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V/I ratio is forced to Z0
On Tue, 27 Sep 2005 02:04:46 GMT, Cecil Moore wrote:
The transmission line length must only be long enough such that the V/I ratio is forced to the Z0 value. According to some pretty smart guys I asked, that's about 2% of a wavelength. Cecil, do you have some quantitative explanation / support for this? The treatments that I have seen of transmission line tuners where different Zo lines are directly connected do not suggest corrections / tolerances of the type you imply. (IIRC, Terman discusses a fringing capacitance as a means of allowing for a physical discontinuity.) I am not asking whether or not field conditions (and V/I on the conductors) immediate to the discontinuity are not Zo of either of the lines, just where has the 2% of a wavelength come from? Owen -- |
Owen Duffy wrote: On Tue, 27 Sep 2005 02:04:46 GMT, Cecil Moore wrote: The transmission line length must only be long enough such that the V/I ratio is forced to the Z0 value. According to some pretty smart guys I asked, that's about 2% of a wavelength. Cecil, do you have some quantitative explanation / support for this? The treatments that I have seen of transmission line tuners where different Zo lines are directly connected do not suggest corrections / tolerances of the type you imply. (IIRC, Terman discusses a fringing capacitance as a means of allowing for a physical discontinuity.) I am not asking whether or not field conditions (and V/I on the conductors) immediate to the discontinuity are not Zo of either of the lines, just where has the 2% of a wavelength come from? Owen As I recall it came from someone on sci.physics.electromag. But think about it. The surge impedance (Zo) is basically just the ratio of the capacitance per unit length to the inductance per unit length. Those quantities might vary a little bit from one place to another, but probably not by much. And there are undoubtedly end effects which locally pull the capacitance and inductance values away from the ideal. So the length really need only be long enough for the variations to average out and for the total values to become large enough to swamp the end effects. ac6xg |
On Fri, 07 Oct 2005 15:09:25 -0700, Jim Kelley
wrote: But think about it. The surge impedance (Zo) is basically just the ratio of the capacitance per unit length to the inductance per unit length. Those quantities might vary a little bit from one place to another, but probably not by much. And there are undoubtedly end effects which locally pull the capacitance and inductance values away from the ideal. So the length really need only be long enough for the variations to average out and for the total values to become large enough to swamp the end effects. I don't doubt there is a discontinuity that disturbs the fields and V/I ratio. What I am asking about is the basis for the 2% of wavelength factor. If I use RG58C/U on 160m, I read that Cecil is suggesting that the V/I ratio is significiantly different to Zo for 2% * 160m or 3.2m (125")from the end of the cable, which seems large when the physical distance between the inner and outer conductor is 0.001m (0.04"). I am looking for quantitative support for Cecil's 2%. Owen -- |
Definition of what you are all waffling about :
"The input impedance Zo applies only for the duration of time taken for an echo to be received back from the point where the line impedance Zo first changes to another value.". Distance can be measured either in metres or, if you like, fractions of a wavelength. Wavelength involves frequency which is rather meaningless because time is already a variable but on a different arbitrary scale. Only Cecil could dream up a use for such an effect. ---- Reg, G4FGQ |
Owen Duffy wrote:
Cecil, do you have some quantitative explanation / support for this? Nope, but there were no disagreeing postings. I am not asking whether or not field conditions (and V/I on the conductors) immediate to the discontinuity are not Zo of either of the lines, just where has the 2% of a wavelength come from? As I remember it came from the spacing between conductors Vs wavelength. The spacing between conductors is about 0.1 inches for RG-58. How many times that value would you think it would take for a transmission line to force its Z0 upon the signals? At 10 MHz, 2% of a wavelength (24 inches) is about 250 times the spacing between conductors. -- 73, Cecil http://www.qsl.net/w5dxp |
Reg Edwards wrote:
Only Cecil could dream up a use for such an effect. Sorry Reg, I only dream of six foot tall blonds with big boobs. The 2% WL value came from sci.physics.electromag. -- 73, Cecil http://www.qsl.net/w5dxp |
Only Cecil could dream up a use for such an effect.
