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Caculating VSWR from rho and rho from VSWR
I note that any textbook I pick up shows that VSWR=(1+rho)/(1-rho) where rho is the magnitude of Gamma (Gamma=(Z-Zo)/(Z+Zo)); rho=abs(Gamma)). Now, reading TL theory texts can be confusing because of the sometimes subtle swithes to and from an assumption of lossless line (under which rho cannot exceed 1). Since VSWR is the ratio of the magnitude of the voltage at a maximum in the standing wave pattern to the magnitude of the voltage at a minimum in the standing wave pattern, if we are to infer SWR at a point on a line (if that makes sense anyway) from rho (which is a property of a point on a lossy line), isn't the formula VSWR=abs(1+rho)/abs(1-rho) correct in the general case (lossy or lossless line)? Given that rho cannot be negative (since it is the magnitude of a complex number), the general formula can be simplified to VSWR=(1+rho)/abs(1-rho). Seems to me that texts almost universally omit the absolute operation on the denominator without necessarily qualifying it with the assumption of lossless line. If VSWR=(1+rho)/abs(1-rho), then doesn't it follow that rho is not a function of VSWR (except in the lossless line case where VSWR=(1+rho)/(1-rho) and therefore rho=(VSWR-1)/(VSWR+1))? Thoughts? Owen |
Owen wrote:
I note that any textbook I pick up shows that VSWR=(1+rho)/(1-rho) where rho is the magnitude of Gamma (Gamma=(Z-Zo)/(Z+Zo)); rho=abs(Gamma)). Now, reading TL theory texts can be confusing because of the sometimes subtle swithes to and from an assumption of lossless line (under which rho cannot exceed 1). To complicate matters, roughly half the textbooks use rho instead of Gamma for the complex reflection coefficient. You've got to be careful. Since VSWR is the ratio of the magnitude of the voltage at a maximum in the standing wave pattern to the magnitude of the voltage at a minimum in the standing wave pattern, if we are to infer SWR at a point on a line (if that makes sense anyway) from rho (which is a property of a point on a lossy line), isn't the formula VSWR=abs(1+rho)/abs(1-rho) correct in the general case (lossy or lossless line)? The whole concept of VSWR gets flakey on a lossy line, and really loses its meaning. It's often analytically convenient to define a quantity at a point and call it "VSWR", although in the presence of loss it no longer means the ratio of maximum to minimum voltage on the line. Since it's lost its original meaning, it comes to mean just about anything you'd like. And the generally accepted definition then is the equation you gave in your first paragraph. That is, in the presence of loss, VSWR is something which is *defined* by that equation, rather than the equation being a means of calculating some otherwise defined property. Under the right conditions and if loss is large enough, rho can be greater than 1, in which case the VSWR as defined by the equation in the first paragraph becomes negative. Again, this is no longer a ratio of voltages along a line, but a quantity defined by an equation. If you alter the equation, you're defining a different quantity. Now, there's no reason that your "VSWR" definition isn't just as good as the conventional one (first paragraph equation). But the conventional one is pretty universally used, and yours is different, so if you're interested in communicating, it would be wise to give it a different name or at least carefully show what you mean when you use it. Given that rho cannot be negative (since it is the magnitude of a complex number), the general formula can be simplified to VSWR=(1+rho)/abs(1-rho). But it can be greater than one. See above. Seems to me that texts almost universally omit the absolute operation on the denominator without necessarily qualifying it with the assumption of lossless line. If VSWR=(1+rho)/abs(1-rho), then doesn't it follow that rho is not a function of VSWR (except in the lossless line case where VSWR=(1+rho)/(1-rho) and therefore rho=(VSWR-1)/(VSWR+1))? Rho is never a function of VSWR. VSWR is a function of rho. Unlike actual VSWR (that is, the ratio of maximum to minimum voltage along a line), the reflection coefficient can be and is rigorously and meaningfully defined at any point along a line, lossy or not. Roy Lewallen, W7EL |
Roy Lewallen wrote:
