On 05/07/2011 09:51 AM, fisico32 wrote: (top posting fixed)> >> On 17.03.2011 20:24, glen herrmannsfeldt wrote: >>> Alexander Sotnikov<alex.sotnikov@qip.ru> wrote: >>>> Just a little add-on to previous replies >>>> >>>> On 16.03.2011 14:33, fisico32 wrote: >>>>> In a balanced transmission line the voltage on one wire is >>>>> always as big as the voltage on the other wire but 180 out of phase. >>> >>>> Strictly speaking, a balanced line is the line that has equal > impedances >>>> of conductors along their length and referring to ground. >>> >>> Such that, given equal and opposite AC voltage, the currents are >>> also equal and opposite. And note that as far as the "equal and >>> opposite" is concerned, it is only the AC part that needs to >>> be considered. It is not unusual to have DC currents and voltages >>> on a balanced line, and the DC source/load resistance may be different >>> from the AC (high frequency) impedance. >>> >>> It used to be common to use 300 ohm balanced line for TV lead in, >>> where now 75 ohm coax is usually used. To keep the line balanced >>> at TV frequencies, it is necessary to keep it away from metal objects, >>> or at least such that the distance is equal for both wires. >>> >>>> What you're >>>> describing is a transmission line driven by a symmetric signal. The >>>> balanced line can be driven by asymmetric signals, as well. >>> >>>> The Wikipedia article about balanced lines >>>> http://en.wikipedia.org/wiki/Balanced_line, which i think Glen was >>>> referring to, explains this in details. >>> >>> Also, for coax, which is considered to be unbalanced, one usually >>> wants equal (AC) currents in the shield and center conductor when >>> it is desired not to radiate. (There are come antenna designs >>> based on radiating from coax.) Keeping the currents equal and >>> opposite means that the (AC) current runs on the inside of >>> the shield, instead of the outside. >>> >>> -- glen >> >> Thank you for correction, Glen. >> >> -- >> >> Alexander >>> Thanks eveyone for your inputs. > > Just to make sure I get the right idea, let me summarize: > > after leaving the generator, the forward direct current I1 travels > through the inner conductor and reaches the load. It eventually needs > to return to the generator via the coax shield. At that point the > current has 3 choices: 1)to flow completely on the shield > inner surface, 2) to flow on the shield outer surface or 3) to > flow on both shield surfaces. In the ideal case, what is the > physics reason the return current I2 chooses the inner surface only? > Also I1=-I2. That's not really how it works. At any point along the length of the coax, the physics of the thing 'wants' to make the current on the inner skin of the outer conductor equal and opposite of the current on the inner conductor. The current doesn't 'go to the end and come back' -- it's more that a loop of current between outer and inner conductors gets launched from the source down the cable. That loop of current may not all get absorbed by the load, and if it doesn't then it'll reflect. The hand-waving physics reason for the current 'choosing' to flow on the inner surface of the outer conductor is that the current on the inner conductor induces an inverse current on the inside skin of outer conductor, and rejects any other current (which then, perforce, has to flow on the outside skin). The precise physics reason is because that's how the math works, when you plug this particular problem into Maxwell's equations. This is one of those times when it's dangerous to try to find an intuitive reason: the math works, so that's what's happening. There are a bazillion different intuitive answers that all sound good, and most of them are wrong. > In the nonideal, case, some current (how much? Less than I1?) flows > on the outer shield surface (the common mode current). Why? Does it > have to do with the concept of "ground loop". Yes, if you lump radiated energy into your concept of "ground loop". > If we ground the coax shield the return current has an extra path it > can take to go back to the generator, the one through ground (if the > generator is connect to ground. It may be that the generator is not > connected to ground). > > thanks > fisico32 -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
balanced and unbalanced lines...
