Roy Lewallen, W7EL wrote:
"Those traveling waves, and hence their sum, cannot cause a reflection
of other waves, or alter those waves in any way."

Let`s reason together on the situation in a quarter-wavelength
short-circuited transmission line stub. I maintain it has a hard short
on its far end and a high impedance on its near end.

A high impedance means just what it says. You can put a high voltage on
it and the resulting current is small.

Reflection from a short-circuit results in a 180-degree voltage phase
reversal at the short.

A round-trip on a 1/4-wave stub produces an additional 180-degree phase
reversal.

This means thats volts returning to the open-circuit end of the stub are
about of the same phase and magnitude as when they started out.

Nearly identical voltages appear at the same pair of terminals from
opposite directions. Significant current flows in either direction? I
think it does not.

Where voltage causes insignificant current flow, we have a high
impedance.

That is why King, Mimno, and Wing on page 30 of Transmission Lines,
Antennas and Wave Guides say:
"Since the input impedance of a short-circuited quarter-wavelength
section of transmission line is a very high resistance, short-circuited
stubs may be used to support the line."

Best regards, Richard Harrison, KB5WZI

On Sat, 14 Apr 2007 10:59:17 -0500, (Richard
Harrison) wrote:

That is why King, Mimno, and Wing on page 30 of Transmission Lines,
Antennas and Wave Guides say:
"Since the input impedance of a short-circuited quarter-wavelength
section of transmission line is a very high resistance, short-circuited
stubs may be used to support the line."

Hi Richard,

One example begs another, symmetrical one; and here is the example of
my asking you why you would struggle through building an open line
reflected as a short (the quarterwave section).

For argument's sake, let's amend KM&W to say:
"Since the input impedance of a open-circuited quarter-wavelength
section of transmission line is a very low resistance, open-circuited
stubs connected in series with the line elements may be used to
support the line."

This statement is equivalent in the transmission line mechanic's tool
kit, certainly. However, it brings with it the difficulties of
maintaining a "good" open (especially when that open is in close
proximity to earth, or at least to the press of humanity). Casting
those problems into the line obviously presents loss issues of a poor
virtual short.

This conforms to my experience with many plumbing designs on the
microwave bench. The absence of open quarterwave sections was nearly
universal due to the problems of their implementation. It takes
little effort to realize why KM&W did not offer this alternative. If
you browse Terman, he also avoids such constructions.

73's
Richard Clark, KB7QHC

On Apr 13, 11:49 pm, Ian White GM3SEK wrote:
Roy Lewallen wrote:

Please let me emphasize again that not I or anyone else who has posted
is disputing the validity of your matching methods or the utility of
the "virtual short" concept. The only disagreement is in the contention
that the "virtual short" actually *effects* reflections rather than
being solely a consequence of them.

The key word there is "utility" - the virtual short/open concept is
*useful* as a short-cut in our thinking. But concepts are only useful if
they help us to think more clearly about physical reality; and
short-cuts are dangerous if they don't reliably bring us back onto the
main track.
....

Indeed. I was thinking about this in terms of short-cuts before
reading Ian's post. What if you take a short-cut and it just takes
you off into the woods? I'm not sure my posting about this made it
into the thread in an intelligible way. (I fear Google may have sent
it off on a "short-cut.")

The gist of it was that, although there are examples where considering
points an even number of half-waves from a short as being shorts
themselves work fine, there are plenty of counter examples too. I
fear that people new to the use of stubs will be lulled into a false
sense of security using that concept, when indeed it fails miserably
at times. Especially in this age of computers and readily available
programs to deal with lines, INCLUDING their loss, why would I use a
concept that may take me on a short-cut that turns out to be the long
way around?

What IS useful to me about the concept is NOT the calculation of the
performance of a particular network of stubs, but rather in coming up
with the trial design to test with full calculations. My example was
the use of two stubs to give me a null on one frequency and pass
another frequency; I can get a null by putting a "virtual short" at
that frequency, and that's a line that's a half wave long on that
frequency, shorted at the other end. But on a slightly lower
frequency, it looks capacitive, so I can put another stub that's
inductive in parallel with it to create an open circuit at the
frequency I want to let pass. THEN I pull out the calculations with
line attenuation included, and discover that in some situations it
works fine, and in others, the performance is terrible.

It's a useful visualization tool and design aid; it's a poor analysis
tool at best. At worst, it will lull you into building something that
just won't work, wasting time and resources.