Owen wrote: I note that any textbook I pick up shows that VSWR=(1+rho)/(1-rho) where rho is the magnitude of Gamma (Gamma=(Z-Zo)/(Z+Zo)); rho=abs(Gamma)). .... property of a point on a lossy line), isn't the formula VSWR=abs(1+rho)/abs(1-rho) correct in the general case (lossy or lossless line)? The whole concept of VSWR gets flakey on a lossy line, and really loses its meaning. It's often analytically convenient to define a quantity at a point and call it "VSWR", although in the presence of loss it no longer means the ratio of maximum to minimum voltage on the line. Since Indeed. It occurs to me that if one was to try to measure VSWR (say using a slotted line with probe) on a lossy line operating at high VSWR, the best estimate would come from finding a minimum, measuring it and the adjacent maxima and calculating VSWR=(Vmax1+Vmax2)/Vmin/2 so that the measurement is biassed towards the notional VSWR at the point of the minimum (the minimum being much more sensitive to line attenuation than the maxima). it's lost its original meaning, it comes to mean just about anything you'd like. And the generally accepted definition then is the equation you gave in your first paragraph. That is, in the presence of loss, VSWR is something which is *defined* by that equation, rather than the equation being a means of calculating some otherwise defined property. Noted Under the right conditions and if loss is large enough, rho can be greater than 1, in which case the VSWR as defined by the equation in the first paragraph becomes negative. Again, this is no longer a ratio of voltages along a line, but a quantity defined by an equation. If you alter the equation, you're defining a different quantity. Now, there's no reason that your "VSWR" definition isn't just as good as the conventional one (first paragraph equation). But the conventional one is pretty universally used, and yours is different, so if you're interested in communicating, it would be wise to give it a different name or at least carefully show what you mean when you use it. Given that rho cannot be negative (since it is the magnitude of a complex number), the general formula can be simplified to VSWR=(1+rho)/abs(1-rho). But it can be greater than one. See above. Agreed. Seems to me that texts almost universally omit the absolute operation on the denominator without necessarily qualifying it with the assumption of lossless line. If VSWR=(1+rho)/abs(1-rho), then doesn't it follow that rho is not a function of VSWR (except in the lossless line case where VSWR=(1+rho)/(1-rho) and therefore rho=(VSWR-1)/(VSWR+1))? Rho is never a function of VSWR. VSWR is a function of rho. Unlike I mean't function in the sense that there is one and only one value of f(x) for x, rather than in the causual sense. actual VSWR (that is, the ratio of maximum to minimum voltage along a line), the reflection coefficient can be and is rigorously and meaningfully defined at any point along a line, lossy or not. Agreed. Thanks for your exhaustive reply Roy, it is appreciated. What I glean from this is that although Gamma and therefore rho are well defined, and both are a function of position on a line in the general case, the "accepted mathematical definition" of VSWR in terms of rho does not behave well (eg producing a negative value) in some cases and is not a good estimator of real VSWR in those cases. It seems fair to say that the reason that the "accepted mathematical definition" of VSWR does not behave well is that it depends on an assumption of a distortionless line (Xo=0). (Lossless lines are distortionless but the converse is not necessarily true). I take your point about the need to qualify a different algorithm by a different name. Thanks again. Owen |
"Roy Lewallen" wrote in message ... Owen wrote: I note that any textbook I pick up shows that VSWR=(1+rho)/(1-rho) where rho is the magnitude of Gamma (Gamma=(Z-Zo)/(Z+Zo)); rho=abs(Gamma)). Now, reading TL theory texts can be confusing because of the sometimes subtle swithes to and from an assumption of lossless line (under which rho cannot exceed 1). To complicate matters, roughly half the textbooks use rho instead of Gamma for the complex reflection coefficient. You've got to be careful. Since VSWR is the ratio of the magnitude of the voltage at a maximum in the standing wave pattern to the magnitude of the voltage at a minimum in the standing wave pattern, if we are to infer SWR at a point on a line (if that makes sense anyway) from rho (which is a property of a point on a lossy line), isn't the formula VSWR=abs(1+rho)/abs(1-rho) correct in the general case (lossy or lossless line)? The whole concept of VSWR gets flakey on a lossy line, and really loses its meaning. It's often analytically convenient to define a quantity at a point and call it "VSWR", although in the presence of loss it no longer means the ratio of maximum to minimum voltage on the line. Since it's lost its original meaning, it comes to mean just about anything you'd like. And the generally accepted definition then is the equation you gave in your first paragraph. That is, in the presence of loss, VSWR is something which is *defined* by that equation, rather than