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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 |
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