Jim Kelley wrote:

Cecil Moore wrote:
Isn't the 50 ohms that causes rho=0 on the 50 ohm feedline
simply the V/I ratio at point '+'?

I see now. Your interested in something else here, I think. The rho
for the whole network which includes both impedance discontinuities is
indeed zero. We've talked about that before. But the rho for the
single discontinuity at '+' is not equal to zero. The reflected
impedance (the load impedance, repeated a half wavelength away) is not
considered in the evaluation of rho at '+'. It is the characteristic
impedance of the line that is considered. You would agree, no?

I would agree if you were talking about s11. But rho on the coax is zero.
The impedance at '+' is 50 ohms. rho = (50-50)/(50+50) = sqrt(0/Pfwd) = 0
at point '+'. And that 50 ohms is a V/I ratio which, I assume, you would
agree cannot cause a voltage.
--
73, Cecil http://www.qsl.net/w5dxp



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wrote:

Cecil Moore wrote:

All (rho = -1) requires is a
short at the end of a transmission line as explained in _Transmission_
Lines_and_Networks_, by Walter C. Johnson when he was chairman of the
Princeton EE Dept. Here's how he calculated rho for a short:

rho = (Z1-Z0)/(Z1+Z0) = (0-Z0)/(0+Z0) = -Z0/Z0 = -1

So your argument is with Dr. Johnson whom I am merely quoting. The
(rho = -1) simply indicates a 180 degree phase shift in the reflected
voltage at the short.

Quite false. Negation is not simply a 180 degree phase shift.

Yes, it is, in the reflected voltage. A short has a voltage rho of -1
indicating a 180 degree phase shift between the incident voltage and
the reflected voltage. An open has a voltage rho of +1 indicating no
phase shift between the incident voltage and reflected voltage. The
rho for current has the opposite sign from the rho for voltage at a
short and open. And sure enough, the reflected current is 180 degrees
out of phase with the reflected voltage.

And let's see about a few values. How about?

0.5 + j0.5 Vs -(0.5 + j0.5) or -0.5 + j0.5 Vs 0.5 - j0.5

Sure enough. The first value is at 45 degrees and the second value
is at 180+45 degrees. The third value is at 135 degrees and the
fourth value is at 180+135 degrees.

And if Walter C. Johnson is worthy of the respect he receives
here, he has certainly never said it is.

Refer to Fig. 1.13 on page 20. The voltage reflection coefficient
at the open end is +1. The current reflection coefficient at the
open end is -1.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:
But rho on the coax is zero.

Coax doesn't have a 'rho' - unless it's broken or damaged coax.

The impedance at '+' is 50 ohms.

By virtue of reflection from the 50 ohm load.

rho = (50-50)/(50+50) = sqrt(0/Pfwd) = 0
at point '+'.

The same is true at points to the left of point '+' as well. So what?

And that 50 ohms is a V/I ratio which, I assume, you would
agree cannot cause a voltage.

As far as I know, V/I ratios don't "cause" anything.

73, Jim AC6XG

Cecil Moore wrote:

Jim Kelley wrote:
The nature of things a point '+' are undefined,

Nope, they are not.

There can exist no real point where the characteristic line impedance is
both 50 ohms and 150 ohms.

There are no reflections on the 50 ohm feedline because it "sees" 50
ohms at point '+'.

Oh my gawd, somebody has brainwashed our Cecil! Maybe the pod people
have invaded Texas! :-)

What about cancelled reflected waves, destructive interference and all
that?

73, Jim AC6XG

Jim, AC6XG wrote:
"As far as I know, V/I ratios don`t cause anything."

As symbols for load impedance, V/I ratios cause reflections when they
differ from the r-f transmission line surge impedance which feeds them.

Best rergards, Richard Harrison, KB5WZI

Richard Harrison wrote:

Jim, AC6XG wrote:
"As far as I know, V/I ratios don`t cause anything."

As symbols for load impedance, V/I ratios cause reflections when they
differ from the r-f transmission line surge impedance which feeds them.

Best rergards, Richard Harrison, KB5WZI

Hi Richard,

Would it be fair to say then that you don't agree that only physical
boundaries cause reflections?

73, Jim AC6XG

Jim Kelley wrote:
As far as I know, V/I ratios don't "cause" anything.

They sometimes cause 'rho' which then becomes an end
result and not the cause of anything. At a two-port
network with reflections, rho usually cannot be calculated
from the physical impedances involved. The moral is be
careful about saying that rho causes anything. Rho may
be only the end result of everything. The s11 reflection
coefficient doesn't suffer from that characteristic.
--
73, Cecil http://www.qsl.net/w5dxp



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Jim Kelley wrote:

There can exist no real point where the characteristic line impedance is
both 50 ohms and 150 ohms.

I agree, and using the same logic, there also can be no such thing as real
steady-state conditions. That doesn't keep us from using it as a real concept.

+----------------
|
---------+
50 ohms 150 ohms
---------+
|
+----------------

You can draw a vertical line through the transition point. That vertical
line has zero width. Ergo, the 50 ohm to 150 ohm transition requires zero
length, conceptually, of course. But you knew that already. While you were
at it, why didn't you point out that there cannot exist a real characteristic
impedance exactly equal to 50 ohms or 150 ohms?

What about cancelled reflected waves, destructive interference and all
that?

I should have said there are no net reflections on the 50 ohm feedline.
There are two component reflections at point '+' that cancel each other
as illustrated by the s-parameter equation, b1 = s11*a1 + s12*a2 = 0
s11 is a reflection coefficient for a1 and s12 is a transmission coefficient
for a2, the voltage reflected from the load. Those two components have to
cancel for the b1 net reflections to equal zero.
--
73, Cecil http://www.qsl.net/w5dxp



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Richard Harrison wrote:

Jim, AC6XG wrote:
"As far as I know, V/I ratios don`t cause anything."

As symbols for load impedance, V/I ratios cause reflections when they
differ from the r-f transmission line surge impedance which feeds them.

So can a V/I ratio cause reflections in the absence of a physical
impedance? For instance:

Source---50 ohm feedline---+---150 ohm feedline---450 ohms

The V/I ratio at point '+' is equal to 1350 ohms and the reflections
on the coax are the same as if a 1350 ohm resistor existed at point '+'.
So does the V/I=1350 ohms ratio cause the reflections at point 'x'?
Or is the reflected voltage, b1, equal to s11*a1 + s12*a2?
--
73, Cecil http://www.qsl.net/w5dxp



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Jim, AC6XG wrote:
"Would it be fair to say that you don`t agree that only physical
boundaries cause reflections?"

A transmission line may be terminated in its surge impedance, yet
somewhere along the line suffer a discontinuity producing an impedance
irregularity and subsequent rteflection.

The line may be matched to its final load, yet have reflections on the
transmitter side of the irregularity. Terman illustrates this on page
118 of his 1955 edition.

Terman suggests several reflection possibilities::
"sharp bends, insulating supports, resistive joints, coupled circuits
and extraneous objects" as typical irregularities.

I am not persuaded that reflection is caused by anything other than a
physical discontinuity, but Terman includes "coupled circuits" in his
list, and I think this indicates a physical discontinuity may be
referred to a reflection point from elsewhere.

Best regards, Richard Harrison, KB5WZI