Started by ●March 16, 2011
Reply by ●May 7, 20112011-05-07
Reply by ●May 7, 20112011-05-07
Hello Tim, please be patient. I really like your reply. I will ask you to be You say: " The current doesn't 'go to the end and come back' -- it's more that a loop of current between outer and inner conductors gets launched from the source down the cable. That loop of current may not all get absorbed by the load, and if it doesn't then it'll reflect". I am so used to think to think that a circuit must be formed by a signal conductor and a return conductor forming a closed loop where current flows. Your explanation that a loop of current is launched from the source down the cable is a new perspective to me....how did you come up with it and how did the math help you with that concept? I have recently read about single-wire transmission lines like the Goubau line, where only a single conductor is used to transfer energy. That is different from a waveguide where the wave travels inside the guide by reflection. The Goubau line seems to have the wave traveling outside the wire.... So the "return conductor" carrying the "return current" does not seem to be indispensable for the functioning of a t-line....What is its purpose then? I guess that also in a single conductor t-line the current must form a loop, but since we have just one conductor, the longitudinal current must be zero inside it since there is not other return current closing the loop... What do you think? Can you provide some more "handwavy" ideas of it works :)? You also say that: "The hand-waving physics reason for the current 'choosing' to flow on the inner surface of the outer conductor is that the current on the inner conductor induces an inverse current on the inside skin of outer conductor, and rejects any other current (which then, perforce, has to flow on the outside skin)".... Why is there such rejection exactly? Even when I look at some of the math of a coax line I miss to get these intuitive and qualitative aspects....my bad of course. Thanks, fisico32>On 05/07/2011 09:51 AM, fisico32 wrote: >(top posting fixed) >> >>> On 17.03.2011 20:24, glen herrmannsfeldt wrote: >>>> Alexander Sotnikov<alex.sotnikov@qip.ru> wrote: >>>>> Just a little add-on to previous replies >>>>> >>>>> On 16.03.2011 14:33, fisico32 wrote: >>>>>> In a balanced transmission line the voltage on one wire is >>>>>> always as big as the voltage on the other wire but 180 out ofphase.>>>> >>>>> Strictly speaking, a balanced line is the line that has equal >> impedances >>>>> of conductors along their length and referring to ground. >>>> >>>> Such that, given equal and opposite AC voltage, the currents are >>>> also equal and opposite. And note that as far as the "equal and >>>> opposite" is concerned, it is only the AC part that needs to >>>> be considered. It is not unusual to have DC currents and voltages >>>> on a balanced line, and the DC source/load resistance may bedifferent>>>> from the AC (high frequency) impedance. >>>> >>>> It used to be common to use 300 ohm balanced line for TV lead in, >>>> where now 75 ohm coax is usually used. To keep the line balanced >>>> at TV frequencies, it is necessary to keep it away from metalobjects,>>>> or at least such that the distance is equal for both wires. >>>> >>>>> What you're >>>>> describing is a transmission line driven by a symmetric signal. The >>>>> balanced line can be driven by asymmetric signals, as well. >>>> >>>>> The Wikipedia article about balanced lines >>>>> http://en.wikipedia.org/wiki/Balanced_line, which i think Glen was >>>>> referring to, explains this in details. >>>> >>>> Also, for coax, which is considered to be unbalanced, one usually >>>> wants equal (AC) currents in the shield and center conductor when >>>> it is desired not to radiate. (There are come antenna designs >>>> based on radiating from coax.) Keeping the currents equal and >>>> opposite means that the (AC) current runs on the inside of >>>> the shield, instead of the outside. >>>> >>>> -- glen >>> >>> Thank you for correction, Glen. >>> >>> -- >>> >>> Alexander >>> > > Thanks eveyone for your inputs. > > > > Just to make sure I get the right idea, let me summarize: > > > > after leaving the generator, the forward direct current I1 travels > > through the inner conductor and reaches the load. It eventually needs > > to return to the generator via the coax shield. At that point the > > current has 3 choices: 1)to flow completely on the shield > > inner surface, 2) to flow on the shield outer surface or 3) to > > flow on both shield surfaces. In the ideal case, what is the > > physics reason the return current I2 chooses the inner surface only? > > Also I1=-I2. > >That's not really how it works. At any point along the length of the >coax, the physics of the thing 'wants' to make the current on the inner >skin of the outer conductor equal and opposite of the current on the >inner conductor. The current doesn't 'go to the end and come back' -- >it's more that a loop of current between outer and inner conductors gets >launched from the source down the cable. That loop of current may not >all get absorbed by the load, and if it doesn't then it'll reflect. > >The hand-waving physics reason for the current 'choosing' to flow on the >inner surface of the outer conductor is that the current on the inner >conductor induces an inverse current on the inside skin of outer >conductor, and rejects any other current (which then, perforce, has to >flow on the outside skin). > >The precise physics reason is because that's how the math works, when >you plug this particular problem into Maxwell's equations. This is one >of those times when it's dangerous to try to find an intuitive reason: >the math works, so that's what's happening. There are a bazillion >different intuitive answers that all sound good, and most of them arewrong.