the equation being a means of calculating some otherwise defined property. Under the right conditions and if loss is large enough, rho can be greater than 1, in which case the VSWR as defined by the equation in the first paragraph becomes negative. Again, this is no longer a ratio of voltages along a line, but a quantity defined by an equation. If you alter the equation, you're defining a different quantity. Now, there's no reason that your "VSWR" definition isn't just as good as the conventional one (first paragraph equation). But the conventional one is pretty universally used, and yours is different, so if you're interested in communicating, it would be wise to give it a different name or at least carefully show what you mean when you use it. Given that rho cannot be negative (since it is the magnitude of a complex number), the general formula can be simplified to VSWR=(1+rho)/abs(1-rho). But it can be greater than one. See above. Seems to me that texts almost universally omit the absolute operation on the denominator without necessarily qualifying it with the assumption of lossless line. If VSWR=(1+rho)/abs(1-rho), then doesn't it follow that rho is not a function of VSWR (except in the lossless line case where VSWR=(1+rho)/(1-rho) and therefore rho=(VSWR-1)/(VSWR+1))? Rho is never a function of VSWR. VSWR is a function of rho. Unlike actual VSWR (that is, the ratio of maximum to minimum voltage along a line), the reflection coefficient can be and is rigorously and meaningfully defined at any point along a line, lossy or not. Roy Lewallen, W7EL Good response, Roy, but concerning rho and gamma to represent reflection coefficient, I refer you to Reflections, Sec 3.1, "Prior to the 1950s rho and sigma, and sometimes 'S' were used to represent standing wave ratio. The symbol of choice to represent reflection coefficient during that era was upper case gamma. However, in 1953 the American Standards Association (now the NIST) announced in its publication ASA Y10.9-1953, that rho is to replace gamma for reflection coefficient, with SWR to represent standing wave ratio (for either voltage or current), and VSWR specifically for voltage standing wave ratio. Most of academia responded to the change, but some individuals did not. Consequently, gamma is occasionally seen representing reflection coefficent, but only rarely." Hope this clarifies any misunderstanding concerning the use of gamma for reflection coefficient. Incidentally, Roy, I recently mailed you a CD containing Laport's book, "Radio Antenna Engineering." I'm wondering if you received it, or did it go astray? Walt, W2DU |
Owen wrote:
Thanks for your exhaustive reply Roy, it is appreciated. It's time to barbecue the virtual reflection coefficient sacred cow or at least understand that it deviates from the field of optics and S-parameter analysis. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
Walter Maxwell wrote:
Good response, Roy, but concerning rho and gamma to represent reflection coefficient, I refer you to Reflections, Sec 3.1, "Prior to the 1950s rho and sigma, and sometimes 'S' were used to represent standing wave ratio. The symbol of choice to represent reflection coefficient during that era was upper case gamma. However, in 1953 the American Standards Association (now the NIST) announced in its publication ASA Y10.9-1953, that rho is to replace gamma for reflection coefficient, with SWR to represent standing wave ratio (for either voltage or current), and VSWR specifically for voltage standing wave ratio. Most of academia responded to the change, but some individuals did not. Consequently, gamma is occasionally seen representing reflection coefficent, but only rarely." Most interesting. I have on my library shelf 14 texts which deal primarily or in a major way with electromagnetic waves and/or transmission lines, and two with microwave circuit design. Of those, 8 (including both microwave design texts) use Gamma 4 use rho 2 use K 2 use k If NIST's pronouncement had any effect at all, it was the opposite of what was intended -- the four texts using rho were copyrighted in 1951, 53, 63, and 65; the 8 using Gamma were copyrighted in 1960 - 2000, 6 of them after 1965. So it appears from my sampling that Gamma is becoming more, not less, prevalant. Hope this clarifies any misunderstanding concerning the use of gamma for reflection coefficient. I'm afraid it doesn't, unless my collection is very atypical. I don't think it is, because it includes many of the classics. Incidentally, Roy, I recently mailed you a CD containing Laport's book, "Radio Antenna Engineering." I'm wondering if you received it, or did it go astray? I did indeed, Walt, and please forgive me for not acknowledging your very kind and thoughtful gift more promptly. Roy Lewallen, W7EL |
