Analyzing Stub Matching with Reflection Coefficients
 

Jim Kelley wrote:

Cecil Moore wrote:
Yes, and HF also observes that, e.g.:

|s11(a1)|^2 = 1 joule/sec

|s12(a2)|^2 = 1 joule/sec

I don't think they do.

Take a look at pages 16 & 17. Quoting:

"Another advantage of s-parameters springs from
the simple relationship between the variables
a1, a2, b1, and b2, and VARIOUS POWER WAVES."

Definitions of |a1|^2, |a2|^2, |b1|^2, & |b2|^2
follow.

"The previous four equations show that s-parameters
are simply related to power gain and mismatch loss,
quantities which are often of more interest than
the corresponding voltage functions."

Definitions of |s11|^2, |s22|^2, |s21|^2, & |s12|^2
follow.

On page 10 it says: "The s-parameters s11 and s22
are the same as optical reflection coefficients;
s12 and s21 are the same as optical transmission
coefficients."

In fact, the intensity equation from optics can
be had from squaring both sides of:

b1 = s11(a1) + s12(a2)
--
73, Cecil, http://www.qsl.net/w5dxp

Analyzing Stub Matching with Reflection Coefficients
 

Dr. Honeydew wrote:
So you don't believe in any of the S-parameter analysis done on linear
small-signal amplifiers, then.

They certainly work for systems designed to be linear
as are *linear* small-signal amplifiers, sources with
circulators, and padded sources as I have said over
and over.

Amateur transmitters are not designed to be linear
and they are NOT linear. Adding a ten cent resistor
to them is not going to make them linear.
--
73, Cecil, http://www.qsl.net/w5dxp

Analyzing Stub Matching with Reflection Coefficients
 

On Apr 23, 1:18 pm, Cecil Moore wrote:
Keith Dysart wrote:
Were one to solve the differential equation for power
on the open circuited line at a location of a voltage
null, the answer would be zero. And all computed without
the need for forward and backward waves.

The *NET* power at any point on a lossless stub is
zero so that is no big deal. The standing wave
current is always 90 degrees out of phase with
the standing wave current so cos(90) = 0. At a
voltage node, all of the energy has simply moved
into the magnetic field.

You really don't WANT to think outside of the box that
you have built for yourself, do you. Anyway, reread
what I wrote and work out why your response is a
non-sequitor.

Calculate the number of electromagnetic joules in
the line and get back to us on how they are reflected
from your above purely virtual impedance at a
voltage "null" and how they can even exist without
a velocity of c(VF).

You really are stuck in a box. Consider a capacitor.
It has no difficulty storing energy without velocity.
And lest you say, "but that is not a transmission
line", consider an open circuited transmission line
excited with a step function. After settling: constant
voltage, no current, energy stored in capacitance,
and the DC coupled directional wattmeter will
tell you there are the same number of 'Bird watts'
going east as there are 'Bird watts' going west

....Keith

Analyzing Stub Matching with Reflection Coefficients
 

Cecil Moore wrote:
Jim Kelley wrote:


Cecil Moore wrote:

Yes, and HF also observes that, e.g.:

|s11(a1)|^2 = 1 joule/sec

|s12(a2)|^2 = 1 joule/sec


I don't think they do.


Take a look at pages 16 & 17. Quoting:

"Another advantage of s-parameters springs from
the simple relationship between the variables
a1, a2, b1, and b2, and VARIOUS POWER WAVES."

Definitions of |a1|^2, |a2|^2, |b1|^2, & |b2|^2
follow.

"The previous four equations show that s-parameters
are simply related to power gain and mismatch loss,
quantities which are often of more interest than
the corresponding voltage functions."

Definitions of |s11|^2, |s22|^2, |s21|^2, & |s12|^2
follow.

On page 10 it says: "The s-parameters s11 and s22
are the same as optical reflection coefficients;
s12 and s21 are the same as optical transmission
coefficients."

In fact, the intensity equation from optics can
be had from squaring both sides of:

b1 = s11(a1) + s12(a2)

But I still don't see where HP said that either s11(a1) or s12(a2) is
equal to the square root of 1 joule/sec. Which pretty much confirms
what I said. I think you might have made that part up to try to sound
authoratative.

