data to the current pointer then moved the pointer down the array, when we reached the bottom, we simply moved the pointer to the top and repeat, also we increment our page fault counter. Something interesting to note, supposing we have the next element in the array that matches so we don't have a page count increment, well the next go around, and that one will still get written over. Since this element wasn't recently written, it doesn't get to keep it's status as "new", and still can be written over in the next iteration. This seems a little less then efficient, however perhaps changing it a small bit so that it skips over to the next element would be beneficial.

LRU Page Replacement
The Least Recently Used algorithm basically makes an association for each array with a clock. The book suggests this is easy to implement using a doubly linked list, the other option is counters.

The LRU page replacement algorithm in theory was pretty easy too, and really looking back on it, it really was. I simply decided to make a second "timing" array that matched up with the frame array. This way whenever we did a frame check, we refreshed the matching frame's time, and if the frame wasn't there, we put the new one into the frame array, and then updated our clock. Really I was able to keep a lot of my code from the fifo page replacement and reuse it, as a lot of it was simply "setup" and structure stuff that I could easily reuse.

Caveats
This sounds really simple, but I did have a few problems along the way one of the main problems I ran into was operator precedence. This occurs when you have + and * in a equation, or a ( or something like that, the compiler/processor has to know what to evaluate first, if for example you try to increment a pointer that is routed through an address, like so *ptr, and you attempt to increment *ptr++, you are actually incrementing your address, effectively losing the pointer to your pointer. This is simply fixed by (*ptr)++; since the ( has higher precedence then ++. Another issue I ran into was that using the function clock() just didn't want to cooperate, clock ran just way too fast, so I had to put wait's in there so that it would adjust the clock even minutely so that when we assigned it to the arrays we didn't use the same time on numerous ones. That makes it hard to find the oldest one if they are all matching times.

Good luck and let me know if you have any problems!