Walter Maxwell wrote:
Good response, Roy, but concerning rho and gamma to represent reflection coefficient, I refer you to Reflections, Sec 3.1, "Prior to the 1950s rho and sigma, and sometimes 'S' were used to represent standing wave ratio. The symbol of choice to represent reflection coefficient during that era was upper case gamma. However, in 1953 the American Standards Association (now the NIST) announced in its publication ASA Y10.9-1953, that rho is to replace gamma for reflection coefficient, with SWR to represent standing wave ratio (for either voltage or current), and VSWR specifically for voltage standing wave ratio. Most of academia responded to the change, but some individuals did not. Consequently, gamma is occasionally seen representing reflection coefficent, but only rarely." Thanks for the information Walter. I must have a few "rare" texts that use Gamma (Gamma to mean uppercase gamma) for the voltage reflection coefficient. I wonder if the recommendation / standard to which you refer is taken up in any international standard? I do note that my ARRL Antenna Handbook (18th edition) and ARRL Handbook (2000) both use rho, however they reckon that rho=(Za-Zo*)/(Za+Zo) (where Zo* means the conjugate of Zo). They do this without derivation, and seem to be in conflict with the derivation in most texts. I suppose the derivation is buried in some article in QST and in the members only section of the ARRL website. Back to notation, accepting that the preferred pronumeral for the voltage reflection coefficient is rho, is there a pronumeral used for abs(rho)? Owen |
"Owen" wrote in message ... Walter Maxwell wrote: Good response, Roy, but concerning rho and gamma to represent reflection coefficient, I refer you to Reflections, Sec 3.1, "Prior to the 1950s rho and sigma, and sometimes 'S' were used to represent standing wave ratio. The symbol of choice to represent reflection coefficient during that era was upper case gamma. However, in 1953 the American Standards Association (now the NIST) announced in its publication ASA Y10.9-1953, that rho is to replace gamma for reflection coefficient, with SWR to represent standing wave ratio (for either voltage or current), and VSWR specifically for voltage standing wave ratio. Most of academia responded to the change, but some individuals did not. Consequently, gamma is occasionally seen representing reflection coefficent, but only rarely." Thanks for the information Walter. I must have a few "rare" texts that use Gamma (Gamma to mean uppercase gamma) for the voltage reflection coefficient. I wonder if the recommendation / standard to which you refer is taken up in any international standard? I do note that my ARRL Antenna Handbook (18th edition) and ARRL Handbook (2000) both use rho, however they reckon that rho=(Za-Zo*)/(Za+Zo) (where Zo* means the conjugate of Zo). They do this without derivation, and seem to be in conflict with the derivation in most texts. I suppose the derivation is buried in some article in QST and in the members only section of the ARRL website. Back to notation, accepting that the preferred pronumeral for the voltage reflection coefficient is rho, is there a pronumeral used for abs(rho)? Owen Hi Owen, From the general use I'm familiar with, rho alone refers to the abs value, while the two vertical bars on each side of rho indicates the magnitude alone. However, following Hewlett-Packard's usage in their AP notes, in Reflections I use a bar over rho for the absolute, and rho alone for the magnitude. However, I explain the term in the book to avoid confusion. Walt, W2DU |
"Owen" wrote in message ... ............................. I do note that my ARRL Antenna Handbook (18th edition) and ARRL Handbook (2000) both use rho, however they reckon that rho=(Za-Zo*)/(Za+Zo) (where Zo* means the conjugate of Zo). They do this without derivation, and seem to be in conflict with the derivation in most texts. I suppose the derivation is buried in some article in QST and in the members only section of the ARRL website. Owen, There was a big discussion about this last year, and somebody posted that the ARRL was going to eliminate the conjugate reference. Tam/WB2TT Back to notation, accepting that the preferred pronumeral for the voltage reflection coefficient is rho, is there a pronumeral used for abs(rho)? Owen |
Tam/WB2TT wrote:
There was a big discussion about this last year, and somebody posted that the ARRL was going to eliminate the conjugate reference. Ok, I take that to mean the ARRL handbooks are in error in stating rho=(Za-Zo*)/(Za+Zo) (where Zo* means the conjugate of Zo), and that they will now use rho=(Za-Zo)/(Za+Zo). Kirchoff lives! I guess we wait and see if it comes to print. Thanks Tam. Owen |