> > > In the nonideal, case, some current (how much? Less than I1?) flows > > on the outer shield surface (the common mode current). Why? Does it > > have to do with the concept of "ground loop". > >Yes, if you lump radiated energy into your concept of "ground loop". > > > If we ground the coax shield the return current has an extra path it > > can take to go back to the generator, the one through ground (if the > > generator is connect to ground. It may be that the generator is not > > connected to ground). > > > > thanks > > fisico32 > > >-- > >Tim Wescott >Wescott Design Services >http://www.wescottdesign.com > >Do you need to implement control loops in software? >"Applied Control Theory for Embedded Systems" was written for you. >See details at http://www.wescottdesign.com/actfes/actfes.html >
Reply by ●May 7, 20112011-05-07
On May 7, 3:16�pm, "fisico32" <marcoscipioni1@n_o_s_p_a_m.gmail.com> wrote:> Hello Tim, > > please be patient. I really like your reply. I will ask you to be > > You say: > " The current doesn't 'go to the end and come back' -- > it's more that a loop of current between outer and inner conductors gets > launched from the source down the cable. �That loop of current may not > all get absorbed by the load, and if it doesn't then it'll reflect". > > I am so used to think to think that a circuit must be formed by a signal > conductor and a return conductor forming a closed loop where current > flows.That's basically right.> Your explanation that a loop of current is launched from the source down > the cable is a new perspective to me....how did you come up with it and how > did the math help you with that concept?The math demands that concept. When the connection to the battery is made, step waves of current and voltage are launched down the cable both in the center conductor and inside the shield. The ratio of voltage to current is the characteristic impedance of the line, sqrt(l/ c) where l and c are the per-unit-distance inductance and capacitance of the cable. (I assume for now that the cable is lossless.) The step propagates down the cable at a velocity of sqrt(1/lc). between the step and the load, there is no current in either conductor. If the load matches the characteristic impedance of the cable, all the energy is absorbed and the system is immediately in steady state. If the impedances don't match, not all of the power can be absorbed, so some is reflected back. Another reflection happens when the step, no reduced because not all energy is reflected from the load. The step bounces back and forth, diminishing in amplitude each time, while the steady part of the energy flow becomes dominant. After maybe 20 round trips, the step becomes too small to distinguish from noise. Unless the cable is extremely long or special measures are taken to achieve slow propagation, the process is so fast that the process seems instantaneous to a casual observer. Intuition gained from casual observation is useless when reasoning about fine detail: do the math.> I have recently read about single-wire transmission lines like the Goubau > line, where only a single conductor is used to transfer energy. That is > different from a waveguide where the wave travels inside the guide by > reflection. The Goubau line seems to have the wave traveling outside the > wire....This is a case where the "return conductor" is infinitely far away. Are you familiar with the launchers used at the ends of the lines? (Goubau lines, also called G-lines or G-strings, not only need launcher cones to establish the energy-carrying fields outside the wire, but an dielectric coating on the wire itself. This can often be an easily overlooked thin oxide coating, but it must be there.> So the "return conductor" carrying the "return current" does not seem to be > indispensable for the functioning of a t-line....What is its purpose then? > I guess that also in a single conductor t-line the current must form a > loop, but since we have just one conductor, the longitudinal current must > be zero inside it since there is not other return current closing the > loop... > What do you think? Can you provide some more "handwavy" ideas of it worksA radio antenna launches a wave that carries power with no conductor at all. A G-line guides the otherwise free-space wave created by the launcher cone, a sort of antenna. There is no easy answer. Without the math, you need to take it on faith. (Look up TEM-0 propagation mode.)> :)? > > You also say that: > "The hand-waving physics reason for the current 'choosing' to flow on the > inner surface of the outer conductor is that the current on the inner > conductor induces an inverse current on the inside skin of outer > conductor, and rejects any other current (which then, perforce, has to > flow on the outside skin)".... > > Why is there such rejection exactly? Even when I look at some of the math > of a coax line I miss to get these intuitive and qualitative aspects....my > bad of course.Even on twisted pair, a current on one conductor will induce an opposite current on the other. In a co-ax, the current is induced on the inside only because the outer conductor's being a shield prevents the outer surface from "seeing" anything that happens in the inner conductor. ... Jerry -- Engineering is the art of making what you want from things you can get.