Sorry Reg, I only dream of six foot tall blonds with big boobs. The 2% WL value came from sci.physics.electromag. ======================================== Yes Cec, you've told us before. I read the thread. That newsgroup has more highly-convincing old-wives than this one has. They are just a little harder to detect. 2% of wavelength is meaningless unless you also state by how much input impedance has diverged from Zo after a time T has elapsed. Wavelength also implies a frequency but what THAT has to do with it is anybody's guess. It merely adds to the confusion. ---- Reg, G4FGQ |
On Sat, 08 Oct 2005 00:25:46 GMT, Cecil Moore wrote:
Owen Duffy wrote: Cecil, do you have some quantitative explanation / support for this? Nope, but there were no disagreeing postings. I am not asking whether or not field conditions (and V/I on the conductors) immediate to the discontinuity are not Zo of either of the lines, just where has the 2% of a wavelength come from? As I remember it came from the spacing between conductors Vs wavelength. The spacing between conductors is about 0.1 inches for RG-58. How many times that value would you think it would take for a transmission line to force its Z0 upon the signals? At 10 MHz, 2% of a wavelength (24 inches) is about 250 times the spacing between conductors. It seems different people have this conceptual model of "a transmission line forcing its Z0 upon the signals" in a gradual way, though differing propositions for the length of line that does not behave as predicted by Zo. An extension of that thinking is in the proposition that I have seen that a Bird 43 cannot give valid readings unless there is at least a quarter wave of 50 ohm line on each side of itself. In this case, the magnitude of significantly affected line seems to be 25%, someone else's is 2%, can they both be correct? It seems to me that apart from the region of the significant distortion of the fields local to some kind of discontinuity, that the fields further along the line at a distance from the discontinuity large compared to the dimension of the discontinuity (which will often be the conductor spacing) should be as constrained by the physical parameters of the line (V/I=Zo for each travelling wave). In the case of the Bird 43, I suggest that if had, say, at 1MHz, 75 ohm line and a 75 ohm load on the load side, that the V/I raio for the travelling waves in the region of the sampling element would be so close to 50 ohms as to not materially affect the accuracy of measurements on the 50 ohms coupler section, irrespective of the fact that the sampling element has only 0.02% of a wavelength of 50 ohm line on its load side. (For avoidance of doubt, nothing in the foregoing is to imply the Bird 43 would be directly measuring or indicating the conditions on the 75 ohm line.) Owen -- |
Cecil Moore wrote:
Owen Duffy wrote: Cecil, do you have some quantitative explanation / support for this? Nope, but there were no disagreeing postings. I am not asking whether or not field conditions (and V/I on the conductors) immediate to the discontinuity are not Zo of either of the lines, just where has the 2% of a wavelength come from? As I remember it came from the spacing between conductors Vs wavelength. The spacing between conductors is about 0.1 inches for RG-58. How many times that value would you think it would take for a transmission line to force its Z0 upon the signals? At 10 MHz, 2% of a wavelength (24 inches) is about 250 times the spacing between conductors. Maybe the electromagnetics people have a useful way to visualize it... Deep inside the coax, the electric field lines between the inner and outer of the coax are exactly at right-angles to the main axis. Where that is exactly true, you have a pure TE10 mode so it's also valid to assume that V/I is exactly equal to Zo. Very close to the end of the coax, the electric field lines from the center conductor start to reach out and connect with whatever is out there beyond the end of the shield. Then you no longer have pure TE10 and can no longer assume that V/I=Zo. Coming at it from the other direction, the question would be: how far into the coax must you go before the field lines become accurately at right-angles? We can be sure that the field lines won't suddenly snap from being divergent to being accurately at right-angles, so what we're really asking is: how far before the field lines are near-enough at right angles to make V/I=Zo a good engineering approximation? Intuitively, the diverging field lines only seem likely to occur within a few diameters of the end of the shield. Field lines always connect with highly conducting surfaces at right-angles, and they won't like to be sharply bent to run along the axis of the coax. In other words, the effect would seem to be mainly a function of shield diameter D. Again intuitively, I can't see where wavelength would come into it, unless D itself is a significant fraction of the wavelength (which is normally never true, and even microwave engineers try to avoid it). Following this picture of diverging field lines, there should also be a secondary effect depending on how the inner and shield of the coax are connected to the circuit outside. All of this suggests that it's impossible to give a single answer that would be valid for all cases (unless you choose a number that's so big, it can't fail to be correct... like "120 radials" :-) However, none of this speculation is of any practical consequence. All practical experience indicates that if a line is so short that V/I is not quite equal to Zo, the impedance transformation along that line will be so small that the effect of any Zo error is lost in the noise. -- 73 from Ian G/GM3SEK 'In Practice' columnist for RadCom (RSGB) http://www.ifwtech.co.uk/g3sek |
