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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 |
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