Reply by ●May 7, 20112011-05-07
On 5/7/2011 11:28 AM, Tim Wescott wrote:> On 05/07/2011 09:51 AM, fisico32 wrote: > (top posting fixed) >> >>> On 17.03.2011 20:24, glen herrmannsfeldt wrote: >>>> Alexander Sotnikov<alex.sotnikov@qip.ru> wrote: >>>>> Just a little add-on to previous replies >>>>> >>>>> On 16.03.2011 14:33, fisico32 wrote: >>>>>> In a balanced transmission line the voltage on one wire is >>>>>> always as big as the voltage on the other wire but 180 out of phase. >>>> >>>>> Strictly speaking, a balanced line is the line that has equal >> impedances >>>>> of conductors along their length and referring to ground. >>>> >>>> Such that, given equal and opposite AC voltage, the currents are >>>> also equal and opposite. And note that as far as the "equal and >>>> opposite" is concerned, it is only the AC part that needs to >>>> be considered. It is not unusual to have DC currents and voltages >>>> on a balanced line, and the DC source/load resistance may be different >>>> from the AC (high frequency) impedance. >>>> >>>> It used to be common to use 300 ohm balanced line for TV lead in, >>>> where now 75 ohm coax is usually used. To keep the line balanced >>>> at TV frequencies, it is necessary to keep it away from metal objects, >>>> or at least such that the distance is equal for both wires. >>>> >>>>> What you're >>>>> describing is a transmission line driven by a symmetric signal. The >>>>> balanced line can be driven by asymmetric signals, as well. >>>> >>>>> The Wikipedia article about balanced lines >>>>> http://en.wikipedia.org/wiki/Balanced_line, which i think Glen was >>>>> referring to, explains this in details. >>>> >>>> Also, for coax, which is considered to be unbalanced, one usually >>>> wants equal (AC) currents in the shield and center conductor when >>>> it is desired not to radiate. (There are come antenna designs >>>> based on radiating from coax.) Keeping the currents equal and >>>> opposite means that the (AC) current runs on the inside of >>>> the shield, instead of the outside. >>>> >>>> -- glen >>> >>> Thank you for correction, Glen. >>> >>> -- >>> >>> Alexander >>> > > Thanks eveyone for your inputs. > > > > Just to make sure I get the right idea, let me summarize: > > > > after leaving the generator, the forward direct current I1 travels > > through the inner conductor and reaches the load. It eventually needs > > to return to the generator via the coax shield. At that point the > > current has 3 choices: 1)to flow completely on the shield > > inner surface, 2) to flow on the shield outer surface or 3) to > > flow on both shield surfaces. In the ideal case, what is the > > physics reason the return current I2 chooses the inner surface only? > > Also I1=-I2. > > That's not really how it works. At any point along the length of the > coax, the physics of the thing 'wants' to make the current on the inner > skin of the outer conductor equal and opposite of the current on the > inner conductor. The current doesn't 'go to the end and come back' -- > it's more that a loop of current between outer and inner conductors gets > launched from the source down the cable. That loop of current may not > all get absorbed by the load, and if it doesn't then it'll reflect. > > The hand-waving physics reason for the current 'choosing' to flow on the > inner surface of the outer conductor is that the current on the inner > conductor induces an inverse current on the inside skin of outer > conductor, and rejects any other current (which then, perforce, has to > flow on the outside skin). > > The precise physics reason is because that's how the math works, when > you plug this particular problem into Maxwell's equations. This is one > of those times when it's dangerous to try to find an intuitive reason: > the math works, so that's what's happening. There are a bazillion > different intuitive answers that all sound good, and most of them are > wrong. > > > In the nonideal, case, some current (how much? Less than I1?) flows > > on the outer shield surface (the common mode current). Why? Does it > > have to do with the concept of "ground loop". > > Yes, if you lump radiated energy into your concept of "ground loop". > > > If we ground the coax shield the return current has an extra path it > > can take to go back to the generator, the one through ground (if the > > generator is connect to ground. It may be that the generator is not > > connected to ground). > > > > thanks > > fisico32 > >Yabbut.... "launch" and "DC" are different things. So where is the math being applied? To the entire epoch or just to the "launch?. I don't do math unless I understand what I'm doing in the first place. If the math generates counterintuitive results then we'll all be well aware of that I'm sure. Fred
