Hello Walt,

On Monday, 13 June 2011 05:05:50 UTC+10, walt wrote:
I have a question for W5DXP, KB7QHC, W7ITM and K0TAR relating to
transmission lines.This is not a trick question. This is a straight-
forward question for which I 'm dead serious. The answer will involve
a resulting reflection coefficient.

Assume a 50-ohm line terminated in a purely resistive 150-ohm load,
yielding a 3:1 mismatch with a voltage reflection coefficient V� equal
to 0.5 at 0°. We want to place a stub on the line at the position
relating to the unit-resistance circle on the Smith Chart. For a 3:1
mismatch this position is exactly 30° rearward from the load, and has
a voltage reflection coefficient V� equal to 0.5 at -60°. The
normalized line resistance at this point is 1.1547 chart ohms, or

This is wrong...

57.735 ohms on the 50-ohm line, for a line impedance of 50 - j57.735
ohms.

as your know!


Placing a short-circuited stub having an inductive reactance of
+j57.735 ohms on the line cancels the line reactance, establishing a
new line impedance at this matching point of 50 + j0. However, a lot
more is involved than just canceling the reactances to establish the
impedance match.

This is vintage "Reflections" stuff Walt, where you try to explain what is happening with one foot in the time domain and the other in the frequency domain.

To begin, the stub generates a new, secondary,
canceling reflection that cancels the primary reflected wave generated
by the mismatched line termination.

I can see you heading to re-re-reflections etc.

Thing is that in the steady state (and your Smith chart works ONLY in the steady state), the reflection on the first mentioned (lossless) line (with the 150 ohm load) is determined entirely by its characteristic impedance and the load impedance. The ratio of V/I looking into that line section is given by those parameters and line electrical length.

Sounds boring, but the reflection on the second mentioned (lossless) line (with s/c termination) is determined entirely by its characteristic impedance and the load impedance. The ratio of V/I looking into that line section is given by those parameters and line electrical length.

The ratio of V/I at the node where both lines are joined is given for a shunt stub by the inverse of the sum of the inverses of V/I for each.

You can think about reflections from on line entering the other, and energy sloshing around, but the Smith chart is not going to help you because is is simply a graphic computer for finding Gamma, calculating Gamma along the line, adding B, G etc by following lines of constant G, B etc.

If we tried to explain the common microwave three screw tuner using re-reflections, virtual open circuits to rearward traveling waves, conjugate mirrors etc, total re-reflection etc, we would complicate something that is easily explained using a Smith chart (and the underlying maths - The Telegrapher's Equation) and quite conventional AC circuit theory.

In fact, a three stub tuner with practical (ie lossy) lines can be explained fairly easily using a Smith chart, even if a little fiddly iteration is needed to deal with lossy elements.

If the total re-reflection etc concepts are hard to translate from mental models to mathematical expressions for simple scenarios, they become near impossible for more complicated topologies and 'real' components.


Owen

On Jun 17, 9:38Â*pm, Owen Duffy wrote:
Hello Walt,

On Monday, 13 June 2011 05:05:50 UTC+10, walt Â*wrote:
I have a question for W5DXP, KB7QHC, W7ITM and K0TAR relating to
transmission lines.This is not a trick question. Â*This is a straight-
forward question for which I 'm dead serious. Â*The answer will involve
a resulting reflection coefficient.

Assume a 50-ohm line terminated in a purely resistive 150-ohm load,
yielding a 3:1 mismatch with a voltage reflection coefficient V� equal
to 0.5 at 0°. We want to place a stub on the line at the position
relating to the unit-resistance circle on the Smith Chart. For a 3:1
mismatch this position is exactly 30° rearward from the load, and has
a voltage reflection coefficient V� equal to 0.5 at -60°. The
normalized line resistance at this point is 1.1547 chart ohms, or

This is wrong...

Sorry Owen, my data above that you say is wrong is not wrong. I'm
absolutely astounded that you would say so. You have misconstrued my
data to make such an assertion. I calculated the data and then plotted
it on the Smith Chart, which verified the data as correct.

So that other readers will understand how I calculated the data I'll
begin with the 150-ohm load terminating the 50-ohm line, yielding a
normalized impedance of Z = 3 + j0. Moving along the line toward the
source the line impedance changes, arriving at a point where
normalized line impedance is 1 - jX. The reactance X is determined by
the SWR on the line which is 3:1. jX = 1/(sqrt SWR/ SWR - 1), which
yields jX = -1.1547 ohms. This value of jX is verified by drawing a
radial of 3:1 SWR to intersect the unit-resistance circle on the Smith
Chart. By inserting an inductive reactance jX = +1.1547 ohms in series
with the line at this point cancels the line reactance, making the
normalized line impedance 1 + j0, where it was previously 1 -
j1.1547.