You've noted what I've noted before in this forum, if you say that
"(V)SWR" is the SWR that would result if the forward and reverse waves at a particular point on the line (the point at which you've measured rho) were allowed to develop a voltage maximum and a voltage minimum (or a current maximum and minimum). I happen to believe that is more in keeping with the original meaning of (V)SWR than the formula which says VSWR=(1+rho)/(1-rho). Of course, this all gets quickly into people trying to assign physical significance to rho which is not really there; they get confused when rho1. If you are clear and consistent with your definitions, I don't see that any problems result either way. I happen to believe that the formula you came up with is a better one than the one you commonly see in texts, but I also agree with Roy that you'd better be clear about it if you use it, because it goes against the commonly accepted grain. On the other hand, I've seen texts that derive it in the same way you have, which come up with the "wrong" answer (without the abs) based on their initial premises. Cheers, Tom |
In article ,
Owen wrote: Indeed. It occurs to me that if one was to try to measure VSWR (say using a slotted line with probe) on a lossy line operating at high VSWR, the best estimate would come from finding a minimum, measuring it and the adjacent maxima and calculating VSWR=(Vmax1+Vmax2)/Vmin/2 so that the measurement is biassed towards the notional VSWR at the point of the minimum (the minimum being much more sensitive to line attenuation than the maxima). If you think about exponential decay along the line, the geometric mean sqrt(Vmax1 * Vmax2) would be an even better replacement for Vmax, to use in the numerator, than the arithmetic mean (Vmax1 * Vmax2)/2. David, ex-W8EZE -- David Ryeburn To send e-mail, use "ca" instead of "caz". |
Tam/WB2TT wrote:
There was a big discussion about this last year, and somebody posted that the ARRL was going to eliminate the conjugate reference. Another problem that needs to be fixed is the difference between the "virtual" rho and the physical rho. What is rho looking into point '+' from the XMTR side? XMTR---50 ohm coax---+---1/4WL 75 ohm coax---112.5 ohm load The physical rho is (75-50)/(75+50) = 0.2 which is the same as 's11' in an S-parameter analysis. The "virtual" rho is SQRT(Pref/Pfor) which, in a Z0-matched system is zero. (The 50 ohm coax "sees" a V/I ratio of 50 ohms) Rho, looking into the load, is (112.5-75)/(112.5+75) = 0.2. The virtual rho, looking back at point '+' from the load side is |1.0| but that same reflection coefficient, s22 for an S-parameter analysis, is (50-75)/(50+75) = -0.2. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
On Fri, 17 Jun 2005 22:44:13 -0400, "Walter Maxwell"
wrote: [snip] Hi Owen, From the general use I'm familiar with, rho alone refers to the abs value, while the two vertical bars on each side of rho indicates the magnitude alone. However, following Hewlett-Packard's usage in their AP notes, in Reflections I use a bar over rho for the absolute, and rho alone for the magnitude. However, I explain the term in the book to avoid confusion. Confusion reigns. Four years ago in another thread I posted thus: Quote On Mon, 12 Feb 2001 15:25:28 -0800, Roy Lewallen wrote: Just a point of clarification. Rho in these equations is the magnitude of the reflection coefficient, not the reflection coefficient itself. The reflection coefficient is actually a complex number. Rho is unfortunately used to sometimes represent the (complex) reflection coefficient and sometimes (like here) its magnitude, although some people (me included) prefer to use uppercase gamma for the complex reflection coefficient and lowercase rho for its magnitude. Roy raises a good point. Tom Bruhns already took me to task for a somewhat careless use of rho. Although I did define it below, as Roy and Tom said, it is often used as a complex number. I too prefer upper case Gamma for the complex number and rho for the magnitude but unfortunately the literature is full of confusing usage. Some of the literature was even published by Tom's employer, the former H-P, now Agilent (how do you pronounce that again?) My autographed copy of Steve Adam's, book "Microwave Theory and Applications", published by H-P, shows on page 23: " |Gamma| = rho " Similarly, my handy dandy H-P "Reflectometer calculator" sliderule says that SWR = (1 + rho) / (1- rho) which bears a striking resemblence to what I wrote below. But then in H-P's App Note 77-3, "Measurement of Complex Impedance 1-1000 MHZ", it says that rho is a vector quantity and it shows: SWR = (1 +|rho| ) / (1 - |rho| ) Finally, the best reference I have is General Radio's "Handbook of Microwave Measurements" (out of print but reissued by Gilbert Engineering) and it says that Gamma is complex and rho isn't. End quote. |