Reply by ●May 7, 20112011-05-07
On May 7, 4:05�pm, Jerry Avins <j...@ieee.org> wrote: ...> propagates down the cable at a velocity of sqrt(1/lc)....Don't misread that. it's the same as 1/sqrt(lc). Jerry -- Engineering is the art of making what you want from things you can get.
Reply by ●May 7, 20112011-05-07
On 05/07/2011 02:55 PM, Fred Marshall wrote:> On 5/7/2011 11:28 AM, Tim Wescott wrote: >> On 05/07/2011 09:51 AM, fisico32 wrote: >> (top posting fixed) >>> >>>> On 17.03.2011 20:24, glen herrmannsfeldt wrote: >>>>> Alexander Sotnikov<alex.sotnikov@qip.ru> wrote: >>>>>> Just a little add-on to previous replies >>>>>> >>>>>> On 16.03.2011 14:33, fisico32 wrote: >>>>>>> In a balanced transmission line the voltage on one wire is >>>>>>> always as big as the voltage on the other wire but 180 out of phase. >>>>> >>>>>> Strictly speaking, a balanced line is the line that has equal >>> impedances >>>>>> of conductors along their length and referring to ground. >>>>> >>>>> Such that, given equal and opposite AC voltage, the currents are >>>>> also equal and opposite. And note that as far as the "equal and >>>>> opposite" is concerned, it is only the AC part that needs to >>>>> be considered. It is not unusual to have DC currents and voltages >>>>> on a balanced line, and the DC source/load resistance may be different >>>>> from the AC (high frequency) impedance. >>>>> >>>>> It used to be common to use 300 ohm balanced line for TV lead in, >>>>> where now 75 ohm coax is usually used. To keep the line balanced >>>>> at TV frequencies, it is necessary to keep it away from metal objects, >>>>> or at least such that the distance is equal for both wires. >>>>> >>>>>> What you're >>>>>> describing is a transmission line driven by a symmetric signal. The >>>>>> balanced line can be driven by asymmetric signals, as well. >>>>> >>>>>> The Wikipedia article about balanced lines >>>>>> http://en.wikipedia.org/wiki/Balanced_line, which i think Glen was >>>>>> referring to, explains this in details. >>>>> >>>>> Also, for coax, which is considered to be unbalanced, one usually >>>>> wants equal (AC) currents in the shield and center conductor when >>>>> it is desired not to radiate. (There are come antenna designs >>>>> based on radiating from coax.) Keeping the currents equal and >>>>> opposite means that the (AC) current runs on the inside of >>>>> the shield, instead of the outside. >>>>> >>>>> -- glen >>>> >>>> Thank you for correction, Glen. >>>> >>>> -- >>>> >>>> Alexander >>>> >> > Thanks eveyone for your inputs. >> > >> > Just to make sure I get the right idea, let me summarize: >> > >> > after leaving the generator, the forward direct current I1 travels >> > through the inner conductor and reaches the load. It eventually needs >> > to return to the generator via the coax shield. At that point the >> > current has 3 choices: 1)to flow completely on the shield >> > inner surface, 2) to flow on the shield outer surface or 3) to >> > flow on both shield surfaces. In the ideal case, what is the >> > physics reason the return current I2 chooses the inner surface only? >> > Also I1=-I2. >> >> That's not really how it works. At any point along the length of the >> coax, the physics of the thing 'wants' to make the current on the inner >> skin of the outer conductor equal and opposite of the current on the >> inner conductor. The current doesn't 'go to the end and come back' -- >> it's more that a loop of current between outer and inner conductors gets >> launched from the source down the cable. That loop of current may not >> all get absorbed by the load, and if it doesn't then it'll reflect. >> >> The hand-waving physics reason for the current 'choosing' to flow on the >> inner surface of the outer conductor is that the current on the inner >> conductor induces an inverse current on the inside skin of outer >> conductor, and rejects any other current (which then, perforce, has to >> flow on the outside skin). >> >> The precise physics reason is because that's how the math works, when >> you plug this particular problem into Maxwell's equations. This is one >> of those times when it's dangerous to try to find an intuitive reason: >> the math works, so that's what's happening. There are a bazillion >> different intuitive answers that all sound good, and most of them are >> wrong. >> >> > In the nonideal, case, some current (how much? Less than I1?) flows >> > on the outer shield surface (the common mode current). Why? Does it >> > have to do with the concept of "ground loop". >> >> Yes, if you lump radiated energy into your concept of "ground loop". >> >> > If we ground the coax shield the return current has an extra path it >> > can take to go back to the generator, the one through ground (if the >> > generator is connect to ground. It may be that the generator is not >> > connected to ground). >> > >> > thanks >> > fisico32 >> >> > > Yabbut.... "launch" and "DC" are different things. So where is the math > being applied? To the entire epoch or just to the "launch?. I don't do > math unless I understand what I'm doing in the first place. If the math > generates counterintuitive results then we'll all be well aware of that > I'm sure.If you're asking about reflections on transmission lines, then you have to concern yourself with the entire event, not just what happens when everything settles out. I'm not sure what you mean by "entire epoch", but if you mean "from the point where we disturb the system from steady state until the point where we declare it settled out", then yes, the math is being applied to the entire epoch. If you were to take a length of lossless transmission line, terminate it with a resistor at it's characteristic impedance, and connect a battery to the other end, you'd 'see' a current step starting at the battery end, with no current ahead of it and V/Z_o (Z_o = characteristic impedance) behind. This would propagate all the way to the load resistor at the cable's propagation speed (which is usually better than 65% of the speed of light, so you have to be on your toes), at which point the system would be in steady state once again. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
Reply by ●May 7, 20112011-05-07
Hi Jerry, thanks for the info. I have discovered a few things this week and I would like to hear your opinion on them: Most introductory physics and EE books(except Chabay, R. and Sherwood, B. A., 2002, Matter and Interactions II) state that a DC current carrying wire only generates a magnetic H field outside it. No electric field E. That is not true. There is also a (weak) electric field E caused not by volume charges in the conductor but by surface charges developed on the wire surface by the battery. Energy flows outside the wires towards the load (and also inside the wire from the outside, Joule effect). There is a short article on current in DC circuits by Ian Sefton, physics educator:Understanding Electricity and Circuits: What the Text Books Don’t Tell You. It is available at http://sydney.edu.au/science/uniserve_science/school/curric/stage6/phys/stw2002/sefton.pdf I later found this book entitled "Electromagnetics Explained" by Ron Schmitt. It gives a good explanation of what is going on in a circuit powered by a RF source. When the circuit is closed, surface charges develop on the two conducting wires of transmission line (for example the twin lead line). Such surface charges on, say, the top conductor, undergo an acceleration producing a field kink. The charges on the other conductor, the bottom one, experience a pulse too being so near but in the opposite direction and get accelerated too. As they accelerate they send out their own field kink which pushes on the charges on the top wire. There is some sort of feedback system. The charge movement in the two wires become coupled. The pulse goes down the transmission line towards the load. This pulse carries energy on the surface of the wires at a speed very close to that of light. Eventually things reach steady state and two coupled electromagnetic surface waves run on each conductor (the signal and return conductor) in opposite directions. Somehow (here I am not very clear how) these two oppositely propagating cylindrical surface waves form the TEM wave that from the generator towards the load and travels outside and in between the wires.Energy gets then dissipated in the load. Books talk about current and voltage waves moving along the transmission line. Those current and voltage waves are derivable from the fields E and H. There are surely longitudinal currents along the wires. But energy is carried outside the wires. In Ian Sefton article we find that "...Arnold Sommerfeld (1952) has pointed out, metals are good conductors of current but