Inserting the inductive reactance at that point causes a discontinuity
in the line that generates a second reflection, a canceling reflection
that cancels the primary reflection, thus establishing an impedance
match at that point.

Now where is the impedance-matching point on the line? First we find
the voltage reflection coefficient of the normalized impedance at that
point, which is rho = (Zo - Zx)/(Zo + Zx) = 0.25 - j0.43301, which
when converted to polar form, rho = 0.50 @ -60°. Extending the radius
through the 3:1 point on the unity-resistance circle on to the
periphery of the Smith Chart, it lands on the -60° point, which again
confirms the validity of my data that you assert is wrong.

This is vintage "Reflections" stuff Walt, where you try to explain what is happening with one foot in the time domain and the other in the frequency domain.

To begin, the stub generates a new, secondary,
canceling reflection that cancels the primary reflected wave generated
by the mismatched line termination.

I can see you heading to re-re-reflections etc.

Owen, you are a brilliant and an excellent lecturer. But I don't
believe I need a lecture on impedance matching with transmission
lines. So just on one point, where have I inferred that I was not
working in the steady state?

Walt, W2DU

Thing is that in the steady state (and your Smith chart works ONLY in the steady state), the reflection on the first mentioned (lossless) line (with the 150 ohm load) is determined entirely by its characteristic impedance and the load impedance. The ratio of V/I looking into that line section is given by those parameters and line electrical length.

Sounds boring, but the reflection on the second mentioned (lossless) line (with s/c termination) is determined entirely by its characteristic impedance and the load impedance. The ratio of V/I looking into that line section is given by those parameters and line electrical length.

The ratio of V/I at the node where both lines are joined is given for a shunt stub by the inverse of the sum of the inverses of V/I for each.

You can think about reflections from on line entering the other, and energy sloshing around, but the Smith chart is not going to help you because is is simply a graphic computer for finding Gamma, calculating Gamma along the line, adding B, G etc by following lines of constant G, B etc.

If we tried to explain the common microwave three screw tuner using re-reflections, virtual open circuits to rearward traveling waves, conjugate mirrors etc, total re-reflection etc, we would complicate something that is easily explained using a Smith chart (and the underlying maths - The Telegrapher's Equation) and quite conventional AC circuit theory.

In fact, a three stub tuner with practical (ie lossy) lines can be explained fairly easily using a Smith chart, even if a little fiddly iteration is needed to deal with lossy elements.

If the total re-reflection etc concepts are hard to translate from mental models to mathematical expressions for simple scenarios, they become near impossible for more complicated topologies and 'real' components.

Owen

On Jun 18, 11:49Â*am, walt wrote:
On Jun 17, 9:38Â*pm, Owen Duffy wrote:









Hello Walt,

On Monday, 13 June 2011 05:05:50 UTC+10, walt Â*wrote:
I have a question for W5DXP, KB7QHC, W7ITM and K0TAR relating to
transmission lines.This is not a trick question. Â*This is a straight-
forward question for which I 'm dead serious. Â*The answer will involve
a resulting reflection coefficient.

Assume a 50-ohm line terminated in a purely resistive 150-ohm load,
yielding a 3:1 mismatch with a voltage reflection coefficient V� equal
to 0.5 at 0°. We want to place a stub on the line at the position
relating to the unit-resistance circle on the Smith Chart. For a 3:1
mismatch this position is exactly 30° rearward from the load, and has
a voltage reflection coefficient V� equal to 0.5 at -60°. The
normalized line resistance at this point is 1.1547 chart ohms, or

This is wrong...

Sorry Owen, my data above that you say is wrong is not wrong. I'm
absolutely astounded that you would say so. You have misconstrued my
data to make such an assertion. I calculated the data and then plotted
it on the Smith Chart, which verified the data as correct.

So that other readers will understand how I calculated the data I'll
begin with the 150-ohm load terminating the 50-ohm line, yielding a
normalized impedance of Z = 3 + j0. Moving along the line toward the
source the line impedance changes, arriving at a point where
normalized line impedance is 1 - jX. The reactance X is determined by
the SWR on the line which is 3:1. jX = 1/(sqrt Â*SWR/ SWR - 1), which
yields jX = -1.1547 ohms. This value of jX is verified by drawing a
radial of 3:1 SWR to intersect the unit-resistance circle on the Smith
Chart. By inserting an inductive reactance jX = +1.1547 ohms in series
with the line at this point cancels the line reactance, making the
normalized line impedance 1 + Â*j0, where it was previously 1 -
j1.1547.