"Cecil Moore" wrote in message . com... Reg Edwards wrote: Only Cecil could dream up a use for such an effect. Sorry Reg, I only dream of six foot tall blonds with big boobs. The 2% WL value came from sci.physics.electromag. -- 73, Cecil http://www.qsl.net/w5dxp then please take it back there. it makes no sense as it would force the effect to get longer and longer at lower frequencies. the more logical effect is fringing effects from edges of the shield if it isn't properly connected to a proper termination... this is something that i can measure with my tdr, and it is definately a very short range effect. just think, if i pulse a line with a 1khz square wave and the effect got longer with lower frequency what would the return pulse look like? |
Dave wrote:
The 2% WL value came from sci.physics.electromag. then please take it back there. it makes no sense as it would force the effect to get longer and longer at lower frequencies. So how long does the coax have to be to force V/I to Z0 when the applied signal is DC? -- 73, Cecil http://www.qsl.net/w5dxp |
Owen Duffy wrote:
An extension of that thinking is in the proposition that I have seen that a Bird 43 cannot give valid readings unless there is at least a quarter wave of 50 ohm line on each side of itself. In this case, the magnitude of significantly affected line seems to be 25%, someone else's is 2%, can they both be correct? I think if you will recheck that posting you will find the assertion was that a Bird 43 cannot give valid readings by sampling at a point. The line must be at least 1/4WL, and preferably 1/2WL, so that voltage maximums and minimums will exist and can be measured. And that 2% of a wavelength is from my faulty memory. I'll try to Google and find the exact quotation. -- 73, Cecil http://www.qsl.net/w5dxp |
"Cecil Moore" wrote in message t... Dave wrote: The 2% WL value came from sci.physics.electromag. then please take it back there. it makes no sense as it would force the effect to get longer and longer at lower frequencies. So how long does the coax have to be to force V/I to Z0 when the applied signal is DC? -- 73, Cecil http://www.qsl.net/w5dxp just enough for any fringe effects to become negligible... no more than a couple diameters if the coax as a rough guess. |
"Cecil Moore" wrote in message t... Owen Duffy wrote: An extension of that thinking is in the proposition that I have seen that a Bird 43 cannot give valid readings unless there is at least a quarter wave of 50 ohm line on each side of itself. In this case, the magnitude of significantly affected line seems to be 25%, someone else's is 2%, can they both be correct? I think if you will recheck that posting you will find the assertion was that a Bird 43 cannot give valid readings by sampling at a point. The line must be at least 1/4WL, and preferably 1/2WL, so that voltage maximums and minimums will exist and can be measured. And that 2% of a wavelength is from my faulty memory. I'll try to Google and find the exact quotation. -- 73, Cecil http://www.qsl.net/w5dxp i want to see a quote from a manufacturer's or good laboratory manual for that 1/4 or 1/2 wave thing on the bird also. |
Dave wrote:
just enough for any fringe effects to become negligible... no more than a couple diameters of the coax as a rough guess. Apparently, the 2% of a wavelength that I was remembering was at 10 MHz. 1'/(98.4*0.66) rounded to 2%. All I was interested in at the time was proving to Reg that one foot of coax forces Vfor/Ifor = Vref/Iref = Z0, the boundary conditions assumed by Bird Wattmeter designers. Since Kevin was not familiar with PL-239's, I erred on the side of caution with the 2% estimate. ****Quote**** Newsgroups: sci.physics.electromag From: "Kevin G. Rhoads" Date: Tue, 07 Oct 2003 12:49:14 -0400 Subject: Transmission Line Question Cecil wrote: It addresses it adequately but doesn't answer any particulars. Given PL-239 connectors and RG-213 coax, I wonder what the answer would be for 10 MHz? I'm not familiar with the connectors in question. Assuming they are properly attached, they should not introduce much mode diversion. For 10 MHz I would expect that all other modes would be non-propagating (i.e., evanescent) even though RG-213 is a large coax (improved RG-8 apparently). The speed of propagation is listed as 66%, so the nominal wavelength is 3/2 times the free space wavelength for the TEM mode. 3/2 x 30m = 45m, which implies the decay rate in space for non-TEM modes is going to be large as the cable diameter is .405" (jacket) which implies the spacing from inner to outer conductors will be less than .203". For order of magnitude estimate, assuming the lowest non-TEM mode can be approximated using a characteristic equation that really is only applicable in Cartesian geometries: (1/45m)**2 = (1/.203")**2 + kz**2 Clearly, kz must be imaginary to make this work. thus an evanescant, non-propagating wave: kz**2 = (1/45m)**2 - (1/.203")**2 To the accuracy used to date, the first term on the right is negligible, so