I'm writing a network simulator using the the FreeSpace propogation model which says that Pr = Pt*Gt*Gr * (lambda / (4*pi*d))^n Where Pt is the power transmitted Gt,Gr are the Gain of the transmitter and reciever d is the distance between the antennas. My question is that if d < 1.0m then it is possible that Pr > Pt. In other words if d is small than the reciever recieves more power than the transmitted power. How do you handle smaller distance less than a meter? Regards, John
Free Space Propogation Model and distance less than a meter.
Started by ●November 12, 2004
Reply by ●November 12, 20042004-11-12
Are your devices really going to be closer than a meter apart? The 1m reference is usually just that, a reference. There's nothing magical about antennas or propagation that makes systems behave differently at less than a meter, and you're correct that the receive power will never be greater than the transmit power. You can always rescale everything for different units so that the effects of the exponent make more sense to you. e.g., use cm instead of m. You'll need Gt and Gr to be valid for the respective antennas at the distances considered, but other than that things should work out okay. Some antenna types will do strange things in very near field, but just getting less than a meter isn't problematic if the power levels are reasonable. If the transmit power is very substantial then practical effects such as coupling through means other than the antennas starts to become a concern at such short ranges. With decent shielding this often isn't a problem, but it's something to think about. I assume you're using an exponent of n = 2? On 11 Nov 2004 21:01:19 -0800, beerbaron@gmail.com (Beer Baron) wrote:>I'm writing a network simulator using the the FreeSpace propogation >model which says that > >Pr = Pt*Gt*Gr * (lambda / (4*pi*d))^n > >Where > >Pt is the power transmitted >Gt,Gr are the Gain of the transmitter and reciever >d is the distance between the antennas. > > >My question is that if d < 1.0m then it is possible that Pr > Pt. In >other words if d is small than the reciever recieves more power than >the transmitted power. How do you handle smaller distance less than a >meter?Eric Jacobsen Minister of Algorithms, Intel Corp. My opinions may not be Intel's opinions. http://www.ericjacobsen.org
Reply by ●November 12, 20042004-11-12
Beer Baron wrote:> I'm writing a network simulator using the the FreeSpace propogation > model which says that > > Pr = Pt*Gt*Gr * (lambda / (4*pi*d))^n > > Where > > Pt is the power transmitted > Gt,Gr are the Gain of the transmitter and reciever > d is the distance between the antennas. > > > My question is that if d < 1.0m then it is possible that Pr > Pt. In > other words if d is small than the reciever recieves more power than > the transmitted power. How do you handle smaller distance less than a > meter? > > Regards, > > JohnIt's a far-field model, (or at least a moderate-field model given that n is not pegged at 2); if your wavelength or any antenna dimension is a significant fraction of a meter then you're in near-field and it won't apply. For near-field conditions you can't just model the antennas as directional emitters and receivers of radiation, you actually have to pay attention to distance variations as well. If you really have to model this in the near field then you'll need a modeling package that takes the effects into account. This would be a good question for nearly any newsgroup with the word "antenna" in it, including rec.radio.amateur.antenna -- there are some pretty sharp RF professionals who don't get enough of it at work so they go home and do more there. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
Reply by ●November 12, 20042004-11-12
"Beer Baron" <beerbaron@gmail.com> wrote in message news:c9a42e73.0411112013.2609242b@posting.google.com...> I'm writing a network simulator using the the FreeSpace propogation > model which says that > > Pr = Pt*Gt*Gr * (lambda / (4*pi*d))^n > > Where > > Pt is the power transmitted > Gt,Gr are the Gain of the transmitter and reciever > d is the distance between the antennas. > > > My question is that if d < 1.0m then it is possible that Pr > Pt. In > other words if d is small than the reciever recieves more power than > the transmitted power. How do you handle smaller distance less than a > meter? > > Regards, > > JohnHello John, If you study EM, you will find that you will have mostly 1/r and 1/r^2 types of fields. There are other orders - 1/r^n , but these only mater in the induction zone. Normally the classical RF propagation model stems from the far field situation where 1/r^n for n>=2 components have decayed to essentially zero relative to the 1/r field. The near field is quite different - it has significant components all orders. When you have antennas less than a meter apart, the assumptions for the derivation for your formula are no longer met, so the formula doesn't apply in that case. Fortunately conservation of energy tells you that the best a receiver could do is simply capture all of the transmitter's power. Normally the receiver captures a tiny fraction of the transmitter's energy. And you were correct in your questioning of the obsevation that your receiver generates energy if your formula is correct. Typically in E-Mag, we describe three general regions which relate wavelength (lambda) , the antenna size(d), and the distance(r). Near (Static) zone: d << r << lambda Intermediate (Induction) zone d << r ~ lambda Far (Radiation) zone d << lambda << r Your formula assumes the radiation zone. And when your antennas are something like a meter apart with antennas only a few cm in size and your wavelength of 12.5 cm (f=2.4GHz), you are in the induction zone. The computations are more difficult in this region. IHTH, Clay