Inserting the inductive reactance at that point causes a discontinuity
in the line that generates a second reflection, a canceling reflection
that cancels the primary reflection, thus establishing an impedance
match at that point.

Now where is the impedance-matching point on the line? First we find
the voltage reflection coefficient of the normalized impedance at that
point, which is rho = (Zo - Zx)/(Zo + Zx) = 0.25 - j0.43301, which
when converted to polar form, rho = 0.50 @ -60°. Extending the radius
through the 3:1 point on the unity-resistance circle on to the
periphery of the Smith Chart, it lands on the -60° point, which again
confirms the validity of my data that you assert is wrong.

This is vintage "Reflections" stuff Walt, where you try to explain what is happening with one foot in the time domain and the other in the frequency domain.

To begin, the stub generates a new, secondary,
canceling reflection that cancels the primary reflected wave generated
by the mismatched line termination.

I can see you heading to re-re-reflections etc.

Owen, you are a brilliant and an excellent lecturer. But I don't
believe I need a lecture on impedance matching with transmission
lines. So just on one point, where have I inferred that I was not
working in the steady state?

Walt, W2DU







Thing is that in the steady state (and your Smith chart works ONLY in the steady state), the reflection on the first mentioned (lossless) line (with the 150 ohm load) is determined entirely by its characteristic impedance and the load impedance. The ratio of V/I looking into that line section is given by those parameters and line electrical length.

Sounds boring, but the reflection on the second mentioned (lossless) line (with s/c termination) is determined entirely by its characteristic impedance and the load impedance. The ratio of V/I looking into that line section is given by those parameters and line electrical length.

The ratio of V/I at the node where both lines are joined is given for a shunt stub by the inverse of the sum of the inverses of V/I for each.

You can think about reflections from on line entering the other, and energy sloshing around, but the Smith chart is not going to help you because is is simply a graphic computer for finding Gamma, calculating Gamma along the line, adding B, G etc by following lines of constant G, B etc.

If we tried to explain the common microwave three screw tuner using re-reflections, virtual open circuits to rearward traveling waves, conjugate mirrors etc, total re-reflection etc, we would complicate something that is easily explained using a Smith chart (and the underlying maths - The Telegrapher's Equation) and quite conventional AC circuit theory.

In fact, a three stub tuner with practical (ie lossy) lines can be explained fairly easily using a Smith chart, even if a little fiddly iteration is needed to deal with lossy elements.

If the total re-reflection etc concepts are hard to translate from mental models to mathematical expressions for simple scenarios, they become near impossible for more complicated topologies and 'real' components.

Owen

I forgot to mention in the above post that the -60° on the Smith Chart
periphery is the angle of the reflection coefficient, which is twice
the value of the electrical length. Therefore, the matching point is
30° from the load toward the source.

Walt

On Jun 18, 10:49*am, walt wrote:
Now where is the impedance-matching point on the line?

By definition, the Z0-match point is where the reflections toward the
source are first eliminated, i.e. Vtot/Itot=Z0, which is on the source
side of the inductive stub. If reflections were eliminated anywhere
else (closer to the load) then there would be no need for that section
of feedline. An energy analysis is possible but would no doubt be
messy. For instance, what is the SWR at the stub short? Where is the
ground reference at the stub short? Since the series stub is a certain
number of degrees long, the two currents at the mouth of the stub
cannot possibly be 100% differential, i.e. common-mode currents exist
on the series stub. Since there is a phase shift in the series stub
that doesn't exist in the main coax braid, common-mode current must be
flowing in the rest of the system. Electrically, how long is the
series stub in degrees? Does an impedance discontinuity exist at the
short at the end of a series stub like it does at the end of a shorted
stub - or not? What is the characteristic impedance of a series stub?
(It is probably not the Z0 of the transmission line being used.)

There is obviously a multiple-port impedance discontinuity at the Z0-
match point but I don't know what it is or how to measure it. I don't
even know how to measure the multiple-port s11 at the Z0-match point.
IMO, this example is just too complicated for 99.9% of the readers
here, including myself.

If we put the series stub (or inductive coil) into a two-port black
box and performed s-parameter measurements on it, we could come up
with the s-parameter equations.

Source--50 ohm line--x--black box--y--150 ohm load

Reckon what s11 we would measure if we replaced the 150 ohm load with
a 50 ohm resistor?
--
73, Cecil, w5dxp.com