the decay rate, alpha, can be estimated: alpha**2 = - (kz)**2 = (1/2.03")**2 Or, the lowest order undesired mode should reduce intensity by a factor of 1/e (0.37) in about 2.03"; power will reduce by that factor squared in the same distance (.135). In about four inches, undesired mode power is down to about 0.018, in six inches, .00248, and after a foot, 6.14x10-6 You should double check my algebra, but I think the estimate is reasonable. To put it into other terms, since the wavelength in the coax dielectric is 45m and the conductor to conductor spacing is about 2", any non-TEM mode will suffer attenuation in E-field intensity with a space-rate constant rounghly equal to the conductor to conductor spacing. INtensity drops by 1/e = 1/2.71828 every 2 inches. Power availalbe drops faster, being square of intensity. So unless almost all the power diverts into an undesireable mode (by a factor of more than a million to one), one foot of cable should see pure TEM at the end. HTH Kevin ****End Quote**** -- 73, Cecil http://www.qsl.net/w5dxp |
Dave wrote:
"Cecil Moore" wrote in message et... I think if you will recheck that posting you will find the assertion was that a Bird 43 cannot give valid readings by sampling at a point. The line must be at least 1/4WL, and preferably 1/2WL, so that voltage maximums and minimums will exist and can be measured. i want to see a quote from a manufacturer's or good laboratory manual for that 1/4 or 1/2 wave thing on the bird also. Cecil was quoting someone else there, and is completely innocent :-) Here's how the Bird 43 measures VSWR. It contains a pair of needle-fine voltage probes, powered by small explosive charges. When coax is connected at either side, it fires those probes out into the coax until it finds a voltage maximum and a voltage minimum. Then it computes the Voltage Standing Wave Ratio and a recoil mechanism reels the probes back in. It's so slick, it all happens before you even know it. Warning: when handling a Bird 43, keep all sensitive parts more than 1/2WL from those sockets! An alternative possibility is that the Bird 43 does give valid readings by sampling at the point where it physically is. -- 73 from Ian G/GM3SEK 'In Practice' columnist for RadCom (RSGB) http://www.ifwtech.co.uk/g3sek |
On Sat, 8 Oct 2005 14:20:46 -0000, "Dave" wrote:
i want to see a quote from a manufacturer's or good laboratory manual for that 1/4 or 1/2 wave thing on the bird also. Hi Dave, Don't hold your breath waiting for that Baloney sandwich to be made. 73's Richard Clark, KB7QHC |
"Richard Clark" wrote in message ... On Sat, 8 Oct 2005 14:20:46 -0000, "Dave" wrote: i want to see a quote from a manufacturer's or good laboratory manual for that 1/4 or 1/2 wave thing on the bird also. Hi Dave, Don't hold your breath waiting for that Baloney sandwich to be made. 73's Richard Clark, KB7QHC i'm not, just bored sitting in here watching it rain. over 2" so far today. |
On Sat, 08 Oct 2005 14:13:02 GMT, Cecil Moore wrote:
Owen Duffy wrote: An extension of that thinking is in the proposition that I have seen that a Bird 43 cannot give valid readings unless there is at least a quarter wave of 50 ohm line on each side of itself. In this case, the magnitude of significantly affected line seems to be 25%, someone else's is 2%, can they both be correct? I think if you will recheck that posting you will find the assertion Cecil, it is someone else who has on a number of occasions suggested the quarter wave thing in email correspondence, and here in postings. My suggestion is that the sampler inside a Bird 43 coupler section is sufficiently far inside the 50 ohm coupler line to provide measurements within the instrument's stated accuracy of what is happening within the 50 ohm coupler, irrespective of whether, for instance, a 75 ohm line is attached to the coupler on the load side. The measurements of what is happening within the Bird 43 coupler could then be used to model what is happening on the adjacent line, having regard for any Zo changes, loss, length etc. Owen -- |
Owen Duffy wrote:
Cecil, it is someone else who has on a number of occasions suggested the quarter wave thing in email correspondence, and here in postings? Yep, it's not me, it's Reg. I have defended the Bird wattmeter design. Reg sez one needs at least 1/4WL and preferably 1/2WL in order to accurately ascertain the "real" SWR. -- 73, Cecil http://www.qsl.net/w5dxp |
On Sat, 08 Oct 2005 20:33:14 GMT, Cecil Moore wrote:
Owen Duffy wrote: Cecil, it is someone else who has on a number of occasions suggested the quarter wave thing in email correspondence, and here in postings? Yep, it's not me, it's Reg. I have defended the Bird wattmeter design. Reg sez one needs at least 1/4WL and preferably 1/2WL in order to accurately ascertain the "real" SWR. Not it was not Reg... end of the guessing game. Having regard to the definition of VSWR (SWR), I can understand Reg's point that the direct way to measure VSWR requires sampling voltage or current over a quarter wave of line where you can find / observe the actual minimum and maximum. Having said that, there are other measurements that one can make that allow one to reasonably predict what the VSWR would be. Owen -- |
Ian White, G/GM3SEK wrote:
