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Smith Chart Quiz
Hi,
I've just become aware of the answers to these questions, but i thought i might see if anyone agrees/disagrees with me. What is the center of the Smith Chart when characteristic impedance Zo= 50 + j50? And what will the complex series impedance be for Rho = -1 in this case? (Hint: it will NOT be a short!) Slick (Garvin) |
Radio913 wrote:
What is the center of the Smith Chart when characteristic impedance Zo= 50 + j50? I suspect that the standard Smith Chart is incapable of handling a Z0 = 50 + j50. On the standard Smith Chart, would the center of the SWR spiral be 1 + j for a Z0 = 50 +j50? If so, that would make for some interesting impedance transformations. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
It's even easier than that. Using the _definition_ of reflection
coefficient, rho = Vr/Vf, we see that Vr=-Vf and therefore the net line voltage at that point is zero (that is, Vf+Vr, or Vf-Vf, or zero). That's either a real short or a virtual short. Cheers, Tom "David Robbins" wrote in message ... lets see... rho = (Z - Zo)/(Z + Zo) if rho = -1 then -1=(Z-Zo)/(Z+Zo) or -1*(Z+Zo)=(Z-Zo) or -Z - Zo = Z - Zo or -Z = Z the only value i know that satisfies that is zero. lets substitute it back in to be sure... -1=(0-(50+j50))/(0+(50+j50)) -1=(-50-j50)/(50+j50) -1=-1(50+j50)/(50+j50) -1=-1 qed. |
I trust we can all agree that the definition of rho is rho=Vr/Vf, and
I trust we can all agree that on a TEM line, the net voltage Vnet=Vr+Vf. If the net voltage is zero (i.e. at a short circuit), then clearly Vr=-Vf, and rho=-Vf/Vf=-1. Then whatever formula you chose to use to find rho at a load from the load impedance, Zload, and the line characteristic impedance, Zo, better work out right for a short circuit termination. Note that "rho=(Zload-Zo*)/(Zload+Zo)" does NOT work for that simple case, unless Zo*=Zo. But "rho=(Zload-Zo)/(Zload+Zo)" does work. You can easily go through something similar for an open circuit load. Note that rho=-1 for a short circuit load, and rho=+1 for an open circuit load, independent of line impedance. Cheers, Tom (Tom Bruhns) wrote in message ... It's even easier than that. Using the _definition_ of reflection coefficient, rho = Vr/Vf, we see that Vr=-Vf and therefore the net line voltage at that point is zero (that is, Vf+Vr, or Vf-Vf, or zero). That's either a real short or a virtual short. Cheers, Tom "David Robbins" wrote in message ... lets see... rho = (Z - Zo)/(Z + Zo) if rho = -1 then -1=(Z-Zo)/(Z+Zo) or -1*(Z+Zo)=(Z-Zo) or -Z - Zo = Z - Zo or -Z = Z the only value i know that satisfies that is zero. lets substitute it back in to be sure... -1=(0-(50+j50))/(0+(50+j50)) -1=(-50-j50)/(50+j50) -1=-1(50+j50)/(50+j50) -1=-1 qed. |
Hmmm ... Tom rho also equals Ir/If. Now if the line is open circuited
.... the net current is zero (i.e. at an open circuit). Therefore Ir = -If, and rho is a -1. Comment? Tom Bruhns wrote: I trust we can all agree that the definition of rho is rho=Vr/Vf, and I trust we can all agree that on a TEM line, the net voltage Vnet=Vr+Vf. If the net voltage is zero (i.e. at a short circuit), then clearly Vr=-Vf, and rho=-Vf/Vf=-1. Then whatever formula you chose to use to find rho at a load from the load impedance, Zload, and the line characteristic impedance, Zo, better work out right for a short circuit termination. Note that "rho=(Zload-Zo*)/(Zload+Zo)" does NOT work for that simple case, unless Zo*=Zo. But "rho=(Zload-Zo)/(Zload+Zo)" does work. You can easily go through something similar for an open circuit load. Note that rho=-1 for a short circuit load, and rho=+1 for an open circuit load, independent of line impedance. Cheers, Tom (Tom Bruhns) wrote in message ... It's even easier than that. Using the _definition_ of reflection coefficient, rho = Vr/Vf, we see that Vr=-Vf and therefore the net line voltage at that point is zero (that is, Vf+Vr, or Vf-Vf, or zero). That's either a real short or a virtual short. Cheers, Tom "David Robbins" wrote in message ... lets see... rho = (Z - Zo)/(Z + Zo) if rho = -1 then -1=(Z-Zo)/(Z+Zo) or -1*(Z+Zo)=(Z-Zo) or -Z - Zo = Z - Zo or -Z = Z the only value i know that satisfies that is zero. lets substitute it back in to be sure... -1=(0-(50+j50))/(0+(50+j50)) -1=(-50-j50)/(50+j50) -1=-1(50+j50)/(50+j50) -1=-1 qed. |
Hi Dave (and lurkers),
