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Standing Wave Phase
On Dec 9, 10:59 pm, "AI4QJ" wrote:
I do not understand why people will not accept that, when one connects a length of transmission line of Zo1 to another length of transmission line with Zo2, there can/will be a phase change at the discontinuity. If there is a phase change on the Zo1 line by itself when the discontinuity is 0 + j0 impedance (short) or 0 +j0 conductance (open), shouldn't there be a phase shift between 0 and pi/2 when the impedance (or as you say, "impedor") is added somewhere in between, i.e. when connecting the line of Zo2, such as -j567 ohms at the electrical degree point of the termination? This argument does seem to have a certain attractiveness. It is partially understood in terms of a lumped component of -j567 but that model describes the mismatch at the discontinuity only. It must be corrected at the system level to also accouint for the known 10 degree additional transmission line contribution of the 100 ohm line. It seems to me that if connecting a -j567 impedance produces a certain response, it should not matter how that -j567 impedance is produced. Consider the following cases: - produce the -j567 with a lumped capacitor - produce the -j567 with 10 degrees of 100 ohm line - produce the -j567 with 46.6 degrees of 600 ohm line The claim appears to be that despite the 43.4 degrees of 600 ohm line being terminated in the exact same -j567 impedance in all cases, the phase shift experienced at the interface is different in each case. Does this really make sense? ....Keith |
Standing Wave Phase
Keith Dysart wrote:
It seems to me that if connecting a -j567 impedance produces a certain response, it should not matter how that -j567 impedance is produced. Consider the following cases: (1) - produce the -j567 with a lumped capacitor (2) - produce the -j567 with 10 degrees of 100 ohm line (3) - produce the -j567 with 46.6 degrees of 600 ohm line The claim appears to be that despite the 43.4 degrees of 600 ohm line being terminated in the exact same -j567 impedance in all cases, the phase shift experienced at the interface is different in each case. They are terminated in the same value of impedance but they are not terminated in identical impedances. Please see the IEEE Dictionary for the three different definitions of impedance. Each case involves a different reflection coefficient at the -j567 point. (I added a number to each case above.) (1) involves an *impedor* with a reflection coefficient of 1.0 at -93.2 degrees when connected to the 600 ohm line. The reflected wave at the feedpoint lags the forward wave by 43.4 + 93.2 + 43.4 = 180 degrees. The impedance is a real impedor, not a virtual impedance. (2) involves an impedance discontinuity with a reflection coefficient of -0.7143. With 10 degrees of 100 ohm line, the reflected wave lags the forward wave by 2(43.4 + 36.6 + 10) = 180 degrees. The phase shift at the discontinuity is 36.6 degrees each way. The impedance is caused by superposition of component waves at the impedance discontinuity. (3) with 46.6 degrees of 600 ohm line, the reflected wave lags the forward wave by 2(43.4 + 0 + 46.6) = 180 degrees. The phase shift at the 600 to 600 ohm junction is zero. The reflection coefficient at the 600 to 600 ohm junction is 0, different from (1) and (2) above. The impedance is completely virtual and thus cannot cause anything. -- 73, Cecil http://www.w5dxp.com |
Standing Wave Phase
On Mon, 10 Dec 2007 23:43:58 -0500, "AI4QJ" wrote:
A lumped component is not enough to make the model correct. Comments welcome. Hi Dan, It works both ways. However, lines can appear to be equivalent to lumped components until you, say, double the frequency. The two models clearly diverge. Where you could place a series resonant lumped circuit in place of a resonant line (or versa vice) at a fundamental frequency, the lumped series circuit would rarely demonstrate parallel resonance at twice that frequency, but the line would. But it appears the model hasn't been constrained to less than an octave bandwidth - until after Keith submitted the winning ticket to the lottery commission. Cecil's lotteries are like that. To date (more than 12 years now) you are the only winner, and I haven't seen your prize awarded yet. 73's Richard Clark, KB7QHC |
Standing Wave Phase
AI4QJ wrote:
Once you get to -j567 at the discontinuity (travelling 10 degrees along the 100 ohm line), now you interface with the 600 ohm line. At that point you have to normalize the -j567 ohms to -j(567/600) = -j0.945 on the smith chart (you normalize to Zo for it to calculate properly). This abrupt switch increases the angle from 10 degrees to arctan (0.945)) or 43 degrees. I think the effect to look for is that the abrupt impedance change when Zo changes. Exactly. Plot -j567/100 and -j567/600 on a Smith Chart and read the phase shift directly from the wavelength scale around the outside of the chart. -- 73, Cecil http://www.w5dxp.com |
Standing Wave Phase
Cecil Moore wrote:
Exactly. Plot -j567/100 and -j567/600 on a Smith Chart and read the phase shift directly from the wavelength scale around the outside of the chart. Well, not exactly directly. One must multiply the delta-wavelength by 360 degrees to get the phase shift. If the angle of reflection coefficient degree scale is used, it must be divided by two to get the actual phase shift. -- 73, Cecil http://www.w5dxp.com |
Standing Wave Phase
On Dec 10, 11:43 pm, "AI4QJ" wrote:
I may have misspoken. Once you get to -j567 at the discontinuity (travelling 10 degrees along the 100 ohm line), now you interface with the 600 ohm line. At that point you have to normalize the -j567 ohms to -j(567/600) = -j0.945 on the smith chart (you normalize to Zo for it to calculate properly). This abrupt switch increases the angle from 10 degrees to arctan (0.945)) or 43 degrees. I think the effect to look for is that the abrupt impedance change when Zo changes. A lumped component is not enough to make the model correct. Comments welcome I follow the arithmetic, and it still has a certain attractiveness but how can it make such a difference how the -j567 is produced. What if you were offerred 3 black boxes, each labelled -j567? Would it make much difference what was in them? How does one compute the phase shift at the terminals? ....Keith |