Cheers,
Tom

I'm not sure I understand the point you're trying to make. Nothing I've
said disputes in any way that the input impedance of a shorted quarter
wave stub is high. I'm quite able to make transmission line calculations
and arrive at correct results.

You're certainly correct that there's very little net current at the
open end of the stub. Yet there are waves traveling in both directions
right through that point. Don't believe it? Then check the current
anywhere else along the stub. How did it get there without going through
the "open" at the input end?

Roy Lewallen, W7EL

Richard Harrison wrote:
Roy Lewallen, W7EL wrote:
"Those traveling waves, and hence their sum, cannot cause a reflection
of other waves, or alter those waves in any way."

Let`s reason together on the situation in a quarter-wavelength
short-circuited transmission line stub. I maintain it has a hard short
on its far end and a high impedance on its near end.

A high impedance means just what it says. You can put a high voltage on
it and the resulting current is small.

Reflection from a short-circuit results in a 180-degree voltage phase
reversal at the short.

A round-trip on a 1/4-wave stub produces an additional 180-degree phase
reversal.

This means thats volts returning to the open-circuit end of the stub are
about of the same phase and magnitude as when they started out.

Nearly identical voltages appear at the same pair of terminals from
opposite directions. Significant current flows in either direction? I
think it does not.

Where voltage causes insignificant current flow, we have a high
impedance.

That is why King, Mimno, and Wing on page 30 of Transmission Lines,
Antennas and Wave Guides say:
"Since the input impedance of a short-circuited quarter-wavelength
section of transmission line is a very high resistance, short-circuited
stubs may be used to support the line."

Best regards, Richard Harrison, KB5WZI

Roy Lewallen wrote in
:

anywhere else along the stub. How did it get there without going through
the "open" at the input end?

Ah, a total re-reflector!

Owen

Richard Clark wrote:
"This conforms to my experience with many plumbing designs on the
microwave bench."

Good. Then, Richard Clark must also be familiar with the grooved
circular flange used in conjunction with a smooth flange to join
waveguide segments. This groove isn`t just used to hold a neoprene
gasket. It is also used as an electrical choke to keep the microwaves
within the pipe. It is approximately a 1/4-wave choke and its high
impedance across its open-circuit helps foil the wave escape. If virtual
open-circuits didn`t work, the "choke-flange wouldn`t work either.

Best regards, Richard Harrison, KB5WZI

On Fri, 13 Apr 2007 19:10:18 -0700, Roy Lewallen wrote:

Walter Maxwell wrote:

Thank you, Roy, I appreciate your comments, as always. However, I knew that you have always considered that
virtual opens and shorts cannot cause reflections, and I was hoping my discussion would have persuaded you
otherwise.
snip

I'd think that this diode-like property of virtual shorts would be a
major clue that they're not real, but a mathematical convenience. The
virtual short is a point where the sum of the voltages of all waves,
forward and reflected, add to zero. If this condition causes waves to
reflect when struck from one direction, what possible physical
explanation could there be for it to do absolutely nothing to waves
traveling the other way?

So I repeat the question: If a virtual short circuit cannot cause reflections, then what causes the reflection
at the stub point?

My answer is this: There is no total re-reflection at the stub point. It
only looks that way.

As you've observed, the waves (traveling in one direction, anyway)
behave just as though there was such a re-reflection. But the waves
actually are reflecting partially or totally from the end of the stub
and other more distant points of impedance discontinuity, not from a
"virtual short". The sum of the forward wave and those reflections add
up to zero at the stub point to create the "virtual short", and to
create waves which look just like they're totally reflecting from the
stub point. This has some parallels to a "virtual ground" at an op amp
input. From the outside world, the point looks just like ground. But it
isn't really. The current you put into that junction isn't going to
ground, but back around to the op amp output. Turn off the op amp and
the "virtual ground" disappears. Likewise, waves arriving at the virtual
short look just like they're reflecting from it. But they aren't.
They're going right on by -- from either direction --, not having any
idea that there's a "virtual short" there -- that is, not having any
idea what the values or sum of other waves are at that point. They go
right on by, reflect from more distant discontinuities, and the sum of
those reflections arrives at the virtual short with the same phase and
amplitude the wave would have if it had actually reflected from the
virtual short. Like with the op amp, you can "turn off" the virtual
short by altering those distant reflection points such as the stub end.

