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