Standing Wave Phase
On Dec 10, 8:33 am, Cecil Moore wrote:
Keith Dysart wrote: It seems to me that if connecting a -j567 impedance produces a certain response, it should not matter how that -j567 impedance is produced. Consider the following cases: (1) - produce the -j567 with a lumped capacitor (2) - produce the -j567 with 10 degrees of 100 ohm line (3) - produce the -j567 with 46.6 degrees of 600 ohm line The claim appears to be that despite the 43.4 degrees of 600 ohm line being terminated in the exact same -j567 impedance in all cases, the phase shift experienced at the interface is different in each case. They are terminated in the same value of impedance but they are not terminated in identical impedances. Please see the IEEE Dictionary for the three different definitions of impedance. Each case involves a different reflection coefficient at the -j567 point. (I added a number to each case above.) (1) involves an *impedor* with a reflection coefficient of 1.0 at -93.2 degrees when connected to the 600 ohm line. The reflected wave at the feedpoint lags the forward wave by 43.4 + 93.2 + 43.4 = 180 degrees. The impedance is a real impedor, not a virtual impedance. (2) involves an impedance discontinuity with a reflection coefficient of -0.7143. With 10 degrees of 100 ohm line, the reflected wave lags the forward wave by 2(43.4 + 36.6 + 10) = 180 degrees. The phase shift at the discontinuity is 36.6 degrees each way. The impedance is caused by superposition of component waves at the impedance discontinuity. (3) with 46.6 degrees of 600 ohm line, the reflected wave lags the forward wave by 2(43.4 + 0 + 46.6) = 180 degrees. The phase shift at the 600 to 600 ohm junction is zero. The reflection coefficient at the 600 to 600 ohm junction is 0, different from (1) and (2) above. The impedance is completely virtual and thus cannot cause anything. I see how you have done the arithmetic, and if I understand correctly, you would claim these are all systems with 90 "electical degrees". What about the following... Open circuit 46.6 degrees of 600 ohm line attached to 80 degrees of 100 ohm line. The drive point impedance will be 0 ohms. And the impedance at the junction will be the same -j567. Using your arithmetic, there is 90-46.6-80 - -36.6 degrees of phase shift at the junction. The physical length is 126.6 degrees while the system is 90 "electrical degrees". Is the definition of a system with 90 "electrical degrees" becoming any system where the drive point impedance has no reactive component? ....Keith |
Standing Wave Phase
Keith Dysart wrote:
Open circuit 46.6 degrees of 600 ohm line attached to 80 degrees of 100 ohm line. The drive point impedance will be 0 ohms. And the impedance at the junction will be the same -j567. Using your arithmetic, there is 90-46.6-80 - -36.6 degrees of phase shift at the junction. The physical length is 126.6 degrees while the system is 90 "electrical degrees". Yes, you have just discovered that a 100 to 600 ohm impedance discontinuity *loses degrees* when the 600 ohm end is open. That's why a center-loaded mobile antenna takes more coil than a base-loaded mobile antenna. Think about it. Hint: if the stub is open, the low Z0 needs to be placed at the open end to make the stub physically shorter. If the stub is shorted, the high Z0 needs to be placed at the shorted end to make the stub physically shorter. You seem to be surprised at that. One wonders why? -- 73, Cecil http://www.w5dxp.com |
Standing Wave Phase
AI4QJ wrote:
"Keith Dysart" wrote in message ... On Dec 10, 11:43 pm, "AI4QJ" wrote: I may have misspoken. Once you get to -j567 at the discontinuity (travelling 10 degrees along the 100 ohm line), now you interface with the 600 ohm line. At that point you have to normalize the -j567 ohms to -j(567/600) = -j0.945 on the smith chart (you normalize to Zo for it to calculate properly). This abrupt switch increases the angle from 10 degrees to arctan (0.945)) or 43 degrees. I think the effect to look for is that the abrupt impedance change when Zo changes. A lumped component is not enough to make the model correct. Comments welcome I follow the arithmetic, and it still has a certain attractiveness but how can it make such a difference how the -j567 is produced. What if you were offerred 3 black boxes, each labelled -j567? Would it make much difference what was in them? How does one compute the phase shift at the terminals? I use the smith chart in my response. If you have 3 black boxes each labeled "input impedance = -j567", they could contain a number of different things but since I am using the smith chart, they will contain 3 different lengths of open transmission lines. Box 1 contains 10 degrees length at 4MHz of 100 ohm transmission line. Based on VF=1, this line would be 3/4*E10E2*10/360 = 2.08m long. On the smith chart, plot from circle 10 degrees (transmission coefficient) and read -j5.67. Normalize to 1 = 100. The impedance at the input of the line is -j567. Label this box "input impedance = -j567". Box 2 contains 19.2 degrees length at 4MHz of 200 ohm transmission line. Based on VF=1, this line would be 3/4*10E2*19.2/360 = 4.0m long On the smith chart, plot from infinte impedance circle 19.2 degrees (transmission coefficient) and read -j2.84. Normalize to 1 = 200. The impedance at the input of the line is -j(200*2.84) = -j567. Label this box "input impedance = -j567". Box 3 contains 27.2 degrees at 4MHz of 300 ohm transmission line. Based on VF=1, this line would be 3/4*10E2*27.2/360 = 5.67m long. On the smith chart, plot from infinte impedance circle 27.2 degrees (transmission coefficient) and read -j1.89. Normalize to 1 = 300. The impedance at the input of the line is -j(300*1.89) = -j567. Label this box "input impedance = -j567". So, do all boxres labeled "input impedance =-j567 ohm" transmission lines behave the same when connected to the 600 ohm transmission line? No. For each of these, impedance at the discontinuity will be -j567. However, each has different electrical lengths, thus the 600 ohm line connecting to it will have to be cut to different electrical lengths, for all of the degrees to add to 90 total. You can only say that all -j567 of a given Zo will affect phase shifts in connected 600 ohms lines in the same way. So far, I find this very interesting. Not all -j567 impedors are