nonconductors of energy. Metals conduct current but space conducts energy and the best conductor of electromagnetic energy is the vacuum...". Surface charge densities on the surface of the wires cause current and fields E, H and the energy they carry. Does this story make sense? Tonite I am going to be reading the following: Barlow, H.E.M, "Surface Waves: A Proposed Definition", Proc. of the IEE Barlow and Cullen, "Surface Waves", Proc. of IEE Brown, M.m "The Types of Waves that may Exist Near a Guiding Surface", Proc. IEE Goubau, G. "Surface Waves and Their Applications to Transmission lines", J.of applied Physics Rosser, W. G. V., 1970, Magnitudes of surface charge distributions associated with electric current flow, American Journal of Physics, 38, 265 - 266.>On May 7, 4:05=A0pm, Jerry Avins <j...@ieee.org> wrote: > > ... > >> propagates down the cable at a velocity of sqrt(1/lc).... > >Don't misread that. it's the same as 1/sqrt(lc). > >Jerry >-- >Engineering is the art of making what you want from things you can >get. > >
Reply by ●May 7, 20112011-05-07
fisico32 <marcoscipioni1@n_o_s_p_a_m.gmail.com> wrote: (snip)> I am so used to think to think that a circuit must be formed by a signal > conductor and a return conductor forming a closed loop where current > flows.> Your explanation that a loop of current is launched from the source down > the cable is a new perspective to me....how did you come up with it and how > did the math help you with that concept?Consider it first from the standpoint of the generator. To make it more obvious, consider that there is a transformer on the generator output. The generator puts some voltage and some current on the center conductor. For the unbalanced case, the shield is considered grounded, so all the voltage goes to the center conductor. When the voltage changes, how much current should flow? That depends on the cable impedance, which depends on the cable capacitance and inductance per unit length. If you consider the cable as a series of inductors, with a capacitor connected between each pair, you can view the signal as charging and discharging capacitors along the way. Now, where does the current come from? For the unbalanced case, it could come from the ground, but if you consider charging and discharging the capacitance of the line you will see that the same amount of current (and opposite sign) has to go into the shield at the same time. So, the "current flows to the other end and then comes back" idea is wrong. (Consider a pulse instead of a continuous wave.) At any instant in time, assuming a properly matched generator and cable, the currents are equal and opposite, and the voltage on the shield is zero. If the shield connects properly to the generator, there is no path for current to the outside. (Assume a generator in a closed metal container.) Now, consider the far end as the signal arrives. If the signal goes either into another closed can, or into a properly designed transformer, the voltage on the shield can stay at zero. Consider, though, that it might go into a dipole antenna without a transformer. The current in the center conductor goes into one side of the antenna, and the shield into the other. Even if properly matched, this will cause a non-zero voltage to appear on the shield. Why was there none before? Current along the shield comes into and out of each small piece of cable at just the right rate not to change its voltage. The current at the end is not properly balanced to keep the voltage at zero, and the reflected signal then goes on the outside. The current on the center conductor is equal and opposite to that on the inside of the shield. That is needed to satisfy the bounary condition of the metal shield. Any extra current, and associated voltage, goes on the outside. The proper termination of a coax line to a (center fed) dipole antenna is through a balun, either a transformer or a piece of coaxial line, which keeps the current and voltage right.> I have recently read about single-wire transmission lines like the Goubau > line, where only a single conductor is used to transfer energy. That is > different from a waveguide where the wave travels inside the guide by > reflection. The Goubau line seems to have the wave traveling outside the > wire....Well, it shouldn't be so surprising. You know that it can be done with zero conductors, so one is one more than is needed.