Reply by ●November 12, 20042004-11-12
I knew I shoulda waited to post. ;) As Clay and Tim pointed out, the big thing to keep in mind is the antenna size and the wavelength and how that affects the situation. I would have expected that for 2.4Ghz and a small antenna 1m would still have been in the far field, but Clay indicates otherwise. This makes me curious how Bluetooth or UWB systems, where the system range is not expected to functionally exceed about 3m, are effectively analyzed, since many links will be <1m. On Fri, 12 Nov 2004 13:58:31 -0500, "Clay Turner" <physics@bellsouth.net> wrote:> >"Beer Baron" <beerbaron@gmail.com> wrote in message >news:c9a42e73.0411112013.2609242b@posting.google.com... >> I'm writing a network simulator using the the FreeSpace propogation >> model which says that >> >> Pr = Pt*Gt*Gr * (lambda / (4*pi*d))^n >> >> Where >> >> Pt is the power transmitted >> Gt,Gr are the Gain of the transmitter and reciever >> d is the distance between the antennas. >> >> >> My question is that if d < 1.0m then it is possible that Pr > Pt. In >> other words if d is small than the reciever recieves more power than >> the transmitted power. How do you handle smaller distance less than a >> meter? >> >> Regards, >> >> John > >Hello John, > >If you study EM, you will find that you will have mostly 1/r and 1/r^2 types >of fields. There are other orders - 1/r^n , but these only mater in the >induction zone. Normally the classical RF propagation model stems from the >far field situation where 1/r^n for n>=2 components have decayed to >essentially zero relative to the 1/r field. The near field is quite >different - it has significant components all orders. When you have antennas >less than a meter apart, the assumptions for the derivation for your formula >are no longer met, so the formula doesn't apply in that case. Fortunately >conservation of energy tells you that the best a receiver could do is simply >capture all of the transmitter's power. Normally the receiver captures a >tiny fraction of the transmitter's energy. And you were correct in your >questioning of the obsevation that your receiver generates energy if your >formula is correct. Typically in E-Mag, we describe three general regions >which relate wavelength (lambda) , the antenna size(d), and the distance(r). > >Near (Static) zone: d << r << lambda > >Intermediate (Induction) zone d << r ~ lambda > >Far (Radiation) zone d << lambda << r > >Your formula assumes the radiation zone. And when your antennas are >something like a meter apart with antennas only a few cm in size and your >wavelength of 12.5 cm (f=2.4GHz), you are in the induction zone. The >computations are more difficult in this region. > >IHTH, > >Clay > > > > >Eric Jacobsen Minister of Algorithms, Intel Corp. My opinions may not be Intel's opinions. http://www.ericjacobsen.org
Reply by ●November 12, 20042004-11-12
Eric Jacobsen wrote:> I knew I shoulda waited to post. ;) > > As Clay and Tim pointed out, the big thing to keep in mind is the > antenna size and the wavelength and how that affects the situation. > I would have expected that for 2.4Ghz and a small antenna 1m would > still have been in the far field, but Clay indicates otherwise. > > This makes me curious how Bluetooth or UWB systems, where the system > range is not expected to functionally exceed about 3m, are effectively > analyzed, since many links will be <1m. > >Just a point to ponder when you get ready to ask why the antenna size makes a difference, too: I do work from time to time for a company that does IR imaging, in the 75THz (3-5 micron) band. The highest magnification uses an objective lens that is over 200mm in diameter, or about 40,000 wavelengths. When you focus on an object at infinity (far field), anything less than about 20 meters away is out of focus -- so your far field approximations (at least for antenna directivity) don't really hold water until you're 4 million wavelengths out. Fortunately with most communications-type antennas you can start using the far-field approximation much sooner. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
Reply by ●November 12, 20042004-11-12
In article <419556bc.1185204437@news.west.cox.net>, eric.jacobsen@delete.ieee.org wrote:>This makes me curious how Bluetooth or UWB systems, where the system >range is not expected to functionally exceed about 3m, are effectively >analyzed, since many links will be <1m. >I'm not sure I understand why people are having trouble with this. The one meter distance is a standard reference distance. It is, in a sense, an engineering fiction in that even signals where the wavelength is many meters in length have the same reference distance. The same fiction is used in sonar and other fields.