equal when it comes to transmission lines. Wait a second. You have three black boxes, all of which present an impedance of -j567 ohms at some frequency. And you say that you can tell the difference between these boxes by connecting a 600 ohm line to each one? If so, please detail exactly what you would measure, and where, which would be different for the three boxes. I maintain that there is no test you can devise at steady state at one frequency (where they're all -j567 ohms) which would enable you to tell them apart. I'll add a fourth box containing only a capacitor and include that in the challenge. If you agree with me that you can't tell them apart but that connecting a 600 ohm line gives different numbers of "electrical degrees", then I'd like to hear your definition of "electrical degrees" because it would have to be something that's different for the different boxes yet you can't tell the difference by any measurement or observation. Roy Lewallen, W7EL |
Standing Wave Phase
Roy Lewallen wrote:
... You have three black boxes, ... ... Roy Lewallen, W7EL Hey! Those ain't illegal citizens band (or, Chicken Band--for you wanna-be-amateurs) linears (or leen-e-airs--as those properly schooled in Chicken Banding would say/pronounce) are they? ;-) And, three? Crud, I'd think a single 5kw would get ya by! Just turn off the "foot warmer" and use the exciter for QRP! wink Regards, JS |
Standing Wave Phase
AI4QJ wrote:
So far, I find this very interesting. Not all -j567 impedors are equal when it comes to transmission lines. An impedor has an impedance but all impedances are not impedors. The IEEE Dictionary gives three different definitions for "impedance". -- 73, Cecil http://www.w5dxp.com |
Standing Wave Phase
On Dec 11, 11:24 pm, "AI4QJ" wrote:
"Keith Dysart" wrote in message ... On Dec 10, 11:43 pm, "AI4QJ" wrote: I may have misspoken. Once you get to -j567 at the discontinuity (travelling 10 degrees along the 100 ohm line), now you interface with the 600 ohm line. At that point you have to normalize the -j567 ohms to -j(567/600) = -j0.945 on the smith chart (you normalize to Zo for it to calculate properly). This abrupt switch increases the angle from 10 degrees to arctan (0.945)) or 43 degrees. I think the effect to look for is that the abrupt impedance change when Zo changes. A lumped component is not enough to make the model correct. Comments welcome I follow the arithmetic, and it still has a certain attractiveness but how can it make such a difference how the -j567 is produced. What if you were offerred 3 black boxes, each labelled -j567? Would it make much difference what was in them? How does one compute the phase shift at the terminals? I use the smith chart in my response. If you have 3 black boxes each labeled "input impedance = -j567", they could contain a number of different things but since I am using the smith chart, they will contain 3 different lengths of open transmission lines. Box 1 contains 10 degrees length at 4MHz of 100 ohm transmission line. Based on VF=1, this line would be 3/4*E10E2*10/360 = 2.08m long. On the smith chart, plot from circle 10 degrees (transmission coefficient) and read -j5.67. Normalize to 1 = 100. The impedance at the input of the line is -j567. Label this box "input impedance = -j567". Box 2 contains 19.2 degrees length at 4MHz of 200 ohm transmission line. Based on VF=1, this line would be 3/4*10E2*19.2/360 = 4.0m long On the smith chart, plot from infinte impedance circle 19.2 degrees (transmission coefficient) and read -j2.84. Normalize to 1 = 200. The impedance at the input of the line is -j(200*2.84) = -j567. Label this box "input impedance = -j567". Box 3 contains 27.2 degrees at 4MHz of 300 ohm transmission line. Based on VF=1, this line would be 3/4*10E2*27.2/360 = 5.67m long. On the smith chart, plot from infinte impedance circle 27.2 degrees (transmission coefficient) and read -j1.89. Normalize to 1 = 300. The impedance at the input of the line is -j(300*1.89) = -j567. Label this box "input impedance = -j567". So, do all boxres labeled "input impedance =-j567 ohm" transmission lines behave the same when connected to the 600 ohm transmission line? No. For each of these, impedance at the discontinuity will be -j567. However, each has different electrical lengths, thus the 600 ohm line connecting to it will have to be cut to different electrical lengths, for all of the degrees to add to 90 total. You can only say that all -j567 of a given Zo will affect phase shifts in connected 600 ohms lines in the same way. So far, I find this very interesting. Not all -j567 impedors are equal when it comes to transmission lines.- Hide quoted text - To find the length of the 600 ohm line, do I not just plot -j567/600 and work from there? It seems to me that it yields the same answer regardless of the content of the black box. ....Keith |
Standing Wave Phase
Keith Dysart wrote:
To find the length of the 600 ohm line, do I not just plot -j567/600 and work from there? It seems to me that it yields the same answer regardless of the content of the black box. Double the frequency and see what happens. -- 73, Cecil http://www.w5dxp.com |
Standing Wave Phase
On Dec 12, 8:21 am, Cecil Moore wrote:
Keith Dysart wrote: To find the length of the 600 ohm line, do I not just plot -j567/600 and work from there? It seems to me that it yields the same answer regardless of the content of the black box. Double the frequency and see what happens. There are certainly many experiments one can conduct to deduce an equivalent circuit for the interior of the black box. But for this sub-thread, which started with 90 "electical degrees", a single frequency seems to me to be strongly implied. ....Keith |
Standing Wave Phase
On Dec 11, 10:48 pm, Cecil Moore wrote:
Keith Dysart wrote: Open circuit 46.6 degrees of 600 ohm line attached to 80 degrees of 100 ohm line. The drive point impedance will be 0 ohms. And the impedance at the junction will be the same -j567. Using your arithmetic, there is 90-46.6-80 - -36.6 degrees of phase shift at the junction. The physical length is 126.6 degrees while the system is 90 "electrical degrees". Yes So this means I could take an antenna element which is longer than 90 physical degrees, i.e. greater than 1/4WL, and with the appropriate matching network make a system that was 90 "electrical degrees". I detect progress on the definition. ....Keith |