"An alternative possibility is that the Bird 43 does give valid readings by sampling at the point where it physically is." Bird claims + or - 5% of Full Scale accuracy for the Model 43. Why is there power from the reverse direction for a Bird Model 43 to indicate? There is no second generator sending power in the peverse direction. The reverse r-f comes from a reflection. The coax enforces a voltage to current ratio equal to Zo in each direction of flow. Zo is 50 ohms in the Model 43. Reflection does a peculiar thing. It produces a 180-degree phase reversal between a wave`s voltage and current. Bird uses the fact that the current is in-phase with the voltage in one direction of travel and out-of-phase in the opposite direction of travel to distinguish between the two directions. To distinguish, Bird takes a voltage sample and a current sample at the same point in a 50 ohm line. These two samples are scaled and calibrated to produce identical deflections of the power indicator. Out-of-phase samples thus cancel leaving the in-phase samples to produce double the deflection either would produce alone. This deflection is carefully calibrated in watts. Reversing the direction of the wattmeter element, reverses the sense of the direction indicated and reverses the direction in which the samples of voltage and current cancel. The Bird has been satisfactory for about a half century. Best regards, Richard Harrison, KB5WZI |
Owen wrote:
"---I can understand Reg`s point that the direct way to measure VSWR requires sampling voltage or current over a quarter wave of line where you can find / observe the actual minimum and maximum." Yes, and the direct way to measure MPH would be to measure the number of miles and divide by the number of hours to get an average value. It`s not often done that way. Maybe about as often as people find maxima and minima on a transmission line and compute their ratio. It is more convenient and sufficiently accurate to use indirect methods for MPH and SWR. Best regards, Richard Harrison, KB5WZI |
In all transmission lines, including coax, there are various shapes of
transverse electric and magnetic fields that can exist for the particular transmission line geometry. For each shape, the "propagation constant" can be calculated. Many transmission lines (at lower frequencies) have only one shape with propagates with low attenuation. The other shapes can exist, but their "propagation constant" is such that they decrease exponentially with distance. The propagation constant for each shape can be calculated, and is often a function of frequency. When there is a discontinuity in a line, other shapes than the usual one must exist at the point of the discontinuity. (for example, in order to ensure that the transverse electric field is zero the surface of a conducting shape that is part of the line discontinuity). Thus, these other shapes exist (at a certain amplitude) at the point of discontinuity. The amplitude of the other shapes decreases exponentially at distances away from the discontinuity. The rate of the fall-off will depend on the particular shape, according to its propagation constant. Thus, the distance needed to be back to regular old TEM propagation in a coax will depend on the particular discontinuity, and the propagation constants of the "higher order modes" or different field shapes, of a coax line. I have seen examples worked out for waveguide propagation and a step change in waveguide width. There are probably worked examples of coax discontinuities in the literature, also. These non-propagating shapes are usually called " evanescent modes", and this would be a good search term to use to investigate this further. Cliff Curry "Ian White G/GM3SEK" wrote in message ... Cecil Moore wrote: Owen Duffy wrote: Cecil, do you have some quantitative explanation / support for this? Nope, but there were no disagreeing postings. I am not asking whether or not field conditions (and V/I on the conductors) immediate to the discontinuity are not Zo of either of the lines, just where has the 2% of a wavelength come from? As I remember it came from the spacing between conductors Vs wavelength. The spacing between conductors is about 0.1 inches for RG-58. How many times that value would you think it would take for a transmission line to force its Z0 upon the signals? At 10 MHz, 2% of a wavelength (24 inches) is about 250 times the spacing between conductors. Maybe the electromagnetics people have a useful way to visualize it... Deep inside the coax, the electric field lines between the inner and outer of the coax are exactly at right-angles to the main axis. Where that is exactly true, you have a pure TE10 mode so it's also valid to assume that V/I is exactly equal to Zo. Very close to the end of the coax, the electric field lines from the center conductor start to reach out and connect with whatever is out there beyond the end of the shield. Then you no longer have pure TE10 and can no longer assume that V/I=Zo. Coming at it from the other direction, the question would be: how far into the coax must you go before the field lines become accurately at right-angles? We can be sure that the field lines won't suddenly snap from being divergent to being accurately at right-angles, so what we're really asking is: how far before the field lines are near-enough at right angles to make V/I=Zo a good engineering approximation? Intuitively, the diverging field lines only seem likely to occur within a few diameters of the end of the shield. Field lines