Ed,
The Smith Charts with R=1 in the center are normalized charts. If you
have 50 ohm coaxial cable, just multiply the numbers by 50. You may
occasionally use 75 ohm cable. While you can buy charts with R=75 in the
center, all you have to do with a normalized chart is to multiply the
values by 75. I doubt if you can even buy charts specifically for 93 ohm
(RG-62) cable, 300 ohm twinlead, or 450 and 440 ohm "window" line.
I always compute the reflection coefficient from a complex load before I
calculate the VSWR.
? (Greek capital letter Gamma) = (Zmeas - Zline)/(Zmeas + Zline)
We usually think that we can replace the line impedance with 50 + j0, but
as my friend Tom, K7ITM, often points out, so-called 50 ohm cable no
longer has a purely resistive characteristic impedance at low
frequencies. So we need to calculate the reflection coefficient as a
complex number. Once we have this, we need to determine the magnitude of
the reflection coefficient.
? (Greek lower case rho) = |?|
= sqrt(Real*Real + Imag*Imag)
Once we know the magnitude of the reflection coefficient, which is always
less than or equal to unity, the voltage standing wave ratio is simply
VSWR = (1 + ?)/(1 - ?)
If you have Windoze, all of these calculations are easily done in Excel.
You may have to install the Analysis Toolpack to get Excel to handle
complex numbers.
I strongly suggest that you continue to use Smith charts. The first
large program for my TI-59 calculator nearly forty years ago was an
electronic Smith chart. It was handy. but I continued to use Smith
charts because they offer graphically significant insight into what
happens along the length of a transmission line. Additionally, they help
you catch mistakes.
73, Barry WA4VZQ
Postscript: I used rich text format solely because I could use Greek
letters. Gamma and rho are almost universally used to represent the
complex reflection coefficient and its magnitude respectively in
engineering publications.
When I read the posting, both Gamma and rho are displayed as a '?'.
--
73, Cecil, w5dxp.com
> We usually think that we can replace the line impedance with 50 + j0, but
> as my friend Tom, K7ITM, often points out, so-called 50 ohm cable no
> longer has a purely resistive characteristic impedance at low
> frequencies. So we need to calculate the reflection coefficient as a
> complex number. Once we have this, we need to determine the magnitude of
> the reflection coefficient.
Hello, and I believe some clarification is needed here. First, for most
ham radio applications we can assume a real characteristic impedance for
the transmission line (e.g. Zo = Ro = 50 ohms). The only time this
leads to large errors when calculating the line-transformed load
impedance (e.g. an antenna at a given frequency) occurs when the VSWR of
the load is greater than 10 or 20. (This was pointed out in a 1953 AIEE
paper by W. W. Macalpine.)
However, just because the reference impedance, Zo, is assumed to be real
doesn't mean that the reflection coefficient has to be real. Sincerely,
and 73s from N4GGO,
--
John Wood (Code 5520) e-mail: wo...@itd.nrl.navy.mil
Naval Research Laboratory
4555 Overlook Avenue, SW
Washington, DC 20375-5337
Barry's assumption of resistance being the same as the DC resistance
and independent of frequency will make the characteristic impedance
look closer to resistive than it actually is, at frequencies where
skin depth is small compared with conductor thickness. Since in
copper, skin depth at 1MHz is only about 2.6 mils (and is inversely
proportional to the square root of the frequency), the resistance of
that particular line's conductors at 1MHz will be considerably higher
than the DC resistance.
Dare I mention that when the line impedance is reactive, you can get
reflection coefficients with magnitude greater than unity? Seems like
every time that comes up here, someone gets their knickers in a knot
about it. Reflection coefficient = rho = (Z-Zo)/(Z+Zo), so if Z (the
load impedance) is a pure inductance, say +j50 ohms, and the line
impedance is (from Barry's chart) 50.4-j4.63 ohms, then rho =
-0.0134+j1.096, and |rho| = 1.096. But to calm your nerves, calculate
what the impedance is looking into any length of that line you wish,
terminated that way, and you'll see that the impedance never has a
negative real part.
Cheers,
Tom
For a Smith Chart which allows any Zo you wish, try:
http://www.fritz.dellsperger.net/PartInterests.htm
I love this chart. I used the free version for years.
Cheers,
John