Standing Wave Phase
On Wed, 12 Dec 2007 07:21:21 -0600, Cecil Moore
wrote: Double the frequency and see what happens. Hi Dan, The Lottery Commission is asking to re-examine your Ticket. Not to worry, winners WILL be notified at a later date. All that is asked is that you be patient as some have been waiting longer than you. 73's Richard Clark, KB7QHC |
Standing Wave Phase
Keith Dysart wrote:
But for this sub-thread, which started with 90 "electical degrees", a single frequency seems to me to be strongly implied. Of course, but handicapping oneself to a steady-state single sinusoidal frequency is like target shooting while blindfolded and spinning on a Lazy Susan. Why enforce a silly arbitrary handicap? -- 73, Cecil http://www.w5dxp.com |
Standing Wave Phase
Keith Dysart wrote:
So this means I could take an antenna element which is longer than 90 physical degrees, i.e. greater than 1/4WL, and with the appropriate matching network make a system that was 90 "electrical degrees". Of course, if you reverse the position of the 600 ohm line and 100 ohm line in the previous open-stub example, you will *lose* 8.3 degrees of phase shift at the impedance discontinuity. The transmission line will have to be physically 98.3 degrees long to get an electrical 90 degree phase shift. In like manner, if you have a straight (Hustler) base rod and make the rest of the antenna a helical coil with no stinger, you will wind up with a resonant mobile antenna that is more than 90 degrees long physically. If the Z0 of the base rod is 400 ohms, the Z0 of the loading coil is 4000 ohms, and the length of the base rod is 10 degrees, you will lose 9 degrees at the base rod-to-coil impedance discontinuity. The coil will need to be 89 degrees long, i.e. almost self-resonant. The antenna will be 99 degrees long physically. -- 73, Cecil http://www.w5dxp.com |
Standing Wave Phase
Keith Dysart wrote:
To find the length of the 600 ohm line, do I not just plot -j567/600 and work from there? It seems to me that it yields the same answer regardless of the content of the black box. ...Keith It darn well better. Roy Lewallen, W7EL |
Standing Wave Phase
Keith Dysart wrote:
I detect progress on the definition. Are you ready to find out why center-loading takes more coil in a mobile antenna than does base loading? Take the 10 degrees of 100 ohm line and move 5 degrees of it to the other end of the stub. From this: ---43.4 deg 600 ohm line---+---10 deg 100 ohm line---open To this: --5 deg 100 ohm line--+--600 ohm line--+--5 deg 100 ohm line--open Now how many physical degrees of 600 ohm line is needed to make the stub electrically 1/4 wavelength? -- 73, Cecil http://www.w5dxp.com |
Standing Wave Phase
On Dec 12, 2:21 am, Roy Lewallen wrote:
AI4QJwrote: "Keith Dysart" wrote in message ... On Dec 10, 11:43 pm, "AI4QJ" wrote: I may have misspoken. Once you get to -j567 at the discontinuity (travelling 10 degrees along the 100 ohm line), now you interface with the 600 ohm line. At that point you have to normalize the -j567 ohms to -j(567/600) = -j0.945 on the smith chart (you normalize to Zo for it to calculate properly). This abrupt switch increases the angle from 10 degrees to arctan (0.945)) or 43 degrees. I think the effect to look for is that the abrupt impedance change when Zo changes. A lumped component is not enough to make the model correct. Comments welcome I follow the arithmetic, and it still has a certain attractiveness but how can it make such a difference how the -j567 is produced. What if you were offerred 3 black boxes, each labelled -j567? Would it make much difference what was in them? How does one compute the phase shift at the terminals? I use the smith chart in my response. If you have 3 black boxes each labeled "input impedance = -j567", they could contain a number of different things but since I am using the smith chart, they will contain 3 different lengths of open transmission lines. Box 1 contains 10 degrees length at 4MHz of 100 ohm transmission line. Based on VF=1, this line would be 3/4*E10E2*10/360 = 2.08m long. On the smith chart, plot from circle 10 degrees (transmission coefficient) and read -j5.67. Normalize to 1 = 100. The impedance at the input of the line is -j567. Label this box "input impedance = -j567". Box 2 contains 19.2 degrees length at 4MHz of 200 ohm transmission line. Based on VF=1, this line would be 3/4*10E2*19.2/360 = 4.0m long On the smith chart, plot from infinte impedance circle 19.2 degrees (transmission coefficient) and read -j2.84. Normalize to 1 = 200. The impedance at the input of the line is -j(200*2.84) = -j567. Label this box "input impedance = -j567". Box 3 contains 27.2 degrees at 4MHz of 300 ohm transmission line. Based on VF=1, this line would be 3/4*10E2*27.2/360 = 5.67m long. On the smith chart, plot from infinte impedance circle 27.2 degrees (transmission coefficient) and read -j1.89. Normalize to 1 = 300. The impedance at the input of the line is -j(300*1.89) = -j567. Label this box "input impedance = -j567". So, do all boxres labeled "input impedance =-j567 ohm" transmission lines behave the same when connected to the 600 ohm transmission line? No. For each of these, impedance at the discontinuity will be -j567. However, each has different electrical lengths, thus the 600 ohm line connecting to it will have to be cut to different electrical lengths, for all of the degrees to add to 90 total. You can only say that all -j567 of a given Zo will affect phase shifts in connected 600 ohms lines in the same way. So far, I find this very interesting. Not all -j567 impedors are equal when it comes to transmission lines. Wait a second. You have three black boxes, all of which present an impedance of -j567 ohms at some frequency. And you say that you can tell the difference between these boxes by connecting a 600 ohm line to each one? If so, please detail exactly what you would measure, and where, which would be different for the three boxes. I maintain that there is no test you can devise at steady state at one frequency (where they're all -j567 ohms) which would enable you to tell them apart. I'll add a fourth box containing only a capacitor and include that in the challenge. If you agree with me that you can't tell them apart but that connecting a 600 ohm line gives different numbers of "electrical degrees", then I'd like to hear your definition of "electrical degrees" because it would have to be something that's different for the different boxes yet you can't tell the difference by any measurement or observation. Roy Lewallen, W7EL In this case, electrical degrees = physical degrees. VF = 1. |