always connect with highly conducting surfaces at right-angles, and they won't like to be sharply bent to run along the axis of the coax. In other words, the effect would seem to be mainly a function of shield diameter D. Again intuitively, I can't see where wavelength would come into it, unless D itself is a significant fraction of the wavelength (which is normally never true, and even microwave engineers try to avoid it). Following this picture of diverging field lines, there should also be a secondary effect depending on how the inner and shield of the coax are connected to the circuit outside. All of this suggests that it's impossible to give a single answer that would be valid for all cases (unless you choose a number that's so big, it can't fail to be correct... like "120 radials" :-) However, none of this speculation is of any practical consequence. All practical experience indicates that if a line is so short that V/I is not quite equal to Zo, the impedance transformation along that line will be so small that the effect of any Zo error is lost in the noise. -- 73 from Ian G/GM3SEK 'In Practice' columnist for RadCom (RSGB) http://www.ifwtech.co.uk/g3sek |
Cliff Curry wrote:
In all transmission lines, including coax, there are various shapes of transverse electric and magnetic fields that can exist for the particular transmission line geometry. For each shape, the "propagation constant" can be calculated. Many transmission lines (at lower frequencies) have only one shape with propagates with low attenuation. The other shapes can exist, but their "propagation constant" is such that they decrease exponentially with distance. The propagation constant for each shape can be calculated, and is often a function of frequency. When there is a discontinuity in a line, other shapes than the usual one must exist at the point of the discontinuity. (for example, in order to ensure that the transverse electric field is zero the surface of a conducting shape that is part of the line discontinuity). Thus, these other shapes exist (at a certain amplitude) at the point of discontinuity. The amplitude of the other shapes decreases exponentially at distances away from the discontinuity. The rate of the fall-off will depend on the particular shape, according to its propagation constant. Thus, the distance needed to be back to regular old TEM propagation in a coax will depend on the particular discontinuity, and the propagation constants of the "higher order modes" or different field shapes, of a coax line. I have seen examples worked out for waveguide propagation and a step change in waveguide width. There are probably worked examples of coax discontinuities in the literature, also. These non-propagating shapes are usually called " evanescent modes", and this would be a good search term to use to investigate this further. All agreed. Along with the math that Cecil has retrieved and quoted again, everything points towards the distance in question being a function of coax diameter only; and not wavelength. -- 73 from Ian G/GM3SEK 'In Practice' columnist for RadCom (RSGB) http://www.ifwtech.co.uk/g3sek |
Richard Harrison wrote:
Ian White, G/GM3SEK wrote: "An alternative possibility is that the Bird 43 does give valid readings by sampling at the point where it physically is." Sorry, Richard, apparently my attempt at irony fell flat. Let me put it another way: The instrument can only make measurements at the point on the line where it physically IS. Therefore the Bird 43 cannot be measuring "SWR" by sampling the maximum and minimum voltages at locations further up and down the line. Therefore it follows that the instrument must actually be measuring something else... namely, what you described in your follow-up: Why is there power from the reverse direction for a Bird Model 43 to indicate? There is no second generator sending power in the peverse direction. The reverse r-f comes from a reflection. The coax enforces a voltage to current ratio equal to Zo in each direction of flow. Zo is 50 ohms in the Model 43. Reflection does a peculiar thing. It produces a 180-degree phase reversal between a wave`s voltage and current. Bird uses the fact that the current is in-phase with the voltage in one direction of travel and out-of-phase in the opposite direction of travel to distinguish between the two directions. To distinguish, Bird takes a voltage sample and a current sample at the same point in a 50 ohm line. These two samples are scaled and calibrated to produce identical deflections of the power indicator. Out-of-phase samples thus cancel leaving the in-phase samples to produce double the deflection either would produce alone. This deflection is carefully calibrated in watts. Reversing the direction of the wattmeter element, reverses the polarity of the current sample, while not affecting the voltage sample... and reverses the direction in which the samples of voltage and current cancel. Yup. It measures the reflection coefficient of whatever impedance is connected to the port on the opposite side from the transmitter. This measurement is made at one physical point along the line. The subsequent conversion to VSWR is a mathematical relationship only. The Bird has been satisfactory for about a half century. As I've often said before, you don't need to defend the Bird 43 to me. I own and use one, and admire the design. It only needs to be defended from weird notions about how it works. -- 73 from Ian G/GM3SEK 'In Practice' columnist for RadCom (RSGB) http://www.ifwtech.co.uk/g3sek |