Please let me emphasize again that not I or anyone else who has posted
is disputing the validity of your matching methods or the utility of the
"virtual short" concept. The only disagreement is in the contention that
the "virtual short" actually *effects* reflections rather than being
solely a consequence of them.
snip
I maintain that such an
example can't be found, because in fact reflection takes place only at
physical discontinuities and not at "virtual shorts". Waves in a linear
medium simply don't reflect from or otherwise affect each other. I'm not
saying that you can't apply the analytical concept of "virtual shorts"
to arrive at the same, valid, result. Or that the "virtual short"
approach won't be easier. But I am saying that it's not necessary in
order to fully analyze any transmission line problem, simply because
it's not real. Can you come up with such an example?

Roy
Hi R oy,

Consider my two explanations, or definitions of what I consider a virtual short--perhaps it should have a
different name, because of course 'virtual' implies non-existence. The short circuit evident at the input of
the two line examples I presented---do you agree that short circuits appear at the input of the two lines? If
so, what would you call them?

Roy, I'd like for you to take another, but perhaps closer look at the summarizing of the reflection
coefficients below. I originally typed in the wrong value for the magnitude of the resultant coefficients.
With the corrected magnitudes in place, the two paragraphs following the summarization now make more sense,
because the short circuit established at the stub point leads correctly to the wave action that occurs there.

Summarizing reflection coefficient values at stub point with stub in place:
Line coefficients: voltage 0.5 at +120°, current -60° (y = 1 + j1.1547)
Stub coefficients: voltage 0.5 at -120°, current +60° (y = 1 - j1.1547)
Resultant coefficients: voltage 0.5 at 180°, current 0.5 at 0° WRONG
Resultant coefficients: voltage 1.0 at 180°, current 1.0 at 0° CORRECT

Repeating from my original post for emphasis:
These two resultant reflection coefficients resulting from the interference between the load-reflected wave at
the stub point and the reflected wave produced by the stub define a virtual short circuit established at the
stub point.

The following paragraph shows how the phases of the reflected waves become in phase with the source waves so
that the reflected waves add directly to the source waves, establishing the forward power, which we know
exceed the source power when the reflected power is re-reflected. The same concept applies to antena tuners.

Again repeating for emphasis:
Let's now consider what occurs when a wave encounters a short circuit. We know that the voltage wave
encounters a phase change of 180°, while the current wave encounters zero change in phase. Note that the
resultant voltage is at 180°, so the voltage phase changes to 0° on reflection at the short circuit, and is
now in phase with the source voltage wave. In addition, the resultant current is already at 0°, and because
the current phase does not change on reflection at the short circuit, it remains at 0° and in phase with
source current wave. Consequently, the reflected waves add in phase with the source waves, thus increasing the
forward power in the line section between the stub and the load.

Keep in mind that the short at the stub point is a one-way short, diode like, as you say, because in the
forward direction the voltage reflection coefficient rho is 0.0 at 0°, while in the reverse direction, rho at
the stub point is 1.0 at 180°, which is why it's a one-way short.

You say that no total re-reflection occurs at the stub point. However, with a perfect match the power rearward
of the stub is zero, and all the source power goes to the load in the forward direction. Is that not total
reflection? Using the numbers of my bench experiment, assuming a source power of 1 watt, and with the
magnitude rho of 0.04, power going rearward of the stub is 0.0016 w, while the power absorbed by the load is
0.9984 w, the sum of which is 1 w. The SWR seen by the source is 1.083:1, and the return loss in this
experiment is 27.96 dB, while the power lost to the load is 0.0070 dB. From a ham's practical viewpoint the
reflected power is totally re-reflected.

In my example using the 49° stub the capacitive reactance it established at its input is Xc = -57.52 ohms.
Thus its inductive susceptance B = 0.0174 mhos, which cancels the capacitive line susceptance B = -0.0174 mhos
appearing at the stub point.

My point is that the 49° stub can be replaced with a lumped capacitance Xc = -57.52 ohms directly on the line
with the same results as with the stub--with the same reflection coefficients. In this case one cannot say
that the re-reflection results from the physical open circuit terminating the stub line.

Various posters have termed my approach as a 'short cut'. I disagree. I prefer to consider it as the wave
analysis to the stub-matching procedure, in contrast to the traditional method of simply saying that the stub
reactance cancels the line reactance at the point on the line where the line resistance R = Zo. In my mind the
wave analysis presents a more detailed view of what's actually happening to the pertinent waves while the
impedance match is being established.