Standing Wave Phase
On Dec 12, 7:26 am, Keith Dysart wrote:
On Dec 11, 11:24 pm, "AI4QJ" wrote: "Keith Dysart" wrote in message ... On Dec 10, 11:43 pm, "AI4QJ" wrote: I may have misspoken. Once you get to -j567 at the discontinuity (travelling 10 degrees along the 100 ohm line), now you interface with the 600 ohm line. At that point you have to normalize the -j567 ohms to -j(567/600) = -j0.945 on the smith chart (you normalize to Zo for it to calculate properly). This abrupt switch increases the angle from 10 degrees to arctan (0.945)) or 43 degrees. I think the effect to look for is that the abrupt impedance change when Zo changes. A lumped component is not enough to make the model correct. Comments welcome I follow the arithmetic, and it still has a certain attractiveness but how can it make such a difference how the -j567 is produced. What if you were offerred 3 black boxes, each labelled -j567? Would it make much difference what was in them? How does one compute the phase shift at the terminals? I use the smith chart in my response. If you have 3 black boxes each labeled "input impedance = -j567", they could contain a number of different things but since I am using the smith chart, they will contain 3 different lengths of open transmission lines. Box 1 contains 10 degrees length at 4MHz of 100 ohm transmission line. Based on VF=1, this line would be 3/4*E10E2*10/360 = 2.08m long. On the smith chart, plot from circle 10 degrees (transmission coefficient) and read -j5.67. Normalize to 1 = 100. The impedance at the input of the line is -j567. Label this box "input impedance = -j567". Box 2 contains 19.2 degrees length at 4MHz of 200 ohm transmission line. Based on VF=1, this line would be 3/4*10E2*19.2/360 = 4.0m long On the smith chart, plot from infinte impedance circle 19.2 degrees (transmission coefficient) and read -j2.84. Normalize to 1 = 200. The impedance at the input of the line is -j(200*2.84) = -j567. Label this box "input impedance = -j567". Box 3 contains 27.2 degrees at 4MHz of 300 ohm transmission line. Based on VF=1, this line would be 3/4*10E2*27.2/360 = 5.67m long. On the smith chart, plot from infinte impedance circle 27.2 degrees (transmission coefficient) and read -j1.89. Normalize to 1 = 300. The impedance at the input of the line is -j(300*1.89) = -j567. Label this box "input impedance = -j567". So, do all boxres labeled "input impedance =-j567 ohm" transmission lines behave the same when connected to the 600 ohm transmission line? No. For each of these, impedance at the discontinuity will be -j567. However, each has different electrical lengths, thus the 600 ohm line connecting to it will have to be cut to different electrical lengths, for all of the degrees to add to 90 total. You can only say that all -j567 of a given Zo will affect phase shifts in connected 600 ohms lines in the same way. So far, I find this very interesting. Not all -j567 impedors are equal when it comes to transmission lines.- Hide quoted text - To find the length of the 600 ohm line, do I not just plot -j567/600 and work from there? It seems to me that it yields the same answer regardless of the content of the black box. ...Keith I see your point. To end up with -j567 on a 200 ohm line, you must use 19.2 degrees of length. 19.2 degrees of 200 ohm line brings you to - j2.84 on the smith chart. To get -j567 on a 100 ohm line, you only need to go 10 degrees, corresponding to -j5.67 on the smith chart. Although we know that the 200 ohm line is longer, there is no indication that the length of the 600 ohm line must or must not change. I cannot know that unless I know the change in phase angle at the new -j567 discontinuity with the 200 ohm line. I do not have enough information on that unless I measure it, as was assumed to be the case using the 100 ohm line. I was assuming that the phase change at the discontinuity would not change but it is not immediately obvious that it will or will not. Then, if it did not change and stayed at 37 degrees, you would have had a new length of 600 ohm line of 90-37-19.2= 33 degrees which at VF =1, l = 33/360*75 = 6.875m. I do not have the information available to make that assumption. I would have to resort to measurement, same as was done with the 600-100 ohm combination. |
Standing Wave Phase
On Dec 12, 2:21 am, Roy Lewallen wrote:
AI4QJwrote: "Keith Dysart" wrote in message ... On Dec 10, 11:43 pm, "AI4QJ" wrote: I may have misspoken. Once you get to -j567 at the discontinuity (travelling 10 degrees along the 100 ohm line), now you interface with the 600 ohm line. At that point you have to normalize the -j567 ohms to -j(567/600) = -j0.945 on the smith chart (you normalize to Zo for it to calculate properly). This abrupt switch increases the angle from 10 degrees to arctan (0.945)) or 43 degrees. I think the effect to look for is that the abrupt impedance change when Zo changes. A lumped component is not enough to make the model correct. Comments welcome I follow the arithmetic, and it still has a certain attractiveness but how can it make such a difference how the -j567 is produced. What if you were offerred 3 black boxes, each labelled -j567? Would it make much difference what was in them? How does one compute the phase shift at the terminals? I use the smith chart in my response. If you have 3 black boxes each labeled "input impedance = -j567", they could contain a number of different things but since I am using the smith chart, they will contain 3 different lengths of open transmission lines. Box 1 contains 10 degrees length at 4MHz of 100 ohm transmission line. Based on VF=1, this line would be 3/4*E10E2*10/360 = 2.08m long. On the smith chart, plot from circle 10 degrees (transmission coefficient) and read -j5.67. Normalize to 1 = 100. The impedance at the input of the line is -j567. Label this box "input impedance = -j567". Box 2 contains 19.2 degrees length at 4MHz of 200 ohm transmission line. Based on VF=1, this line would be 3/4*10E2*19.2/360 = 4.0m long On the smith chart, plot