But I will concede that if you had been right, it may have lent
support to your contention that waves and energy are moving in the
direction of b1 while not moving in the direction of b1. :-)

73, Jim AC6XG

Analyzing Stub Matching with Reflection Coefficients
 

Keith Dysart wrote:
You really are stuck in a box. Consider a capacitor.
It has no difficulty storing energy without velocity.

Uhhhh Keith, a capacitor cannot store an EM wave which
is photonic energy. If a capacitor stores the energy in
an EM wave, it must first convert the EM wave energy to
some other form of energy, e.g. electronic energy. That
is certainly possible, but the energy ceases to be EM
energy and cannot be recovered as the same EM energy
coherent with that which was stored. If thinking outside
the box requires a complete denial of reality and the
laws of physics, then I think I will pass.

To summarize, converting EM energy to electronic energy
causes the EM energy to lose coherence. It would take
heroic measures to recover that coherency and those
measures simply do not exist inside a transmission line.

It certainly seems that you guys don't understand the
difference between photonic energy and electronic
energy and you treat them as if they were interchangeable
with no conversion process required. Hint: When energy
is converted from one form to another, the previous
form disappears and is generally not recoverable in
a form coherent with the original. Yet, all the energy
inside a transmission line remains coherent. You have
generated a logical and technical contradiction.
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Jim Kelley wrote:
Cecil Moore wrote:
In fact, the intensity equation from optics can
be had from squaring both sides of:

b1 = s11(a1) + s12(a2)

But I still don't see where HP said that either s11(a1) or s12(a2) is
equal to the square root of 1 joule/sec.

Good grief, Jim, are you too lazy to perform the math???

Squaring both sides of the above equation to obtain
joules/sec:

b1^2 = |s11(a1)|^2 + 2*s11(a1)*s12(a2) + |s12(a2)|^2

Does that center term look familiar yet? It should since
it is the interference term in the intensity (power density)
equations from "Optics" by Hecht and from "Principles of
Optics", by Born and Wolf.

Let |s11(a2)|^2 = P1 Then s11(a1) = SQRT(P1)

Let |s12(a2)|^2 = P2 Then s12(a2) = SQRT(P2)

Is your mental light bulb coming on yet? Assuming the phase
angle between a1 and a2 is zero, we can directly write the
equation for total constructive interference.

Ptot = P1 + P2 + 2*SQRT(P1*P2)
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

On Apr 23, 6:07 pm, Cecil Moore wrote:
Uhhhh Keith, a capacitor cannot store an EM wave which
is photonic energy.

Could you clarify; At what frequency does the signal in the
transmission line become "EM" energy? A step function has
high frequency components, and yet the end result is DC.
Did it stop being "EM" energy at some point? Which point?

Or how about a square wave at 0.00001 Hertz? "EM" energy?
Or not? Harmonics are only limited by the rise time. But my
meter will have trouble knowing its not DC.

....Keith

Analyzing Stub Matching with Reflection Coefficients
 

On 23 Apr 2007 15:30:36 -0700, Keith Dysart wrote:

On Apr 23, 6:07 pm, Cecil Moore wrote:
Uhhhh Keith, a capacitor cannot store an EM wave which
is photonic energy.

Could you clarify; At what frequency does the signal in the
transmission line become "EM" energy?

A Question for the Ages!

(another, less profound question: "When did caps stop working in tank
circuits?")

73's
Richard Clark, KB7QHC

Analyzing Stub Matching with Reflection Coefficients
 

Keith Dysart wrote:
Could you clarify; At what frequency does the signal in the
transmission line become "EM" energy?

This is not the proper forum to try to teach you the difference
between a photonic wave and a DC electronic circuit. Please feel
free to return when you have educated yourself of the considerable
technical information available on the subject.
--
73, Cecil http://www.w5dxp.com

Analyzing Stub Matching with Reflection Coefficients
 

Richard Clark wrote:
(another, less profound question: "When did caps stop working in tank
circuits?")

When the tank gets hit by a Scud missile?
--
73, Cecil http://www.w5dxp.com