> So the "return conductor" carrying the "return current" does not seem to be > indispensable for the functioning of a t-line....What is its purpose then? > I guess that also in a single conductor t-line the current must form a > loop, but since we have just one conductor, the longitudinal current must > be zero inside it since there is not other return current closing the > loop...In the case of a balanced line, one must keep the line sufficiently far from other metal object to keep it balanced. In the case of the single wire line, one keep it even farther from other objects. That distance depends on the frequency.> What do you think? Can you provide some more "handwavy" ideas of > it works :)?> You also say that: > "The hand-waving physics reason for the current 'choosing' to flow on the > inner surface of the outer conductor is that the current on the inner > conductor induces an inverse current on the inside skin of outer > conductor, and rejects any other current (which then, perforce, has to > flow on the outside skin)"....> Why is there such rejection exactly? Even when I look at some of the math > of a coax line I miss to get these intuitive and qualitative aspects....my > bad of course.There can be no E field inside a perfect conductor. There can in a less than perfect conductor, but not enough at the frequencies involved. Currents, especially surface currents, are generated to keep E out. Ideally, and close enough for the usual cases, the exact surface current is generated such that E inside is zero. -- glen
Reply by ●May 7, 20112011-05-07
Tim Wescott <tim@seemywebsite.com> wrote: (snip)> If you were to take a length of lossless transmission line, terminate it > with a resistor at it's characteristic impedance, and connect a battery > to the other end, you'd 'see' a current step starting at the battery > end, with no current ahead of it and V/Z_o (Z_o = characteristic > impedance) behind. This would propagate all the way to the load > resistor at the cable's propagation speed (which is usually better than > 65% of the speed of light, so you have to be on your toes), at which > point the system would be in steady state once again.In one undergrad physics lab that I remember, we measured the impedance, attenuation, and propagation velocity of some cables. (On spools, as they were somewhat long.) One cable had a spiral center conductor, which increases its inductance, and so, as Jerry mentioned decreases the velocity as sqrt(1/LC). The velocity also decreases as the square root of the dielectric constant, more commonly called the index of refraction. (The dielectric increases C.) The funny thing, though, is that the velocity is about what you would get if you considered the wave as following along the spiral wound (high inductance) center conductor. So, does the wave go round and round, or just slow and steady straight ahead down the cable? Otherwise, the experiment is pretty easy. There is a pulse genrator with a series resistor, and a variable resistor at the far end. If the far end resistor matches the impedance there is no reflection, as can be seen with an oscilloscope in the cable input. Measure the round trip time to the reflection on the scope to determine velocity. (The length is indicated on the spools.) Attenuation is also measured using the scope, to see how big it is at the far end. -- glen
Reply by ●May 7, 20112011-05-07
On 5/7/2011 3:41 PM, Tim Wescott wrote:> > If you're asking about reflections on transmission lines, then you have > to concern yourself with the entire event, not just what happens when > everything settles out. I'm not sure what you mean by "entire epoch", > but if you mean "from the point where we disturb the system from steady > state until the point where we declare it settled out", then yes, the > math is being applied to the entire epoch. > > If you were to take a length of lossless transmission line, terminate it > with a resistor at it's characteristic impedance, and connect a battery > to the other end, you'd 'see' a current step starting at the battery > end, with no current ahead of it and V/Z_o (Z_o = characteristic > impedance) behind. This would propagate all the way to the load resistor > at the cable's propagation speed (which is usually better than 65% of > the speed of light, so you have to be on your toes), at which point the > system would be in steady state once again. >Tim, OK. Thanks. In other posts it seems to me dangerous to talk about "DC" and then introduce batteries with switches. The "entire epoch" is what you figured - it just isn't steady state. Only a minor point really. Fred