from infinte impedance circle 19.2 degrees (transmission coefficient) and read -j2.84. Normalize to 1 = 200. The impedance at the input of the line is -j(200*2.84) = -j567. Label this box "input impedance = -j567". Box 3 contains 27.2 degrees at 4MHz of 300 ohm transmission line. Based on VF=1, this line would be 3/4*10E2*27.2/360 = 5.67m long. On the smith chart, plot from infinte impedance circle 27.2 degrees (transmission coefficient) and read -j1.89. Normalize to 1 = 300. The impedance at the input of the line is -j(300*1.89) = -j567. Label this box "input impedance = -j567". So, do all boxres labeled "input impedance =-j567 ohm" transmission lines behave the same when connected to the 600 ohm transmission line? No. For each of these, impedance at the discontinuity will be -j567. However, each has different electrical lengths, thus the 600 ohm line connecting to it will have to be cut to different electrical lengths, for all of the degrees to add to 90 total. You can only say that all -j567 of a given Zo will affect phase shifts in connected 600 ohms lines in the same way. So far, I find this very interesting. Not all -j567 impedors are equal when it comes to transmission lines. Wait a second. You have three black boxes, all of which present an impedance of -j567 ohms at some frequency. And you say that you can tell the difference between these boxes by connecting a 600 ohm line to each one? If so, please detail exactly what you would measure, and where, which would be different for the three boxes. I maintain that there is no test you can devise at steady state at one frequency (where they're all -j567 ohms) which would enable you to tell them apart. I'll add a fourth box containing only a capacitor and include that in the challenge. If you agree with me that you can't tell them apart but that connecting a 600 ohm line gives different numbers of "electrical degrees", then I'd like to hear your definition of "electrical degrees" because it would have to be something that's different for the different boxes yet you can't tell the difference by any measurement or observation. Roy Lewallen, W7EL I see your point. I assumed that the phase change specific to the discontinuity was the same for all -j567 impedances. I do not have the information to say it is true (or I lack the tools to prove it). To end up with -j567 on a 200 ohm line, you must use 19.2 degrees of length. 19.2 degrees of 200 ohm line brings you to -j2.84 on the smith chart. Then, 200* -j2.84 = -j567. To get -j567 on a 100 ohm line, you only need to go 10 degrees, corresponding to -j5.67 on the smith chart. The 100* -j5.67 = -j567. Although we know that the 200 ohm line is longer by 9.2 degrees, there is no indication that the length of the 600 ohm line must or must not change. I cannot know that unless I know the change in phase angle at the new -j567 discontinuity with the 200 ohm line. I do not have enough information on that without additional measurements. I would need to measure it to to system resonance (1/4WL = 90 degrees), as was the case using the 100 ohm line. I was assuming that the phase change at the discontinuity would not change but it is not immediately obvious that it will or will not. But, assuming (rightly or wrongly) that the phase angle at the discontinuity did not change and stayed at 37 degrees, you would have had a new length of 600 ohm line of 90-37-19.2= 33 degrees which at VF =1, l = 33/360*75 = 6.875m. I cannot make that assumption right now. I would have to resort to measurement, same as was done with the 600-100 ohm combination. In any case, this does not disprove the basic premise of the existence of a phase change due to the impedance discontinuity. It does prove that my minor point of elaboration may have had a false assumption.Sorry for any confusion. |
Standing Wave Phase
On Dec 12, 11:00 am, Richard Clark wrote:
On Wed, 12 Dec 2007 07:21:21 -0600, Cecil Moore wrote: Double the frequency and see what happens. Hi Dan, The Lottery Commission is asking to re-examine your Ticket. Not to worry, winners WILL be notified at a later date. All that is asked is that you be patient as some have been waiting longer than you. 73's Richard Clark, KB7QHC Those white floating things are way, way off in the distance and nowhere near to approaching us. The chairs for the band remain in place, folded up. The angle of the deck is 0 degrees horizontal. |
Standing Wave Phase
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Standing Wave Phase
Roy Lewallen wrote:
-- IF there's a phase change at the impedance discontinuity (which I assume means the black box terminals), it's the same for all four (your three and my one) black boxes. To see exactly how wrong that statement really is let's take a look at two of the cases. (1) with a -j567 ohm impedor (capacitor) connected inside the black box to the existing 43.4 degrees of 600 ohm line external to the black box. (2) with 46.6 degrees of 600 ohm line connected to the existing 43.4 degrees of 600 ohm line external to the black box. The reflection coefficient for (1) at the black box terminals is 1.0 at -93 degrees, i.e. 100% reflection at that point with an accompanying phase angle shift. The reflection coefficient for (2) at the black box terminals is 0 at 0 degrees, i.e. no reflections at that point and no accompanying phase angle shift at that point. To assert that the phase change at the black box terminals is the same in both of those cases is ridiculous. In case (1) it is 93 degrees. In case (2) it is zero degrees. -- 73, Cecil http://www.w5dxp.com |
Standing Wave Phase
On Dec 12, 5:52 pm, Cecil Moore wrote:
Roy Lewallen wrote: -- IF there's a phase change at the impedance discontinuity (which I assume means the black box terminals), it's the same for all four (your three and my one) black boxes. To see exactly how wrong that statement really is let's take a look at two of the cases. (1) with a -j567 ohm impedor (capacitor) connected inside the black box to the existing 43.4 degrees of 600 ohm line external to the black box. (2) with 46.6 degrees of 600 ohm line connected to the existing 43.4 degrees of 600 ohm line external to the black box. The reflection coefficient for (1) at the black box terminals is 1.0 at -93 degrees, i.e. 100% reflection at that point with an accompanying phase angle shift. The reflection coefficient for (2) at the black box terminals is 0 at 0 degrees, i.e. no reflections at that point and no accompanying phase angle shift at that point. To assert that the phase change at the black box terminals is the same in both of those cases is ridiculous. In case (1) it is 93 degrees. In case (2) it is zero degrees. So, when you are presented with 43.6 degrees of 600 ohm line connected to a black box (about which you know nothing of the internals) which provides an impedance -j567, do you answer the question: What is the impedance at the input of the 600 ohm line? Or do you refuse, because you do not know what phase change is occuring at the terminals? ....Keith |