Well, if rho=Ir/If, then the net current on the line is If-Ir, NOT If+Ir. You have to be careful about directions, and be careful to define things and stick with those definitions. I usually use Zo=Vf/If=-Vr/Ir, so that Ir and If measure "positive" in the same physical direction along the line, and then of course rho=Vr/Vf=-Ir/If. If that's at all hazy, just draw a picture and it should be clear. Anyway, rho=+1 at a point of zero net current, either way you define the current direction, and rho=-1 at a point of zero net voltage. No Smith chart, and no Zo, needed to figure that out. Cheers, Tom Dave Shrader wrote in message news:iBkcb.568264$o%2.253779@sccrnsc02... Hmmm ... Tom rho also equals Ir/If. Now if the line is open circuited ... the net current is zero (i.e. at an open circuit). Therefore Ir = -If, and rho is a -1. Comment? Tom Bruhns wrote: I trust we can all agree that the definition of rho is rho=Vr/Vf, and I trust we can all agree that on a TEM line, the net voltage Vnet=Vr+Vf. If the net voltage is zero (i.e. at a short circuit), then clearly Vr=-Vf, and rho=-Vf/Vf=-1. Then whatever formula you chose to use to find rho at a load from the load impedance, Zload, and the line characteristic impedance, Zo, better work out right for a short circuit termination. Note that "rho=(Zload-Zo*)/(Zload+Zo)" does NOT work for that simple case, unless Zo*=Zo. But "rho=(Zload-Zo)/(Zload+Zo)" does work. You can easily go through something similar for an open circuit load. Note that rho=-1 for a short circuit load, and rho=+1 for an open circuit load, independent of line impedance. Cheers, Tom (Tom Bruhns) wrote in message ... It's even easier than that. Using the _definition_ of reflection coefficient, rho = Vr/Vf, we see that Vr=-Vf and therefore the net line voltage at that point is zero (that is, Vf+Vr, or Vf-Vf, or zero). That's either a real short or a virtual short. Cheers, Tom "David Robbins" wrote in message ... lets see... rho = (Z - Zo)/(Z + Zo) if rho = -1 then -1=(Z-Zo)/(Z+Zo) or -1*(Z+Zo)=(Z-Zo) or -Z - Zo = Z - Zo or -Z = Z the only value i know that satisfies that is zero. lets substitute it back in to be sure... -1=(0-(50+j50))/(0+(50+j50)) -1=(-50-j50)/(50+j50) -1=-1(50+j50)/(50+j50) -1=-1 qed. |
Hi Tom,
Since rho represents the fraction of forward power that is reflected, what does a negative value for rho indicate? Thanks, Jim AC6XG Tom Bruhns wrote: Hi Dave (and lurkers), Well, if rho=Ir/If, then the net current on the line is If-Ir, NOT If+Ir. You have to be careful about directions, and be careful to define things and stick with those definitions. I usually use Zo=Vf/If=-Vr/Ir, so that Ir and If measure "positive" in the same physical direction along the line, and then of course rho=Vr/Vf=-Ir/If. If that's at all hazy, just draw a picture and it should be clear. Anyway, rho=+1 at a point of zero net current, either way you define the current direction, and rho=-1 at a point of zero net voltage. No Smith chart, and no Zo, needed to figure that out. Cheers, Tom Dave Shrader wrote in message news:iBkcb.568264$o%2.253779@sccrnsc02... Hmmm ... Tom rho also equals Ir/If. Now if the line is open circuited ... the net current is zero (i.e. at an open circuit). Therefore Ir = -If, and rho is a -1. Comment? Tom Bruhns wrote: I trust we can all agree that the definition of rho is rho=Vr/Vf, and I trust we can all agree that on a TEM line, the net voltage Vnet=Vr+Vf. If the net voltage is zero (i.e. at a short circuit), then clearly Vr=-Vf, and rho=-Vf/Vf=-1. Then whatever formula you chose to use to find rho at a load from the load impedance, Zload, and the line characteristic impedance, Zo, better work out right for a short circuit termination. Note that "rho=(Zload-Zo*)/(Zload+Zo)" does NOT work for that simple case, unless Zo*=Zo. But "rho=(Zload-Zo)/(Zload+Zo)" does work. You can easily go through something similar for an open circuit load. Note that rho=-1 for a short circuit load, and rho=+1 for an open circuit load, independent of line impedance. Cheers, Tom (Tom Bruhns) wrote in message ... It's even easier than that. Using the _definition_ of reflection coefficient, rho = Vr/Vf, we see that Vr=-Vf and therefore the net line voltage at that point is zero (that is, Vf+Vr, or Vf-Vf, or zero). That's either a real short or a virtual short. Cheers, Tom "David Robbins" wrote in message ... lets see... rho = (Z - Zo)/(Z + Zo) if rho = -1 then -1=(Z-Zo)/(Z+Zo) or -1*(Z+Zo)=(Z-Zo) or -Z - Zo = Z - Zo or -Z = Z the only value i know that satisfies that is zero. lets substitute it back in to be sure... -1=(0-(50+j50))/(0+(50+j50)) -1=(-50-j50)/(50+j50) -1=-1(50+j50)/(50+j50) -1=-1 qed. |
Tom, rho^2 represents the fraction of forward power that is reflected.
