How much of Notes 11 will be on test 3? I heard a week ago Alexander say all of it it wouldn't be covered, but then I never heard anything else.

11/14/2005 12:06:24 PM

i heard all of it from someone

11/14/2005 5:59:14 PM

are we supposed to use eqn4, ch11 for 11-1 ?

if so, is B = 2pi/lambda ?
and then, if so, how do we go about simplifying it to just get the answer he has?

i cant do any of ch 11!!

11/14/2005 6:45:52 PM

how well did any of you guys do on the noteset 9 and 10 homework? i have nothing to really study from :\ i bombed most all of those HW's

i just cut out of work a couple hours early so i could study for this beast

[Edited on November 14, 2005 at 7:14 PM. Reason : .]

11/14/2005 7:13:18 PM

what is the value of mu?

11/14/2005 8:38:13 PM

4pi*10^-7?

not having anything to study from fucking sucks


[Edited on November 14, 2005 at 8:48 PM. Reason : §]

11/14/2005 8:42:47 PM

i think that mu value is correct

for 10-5 i got
Up = .337 ux + .925 uy + .173 uz
Uh = -.939 ux + .342 uy
Ue = .069 ux + .159 uy + .985 uz

can anyone else confirm?

11/14/2005 9:01:03 PM

i got that HW wrong :\

what were the last equations you used for up uh and ue

go back and use the same eq on the problem above and if you get the given answers then it should be right. ... infact if you tell me that much ill be able to do most of 10

11/14/2005 9:03:44 PM

^^ those are correct

11/14/2005 9:05:04 PM

Up is in the direction of the radius vector. To get Uh, map the radius vector in the XY plane then add 90degrees to the vector and the resulting vector is in the direction of Uh. Then to get Ue, set Up equal to the written out form of Ue X Uh. This gives you 3 equations. You'll need another equation to solve it: use 1 = sqrt( Ex^2 + Ey^2 + Ez^2). Basically to do the problem you should know H is orthogonal to P.

11/14/2005 9:20:42 PM

For 11-1, you can not use far-field approximation due to the fact that it is a DC current element. So you have to do it the hard way and find the radius using trig.

There are three radii to work with, call them r-1, r0, and r1. r0 is the easy one to compute, it's basically the spherical conversion of the cartesian point that is given. Turns out it is 8.66. To get this, first do an xy projection and solve for d. d is sqrt(50) via pythogorean (sp) theorem --- to find r, do a x-z projection. One leg of the triangle is 5, the other is sqrt(50). The hyp is r, and it works out to be 5*sqrt(3). For the other two radii, instead of 5 being the length of the leg, use 3 and 7. Respectively, they give you r1=sqrt(59) r-1=sqrt(99). Use these distances in finding the summation of the A vector. So r0=8.66; r1=7.68; r-1=9.94. Plug it all in, you get A=6.93E-8 in the uz direction.

There is also a much, much easier way to find the H and E unit vectors. Say your point is P(r, theta, phi)=(13.3,33.7,110.5). For the H vector, add 90 degrees to phi and neglect the z component: (13.3,33.7,200.5) convert this to cartesian and normalize, and you have (-0.936, -.350,0). This agrees with the other approach of doing an XY projection. For the E vector, add 90 degrees to theta and do that conversion to cartesian. If you don't believe me, all of the inner products come out to be zero, showing their respective orthogonality.

11/14/2005 9:38:20 PM

if P = (13.3,33.7,110.5) spherical

and add 90 to phi (200.5) then convert to cartesian i get

x = rsin(theta)cos(phi) = -6.912
y = rsin(theta)sin(phi) = -2.584
z = rcos(theta) = 11.064

(-6.912, -2.584, 11.064) != (-0.936, -.350,0)
what is this normalize you speak of

[Edited on November 14, 2005 at 9:58 PM. Reason : see i dont understand this stuff ]

11/14/2005 9:56:02 PM

(-6.912, -2.584, 11.064) <---- you did it right
take the vector from the origin to ^ that point. Drop the Z term (-6.912, -2.584, 0). Then find the unit vector in that direction (divide the X and Y term by the mag of the vector).

11/14/2005 10:03:20 PM

oh ok sweet, so if im understanding that right, that can be applied to the problems 10-3,10-4 and 10-5 as well as a much quicker way to solve it?

[Edited on November 14, 2005 at 10:13 PM. Reason : .]

11/14/2005 10:11:35 PM

that is correct

11/14/2005 10:21:13 PM

sweet i just did 10-5 that way, took about 2 minutes and i got the answers posted above

11/14/2005 10:29:06 PM

whats the value for the "sigma like" E in the intrinsic impedance formula? eta = beta/(omega*sigmaE) (Eq 36 in Notes11). Trying to do 11-4

Also whats vp for free space?

[Edited on November 14, 2005 at 10:42 PM. Reason : ]

11/14/2005 10:39:39 PM

thats propogation throuhg free space, 6.626*10^-34 ?

11/14/2005 10:43:16 PM

uh so what are the values for sigmaE and vp?

11/14/2005 10:49:14 PM

eta = beta / omega*6.626*10^-34


? is that what you're asking?

11/14/2005 10:59:48 PM

while we are at it, anyone know how to do 11.4 and 11.5?

11/14/2005 11:01:47 PM

In 11-4, it is straight-up plug and chug to solve.

Use equation 39b, where eta=377Ohms. Set it all equal to 10^-3, and solve for Io. Since we're talking about magnitude, you can completely neglect the complex exponential, as that only affects the phase of the field.

For 11-5, I did it using Euler's Identity. Just do the integral -- that's all there is to it. Takes some algebra, but it works out. The only thing that may hang you up is pulling the sine out of the complex terms -- remember that -j=1/j, and j=-1/j.

//

[Edited on November 14, 2005 at 11:18 PM. Reason : ^]

11/14/2005 11:17:57 PM

^ dude you're good at this shit

11/14/2005 11:18:53 PM

meh, i just have no life

11/14/2005 11:21:20 PM

T minus 2hr 17min

[Edited on November 15, 2005 at 12:03 PM. Reason : ]

11/15/2005 12:03:06 PM

the calm before the storm...i'm amazingly at peace right now

11/15/2005 1:58:49 PM

Problem 5 took me about 30 mins.

11/15/2005 4:03:15 PM

why? just use a calculator and find it at -.00001 and .00001 and you get ~ pi/4

11/15/2005 4:43:48 PM

I don't know how to do that so I just did it by hand.

11/15/2005 4:46:34 PM

ah, i dont know how to do it that way
i wish i did im sure i wont get full credit

11/15/2005 5:24:13 PM

i used the calculator to approx the limit of the function, which gave me .7854, or pi/4 to 4 digits. when you do it by hand you have to do l'hopital's rule twice. the first time isn't so bad, the second time is just like omgwtfwhyme kind of thing. it's really easy to get lost in the chain and product rules -- it sucked. there really isn't an easy way to do it by hand...euler is of no help there. you just have to take it and do it. ah well.

other than that, not that bad of an exam. i messed up a partial derivative now that i look back on it, but i don't think it'll be that bad.

//

[Edited on November 15, 2005 at 6:49 PM. Reason : a]

11/15/2005 6:47:11 PM