Standing Wave Phase
Keith Dysart wrote:
So, when you are presented with 43.6 degrees of 600 ohm line connected to a black box (about which you know nothing of the internals) which provides an impedance -j567, do you answer the question: What is the impedance at the input of the 600 ohm line? Or do you refuse, because you do not know what phase change is occuring at the terminals? I'm not sure I understand the question. Measuring the s22 parameter should be all we need to know. -- 73, Cecil http://www.w5dxp.com |
Standing Wave Phase
AI4QJ wrote:
"Cecil Moore" wrote in message . net... wrote: Although we know that the 200 ohm line is longer, there is no indication that the length of the 600 ohm line must or must not change. The phase shift at the impedance discontinuity depends upon the *ratio of Z0High/Z0Low*. The following two examples have the same phase shift at the impedance discontinuity. Z0High Z0Low ---43.4 deg 600 ohm line---+---10 deg 100 ohm line---open ---43.4 deg 300 ohm line---+---10 deg 50 ohm line---open So how long does the 600 ohm line have to be in the following example for the stub to exhibit 1/4WL of electrical length? ---??? deg 600 ohm line---+---10 deg 50 ohm line---open Based on what you say above, Zo high/Zo low = 6 resul;ts in 37 degrees. Then for phase shift, (Zo high/Zo low)*37 deg = 6*D where D = phase shift. If the above ratio is literally true, then in the above example, 12*37deg/6 = 74 degrees and length of the 600 ohm line = 90-74-10 = 6 degrees. However, I would like to find a reference for the math showing the characteristic impedance ratio relationship with phase shift. I am reluctant to accepting formulas without seeing them derived at least once. :-) Looks like you got the Z0 ratio upside down? Actually I get 25.3 degrees for the 600 ohm line. Arctan(Z0Low/Z0High)cot(10)) = 25.3 degrees -- 73, Cecil http://www.w5dxp.com |
Standing Wave Phase
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Standing Wave Phase
On Dec 12, 8:44 pm, "AI4QJ" wrote:
"Cecil Moore" wrote in message . net... wrote: Although we know that the 200 ohm line is longer, there is no indication that the length of the 600 ohm line must or must not change. The phase shift at the impedance discontinuity depends upon the *ratio of Z0High/Z0Low*. The following two examples have the same phase shift at the impedance discontinuity. Z0High Z0Low ---43.4 deg 600 ohm line---+---10 deg 100 ohm line---open ---43.4 deg 300 ohm line---+---10 deg 50 ohm line---open So how long does the 600 ohm line have to be in the following example for the stub to exhibit 1/4WL of electrical length? ---??? deg 600 ohm line---+---10 deg 50 ohm line---open Based on what you say above, Zo high/Zo low = 6 resul;ts in 37 degrees. Then for phase shift, (Zo high/Zo low)*37 deg = 6*D where D = phase shift. If the above ratio is literally true, then in the above example, 12*37deg/6 = 74 degrees and length of the 600 ohm line = 90-74-10 = 6 degrees. However, I would like to find a reference for the math showing the characteristic impedance ratio relationship with phase shift. I am reluctant to accepting formulas without seeing them derived at least once. In another thread, "Calculating a (fictitious) phase shift; was : Loading Colis", David Ryeburn has provided the arithmetic for the general case which includes the following expression: -j*(Z_2)*cot(alpha + beta) = -j*(Z_1)*cot(alpha) "alpha" is the length of the open line "beta" is the "phase shift" at the joint "Z_2" is the impedance of the open line "Z_1" is the impedance of the driven line It can be seen, as noted by Cecil, that beta will be the same if the impedances of the two lines are scaled proportionally. But beta is dependant not only on the two impedances, but also on the length of the open line. There are no simple relationships here. It does not seem to be a concept that is particularly useful for the solving of problems. ....Keith |
Standing Wave Phase
It does not seem to be a concept that is
particularly useful for the solving of problems. Looks like you haven't thought it through. If one wants to create a shortened dual-Z0 stub with equal lengths of each section of Z0High and Z0Low, here is the corresponding chart for different ratios of Z0High/Z0Low. http://www.w5dxp.com/DualZ0.gif As an example, one can create an electrical 1/4WL stub that is 1/3 the normal physical length by using 600 ohm line and 50 ohm line. -- 73, Cecil http://www.w5dxp.com |
Standing Wave Phase
On Dec 13, 5:57 am, Cecil Moore wrote:
It does not seem to be a concept that is particularly useful for the solving of problems. Looks like you haven't thought it through. If one wants to create a shortened dual-Z0 stub with equal lengths of each section of Z0High and Z0Low, here is the corresponding chart for different ratios of Z0High/Z0Low. http://www.w5dxp.com/DualZ0.gif As an example, one can create an electrical 1/4WL stub that is 1/3 the normal physical length by using 600 ohm line and 50 ohm line. There are many ways to create the impedances for matching, each with different advantages. As you point out, one of the benefits of using two different impedance lines is a reduction in material, though, you could go all the way to just using a lumped capacitor and save even more. This reality, however, does not demonstrate any value for the *concept* of phase shift at a discontinuity. For the concept to be useful it should facilitate understanding or problem solving. As far as I can tell, you always solve the problem in the conventional way (change the angle on the Smith chart, un-normalize the impedance, re-normalize the impedance to the new Z0, measure the angle to get to the desired impedance) and then work out the "phase shift" at the discontinuity. It sure looks like additional work that adds no value since the important and useful information has already been derived before computing the "phase shift". As such, I declare it "not particularly useful". ....Keith |
