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On 5 Apr 2007 09:49:52 -0700, "K7ITM" wrote:
On Apr 4, 8:38 pm, Walter Maxwell wrote: On Thu, 05 Apr 2007 03:01:11 GMT, Owen Duffy wrote: Walter Maxwell wrote in : ... R/Zo = (1 - rho squared)/(1 + rho squared - 2 rho cos phi) X/Zo = (2 rho sin phi)/(1 + rho squared - 2 rho cos phi) ... These appear to depend on Zo=Ro to be correct, perhaps they would be more correctly expressed using Ro instead of Zo. Owen Owen, for historical accuracy, at least in the US, prior to 1950, rho, sigma, and S were used to represent standing wave ratio. The symbol of choice used to represent reflection coefficient during that early era was upper case lambda. However, in 1953 the American Standards Association (now NIST) announced in its publication ASA Y10.9-1953, that rho is to replace upper case lambda as the standard symbol for reflection coefficient, and SWR to represent standing wave ratio. Most of academia responded to the change, but a few have not. I don't know about Australia, but in the US lambda is rarely seen as the symbol for reflection coefficient. WRT Ro vs Zo, I was simply copying directly from Chipman, where Zo is routinely considered the characteristic impedance of a transmission line, and where it's usually considered sufficiently low loss to the thought of as Ro. Walt About Zo being reasonably approximated by Ro, or not: I made a note to myself some time ago, and I believe it's reasonably accurate, that neglecting dielectric loss, for a TEM line, given Zo = Ro+jXo, then to a good approximation Xo = -0.180*Ro*A*Vf/f where A = line attenuation in dB/100ft Vf = line velocity factor f = frequency in MHz. So for small diameter 50 ohm polyethylene dielectric line at 1.8MHz, the worst case for most ham applications, Xo/Ro is about .18. For that, I used 2.7dB/100ft for RG174 type line. That's getting to be pretty significant, a ten degree phase angle away from purely resistive. As Owen posted, it's so easy these days to deal with complex numbers that you may as well just carry them all along. Given the above formula, it's easy to figure the complex Zo for a line where you know the nominal attenuation, the velocity factor, and the frequency, and of course the nominal high frequency Zo value. Cheers, Tom Can't argue with your comments above, Tom, but what ham in his right mind would use 100 feet of RG174? Walt |
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