The squaring function produces a positive value. Rho represents the percentage of voltage or current. Rho^2 is the power function. Dave Jim Kelley wrote: Hi Tom, Since rho represents the fraction of forward power that is reflected, what does a negative value for rho indicate? Thanks, Jim AC6XG Tom Bruhns wrote: Hi Dave (and lurkers), Well, if rho=Ir/If, then the net current on the line is If-Ir, NOT If+Ir. You have to be careful about directions, and be careful to define things and stick with those definitions. I usually use Zo=Vf/If=-Vr/Ir, so that Ir and If measure "positive" in the same physical direction along the line, and then of course rho=Vr/Vf=-Ir/If. If that's at all hazy, just draw a picture and it should be clear. Anyway, rho=+1 at a point of zero net current, either way you define the current direction, and rho=-1 at a point of zero net voltage. No Smith chart, and no Zo, needed to figure that out. Cheers, Tom Dave Shrader wrote in message news:iBkcb.568264$o%2.253779@sccrnsc02... Hmmm ... Tom rho also equals Ir/If. Now if the line is open circuited ... the net current is zero (i.e. at an open circuit). Therefore Ir = -If, and rho is a -1. Comment? Tom Bruhns wrote: I trust we can all agree that the definition of rho is rho=Vr/Vf, and I trust we can all agree that on a TEM line, the net voltage Vnet=Vr+Vf. If the net voltage is zero (i.e. at a short circuit), then clearly Vr=-Vf, and rho=-Vf/Vf=-1. Then whatever formula you chose to use to find rho at a load from the load impedance, Zload, and the line characteristic impedance, Zo, better work out right for a short circuit termination. Note that "rho=(Zload-Zo*)/(Zload+Zo)" does NOT work for that simple case, unless Zo*=Zo. But "rho=(Zload-Zo)/(Zload+Zo)" does work. You can easily go through something similar for an open circuit load. Note that rho=-1 for a short circuit load, and rho=+1 for an open circuit load, independent of line impedance. Cheers, Tom (Tom Bruhns) wrote in message ... It's even easier than that. Using the _definition_ of reflection coefficient, rho = Vr/Vf, we see that Vr=-Vf and therefore the net line voltage at that point is zero (that is, Vf+Vr, or Vf-Vf, or zero). That's either a real short or a virtual short. Cheers, Tom "David Robbins" wrote in message ... lets see... rho = (Z - Zo)/(Z + Zo) if rho = -1 then -1=(Z-Zo)/(Z+Zo) or -1*(Z+Zo)=(Z-Zo) or -Z - Zo = Z - Zo or -Z = Z the only value i know that satisfies that is zero. lets substitute it back in to be sure... -1=(0-(50+j50))/(0+(50+j50)) -1=(-50-j50)/(50+j50) -1=-1(50+j50)/(50+j50) -1=-1 qed. |
Ur right, thanks Dave. I meant to say voltage rather than power. Let
me ask the question properly. Tom, Since rho represents the fraction of forward voltage that is reflected, what does a negative value for rho indicate? Thanks and 73, Jim AC6XG Dave Shrader wrote: Tom, rho^2 represents the fraction of forward power that is reflected. The squaring function produces a positive value. Rho represents the percentage of voltage or current. Rho^2 is the power function. Dave Jim Kelley wrote: Hi Tom, Since rho represents the fraction of forward power that is reflected, what does a negative value for rho indicate? Thanks, Jim AC6XG Tom Bruhns wrote: Hi Dave (and lurkers), Well, if rho=Ir/If, then the net current on the line is If-Ir, NOT If+Ir. You have to be careful about directions, and be careful to define things and stick with those definitions. I usually use Zo=Vf/If=-Vr/Ir, so that Ir and If measure "positive" in the same physical direction along the line, and then of course rho=Vr/Vf=-Ir/If. If that's at all hazy, just draw a picture and it should be clear. Anyway, rho=+1 at a point of zero net current, either way you define the current direction, and rho=-1 at a point of zero net voltage. No Smith chart, and no Zo, needed to figure that out. Cheers, Tom Dave Shrader wrote in message news:iBkcb.568264$o%2.253779@sccrnsc02... Hmmm ... Tom rho also equals Ir/If. Now if the line is open circuited ... the net current is zero (i.e. at an open circuit). Therefore Ir = -If, and rho is a -1. Comment? Tom Bruhns wrote: I trust we can all agree that the definition of rho is rho=Vr/Vf, and I trust we can all agree that on a TEM line, the net voltage Vnet=Vr+Vf. If the net voltage is zero (i.e. at a short circuit), then clearly Vr=-Vf, and rho=-Vf/Vf=-1. Then whatever formula you chose to use to find rho at a load from the load impedance, Zload, and the line characteristic impedance, Zo, better work out right for a short circuit termination. Note that "rho=(Zload-Zo*)/(Zload+Zo)" does NOT work for that simple case, unless Zo*=Zo. But "rho=(Zload-Zo)/(Zload+Zo)" does work. You can easily go through something similar for an open circuit load. Note that rho=-1 for a short circuit load, and rho=+1 for an open circuit load, independent of line impedance. Cheers, Tom (Tom Bruhns) wrote in message ... It's even easier than that. Using the _definition_ of reflection coefficient, rho = Vr/Vf, we see that Vr=-Vf and therefore the net line voltage at that point is zero (that is, Vf+Vr, or Vf-Vf, or zero). That's either a real short or a virtual short. Cheers, Tom "David Robbins" wrote in message ... lets see... rho = (Z - Zo)/(Z + Zo) if rho = -1 then -1=(Z-Zo)/(Z+Zo) or -1*(Z+Zo)=(Z-Zo) or -Z - Zo = Z - Zo or -Z = Z the only value i know that satisfies that is zero. lets substitute it back in to be sure... -1=(0-(50+j50))/(0+(50+j50)) -1=(-50-j50)/(50+j50) -1=-1(50+j50)/(50+j50) -1=-1 qed. |
The voltage reflection coefficent, sometimes designated uppercase gamma,