"Ian White wrote Yup. It measures the reflection coefficient of whatever impedance is connected to the port on the opposite side from the transmitter. ===================================== No, it doesn't. It measures the MAGNITUDE of the reflection coefficient. It discards the information which is contained in the phase angle of the reflection coefficient. As a consequence the only use which can be made of the magnitude is to calculate the SWR on an imaginary 50-ohm line. The SWR can be used to calculate the magnitude of the reflection coefficient. --- Reg. |
Reg Edwards wrote:
"Ian White wrote Yup. It measures the reflection coefficient of whatever impedance is connected to the port on the opposite side from the transmitter. ===================================== No, it doesn't. It measures the MAGNITUDE of the reflection coefficient. It discards the information which is contained in the phase angle of the reflection coefficient. Sorry, I left that important word out. As a consequence the only use which can be made of the magnitude is to calculate the SWR on an imaginary 50-ohm line. Agreed. SWR has become a number that indicates the general "goodness" of an impedance match. It is almost always determined indirectly, by actually measuring something else and then calculating an SWR value. The only way to measure VSWR truly and directly is to find the points of maximum and minimum voltage along the line, and measure the ratio of those two voltages. That is the classical definition of VSWR, but hardly anyone measures it that way, because it requires physical access to all points along the line. But if they do, then... The SWR can be used to calculate the magnitude of the reflection coefficient. Engineers swap freely between the different available ways of expressing the "goodness" of an impedance match, choosing whichever one is the most convenient (or conventional) for the application. -- 73 from Ian G/GM3SEK 'In Practice' columnist for RadCom (RSGB) http://www.ifwtech.co.uk/g3sek |
On Tue, 11 Oct 2005 10:43:00 +0100, Ian White G/GM3SEK
wrote: Reg Edwards wrote: "Ian White wrote Yup. It measures the reflection coefficient of whatever impedance is connected to the port on the opposite side from the transmitter. ===================================== No, it doesn't. It measures the MAGNITUDE of the reflection coefficient. It discards the information which is contained in the phase angle of the reflection coefficient. Sorry, I left that important word out. To be picky, in most implementations, its response is a function of the forward or reflected power provided that Zo is real, and the magnitude of the complex reflection coefficient can be calculated from those measurements. Owen -- |
"Owen Duffy" wrote To be picky, in most implementations, its response is a function of the forward or reflected power provided that Zo is real, and the magnitude of the complex reflection coefficient can be calculated from those measurements. ================================ Owen, Forward and especially reflected power are even more imaginary than the SWR on a non-existent 50-ohm line. The only use for forward and reflected power is to calculate the magnitude of the reflection coefficient. And the only use for the reflection coefficient is to calculate the imaginary SWR. And the only use . . . . . I understand meter manufacturers provide graphs, which, if you don't know how to use a pocket calculator, will do the calculations for you. But you will still go round in circles. ---- Reg. |
On Tue, 11 Oct 2005 09:11:19 +0100, Ian White G/GM3SEK
wrote: The subsequent conversion to VSWR is a mathematical relationship only. Hi Ian, This seems to be a particularly notable difference - to which absolutely NO ONE has ever deviated from in ANY determination of SWR! That is to say, this "mathematical" distinction that some rely on to differentiate their arguments has not got one scintilla of difference over any other method. The only way to claim you "directly" measure SWR is to find some way to place two probes of a meter along the line such that one probe goes into the trough and the other into the peak and the meter reads SWR directly. Unfortunately for rhetoric's sake, this STILL renders the determination in terms of a mathematical relationship. It cannot be escaped. Why this keeps on being revisited must be to allow the new lurkers to observe my correction. 73's Richard Clark, KB7QHC |
Reg Edwards wrote:
It measures the MAGNITUDE of the reflection coefficient. It discards the information which is contained in the phase angle of the reflection coefficient. As a consequence the only use which can be made of the magnitude is to calculate the SWR on an imaginary 50-ohm line. Reg, I dug up some calculations from sci.physics.electromag from about a year ago that indicate one foot of 50 ohm coax on each side of the Bird is enough to make the line real, i.e. not imaginary, and that's a conservative estimate. -- 73, Cecil http://www.qsl.net/w5dxp |
