RE: GTEM Stuff
The GTEM is a guided wave device and usually we tend to think of polarization in terms of free space (unguided waves). Being a TEM transmission line, the direction of the field in the cell is between the inner and outer conductor, the way most GTEM's are constructed and given the location where the EUT is located you could describe it as vertical polarization Vicente Rodríguez, Ph.D., E.I.T. RF/Electromagnetics Engineer ETS-Lindgren (an ESCO Company) P.O.Box 80589, Austin TX 78708-0589 phone 512.835.4684 x648 fax 512.835.4729 vicente.rodrig...@emctest.com http://www.emctest.com http://home.austintx.com/~vicenter -Original Message- From: Aschenberg, Mat [SMTP:matt.aschenb...@echostar.com] Sent: Friday, January 12, 2001 12:02 PM To: emc-p...@majordomo.ieee.org Subject: GTEM Stuff Here is another specific question dealing with GTEM - OATS correlation. Is a GTEM cell horizontally or vertically polarized? Thanks for your help, Mat Mathew Aschenberg Agency Engineer EchoStar Technologies Corporation 90 Inverness Circle East Englewood, CO 80112 --- This message is from the IEEE EMC Society Product Safety Technical Committee emc-pstc discussion list. To cancel your subscription, send mail to: majord...@ieee.org with the single line: unsubscribe emc-pstc For help, send mail to the list administrators: Jim Bacher: jim_bac...@mail.monarch.com Michael Garretson:pstc_ad...@garretson.org For policy questions, send mail to: Richard Nute: ri...@ieee.org --- This message is from the IEEE EMC Society Product Safety Technical Committee emc-pstc discussion list. To cancel your subscription, send mail to: majord...@ieee.org with the single line: unsubscribe emc-pstc For help, send mail to the list administrators: Jim Bacher: jim_bac...@mail.monarch.com Michael Garretson:pstc_ad...@garretson.org For policy questions, send mail to: Richard Nute: ri...@ieee.org
RE: Zo
Zo= eta(free space)/ (pi * sqrt(epsilon effective) * inversecosh (D/d) where epsilon effective is = 1 + (.25 + 0.0004 theta^2)*(epsilon of insulation-1.) use 0.001 instead of 0.0004 for soft insulation theta= inversetan( twists per length * pi * D) D= diameter of coductor + insulator d = diameter of conductor I am attaching a Powerpoint file that should help twistedpairzo.ppt Vicente Rodríguez, Ph.D., E.I.T. RF/Electromagnetics Engineer ETS-Lindgren (an ESCO Company) P.O.Box 80589, Austin TX 78708-0589 phone 512.835.4684 x648 fax 512.835.4729 vicente.rodrig...@emctest.com http://www.emctest.com http://home.austintx.com/~vicenter -Original Message- From: William D'Orazio [SMTP:dora...@cae.ca] Sent: Thursday, January 11, 2001 8:49 AM To: EMC Posting (E-mail) Subject: Zo Does anybody know the characteristic impedance of a twisted pair? Thanks in advance, ...OLE_Obj... William D'Orazio CAE Electronics Ltd. Electrical System Designer Phone: (514) 341-2000 (X4555) Fax: (514)340-5552 Email: dora...@cae.ca --- This message is from the IEEE EMC Society Product Safety Technical Committee emc-pstc discussion list. To cancel your subscription, send mail to: majord...@ieee.org with the single line: unsubscribe emc-pstc For help, send mail to the list administrators: Jim Bacher: jim_bac...@mail.monarch.com Michael Garretson:pstc_ad...@garretson.org For policy questions, send mail to: Richard Nute: ri...@ieee.org twistedpairzo.ppt Description: MS-Powerpoint presentation
RE: Capacitance calculation
I am using the approximation that this cylinder is very long, longer than the other dimensions, then I can solve it as a two dimensional problem and find the capacitance per unit length. I am also assuming that the plane is much larger than the distance D to the cylinder of radius R, then I can approach the problem as PEC cylinder above a PEC plane, an image plane. then by the image theory you can substitute the flat plane by a cylinder of the same dimensions as the other one but with different polarity. So the equivalent problem is now two cylinders of radius R separated by 2*D. that is an analytical problem that can be found in the literature (among them Cheng fundamentals of engineering Electromagnetics Addison Wesley) and whose solution I posted in a previous message Vicente Rodríguez, Ph.D., E.I.T. RF/Electromagnetics Engineer ETS-Lindgren (an ESCO Company) P.O.Box 80589, Austin TX 78708-0589 phone 512.835.4684 x648 fax 512.835.4729 vicente.rodrig...