sometimes lowercase rho, is the ratio of the reflected voltage to forward voltage. It doesn't represent power at all, and you have to make some assumptions of questionable validity to try and associate it with a power. People who like to speak of "forward power" and "reverse power" calculate that the ratio of "reverse power" to "forward power" is equal to the square of the magnitude of the voltage reflection coefficient, providing that the characteristic impedance of the line is assumed to be purely real. The reflection coefficient is a complex number, so it isn't restricted to "positive" and "negative", but can have any phase angle. Its magnitude, like the magnitude of any complex number, is a real number, so is square is always positive. (Complex numbers can be written with a negative magnitude, but this isn't commonly done. If it were, though, the square of the magnitude would still be positive.) Incidentally, the magnitude of the reflection coefficient is also often designated as lowercase rho, so any question about "rho" is unclear unless you say whether you're talking about the complex reflection coefficient or just its magnitude. The complex reflection coefficient can have a value of exactly -1, or a magnitude of 1 with a phase angle of 180 degrees. This simply means that the reflected voltage is equal in magnitude to the forward voltage, and exactly out of phase with it. The sum of forward and reflected voltage has to equal the total voltage at any point along the line. If you terminate a line with a short circuit, the reflection coefficient is -1 and the forward and reverse voltages at the end of the line are equal and opposite, so they add up to zero. Which is the voltage you have at a short circuit. Roy Lewallen, W7EL Jim Kelley wrote: Ur right, thanks Dave. I meant to say voltage rather than power. Let me ask the question properly. Tom, Since rho represents the fraction of forward voltage that is reflected, what does a negative value for rho indicate? Thanks and 73, Jim AC6XG |
The short answer is that you'd use the magnitude of rho, not its
complex value, to find the magnitude of Vr in terms of the magnitude of Vf, and from that, the relative value of the powers. So I suppose you'll get the same values for rho=+1 and rho=-1, since the magnitude is +1 in both cases. Beware of how you do the calcs: rho=+j, rho=-j, and rho=(1+j)/sqrt(2) all should also give you |rho|=1. But I'll leave the power calcs to you. Resolving things into "forward power" and "reflected power" for steady-state excitation really doesn't do a thing for me. I want to know the load presented to the source, and the power delivered to the line by the source and to the load by the line, and perhaps some other things like power dissipation as a function of distance along the line, but I can't think of any reason why I'd care about "f.p." or "r.p." Now what happens during transient situations is a completely different story. Cheers, Tom Jim Kelley wrote in message ... Hi Tom, Since rho represents the fraction of forward power that is reflected, what does a negative value for rho indicate? Thanks, Jim AC6XG Tom Bruhns wrote: Hi Dave (and lurkers), Well, if rho=Ir/If, then the net current on the line is If-Ir, NOT If+Ir. You have to be careful about directions, and be careful to define things and stick with those definitions. I usually use Zo=Vf/If=-Vr/Ir, so that Ir and If measure "positive" in the same physical direction along the line, and then of course rho=Vr/Vf=-Ir/If. If that's at all hazy, just draw a picture and it should be clear. Anyway, rho=+1 at a point of zero net current, either way you define the current direction, and rho=-1 at a point of zero net voltage. No Smith chart, and no Zo, needed to figure that out. Cheers, Tom Dave Shrader wrote in message news:iBkcb.568264$o%2.253779@sccrnsc02... Hmmm ... Tom rho also equals Ir/If. Now if the line is open circuited ... the net current is zero (i.e. at an open circuit). Therefore Ir = -If, and rho is a -1. Comment? Tom Bruhns wrote: I trust we can all agree that the definition of rho is rho=Vr/Vf, and I trust we can all agree that on a TEM line, the net voltage Vnet=Vr+Vf. If the net voltage is zero (i.e. at a short circuit), then clearly Vr=-Vf, and rho=-Vf/Vf=-1. Then whatever formula you chose to use to find rho at a load from the load impedance, Zload, and the line characteristic impedance, Zo, better work out right for a short circuit termination. Note that "rho=(Zload-Zo*)/(Zload+Zo)" does NOT work for that simple case, unless Zo*=Zo. But "rho=(Zload-Zo)/(Zload+Zo)" does work. You can easily go through something similar for an open circuit load. Note that rho=-1 for a short circuit load, and rho=+1 for an open circuit load, independent of line impedance. .... |
Jim Kelley wrote:
Since rho represents the fraction of forward power that is reflected, what does a negative value for rho indicate? rho represents the fraction of forward voltage that is reflected. |rho|^2 represents the fraction of forward power that is reflected. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Jim Kelley wrote:
Since rho represents the fraction of forward voltage that is reflected, what does a negative value for rho indicate? rho = +1 means there is a zero degree phase shift in the reflected voltage. rho = -1 means there is a 180 degree phase shift in the reflected voltage. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