"Cecil Moore" wrote in message .. . Reg Edwards wrote: It measures the MAGNITUDE of the reflection coefficient. It discards the information which is contained in the phase angle of the reflection coefficient. As a consequence the only use which can be made of the magnitude is to calculate the SWR on an imaginary 50-ohm line. Reg, I dug up some calculations from sci.physics.electromag from about a year ago that indicate one foot of 50 ohm coax on each side of the Bird is enough to make the line real, i.e. not imaginary, and that's a conservative estimate. -- 73, Cecil http://www.qsl.net/w5dxp i like imaginary! |
Reg Edwards wrote:
Forward and especially reflected power are even more imaginary than the SWR on a non-existent 50-ohm line. My old Heathkit HM-15 SWR meter samples the peak forward and reflected currents. A pot is used to set the meter to full scale using a voltage proportional to the peak forward current. When the voltage proportional to the peak reflected current is then switched into the meter circuit, it reads SWR from the precalibrated scale which is linear with |rho|, i.e. at half-scale, SWR=3 and |rho|=(3-1)/(3+1)=0.5 -- 73, Cecil http://www.qsl.net/w5dxp |
Ian White G/GM3SEK wrote:
All agreed. Along with the math that Cecil has retrieved and quoted again, everything points towards the distance in question being a function of coax diameter only; and not wavelength. Please forgive my previous senior moment. It was ~2% of a wavelength at 10 MHz for RG-213. It appears that one foot of coax on each side of a Bird wattmeter is enough to establish Z0 at 50 ohms which forces Vfor/Ifor=Vref/Iref=50, the necessary Bird boundary conditions. -- 73, Cecil http://www.qsl.net/w5dxp |
On Tue, 11 Oct 2005 15:40:55 -0000, "Dave" wrote:
i like imaginary! fantasy is more appropriate. |
On Tue, 11 Oct 2005 15:37:51 GMT, Cecil Moore wrote:
I dug up some calculations from sci.physics.electromag which you recite here; then in sci.physics.electromag you can quote their use by authorities (sic both times) in rec.radio.amateur.antenna.... This appeal is called a circle of friendship - not evidence. |
Richard Clark wrote:
That is to say, this "mathematical" distinction that some rely on to differentiate their arguments has not got one scintilla of difference over any other method. I must have a dozen equations for SWR, all mathematically consistent with each other. A lot of the math is performed by simply calibrating a meter face. For instance, given a linear meter reading for |rho| with full-scale equal to 1.0, SWR values can be applied to the meter face with 3.0 at half-scale. (Ever notice how many SWR meters have 3.0 at half scale?) -- 73, Cecil http://www.qsl.net/w5dxp |
On Tue, 11 Oct 2005 16:37:42 GMT, Cecil Moore wrote:
Ever notice how many SWR meters have 3.0 at half scale? Amateur grade - 1 in 4; Professional grade - none. |
Richard Clark wrote:
On Tue, 11 Oct 2005 15:37:51 GMT, Cecil Moore wrote: I dug up some calculations from sci.physics.electromag which you recite here; then in sci.physics.electromag you can quote their use by authorities (sic both times) in rec.radio.amateur.antenna.... This appeal is called a circle of friendship - not evidence. Actually, it is called an argumentum ad verecundiam, an appeal to authority - a technical authority in this case. I don't know Kevin G. Rhodes at Dartmouth. He merely answered my question on s.p.e. that didn't find an answer on this newsgroup. Exactly what did you find technically wrong with the following evidence? ****Quote**** Newsgroups: sci.physics.electromag From: "Kevin G. Rhoads" Date: Tue, 07 Oct 2003 12:49:14 -0400 Subject: Transmission Line Question For 10 MHz I would expect that all other modes would be non-propagating (i.e., evanescent) even though RG-213 is a large coax (improved RG-8 apparently). The speed of propagation is listed as 66%, so the nominal wavelength is 3/2 times the free space wavelength for the TEM mode. 3/2 x 30m = 45m, which implies the decay rate in space for non-TEM modes is going to be large as the cable diameter is .405" (jacket) which implies the spacing from inner to outer conductors will be less than .203". For order of magnitude estimate, assuming the lowest non-TEM mode can be approximated using a characteristic equation that really is only applicable in Cartesian geometries: (1/45m)**2 = (1/.203")**2 + kz**2 Clearly, kz must be imaginary to make this work. thus an evanescant, non-propagating wave: kz**2 = (1/45m)**2 - (1/.203")**2 To the accuracy used to date, the first term on the right is negligible, so the decay rate, alpha, can be estimated: alpha**2 = - (kz)**2 = (1/2.03")**2 Or, the lowest order undesired mode should reduce intensity by a factor of 1/e (0.37) in about 2.03"; power will reduce by that factor squared in the same distance (.135). In about four inches, undesired mode power is down to about 0.018, in six inches, .00248, and after a foot, 6.14x10-6 You should double check my algebra, but I think the estimate is reasonable. To put it into other terms, since the wavelength in the coax dielectric is 45m and the conductor to conductor spacing is about 2", any non-TEM mode will suffer attenuation in E-field intensity with a space-rate constant rounghly equal to the conductor to conductor spacing. INtensity drops by 1/e = 1/2.71828 every 2 inches. Power availalbe drops faster, being square of intensity. So unless almost all the power diverts into an undesireable mode (by a factor of more than a million to one), one foot of cable should see pure TEM at the end. ***End Quote*** -- 73, Cecil http://www.qsl.net/w5dxp |
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