"Peter O. Brackett" wrote in message thlink.net...
Slick:
[snip]
If you believe that there are theoretically no
reflections in a conjugate match, then with
Zl=50+j10 and Zo=50-j10,
the conjugate equation correctly cancels the reactances
giving no reflections, while the non-conjugate still
incorrectly gives a magitude (non zero) for rho.
Slick
[snip]
Oh yes here are voltage reflections at a conjugate match!
Simply put as waves pass across the transition from an impedance
of Zo to an impedance of conj(Zo) they are crossing a boundary
with an impedance discontinuity. Zo on one side and conj(Zo)
on the other side is definitely discontinuous! Unless of course,
Zo = conj(Zo) which occurs only when Zo is real.
There will always reflections at such an impedance
discontinuity where an impedance faces its' conjugate.
I disagree completely. The theoretical impedance of a resonant
series L and C (which is lossless) is zero. So in a conjugate
match, where they cancel out, in an ideal loss-less world, it
would be equivalent to the series C and L not being there at all,
with the source and load 50 ohms free to pass max. power delivered to
the load.
If an impedance Zo faces its' conjugate conj(Zo) then there will be
no "power reflections", but there will in general be voltage reflections,
??? if the square of the magnitude of the voltage RC is the power RC,
then your statement is incorrect. And rho (magnitude of Voltage RC) is the
square root of the Power RC.
i.e. "classical" rho = (Z - Zo)/(Z + Zo) is not zero at a conjugate match.
That's why it is incorrect for complex Zo.
I also believe that this is "Mother Nature's" reflection coefficient
for it is exactly what she uses as she lets the waves propagate
down her lines of surge impedance Zo following her partial differential
equations at every point along the way. At every infinitesimal
length of line all along it's length the waves are passing from a
infinitesimal region of surge impedance Zo to the next infinitesimal
region of surge impedance Zo and there are no voltage reflections
anywhere along that [uniform] line, although if the line is not lossless
there will be energy lost as the wave progresses.
Maybe "Mother Nature" should take a Les Besser course...
Slick... On another whole level it simply does not matter which defiinition
of the reflection coefficient one uses to make design calculations though,
as
long as the definition is used consistently throughout any calculations.
I totally disagree again:
Did you read Williams' data?
The data follows:
Note: |rho1*| is conjugated rho1, SWR1 is for
|rho1*|, |rho2| is not conjugated and SWR2 applies
to |rho2|
X0.......|rho1*|..SWR1.....|rho2|..SWR2
-250..... 0.935...30.0.....1.865...-3.30
-200..... 0.937...30.8.....1.705...-3.80
-150..... 0.942...33.3.....1.517...-4.87
-100..... 0.948...37.5.....1.320...-7.25
-050..... 0.955...43.3.....1.131...-16.3
-020..... 0.959...47.6.....1.030...-76.5
-015..... 0.960...48.4.....1.010...-204
-012..... 0.960...48.9.....0.997....+/- infinity
-010..... 0.960...49.2.....0.990....+305
-004..... 0.961...76.3.....0.974....+76.3
0000..... 0.961...50.9.....0.961....+50.9
The numbers for not-conjugate rho are all over the
place and lead to ridiculous numbers for SWR. It
is also obvious that for a low-loss line it
doesn't matter much. But values of rho greater
than 1.0, on a Smith chart correspond to negative
values of resistance (see the data).
Excellent work William. You are also showing how
a rho1 leads to ridiculous numbers for the equation:
SWR = (1 + rho)/(1 - rho)
The non-conjugate equation simply cannot handle
complex Zo.
Some people think we should throw out the SWR formula
completely, but this is complete nonsense, of course.
SWR = (1 + rho)/(1 - rho) works for 0=rho=1,
for very good reason, as it applies to passive networks only.
And the conjugate will always give 0=rho=1,
even with a complex Zo.
Slick