"Cecil Moore" wrote in message ... Jim Kelley wrote: Since rho represents the fraction of forward voltage that is reflected, what does a negative value for rho indicate? rho = +1 means there is a zero degree phase shift in the reflected voltage. rho = -1 means there is a 180 degree phase shift in the reflected voltage. So in other word the sign sometimes indicates phase, and other times indicates a direction in propagation depending on which hand is waving.. Thanks. 73, Jim AC6XG |
Jim Kelley wrote:
"Cecil Moore" wrote: rho = +1 means there is a zero degree phase shift in the reflected voltage. rho = -1 means there is a 180 degree phase shift in the reflected voltage. So in other word the sign sometimes indicates phase, and other times indicates a direction in propagation depending on which hand is waving.. Thanks. Consider the equation, rho = (ZL-Z0)/(ZL+Z0). If ZLZ0, then the voltage reflection coefficient is positive and there is no reflected voltage phase shift. If ZLZ0, then the voltage reflection coefficient is negative and there is a 180 degree phase shift in the reflected voltage. The same holds true for the E-field of reflected light. The RF reflected current convention differs from the reflected light H-field convention. Kirchhoff's current convention enters into the sign of the reflected RF current where no such convention exists for light. -- 73, Cecil, W5DXP |
Jim,
Take the sign of rho literally. For instance, if Zl=0, a short circuit, rho=-1; therefore, at the termination, V=V+ +(- V+) = 0. If Zl= open, then rho=+1, and V= V+ + V+ =2V+ Tam/WB2TT "Jim Kelley" wrote in message ... Ur right, thanks Dave. I meant to say voltage rather than power. Let me ask the question properly. Tom, Since rho represents the fraction of forward voltage that is reflected, what does a negative value for rho indicate? Thanks and 73, Jim AC6XG Dave Shrader wrote: Tom, rho^2 represents the fraction of forward power that is reflected. The squaring function produces a positive value. Rho represents the percentage of voltage or current. Rho^2 is the power function. Dave Jim Kelley wrote: Hi Tom, Since rho represents the fraction of forward power that is reflected, what does a negative value for rho indicate? Thanks, Jim AC6XG Tom Bruhns wrote: Hi Dave (and lurkers), Well, if rho=Ir/If, then the net current on the line is If-Ir, NOT If+Ir. You have to be careful about directions, and be careful to define things and stick with those definitions. I usually use Zo=Vf/If=-Vr/Ir, so that Ir and If measure "positive" in the same physical direction along the line, and then of course rho=Vr/Vf=-Ir/If. If that's at all hazy, just draw a picture and it should be clear. Anyway, rho=+1 at a point of zero net current, either way you define the current direction, and rho=-1 at a point of zero net voltage. No Smith chart, and no Zo, needed to figure that out. Cheers, Tom Dave Shrader wrote in message news:iBkcb.568264$o%2.253779@sccrnsc02... Hmmm ... Tom rho also equals Ir/If. Now if the line is open circuited ... the net current is zero (i.e. at an open circuit). Therefore Ir = -If, and rho is a -1. Comment? Tom Bruhns wrote: I trust we can all agree that the definition of rho is rho=Vr/Vf, and I trust we can all agree that on a TEM line, the net voltage Vnet=Vr+Vf. If the net voltage is zero (i.e. at a short circuit), then clearly Vr=-Vf, and rho=-Vf/Vf=-1. Then whatever formula you chose to use to find rho at a load from the load impedance, Zload, and the line characteristic impedance, Zo, better work out right for a short circuit termination. Note that "rho=(Zload-Zo*)/(Zload+Zo)" does NOT work for that simple case, unless Zo*=Zo. But "rho=(Zload-Zo)/(Zload+Zo)" does work. You can easily go through something similar for an open circuit load. Note that rho=-1 for a short circuit load, and rho=+1 for an open circuit load, independent of line impedance. Cheers, Tom (Tom Bruhns) wrote in message ... It's even easier than that. Using the _definition_ of reflection coefficient, rho = Vr/Vf, we see that Vr=-Vf and therefore the net line voltage at that point is zero (that is, Vf+Vr, or Vf-Vf, or zero). That's either a real short or a virtual short. Cheers, Tom "David Robbins" wrote in message ... lets see... rho = (Z - Zo)/(Z + Zo) if rho = -1 then -1=(Z-Zo)/(Z+Zo) or -1*(Z+Zo)=(Z-Zo) or -Z - Zo = Z - Zo or -Z = Z the only value i know that satisfies that is zero. lets substitute it back in to be sure... -1=(0-(50+j50))/(0+(50+j50)) -1=(-50-j50)/(50+j50) -1=-1(50+j50)/(50+j50) -1=-1 qed. |
Yes, Cecil. I have considered it, and I agree. It just bothers me when
people forget what the minus sign means, and try using it to make unrealistic claims. 73, Jim AC6XG Cecil Moore wrote: Cecil Moore wrote: Consider the equation, rho = (ZL-Z0)/(ZL+Z0). If ZLZ0, then the voltage reflection coefficient is positive and there is no reflected voltage phase shift. If ZLZ0, then the voltage reflection coefficient is negative and there is a 180 degree phase shift in the reflected voltage. The same holds true for the E-field of reflected light. The above assumes ZL and Z0 to be real numbers. The light index of refraction is usually a real number. -- 73, Cecil, W5DXP |
Cecil Moore wrote:
Consider the equation, rho = (ZL-Z0)/(ZL+Z0). If ZLZ0, then the voltage reflection coefficient is positive and there is no reflected voltage phase shift. If ZLZ0, then the voltage reflection coefficient is negative and there is a 180 degree phase shift in the reflected voltage. The same holds true for the E-field of reflected light. The above assumes ZL and Z0 to be real numbers. The light index of refraction is usually a real number. -- 73, Cecil, W5DXP |