@emctest.com http://www.emctest.com http://home.austintx.com/~vicenter -Original Message- From: Cortland Richmond [SMTP:72146@compuserve.com] Sent: Monday, December 18, 2000 10:56 AM To: Vince Rodriguez Subject: RE: Capacitance calculation Vince Rodriguez vicente.rodrig...@emctest.com wrote: You can use image theory and just get the capacitance between two cylinders separated by 2D That doesn't sound right. How is the capacitance between two objects separated by some distance going to be equalled by that between two objects (and one of them is smaller than before) separated by TWICE the distance? This is an experiment that can easily be done with common test equipment. Cortland --- This message is from the IEEE EMC Society Product Safety Technical Committee emc-pstc discussion list. To cancel your subscription, send mail to: majord...@ieee.org with the single line: unsubscribe emc-pstc For help, send mail to the list administrators: Jim Bacher: jim_bac...@mail.monarch.com Michael Garretson:pstc_ad...@garretson.org For policy questions, send mail to: Richard Nute: ri...@ieee.org
RE: Capacitance calculation
Another comment,
this is an analytical problem the steps are, first use image theory to get
rid of the plane and end up with two parallel cylinders of radius r at a
distance 2D from each other. then using image theory again the wires can be
reduced to line charges then the problem is easily solved
C= (pi)(epsilon)/ Ln[(D/r)+sqrt{(D/r)^2-1}]
if D/r is bigger than 1 then since Ln[x+sqrt(x^2-1)]=cosh^(-1) x for x1
C=(pi)(epsilon)/cosh^(-1)(D/r)
Vicente Rodríguez, Ph.D., E.I.T.
RF/Electromagnetics Engineer
ETS-Lindgren
(an ESCO Company)
P.O.Box 80589, Austin TX 78708-0589
phone 512.835.4684 x648
fax 512.835.4729
vicente.rodrig...@emctest.com
http://www.emctest.com
http://home.austintx.com/~vicenter
-Original Message-
From: William D'Orazio [SMTP:dora...@cae.ca]
Sent: Friday, December 15, 2000 3:07 PM
To: EMC Posting (E-mail)
Subject: Capacitance calculation
Gents,
I do not have all my resources next to me and I was wondering what
is the capacitance between a cylinder of radius R separated from a plane.
The distance between the center of the cylinder and the plane is D (see
Fig.1).
In addition is the dielectric constant of styrofoam 1.
Thanks in advance,
...
Fig. 1
...
William D'Orazio
CAE Electronics Ltd.
Electrical System Designer
Phone: (514) 341-2000 (X4555)
Fax: (514)340-5552
Email: dora...@cae.ca
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RE: Capacitance calculation
You can use image theory and just get the capacitance between two cylinders separated by 2D Vicente Rodríguez, Ph.D., E.I.T. RF/Electromagnetics Engineer ETS-Lindgren (an ESCO Company) P.O.Box 80589, Austin TX 78708-0589 phone 512.835.4684 x648 fax 512.835.4729 vicente.rodrig...@emctest.com http://www.emctest.com http://home.austintx.com/~vicenter -Original Message- From: William D'Orazio [SMTP:dora...@cae.ca] Sent: Friday, December 15, 2000 3:07 PM To: EMC Posting (E-mail) Subject: Capacitance calculation Gents, I do not have all my resources next to me and I was wondering what is the capacitance between a cylinder of radius R separated from a plane. The distance between the center of the cylinder and the plane is D (see Fig.1). In addition is the dielectric constant of styrofoam 1. Thanks in advance, ... Fig. 1 ... William D'Orazio CAE Electronics Ltd. Electrical System Designer Phone: (514) 341-2000 (X4555) Fax: (514)340-5552 Email: dora...@cae.ca --- This message is from the IEEE EMC Society Product Safety Technical Committee emc-pstc discussion list. To cancel your subscription, send mail to: majord...@ieee.org with the single line: unsubscribe emc-pstc For help, send mail to the list administrators: Jim Bacher: jim_bac...@mail.monarch.com Michael Garretson:pstc_ad...@garretson.org For policy questions, send mail to: Richard Nute: ri...@ieee.org --- This message is from the IEEE EMC Society Product Safety Technical Committee emc-pstc discussion list. To cancel your subscription, send mail to: majord...@ieee.org with the single line: unsubscribe emc-pstc For help, send mail to the list administrators: Jim Bacher: jim_bac...@mail.monarch.com Michael Garretson:pstc_ad...@garretson.org For policy questions, send mail to: Richard Nute: ri...@ieee.org
RE: How does RF travel through outer space?