Tam/WB2TT wrote: "Owen" wrote in message ... ............................. I do note that my ARRL Antenna Handbook (18th edition) and ARRL Handbook (2000) both use rho, however they reckon that rho=(Za-Zo*)/(Za+Zo) (where Zo* means the conjugate of Zo). They do this without derivation, and seem to be in conflict with the derivation in most texts. I suppose the derivation is buried in some article in QST and in the members only section of the ARRL website. Owen, There was a big discussion about this last year, and somebody posted that the ARRL was going to eliminate the conjugate reference. Les Besser agrees with the ARRL Handbook except he uses gamma for the complex reflection coefficient: gamma=(Za-Zo*)/(Za+Zo) (where Zo* means the conjugate of Zo) But for most practical calculations, the Zo is assumed to be purely real, so many texts give gamma=(Za-Zo)/(Za-Zo). rho=[gamma]=absolute value of gamma=magnitude of gamma Rho can never be greater than one going into a passive network. Only when you have an active device, or gain, can you move outside of the unity circle on the Smith Chart. Slick Tam/WB2TT Back to notation, accepting that the preferred pronumeral for the voltage reflection coefficient is rho, is there a pronumeral used for abs(rho)? Owen |
|
Makes things easy, gamma always equal to 1
wrote in message oups.com... - But for most practical calculations, the Zo is assumed to be purely real, so many texts give gamma=(Za-Zo)/(Za-Zo). |
In article ,
David Ryeburn wrote: In article , Owen wrote: Indeed. It occurs to me that if one was to try to measure VSWR (say using a slotted line with probe) on a lossy line operating at high VSWR, the best estimate would come from finding a minimum, measuring it and the adjacent maxima and calculating VSWR=(Vmax1+Vmax2)/Vmin/2 so that the measurement is biassed towards the notional VSWR at the point of the minimum (the minimum being much more sensitive to line attenuation than the maxima). If you think about exponential decay along the line, the geometric mean sqrt(Vmax1 * Vmax2) would be an even better replacement for Vmax, to use in the numerator, than the arithmetic mean (Vmax1 * Vmax2)/2. ? I should have typed (Vmax1 + Vmax2)/2 and not (Vmax1 * Vmax2)/2. Shouldn't post at three minutes before midnight. Also, in article , Wes Stewart wrote: On 18 Jun 2005 14:20:34 -0700, wrote: [snip] Rho can never be greater than one going into a passive network. Only when you have an active device, or gain, can you move outside of the unity circle on the Smith Chart. You better raise your deflection shield. Go ahead Tom, you have the honor. I did my best with this same issue a year or two ago, and in the light of what transpired then I suggest you not waste your time, Tom! The easiest way to see that (under very special circumstances) rho can exceed 1 in a passive network uses a simple geometrical argument*. Unfortunately geometry has even less attention paid to it today than it did when I was in high school. *Tradition has it that words translatable as "Let no one ignorant of geometry enter" were written on the door of Plato's Academy. This is probably not true, though Plato certainly had a high opinion of geometry. See http://plato-dialogues.org/faq/faq009.htm David, ex-W8EZE, retired SFU mathematician, and Google addict -- David Ryeburn To send e-mail, use "ca" instead of "caz". |
Wes Stewart wrote: On 18 Jun 2005 14:20:34 -0700, wrote: [snip] Rho can never be greater than one going into a passive network. Only when you have an active device, or gain, can you move outside of the unity circle on the Smith Chart. You better raise your deflection shield. No need to, i have the laws of physics on my side! Hint: Conservation of Energy! Slick |
Rho can never be greater than one going into a passive
network. Only when you have an active device, or gain, can you move outside of the unity circle on the Smith Chart. Slick ================================== Yes, Rho CAN exceed unity when the termination is a passive network. For example, when on a real line, Zo = Ro - jXo and the termination Zt = Rt + jXt then Rho can exceed unity. Rho has an absolute maximum value which approaches 1 + Sqrt(2) = 2.4142 which occurs when the angle of Zo approaches -45 degrees and Zt is purely inductive. It arises because of a weak resonant effect between -jXo and + jXt. The angle of Zo of real lines always becomes more negative as frequency decreases. Mr Smith's Chart does not recognise this. He did it knowingly and deliberately. There are other departures from reality. But at least it is fit for its few intended purposes. ---- Reg, G4FGQ |