Jim Kelley wrote:
Yes, Cecil. I have considered it, and I agree. It just bothers me when people forget what the minus sign means, and try using it to make unrealistic claims. At least for real Z0's, it should be consistent. Wouldn't a rho of 0.5 at 20 degrees would be the same as a rho of -0.5 at 200 degrees? -- 73, Cecil, W5DXP |
No. The voltage reflection coefficient is the ratio of two voltages, and
has nothing to do with their directions. The reverse voltage wave is always traveling toward the source. The forward voltage wave is always traveling toward the load. The angle of the reflection coefficient is the relative phases of these two at the point the reflection coefficient is being evaluated. It makes sense to speak of the reflection coefficient as having a "sign" only in the two special cases of zero and 180 degree angles. Otherwise, it has an angle, not a positive or negative "sign". The current reflection coefficient is a little more ambiguous because of the freedom of defining which direction represents positive flow of Ir. If Ir is defined to be positive toward the load (the more common definition), then the current reflection coefficient, Ir/If = -Vr/Vf. If it's defined to be positive toward the source, then Ir/If = Vr/Vf. Roy Lewallen, W7EL Jim Kelley wrote: "Cecil Moore" wrote in message ... Jim Kelley wrote: Since rho represents the fraction of forward voltage that is reflected, what does a negative value for rho indicate? rho = +1 means there is a zero degree phase shift in the reflected voltage. rho = -1 means there is a 180 degree phase shift in the reflected voltage. So in other word the sign sometimes indicates phase, and other times indicates a direction in propagation depending on which hand is waving.. Thanks. 73, Jim AC6XG |
On Fri, 26 Sep 2003 12:53:50 -0500, Cecil Moore
wrote: At least for real Z0's, it should be consistent. Wouldn't a rho of 0.5 at 20 degrees would be the same as a rho of -0.5 at 200 degrees? Hi Cecil, The Reflection Coefficient is a characteristic of the Load or Source, not a value projected all along the line. This is the teaching of Chipman that you undoubtedly speed-read past on your way to the cut-and-paste opportunities you sought. When are you going to ride your bike back to the library to fill all these shortfalls of his teachings you so liberally sprinkle your missives with? 73's Richard Clark, KB7QHC |
Richard Clark wrote:
The Reflection Coefficient is a characteristic of the Load or Source, not a value projected all along the line. This is the teaching of Chipman that you undoubtedly speed-read past on your way to the cut-and-paste opportunities you sought. For lossless transmission lines, |rho| = Sqrt(Pref/Pfwd). You don't even need to know the load and/or source impedances. When are you going to ride your bike back to the library to fill all these shortfalls of his teachings you so liberally sprinkle your missives with? Just as soon as I am over my sinusitis and have a day off. -- 73, Cecil, W5DXP |
On Fri, 26 Sep 2003 14:35:33 -0500, Cecil Moore
wrote: For lossless transmission lines, |rho| = Sqrt(Pref/Pfwd). You don't even need to know the load and/or source impedances. Hi Cecil, How did you get a -1 out of your |rho|? Take a box of kleenex on your bike ride to the library - your logic is dribbling. 73's Richard Clark, KB7QHC |
Richard Clark wrote:
wrote: For lossless transmission lines, |rho| = Sqrt(Pref/Pfwd). You don't even need to know the load and/or source impedances. How did you get a -1 out of your |rho|? I probably should have said rho^2 = Pref/Pfwd. When Pref = Pfwd, rho can be plus or minus one. I used |rho| to indicate a magnitude, sans phase angle, not an absolute value. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Jim Kelley wrote in message ...
Ur right, thanks Dave. I meant to say voltage rather than power. Let me ask the question properly. Tom, Since rho represents the fraction of forward voltage that is reflected, what does a negative value for rho indicate? Jim, I suppose your question has been answered sufficiently (Thanks, Roy, Cecil and Tam), but I'd like to offer a bit different viewpoint than is implied by your "fraction that is reflected." I prefer to think of it not as "a fraction that's reflected" but rather as a resolution of a particular voltage and current into two modes. There are two modes of propagation supported by TEM line, one in each direction along the line. If you excite a line to steady-state at one frequency, there will be some sinusoidal current at each point along the line, and some sinusoidal voltage across the line at each point along its length. (You can have a load at one end and a source at the other, or two sources each with its own internal impedance, one at each end, so long as they are on the same frequency.) That set of voltages and currents can be resolved mathematically into two components, one corresponding to the mode of propagation in one direction and one corresponding to the mode in the other direction. Rho is simply a number representing that resolution at the point on the line where that rho is measured (or calculated). It's a complex number because it represents both phase and amplitude. (Note that our resolution of measured voltage and current into the two modes generally assumes that we know the line's Zo, and the degree to which we don't know that will introduce an error in our determination of rho. But that's a whole 'nuther topic...) Cheers, Tom |
Cecil Moore wrote in message ...