Walt

Walter Maxwell wrote in
:

Summarizing reflection coefficient values at stub point with stub in
place: Line coefficients: voltage 0.5 at +120°, current -60° (y = 1
+ j1.1547) Stub coefficients: voltage 0.5 at -120°, current +60° (y
= 1 - j1.1547) Resultant coefficients: voltage 0.5 at 180°, current
0.5 at 0° WRONG Resultant coefficients: voltage 1.0 at 180°, current
1.0 at 0° CORRECT

Walt,

Though admittance or impedance at a point on the mismatched line are
calculated from the underlying Zo and the reflection coefficient
corrected for line loss, they are easier to work in than the raw
reflection coefficient.

It is easier to explain why the stub is located at a position where Yn'=1
+jB than where Gamma=0.5120 (assuming lossless line). It is relatively
obvious that where Yn'=1+jB, a shunt reactance of -jB from a s/c or o/c
stub will leave Yn=1 which is the matched condition.

Re your worked solution (above), I agree that the normalised admittance
looking into 30deg of line with load 16.667+j0 is about 1-j1.1547 (not
the different sign).

I make the normalised admittance looking into the stub about 0+j1.15 (and
the reflection coefficient about 0.5-98, how do you get 1+j1.15?

The addition of the two normalised admittances 1-j1.15 + 0+j1.15 gives 1
+j0 which is the matched condition.

The design is correct, the stub results in a match at the stub connection
point (irrespective of what is connected on the source side of the
point), but I can't understand your maths above (allowing for the sign
error that I think you have made).

Is the reflection coefficient explanation a clearer explanation than
using admittances?

Owen

BTW, my line loss calculator solutions (http://www.vk1od.net/tl/tllc.php)
for Belden 8262 RG58 (you said RG53, but you probably meant RG58) a

(Note some symbols arent supported by plain ascii and appear as '?'.)

Load to Stub connection:

Parameters
Transmission Line Belden 8262 (RG-58C/U)
Code B8262
Data source Belden
Frequency 16.000 MHz
Length 1.030 metres
Zload 16.67+j0.00 ?
Yload 0.059999+j0.000000 ?
Results
Zo 50.00-j0.54 ?
Velocity Factor 0.660
Length 29.97 ?, 0.083 ?
Line Loss (matched) 0.059 dB
Line Loss 0.149 dB
Efficiency 96.63%
Zin 22.12+j24.55 ?
Yin 0.020258-j0.022480 ?
Gamma, rho, theta, VSWR (source end) -2.44e-1+j4.29e-1, 0.493,
119.6?, 2.950
Gamma, rho, theta, VSWR (load end) -5.00e-1+j4.03e-3, 0.500, 179.5?,
3.000
? 6.54e-3+j5.08e-1
k1, k2 1.30e-5, 2.95e-10
Loss model source data lowest frequency 1.000 MHz
Correlation coefficient (r) 0.999884

Stub:

Parameters
Transmission Line Belden 8262 (RG-58C/U)
Code B8262
Data source Belden
Frequency 16.000 MHz
Length 1.685 metres
Zload 100000000.00+j0.00 ?
Yload 0.000000+j0.000000 ?
Results
Zo 50.00-j0.54 ?
Velocity Factor 0.660
Length 49.02 ?, 0.136 ?
Line Loss (matched) 0.096 dB
Line Loss 40.574 dB
Efficiency 0.01%
Zin 0.50-j43.44 ?
Yin 0.000265+j0.023019 ?
Gamma, rho, theta, VSWR (source end) -1.37e-1-j9.69e-1, 0.978, -
98.0?, 90.720
Gamma, rho, theta, VSWR (load end) 1.00e+0+j1.07e-8, 1.000, 0.0?, inf
? 6.54e-3+j5.08e-1
k1, k2 1.30e-5, 2.95e-10
Loss model source data lowest frequency 1.000 MHz
Correlation coefficient (r) 0.999884

Walt, before digging into your recent posting, I'd really like to get
one issue settled. I think it would be helpful in our discussion. The
issue is:

Can you find even one example of any transmission line problem which
cannot be solved, or a complete analysis done, without making the
assumption that waves reflect from a "virtual short" or "virtual open"?
That is, any example where such an assumption is necessary in order to
find the currents, voltages, and impedances, and the magnitude and phase
of forward and reverse voltage and current waves?

Roy Lewallen, W7EL

On Sat, 14 Apr 2007 22:31:37 GMT, Owen Duffy wrote:

Walter Maxwell wrote in
:

Summarizing reflection coefficient values at stub point with stub in
place:
Line coefficients: voltage 0.5 at +120°, current -60° (y = 1 + j1.1547)
Stub coefficients: voltage 0.5 at -120°, current +60° (y = 1 - j1.1547)
Resultant coefficients: voltage 0.5 at 180°, current 0.5 at 0° WRONG
Resultant coefficients: voltage 1.0 at 180°, current 1.0 at 0° CORRECT

Walt,

Though admittance or impedance at a point on the mismatched line are
calculated from the underlying Zo and the reflection coefficient
corrected for line loss, they are easier to work in than the raw
reflection coefficient.

