Roy Lewallen wrote:
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)).
....
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

Indeed. It occurs to me that if one was to try to measure VSWR (say
using a slotted line with probe) on a lossy line operating at high VSWR,
the best estimate would come from finding a minimum, measuring it and
the adjacent maxima and calculating VSWR=(Vmax1+Vmax2)/Vmin/2 so that
the measurement is biassed towards the notional VSWR at the point of the
minimum (the minimum being much more sensitive to line attenuation than
the maxima).

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.

Noted

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.

Agreed.


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

I mean't function in the sense that there is one and only one value of
f(x) for x, rather than in the causual sense.

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.

Agreed.

Thanks for your exhaustive reply Roy, it is appreciated.

What I glean from this is that although Gamma and therefore rho are well
defined, and both are a function of position on a line in the general
case, the "accepted mathematical definition" of VSWR in terms of rho
does not behave well (eg producing a negative value) in some cases and
is not a good estimator of real VSWR in those cases.

It seems fair to say that the reason that the "accepted mathematical
definition" of VSWR does not behave well is that it depends on an
assumption of a distortionless line (Xo=0). (Lossless lines are
distortionless but the converse is not necessarily true).

I take your point about the need to qualify a different algorithm by a
different name.

Thanks again.

Owen