.... Wouldn't a rho of 0.5 at 20 degrees would be the same as a rho of -0.5 at 200 degrees? Yes, and any other complex quantity would likewise be the same expressed either way. But it would certainly be confusing to the readers. It would be much better to stick with rectangular or with polar and not mix them in the same quantity. Of course, sometimes one is easier to work with, or offers more insight, than the other and you're welcome to convert between them at any time. Let's see if we can keep it more along the lines of 0.5 at 20 degrees being (very nearly) the same as .4698+j.1710 Cheers, Tom |
On Fri, 26 Sep 2003 23:15:44 -0500, Cecil Moore
wrote: Richard Clark wrote: wrote: For lossless transmission lines, |rho| = Sqrt(Pref/Pfwd). You don't even need to know the load and/or source impedances. How did you get a -1 out of your |rho|? I probably should have said rho^2 = Pref/Pfwd. When Pref = Pfwd, rho can be plus or minus one. I used |rho| to indicate a magnitude, sans phase angle, not an absolute value. Hi Cecil, You obviously don't respect/know the difference between a dependant variable (rho) and independent variables (P). Rho is a dependency of the interface, not a translatable value you are forcing illogic to perform. You really need to ride your bike to the library more and offer these poor examples less. Since Rho is the dependant variable, even squared (for you to force a -1 into this charade) requires a concurrent observance of a negative in the right hand side (negative power - perhaps if you were in a black hole). The long and short of it is that this confirms Jim's observance of your forced math serving your canards rather than logic. Oh, and please stop offering and polluting Chipman as a resource when you've only copied one page. 73's Richard Clark, KB7QHC |
Richard Clark wrote:
Since Rho is the dependant variable, even squared (for you to force a -1 into this charade) requires a concurrent observance of a negative in the right hand side (negative power - perhaps if you were in a black hole). You seem to have forgotten some junior high math, Richard. There is no requirement for a negative anywhere in order for the square root of a number to be negative. The square root of 100W/100W has two values, plus or minus one, and sure enough, an open or a short will cause 100% reflection. BTW, I copied that page in Chipman with which you are having such a problem and I don't see the problem you described. Absolutely nothing said about reflections from the source. In fact, the source has the same impedance as the transmission line so there are no reflections from the source. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
On Sat, 27 Sep 2003 16:34:43 -0500, Cecil Moore
wrote: Richard Clark wrote: Since Rho is the dependant variable, even squared (for you to force a -1 into this charade) requires a concurrent observance of a negative in the right hand side (negative power - perhaps if you were in a black hole). You seem to have forgotten some junior high math, Richard. There is no requirement for a negative anywhere in order for the square root of a number to be negative. The square root of 100W/100W has two values, plus or minus one, and sure enough, an open or a short will cause 100% reflection. Hi Cecil, If neither powers are negative, the square root of them cannot possibly enclose a negative. There is no possibility of Rho being negative by your description. You should start biking to junior high. BTW, I copied that page in Chipman with which you are having such a problem and I don't see the problem you described. My problem? Quote me rather than give me your tarted up remembrance of what I said. Clearly your head cold cannot answer for such consistently unreliable correspondence. Absolutely nothing said about reflections from the source. In fact, the source has the same impedance as the transmission line so there are no reflections from the source. Duh! Cecil, You are going to run your bike's mileage warrantee out by pedalling to the library for one page copies at a time. Why don't you spend a couple hours there and read it instead? 73's Richard Clark, KB7QHC |
Richard Clark wrote:
If neither powers are negative, the square root of them cannot possibly enclose a negative. Huh?????? The square root of +100W/+100W cannot be negative????? WOW! Sounds like you are letting your personal feelings get in the way of accepted math principles. Hint: If one of those powers is negative, the square root will be imaginary. Cecil, You are going to run your bike's mileage warrantee out by pedalling to the library for one page copies at a time. Why don't you spend a couple hours there and read it instead? I've got page 139, the one you referenced, in front of me. It says absolutely nothing about reflections from the source. All it seems to say is that conjugately matched loads accept more power than non-conjugately matched loads but we knew that already. Incidentally, pages 140-143 discusses "Transmission line sections as two-port networks" using the h-parameter analysis. Who said transmission lines didn't have ports? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
On Sat, 27 Sep 2003 22:41:41 -0500, Cecil Moore
wrote: Richard Clark wrote: If neither powers are negative, the square root of them cannot possibly enclose a negative. Huh?????? The square root of +100W/+100W cannot be negative????? WOW! Sounds like you are letting your personal feelings get in the way of accepted math principles. Personal feelings, hmmm? You are the one enclosing the statement with excessive marks, bucko. And I also note that what is enclosed is a hoot! Hint: If one of those powers is negative, the square root will be imaginary. We can all tell where imagination springs from. Give me better than a hint of negative power - you obviously didn't embrace it between your emotional markings. I've got page 139, the one you referenced, in front of me. It says absolutely nothing about reflections from the source. Duh! Two for Two. You still can't do any better than your tarted up versions of what you "think" I said? Your ability to find a Google copy is no better than your cut-and-paste library skills. You can (and have) spin these fantasies out to 600 postings if you put your mind to it. Could we at least expect you may actually read Chipman at some future date? 73's Richard Clark, KB7QHC |
