On chapter 6, hw set: #4b:

for this one, 10 is the max, so it's obvious it would hit the max exactly between each pair of zeros, but if we were looking for the magnitude equal to any other number, how would you calculate that? is there a mathematical way to solve this?

3/16/2006 7:14:04 PM

so it is ten when they are both +5, or, better when they're equal to each other, so you can have the equation:

5*exp(j70t)=5*exp(j60t)
exp(j70t)=1*exp(j60t)
exp(j70t)=exp(j*2pi*n)*exp(j60t)
exp(j70t)=exp(j(2*pi*n+60t)

equation phase/taking ln of both sides

70t = 2pi*n+60t
10t=2pi*n
t=.2*pi*n

a single time can be n=1, so t=.2*pi.

hope this helps

3/16/2006 9:19:45 PM

and i just realized i completely didn't answer the question.

damn. sorry.

3/17/2006 12:09:59 AM

This is gonna be so much fun....

Did he review PFEs? I missed a day recently, and I think maybe he did. Or maybe he just expected us to know it. I guess I better review it either way.

On the homework that we had to do an inverse Z transform, I used matlab

[Edited on March 17, 2006 at 12:47 AM. Reason : ads]

3/17/2006 12:37:15 AM

3/17/2006 12:44:46 AM

3/17/2006 12:45:10 AM

To answer the previous question:

for part a: odd multiples of pi = 70t - 60t

for part b: even multiples of pi = 70t - 60t

So solve for the first couple (like 1*pi and 3*pi / 2*pi and 4*pi) and then see the period

3/17/2006 1:07:48 AM

how do you do 8-1?

3/17/2006 1:13:53 AM

Yeah, the chapter 8 stuff is definitely the worst.

Solving for the filter coefficients, you use b = (X^TX)^-1*X^Tr

And I have written off to the side that the ^-1 signifies a 'psuedo inverse', but I don't have it written anywhere just what that actually means. I don't have a good example worked out of actually solving a problem like that.

On 8-1 b, I got the dreaded right answer... wrong method. My matrices came out funny, and I'm not sure how they should look.

I had [A;(...)B] = ([6,1;(...)1,3] [6;1;(...)1;3])^-1 [6,1;(...)1,3] [3;4]

and that = [0.2941;(...)1.2353]

I think I had my X^T
I think it should have looked more like [6;1;(...)1;3]

[Edited on March 17, 2006 at 1:22 AM. Reason : asd]

3/17/2006 1:19:50 AM

So anyone wanna meet in the morning?

I'd like to go over the way to solve for the orthagonality condition ala chapter 8 with some kind soul.

3/17/2006 1:50:14 AM

3/17/2006 2:04:42 AM

3/17/2006 2:36:41 AM

3/17/2006 3:19:32 AM

as for the first question, i'm thinking the quesiton is only interesting at the min or max of the function, since it's representing the doppler fading....you'd either want to know where it's at its max, or its min..

3/17/2006 7:20:22 AM

anyone know how to do 8-4?

3/17/2006 8:20:12 AM

that's what i figured, since there's 1 min and max per period, but 2 of any other #

#9-2 & 3...why is this stumping me? i thought there were supposed to be the same # of roots as the degree...for a) i get .216j and -.216j, b) .6 & -.6, c) 2.2, -2.2, 2.2j, -2.2j...for 9-2

for 9-3 i get zeros at +/- .9537j and holes at +/- .79999j

rhea and i are in eb1



[Edited on March 17, 2006 at 8:33 AM. Reason : #]

3/17/2006 8:20:45 AM

I'll be over there in a little bit

3/17/2006 8:26:15 AM

3/17/2006 8:53:05 AM

stupid questions

3/17/2006 12:44:59 PM

Wasn't so bad. Of course, I thought that about the first one.

3/17/2006 4:20:37 PM

wasn't so bad except there is no stable E(z)...damnit all to hell.

i feel ok about it. i'd say i got between a 50 and an 80 on it

3/17/2006 6:56:20 PM

can't wait for the histogram on this one...ohh yeah

3/18/2006 12:06:07 AM

So it was a trick question?

3/18/2006 12:24:42 AM

Yes

3/18/2006 10:48:20 AM

yea. and i wrote "poles @ 1 & 2 - not stable", turned the page, went "wait, it asked for a stable one...i don't know how to get a stable one..." so i erased the line bout the poles and decided i must not know what i was talking about with the pole business. grrrr

3/18/2006 11:37:38 AM

I thought it just asked for a stable E(z) and that was it.

Oh well, I guess half credit is about the best I can hope for on that problem.

3/18/2006 12:26:31 PM

Friday afternoon:

me:


alexander:

3/18/2006 12:27:10 PM

^well it did, but 1/C(z) turned out to be unstable...

haha such a great picture of him

3/18/2006 12:58:08 PM

Yeah, he dressed up



[Edited on March 18, 2006 at 2:13 PM. Reason : I kid Dr. Alexander, you're great!11]

3/18/2006 2:13:08 PM

teehee i was thinking he just looked awfully cheerful but you're right!

it's sad that i won't be able to take him for the 4th time this fall

3/18/2006 4:31:30 PM

3/20/2006 4:06:58 PM