yes you are right on that, the behavior of the energy with distance depends on the source. Vicente Rodríguez, Ph.D., E.I.T. RF/Electromagnetics Engineer ETS-Lindgren (an ESCO Company) P.O.Box 80589, Austin TX 78708-0589 phone 512.835.4684 x648 fax 512.835.4729 vicente.rodrig...@emctest.com http://www.emctest.com http://home.austintx.com/~vicenter -Original Message- From: geor...@lexmark.com [SMTP:geor...@lexmark.com] Sent: Thursday, November 30, 2000 4:11 PM To: george_t...@dell.com Cc: brian_ku...@leco.com; emc-p...@majordomo.ieee.org Subject: RE: How does RF travel through outer space? Geez, this question makes me feel like I'm on Who want to be a millionaire! Brain should have provided multiple choices for his two questions. George, good anaolgy re the pumpkin. Here is some other stuff I remember from my hated electric field classes some 40 years ago. If the radiator is a point source in space, and the energy is equally distributed in a spherical pattern, the energy at any point in space drops off as the cube of the distance from the source. If the radiator is an infinite line, the energy drops off as the square of the distance from the source. If the source is an infinite plane, the energy remains the same regardless of the distance from the plane. Are these right? Now to the tough part about RF travelling in space. Note that all energy we receive from the sun comes in some form of electromagnetic energy, ranging from frequncies which give us light, to infrared which gives us heat, in addition to all kinds of non-structered RF signals. Doesn't electromagnetic energy depend on the sequential exitation of electrons from one adjacent molecule to another to travel anywhere? It is my understanding that space is not a vacuum, but merely a lower pressure than that here on earth. The molecules may be further apart (less dense), but are still out there. The only TRUE vacuum I know of is what they call a black hole in space, which literally sucks all nearby matter into it..but this is still a theory. Brain, the RF energy, or any other part of the EM spectrum, travels through space just as in the water you mentioned, only with different molecules, and spacing of same. These are my FINAL answers. Did I win anything, or must I go home a loser? George george_tang%dell@interlock.lexmark.com on 11/30/2000 03:56:59 PM Please respond to george_tang%dell@interlock.lexmark.com To: brian_kunde%leco@interlock.lexmark.com, emc-pstc%majordomo.ieee@interlock.lexmark.com cc:(bcc: George Alspaugh/Lex/Lexmark) Subject: RE: How does RF travel through outer space? We all just carved pumpkins not too long ago. We can use pumpkins to explain one of these questions. If you put a 5 watt light bulb at the center of your carved pumpkin, then each square inch of the internal pumpkin surface gets the amount of light energy given by the expression: 5W / (internal surface area of pumpkin) = light energy per square inch Now you move your light bulb to a bigger pumpkin and do the same calculation. You find that each square inch of your bigger pumpkin gets less light energy due to a bigger surface area. This is why RF signals drop off at the rate of 1/distance squared, since the pumpkin surface area is proportional to the square of the radius. Light is simply a higher frequency emission than RF, but the same concept applies. As far as How RF travel through vacuum, you can think of it this way: RF is composed of electric field and magnetic field. Electric field is simply the attraction force between the positive charges and the negative charges. And magnetic field is the interaction between 2 current loops. It is not hard to imagine that refrigerator magnets will work in vacuum or protons and electrons will attract in outer space. RF is simply the electric and magnetic fields changing polarity at a very rapid rate. George -Original Message- From: brian_kunde [mailto:brian_ku...@leco.com] Sent: Thursday, November 30, 2000 12:51 PM To: emc-pstc Subject: How does RF travel through outer space? Hello, I'm sorry if this is too simple of question... How does RF travel through outer space?. I will be teaching a class in which this question will come up. I want to be prepared with all the basic science behind this principal. I need an explaination that is simple and easy to understand. People seem to have no problem understanding how waves can travel through mass such as a body of water but can not understand how it can travel where there is no mass. I also understand that there is a lot of debate over how Light travels through space (photons and all). Also, I understand that RF signals degrade at a rate of 1/distance(squared). What force is causing this attenuation? Try to keep it simple for