|
Tom Bruhns wrote:
Now can you children quit bickering? Does it really hurt anything to remind everyone that +1 at 180 degrees equals -1 at zero degrees? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Richard Clark wrote:
wrote: Richard Clark wrote: If neither powers are negative, the square root of them cannot possibly enclose a negative. Huh?????? The square root of +100W/+100W cannot be negative????? WOW! Sounds like you are letting your personal feelings get in the way of accepted math principles. Personal feelings, hmmm? You are the one enclosing the statement with excessive marks, bucko. And I also note that what is enclosed is a hoot! Owls are not really all that intelligent, Richard, even if they are MENSA's mascot. You really should upgrade to parrots if you want an intelligent bird. Give me better than a hint of negative power - you obviously didn't embrace it between your emotional markings. By convention, direction can change power to a negative number. That is positive power flowing in a negative direction. To the best of my knowledge, there is really no such thing as negative energy as would be required for negative power. Could we at least expect you may actually read Chipman at some future date? I've got page 139 in front of me. It doesn't say what you said it said. It says a conjugate match will ensure maximum power transfer. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
On Sun, 28 Sep 2003 14:35:30 -0500, Cecil Moore
wrote: Owls are not really all that intelligent, Richard, even if they are MENSA's mascot. You really should upgrade to parrots if you want an intelligent bird. I didn't think you could deal with the negative power intelligently. Give me better than a hint of negative power - you obviously didn't embrace it between your emotional markings. By convention, direction can change power to a negative number. What convention, Shriners? That is positive power flowing in a negative direction. To the best of my knowledge, there is really no such thing as negative energy as would be required for negative power. Thus it follows your negative that doesn't exist in your own formula, doesn't exist as a figment of someone else's over-indulgent imagination. I already said as much. Could we at least expect you may actually read Chipman at some future date? I've got page 139 in front of me. It doesn't say what you said it said. It says a conjugate match will ensure maximum power transfer. Duh! You are still at a loss to respond to the post, and instead in your own tradition of the unreliable correspondent decide to respond to your own rhetoric. Three for three now. It is painfully obvious you haven't got a clue what you are answering to. This is a curious state of affairs where your imagination works overtime to build these fantasies of power flow signs, and then paint up these supposed quotes of mine with such pale and weak colors. 73's Richard Clark, KB7QHC |
Richard Clark wrote:
Thus it follows your negative that doesn't exist in your own formula, doesn't exist as a figment of someone else's over-indulgent imagination. I already said as much. The sign of the reflection coefficient has absolutely nothing to do with negative power. The sign of the reflection coefficient is simply a math convention where -1 at zero degrees equals +1 at 180 degrees. You are still at a loss to respond to the post, ... Exactly where does Chipman talk about reflections from the source? As far as I can tell, the concept of reflections from the source originated with you, not with Chipman. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
On Mon, 29 Sep 2003 09:18:39 -0500, Cecil Moore
wrote: You are still at a loss to respond to the post, ... Exactly where does Chipman talk about reflections from the source? As far as I can tell, the concept of reflections from the source originated with you, not with Chipman. Hi Cecil, You deliberately isolate my complaint, and then entirely ignore it. Does this give you some insight into you being observed as an unreliable correspondent? Is it so difficult to search Google to find the actual post that you claim you can quote from memory (sic)? If you do find it difficult, what makes me think you have the library research skills to find any printed material to a citation? Your inability to quote or follow Chipman's work is equal in the poor treatment you post in regards to Optics here, a field you are so naive about, and so terribly unfamiliar with, you cannot even perform the simplest of field solutions. This is indeed the price and evidence of your career in binary electronics. 73's Richard Clark, KB7QHC |
Tom Bruhns wrote: (Note that our resolution of measured voltage and current into the two modes generally assumes that we know the line's Zo, and the degree to which we don't know that will introduce an error in our determination of rho. But that's a whole 'nuther topic...) To my way of thinking, rho is entirely dependent upon the impedances, and the voltages (reflected voltages in particular) are dependent upon rho. Not the other way around. 73, Jim AC6XG |
Cecil Moore wrote in message ...
Does it really hurt anything to remind everyone that +1 at 180 degrees equals -1 at zero degrees? No, and I already agreed with that in another posting in this thread. Perhaps you missed it. But it's just flat wrong to claim that the negative value for sqrt(x^2) can be correct when you know that the the original value of x is not negative: x in this case is the magnitude of a complex number, and that magnitude is real and never negative. Not only is that wrong, but it's also potentially confusing to lurkers who may read into it that the only two values of rho which can result in |rho|=1 are rho=+1 and rho=-1, and that's wrong. Just do it right and say that your square root = |rho| = +1 and not -1, because it's a magnitude, and that rho then can be magnitude 1 at ANY phase angle, not just 0 and 180. None of which has anything to do with the two of you continuing to squabble like a couple of young children. Cheers, Tom |
Richard Clark wrote:
You deliberately isolate my complaint, and then entirely ignore it. I give up trying to communicate with you as have most others. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
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