"Reg Edwards" wrote in message ... ............................. Yes, Rho CAN exceed unity when the termination is a passive network. For example, when on a real line, Zo = Ro - jXo and the termination Zt = Rt + jXt then Rho can exceed unity. Rho has an absolute maximum value which approaches 1 + Sqrt(2) = 2.4142 which occurs when the angle of Zo approaches -45 degrees and Zt is purely inductive. It arises because of a weak resonant effect between -jXo and + jXt. The angle of Zo of real lines always becomes more negative as frequency decreases. Mr Smith's Chart does not recognise this. He did it knowingly and deliberately. There are other departures from reality. But at least it is fit for its few intended purposes. ---- Reg, G4FGQ That is exacly what is in a book I have. Tam |
On Sun, 19 Jun 2005 10:58:15 +0000 (UTC), "Reg Edwards"
wrote: Rho can never be greater than one going into a passive network. Only when you have an active device, or gain, can you move outside of the unity circle on the Smith Chart. Slick ================================== Yes, Rho CAN exceed unity when the termination is a passive network. For example, when on a real line, Zo = Ro - jXo and the termination Zt = Rt + jXt then Rho can exceed unity. Thanks Reg, you beat Tom to it. [g] This was beaten to death (well, I guess it *wasn't* beaten to death, here it is again) over and over again. Chipman in section 7.6 "Complex characteristic impedance" deals with this and concurs with what Reg says above and below. Rho has an absolute maximum value which approaches 1 + Sqrt(2) = 2.4142 which occurs when the angle of Zo approaches -45 degrees and Zt is purely inductive. It arises because of a weak resonant effect between -jXo and + jXt. The angle of Zo of real lines always becomes more negative as frequency decreases. Mr Smith's Chart does not recognise this. He did it knowingly and deliberately. There are other departures from reality. But at least it is fit for its few intended purposes. ---- Reg, G4FGQ |
Wes Stewart wrote:
He states in part, "To establish that the principle of conservation of energy is not violated on a transmission line even when the magnitude of the reflection coefficient at a point on the line exceeds unity......." Isn't it the same principle as a capacitor in a resonant circuit being able to develop a higher voltage than the voltage incident upon the resonant circuit? -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
That is exacly what is in a book I have.
Tam ======================== Good! That proves your book is correct. Is it by Kraus or Terman? |
"Reg Edwards" wrote in message ... That is exacly what is in a book I have. Tam ======================== Good! That proves your book is correct. Is it by Kraus or Terman? Adler, Fano & Chu. It is the book Chipman used in his course. Tam |
Wes Stewart wrote:
"Go ahead Tom, you have the honor. " It's not worth the effort, Wes. He wasn't willing (or perhaps he was just unable) to do the math last year, and I suppose he isn't this year either. He was willing to call me a liar instead of getting ahold of Besser himself and verifying that Besser long ago corrected that typo about using the complex conjugate in the formula for reflection coefficient. Anyone willing to start with the very basic idea that forward and reverse waves on uniform TEM lines are independent and the ratio of voltage to current in each of them is equal to the line impedance, plus a few other elementary ideas like the sum of currents into a node is zero can, with a bit of algebraic facility, verify the derivation of reflection coefficient expressed in terms of line and load impedance. Or, they can look in the archives (via Google, etc.) to see it presented in past threads here. Cheers, Tom |
K7ITM wrote:
Anyone willing to start with the very basic idea that forward and reverse waves on uniform TEM lines are independent and the ratio of voltage to current in each of them is equal to the line impedance, ... What? You mean reflected energy doesn't "slosh" around? :-) -- 73, Cecil http://www.qsl.net/w5dxp |
K7ITM wrote: Wes Stewart wrote: "Go ahead Tom, you have the honor. " It's not worth the effort, Wes. He wasn't willing (or perhaps he was just unable) to do the math last year, and I suppose he isn't this year either. He was willing to call me a liar instead of getting ahold of Besser himself and verifying that Besser long ago corrected that typo about using the complex conjugate in the formula for reflection coefficient. And a liar you still are, apparently. Besser has NOT corrected it, because it isn't a typo! The ARRL also agrees. Slick |