Reply by ●November 13, 20042004-11-13
beerbaron@gmail.com (Beer Baron) wrote in message news:<c9a42e73.0411112013.2609242b@posting.google.com>...> I'm writing a network simulator using the the FreeSpace propogation > model which says that > > Pr = Pt*Gt*Gr * (lambda / (4*pi*d))^n > > Where > > Pt is the power transmitted > Gt,Gr are the Gain of the transmitter and reciever > d is the distance between the antennas. > > > My question is that if d < 1.0m then it is possible that Pr > Pt. In > other words if d is small than the reciever recieves more power than > the transmitted power. How do you handle smaller distance less than a > meter?You have already recieved lots of good answers to your question. Just let me add that the difference between the far-field approximation and neart-field physics can cause some trouble. Some people I know who were very intersted in sonars ran into trouble since they had to do certain measurements in the near field of the sonar, and did not get the measurements to match up with the (far field) specs without doing some serious thinking about these issues. Another time, I saw a presentation made by some guy who had designed his own active sonar array. The array consisted of several individual transducer elements that were mounted in a 2D array shape. It turned out that in transmit mode, certain of these transducer elements were actually draining energy away from the acoustic field. They generated voltage from the pressure waves transmitted from the other elements, instead of adding more pressure. The phenomenon was explained by near field effects. So yes, the question you have stumbled upon is important. You really need to go into the detaile equations of the physics of the transmitter to sort it out. Rune
Reply by ●November 13, 20042004-11-13
On Sat, 13 Nov 2004 02:16:26 GMT, george.w.bush@whitehouse.com (George Bush) wrote:>In article <419556bc.1185204437@news.west.cox.net>, eric.jacobsen@delete.ieee.org wrote: > >>This makes me curious how Bluetooth or UWB systems, where the system >>range is not expected to functionally exceed about 3m, are effectively >>analyzed, since many links will be <1m. >> > >I'm not sure I understand why people are having trouble with this. The one >meter distance is a standard reference distance. It is, in a sense, an >engineering fiction in that even signals where the wavelength is many meters >in length have the same reference distance. The same fiction is used in sonar >and other fields.Absolutely, there's nothing magical about 1m. However, there is a difference between far field and near field (induction) with antennas, and Clay indicated that even with a 2.4GHz center frequency and a small antenna that 1m is still near-field, which surprised me. There are a lot of emerging applications (Bluetooth, UWB) where wireless link distances are expected to be very short, from a few centimeters (or less, even immediately adjacent) up to a few meters. Now I'm curious about what affect that will have on the operation of these systems compared to typical analysis. Eric Jacobsen Minister of Algorithms, Intel Corp. My opinions may not be Intel's opinions. http://www.ericjacobsen.org
Reply by ●November 13, 20042004-11-13
Tim Wescott wrote:> Eric Jacobsen wrote: > >> I knew I shoulda waited to post. ;) >> >> As Clay and Tim pointed out, the big thing to keep in mind is the >> antenna size and the wavelength and how that affects the situation. >> I would have expected that for 2.4Ghz and a small antenna 1m would >> still have been in the far field, but Clay indicates otherwise. >> >> This makes me curious how Bluetooth or UWB systems, where the system >> range is not expected to functionally exceed about 3m, are effectively >> analyzed, since many links will be <1m. >> >> > Just a point to ponder when you get ready to ask why the antenna size > makes a difference, too: > > I do work from time to time for a company that does IR imaging, in the > 75THz (3-5 micron) band. The highest magnification uses an objective > lens that is over 200mm in diameter, or about 40,000 wavelengths. When > you focus on an object at infinity (far field), anything less than about > 20 meters away is out of focus -- so your far field approximations (at > least for antenna directivity) don't really hold water until you're 4 > million wavelengths out. > > Fortunately with most communications-type antennas you can start using > the far-field approximation much sooner.I doubt that near-field/far-field consideration has much connection with depth of field. In fact, using wave (rather than ray) theory, if your lens were focused at 20 meters -- the presumed hyperfocal distance -- then everything from 10 meters to infinity will be as sharp as it could get anyway. The Rayleigh limit works at both ends. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