Standing Wave Phase
Cecil Moore wrote:
It does not seem to be a concept that is particularly useful for the solving of problems. Looks like you haven't thought it through. If one wants to create a shortened dual-Z0 stub with equal lengths of each section of Z0High and Z0Low, here is the corresponding chart for different ratios of Z0High/Z0Low. http://www.w5dxp.com/DualZ0.gif As an example, one can create an electrical 1/4WL stub that is 1/3 the normal physical length by using 600 ohm line and 50 ohm line. Cecil, So how do you make that 12:1 connection, say 50 ohm coax to 600 ohm open line? Do you s'pose the connection bits add any phase shift all by themselves? Do you have a model for that extra phase shift? 8-) 73, Gene W4SZ |
Standing Wave Phase in Loaded Mobile Antennas
Keith Dysart wrote:
This reality, however, does not demonstrate any value for the *concept* of phase shift at a discontinuity. It may indeed have little value for stubs. But for loaded mobile antennas the value is obvious. The value is that it explains the phase shift through a loading coil in a loaded mobile antenna and the phase shift at the coil to stinger junction. Using dual-Z0 transmission line stubs we are ready to understand loaded mobile antennas, the phase shift through the loading coil, and the "missing degrees" at the coil to stinger junction. According to Dr. Corum, my 75m Texas Bugcatcher coil has a Z0 of ~4000 and a VF of ~0.02. The stinger has a Z0 in the ballpark of 400 ohms and a VF close to 1.0 Knowing what we know about a dual-Z0 1/4WL stub, we can now use that knowledge to analyze a base- loaded mobile antenna with coil and stinger. ---Z0=4000 ohm coil---+---10 deg 400 ohm stinger Now it's a piece of cake. How many degrees of loading coil do we need to make the configuration 90 electrical degrees long? Arctan((400/4000)*cot(10)) = ~30 degrees What is the impedance at the coil to stinger junction? 400*cot(10) = ~ -j2300 ohms What is the phase shift at the coil to stinger junction? 90 - 30 - 10 = ~50 degrees I stumbled upon the dual-Z0 stub idea in trying to understand the phase shifts and delays in a loaded mobile antenna. The same general principles apply. Using traveling-wave current to measure the delay through my Texas Bugcatcher coil agreed within 15% with these calculated values. One side said the coil had to make up the missing 80 degrees of antenna that necessarily had to be there with a 10 degree stinger. This side did not understand the phase shift at the coil to stinger junction. The other side said the coil, like a lumped inductor, has ~zero phase shift through it. This side did not understand the limitations of the lumped circuit model. The delay through a coil is what it measures and calculates to be within a certain accuracy. It is not 80 degrees and it is not ~zero degrees. -- 73, Cecil http://www.w5dxp.com |
Standing Wave Phase in Loaded Mobile Antennas
Cecil Moore wrote:
... 400*cot(10) = ~ -j2300 ohms ... I am in the middle of a brain fog/block. I am attempting to get an equation to obtain the cotangent of x (or 10 in the above.) I HAVE DONE THIS BEFORE-got something wrong here ... (1/tan(10)) = 5.67128 and (400*(1/tan(10)) = 2268.51273 There is no "built in" cotan function on my ti-86, ti-83, etc. Help me out Cecil, anyone? Regards, JS |
Standing Wave Phase in Loaded Mobile Antennas
John Smith wrote:
There is no "built in" cotan function on my ti-86, ti-83, etc. Poor guy - why can't you do cotangent functions in your head? :-) Help me out Cecil, anyone? How about: cot(x) = tan(90-x) cot(10) = tan(80) = 5.67 -- 73, Cecil http://www.w5dxp.com |
Standing Wave Phase
In article
, Keith Dysart wrote: In another thread, "Calculating a (fictitious) phase shift; was : Loading Colis", David Ryeburn has provided the arithmetic for the general case which includes the following expression: -j*(Z_2)*cot(alpha + beta) = -j*(Z_1)*cot(alpha) "alpha" is the length of the open line "beta" is the "phase shift" at the joint Yes. "Z_2" is the impedance of the open line "Z_1" is the impedance of the driven line Backwards. Z_1 is the characteristic impedance of the open-circuited line (100 ohms, in some of the examples previously discussed). Z_2 is the characteristic impedance of line from the junction point back to the source (600 ohms, in those examples). But beta is dependant not only on the two impedances, but also on the length of the open line. There are no simple relationships here. It does not seem to be a concept that is particularly useful for the solving of problems. I couldn't agree more. The angle alpha + beta is useful; the angle beta is not. But I thought I would provide the formulas in an attempt to set things straight. In article . Keith Dysart wrote: There are many ways to create the impedances for matching, each with different advantages. As you point out, one of the benefits of using two different impedance lines is a reduction in material, though, you could go all the way to just using a lumped capacitor and save even more. Agreed. I've always liked capacitors better than transmission line segments. It takes a pretty crummy capacitor to have as low a Q as a transmission line section is likely to have. Even inductors are often better than transmission line segments. But W5DXP was trying to explain how a loaded mobile antenna worked (using transmission line concepts). However, loaded mobile antennas presumably radiate, at least a little, and my analysis (and W5DXP's discussion of angle lengths of transmission lines and "phase shift" at their junction) is for *LOSSLESS* transmission lines. This makes me wonder. David, ex-W8EZE -- David Ryeburn To send e-mail, use "ca" instead of "caz". |
Standing Wave Phase in Loaded Mobile Antennas
Cecil Moore wrote:
... Poor guy - why can't you do cotangent functions in your head? :-) ... Cecil: The extremly difficult will take me a couple of minutes ... The impossible takes just a bit more time. :-) I AM NOT A GURU! yanno? LOL Regards, JS |
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