Depends on the instrumentation available for obtaining the raw data.

It is easier to explain why the stub is located at a position where Yn'=1
+jB than where Gamma=0.5120 (assuming lossless line). It is relatively
obvious that where Yn'=1+jB, a shunt reactance of -jB from a s/c or o/c
stub will leave Yn=1 which is the matched condition.

Re your worked solution (above), I agree that the normalised admittance
looking into 30deg of line with load 16.667+j0 is about 1-j1.1547 (not
the different sign).

Yes Owen, you're right. I added the y values at the last moment, and didn't catch the errors. Both the line
and stub signs are reversed. Sorry 'bout that.

I make the normalised admittance looking into the stub about 0+j1.15 (and
the reflection coefficient about 0.5-98, how do you get 1+j1.15?

Normalized y looking into the stub directly is y = 0 + 1.1547, but looking at the stub while on the line at
the 30° point is y = 1 + 1.1547. To view the stub separately on the line the line is terminated in 50 ohms,
because the real component of the line impedance at the match point is 50 ohms.

The addition of the two normalised admittances 1-j1.15 + 0+j1.15 gives 1
+j0 which is the matched condition.

Of course.

The design is correct, the stub results in a match at the stub connection
point (irrespective of what is connected on the source side of the
point), but I can't understand your maths above (allowing for the sign
error that I think you have made).

The resultant coefficients are obtained by simply adding the voltage coefficients and the current
coefficients, as in the adding of the line and stub admittances.

Is the reflection coefficient explanation a clearer explanation than
using admittances?

Not at all, Owen, but as I said in the original post, the instrument I acquired early in my time with the RCA
antenna lab was the PRD-219 reflectometer. At that time I considered it the best instrument for measuring
transmission line circuitry at VHF, and because it delivered readings in reflection coefficient I became
somewhat more efficient in my thinking process using that mode.

Walt





















Owen

BTW, my line loss calculator solutions (http://www.vk1od.net/tl/tllc.php)
for Belden 8262 RG58 (you said RG53, but you probably meant RG58) a

(Note some symbols arent supported by plain ascii and appear as '?'.)

Load to Stub connection:

Parameters
Transmission Line Belden 8262 (RG-58C/U)
Code B8262
Data source Belden
Frequency 16.000 MHz
Length 1.030 metres
Zload 16.67+j0.00 ?
Yload 0.059999+j0.000000 ?
Results
Zo 50.00-j0.54 ?
Velocity Factor 0.660
Length 29.97 ?, 0.083 ?
Line Loss (matched) 0.059 dB
Line Loss 0.149 dB
Efficiency 96.63%
Zin 22.12+j24.55 ?
Yin 0.020258-j0.022480 ?
Gamma, rho, theta, VSWR (source end) -2.44e-1+j4.29e-1, 0.493,
119.6?, 2.950
Gamma, rho, theta, VSWR (load end) -5.00e-1+j4.03e-3, 0.500, 179.5?,
3.000
? 6.54e-3+j5.08e-1
k1, k2 1.30e-5, 2.95e-10
Loss model source data lowest frequency 1.000 MHz
Correlation coefficient (r) 0.999884

Stub:

Parameters
Transmission Line Belden 8262 (RG-58C/U)
Code B8262
Data source Belden
Frequency 16.000 MHz
Length 1.685 metres
Zload 100000000.00+j0.00 ?
Yload 0.000000+j0.000000 ?
Results
Zo 50.00-j0.54 ?
Velocity Factor 0.660
Length 49.02 ?, 0.136 ?
Line Loss (matched) 0.096 dB
Line Loss 40.574 dB
Efficiency 0.01%
Zin 0.50-j43.44 ?
Yin 0.000265+j0.023019 ?
Gamma, rho, theta, VSWR (source end) -1.37e-1-j9.69e-1, 0.978, -
98.0?, 90.720
Gamma, rho, theta, VSWR (load end) 1.00e+0+j1.07e-8, 1.000, 0.0?, inf
? 6.54e-3+j5.08e-1
k1, k2 1.30e-5, 2.95e-10
Loss model source data lowest frequency 1.000 MHz
Correlation coefficient (r) 0.999884