Tam/WB2TT wrote: "Owen" wrote in message ... ............................. I do note that my ARRL Antenna Handbook (18th edition) and ARRL Handbook (2000) both use rho, however they reckon that rho=(Za-Zo*)/(Za+Zo) (where Zo* means the conjugate of Zo). They do this without derivation, and seem to be in conflict with the derivation in most texts. I suppose the derivation is buried in some article in QST and in the members only section of the ARRL website. Thank you, Owen. Les Besser agrees with the ARRL. However, in almost all practical calculations, Zo is purely real, so that gamma=(Za-Zo)/(Za+Zo) is used in most texts, and the results are the same. Owen, There was a big discussion about this last year, and somebody posted that the ARRL was going to eliminate the conjugate reference. Tam/WB2TT Going to? It says 2000 on that ARRL Handbook! They are NOT going to eliminate the conjugate reference, because it's correct. Slick |
There was a big discussion about this last year, and somebody posted that the ARRL was going to eliminate the conjugate reference. Tam/WB2TT Going to? It says 2000 on that ARRL Handbook! They are NOT going to eliminate the conjugate reference, because it's correct. Point of fact: in the current 20th edition, dated 2003, it has gone. It had been correct; then they incorrectly revised it; now it has been corrected again. -- 73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB) http://www.ifwtech.co.uk/g3sek |
wrote:
However, in almost all practical calculations, Zo is purely real, ... Nope, "purely real" requires a special expensive line. I forget what they call it. Ordinary feedline is not purely real however close it might be. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
Ian White GM3SEK wrote: There was a big discussion about this last year, and somebody posted that the ARRL was going to eliminate the conjugate reference. Tam/WB2TT Going to? It says 2000 on that ARRL Handbook! They are NOT going to eliminate the conjugate reference, because it's correct. Point of fact: in the current 20th edition, dated 2003, it has gone. It had been correct; then they incorrectly revised it; now it has been corrected again. Point of fact: If Zo is purely real, then Zo*=Zo, and so both forms are correct in this case. S. |
Cecil Moore wrote: wrote: However, in almost all practical calculations, Zo is purely real, ... Nope, "purely real" requires a special expensive line. I forget what they call it. Ordinary feedline is not purely real however close it might be. -- I don't care how much it costs, your line is NEVER going to be 100% purely real! There is always a gap between mathematical ideals and reality. So i totally agree with you. But for most practical calculations, Zo can be considered purely real, and your calculations will be close. S. |
wrote:
I don't care how much it costs, your line is NEVER going to be 100% purely real! There is always a gap between mathematical ideals and reality. Therefore, Z0 will never equal Z0*? One can purchase test leads guaranteed within a certain percentage at a certain frequency. I believe it occurs when G/C = R/L or some such. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
wrote:
I don't care how much it costs, your line is NEVER going to be 100% purely real! There is always a gap between mathematical ideals and reality. Therefore, Z0 will never equal Z0*? One can purchase test leads guaranteed within a certain percentage at a certain frequency. I believe it occurs when G/C = R/L or some such. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
wrote:
I don't care how much it costs, your line is NEVER going to be 100% purely real! There is always a gap between mathematical ideals and reality. Therefore, Z0 will never equal Z0*? One can purchase test leads guaranteed within a certain percentage at a certain frequency. I believe it occurs when G/C = R/L or some such. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
Cecil Moore wrote: wrote: I don't care how much it costs, your line is NEVER going to be 100% purely real! There is always a gap between mathematical ideals and reality. Therefore, Z0 will never equal Z0*? Never exactly, but they will be damn close most of the time, unless your coax is really cheap/bad. One can purchase test leads guaranteed within a certain percentage at a certain frequency. I believe it occurs when G/C = R/L or some such. -- 73, Cecil http://www.qsl.net/w5dxp If Zo=50+j0.000000000001, then it's close enough! S. |
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Richard Clark wrote: On 20 Jun 2005 01:00:51 -0700, wrote: If Zo=50+j0.000000000001, And it is not then it's close enough! hence it follows from this logic, it is not close enough. Even if Zo=50+j2, then the VSWR will still be very close to a 1:1 match. You antenna boys don't build too many power amplifiers, obviously. Slick |
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