On Tue, 7 Sep 2004 10:48:16 -0500, (Richard
Harrison) wrote:

Richard Clark wrote:
In a world of mismatches, how does it happen that the transmitter always
sees an in-phase, resistive load?"

It doesn`t. You can put a capacitor directly across its output
terminals, and the transmitter will energize the capacitor.

Hi Richard,

And so it would seem that a too-short antenna would act likewise.
This, then, returns us to my original question for this, a very simple
example that has a currency that is frequent enough in our experience:
how does a transmitter happen to always be "in-phase" to any
reflection?

73's
Richard Clark, KB7QHC

Richard Harrison wrote:
Cecil, W5DXP wrote:
"The superposed forward voltage and reflected voltage can damage an
unprotected transmitter."

To do so, they would be in-phase and not out-of-phase.

Nope, they don't have to be in-phase. The reflected voltage can
arrive with any phase with respect to the forward voltage. You
trimmed out the example I gave. I will repeat it so you can
study it closer.

Just a for instance - assume the transmitter is putting out 70.7v
in phase with 1.4a at zero deg. The arriving reflected wave is 50v
at 90 deg and 1.0a at -90 deg. The load seen by the transmitter is
86.6v at 35 deg and 1.72a at -35 deg. Over voltage and over current
exist at the transmitter output. The forward power is 100w and the
reflected power is 50w. The net power being delivered to the reactive
"load" seen by the transmitter is 86.6*1.72*cos(70.4) = 50w.

In the above example, the designed for output voltage is 70.7v and
the designed-for output current is 1.4a. The superposed voltage is
86.6v, higher than the designed-for voltage. The superposed current
is 1.72a, higher than the designed-for current. In this example,
we have both over-voltage and over-current occurring *at the same
time* even when the forward voltage and reflected voltage are 90
degrees out of phase. The superposed voltage is not smaller than
the forward voltage until the phase angle between them is in the
neighborhood of 120 degrees. Two superposed voltages with a 90
degree phase angle are *always* larger than either voltage component,
i.e. SQRT(V1^2+V2^2) is always larger than either V1 or V2.

The superposed voltage will be higher than the forward voltage for
any phase angle from zero to 90 degrees. At some phase angle higher
than 90 degrees, in the neighborhood of 120 degrees, the superposed
voltage starts to decrease.

If you sit down and draw these phasors on a piece of paper, you
will discover that you are mistaken. It appears that you are
thinking one-dimensionally instead of two-dimensional phasors.
--
73, Cecil http://www.qsl.net/w5dxp


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Richard Clark wrote:
how does a transmitter happen to always be "in-phase" to any
reflection?

It doesn't. The reflected voltage can obviously be 90 degrees
out of phase with the forward voltage in which case, the
interference term equals zero, and the superposed voltage is
SQRT(Vf^2+Vr^2), i.e. greater than Vf. (The argument reminds
me of Gary Coffman's one-dimensional "spitting up the fire
hose" argument.)
--
73, Cecil http://www.qsl.net/w5dxp


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Richard Harrison wrote:
Standing waves display interference between incident and reflected waves
whiich ideally have in-phase and out-of-phase constituents.

In the following, the interference term equals zero.

XMTR--------1/8WL 50 ohm lossless coax------291.5 ohm load

Forward power = 100w, reflected power = 50w.

Vf = 70.7v at zero degrees, If = 1.4a at zero degrees

Vr = 50v at 90 degrees, Ir = 1.0a at -90 degrees

Superposing:

Vtot = SQRT(Vf^2+Vr^2) = 86.6v over-voltage condition

Itot = SQRT(If^2+Ir^2) = 1.72a over-current condition

The phase angle between Vtot and Itot is about 70.4 degrees,
i.e. the source sees a highly reactive impedance.

Vtot is NOT in phase with Vf, Itot is NOT in phase with If,
Vtot is NOT in phase with Itot. Vf is NOT in phase with Vr.
No two voltages are in phase. No two currents are in phase.
--
73, Cecil http://www.qsl.net/w5dxp


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Cecil, W5DXP wrote:
"Nope, they don`t have to be in-phase."

Cecil is right. The third side of a triangle can be longer or shorter
than one of the sides.

I should have said, it is entirely possible that the superposed forward
and reflected voltages can damage an unprotected transmitter, The
reflected voltage can add to the forward voltage applied to the
transmitter.

The phase of the reflected voltage should be out of phase with the
transmitter source voltage to maximize the volts across the internal
impedance of the transmitter if we want to damage an unprotected
transmitter.

Best regards, Richard Harrisin, KB5WZI

Richard Harrison wrote:
Cecil, W5DXP wrote:
"Nope, they don`t have to be in-phase."

Cecil is right. The third side of a triangle can be longer or shorter
than one of the sides.

I should have said, it is entirely possible that the superposed forward
and reflected voltages can damage an unprotected transmitter, The
reflected voltage can add to the forward voltage applied to the
transmitter.

The phase of the reflected voltage should be out of phase with the
transmitter source voltage to maximize the volts across the internal
impedance of the transmitter if we want to damage an unprotected
transmitter.

Now assuming, as Richard H. logically did, that we don't want to damage
an unprotected transmitter, let's introduce the concept of an active
match Vs a passive match.

The match at the transmitter output is passive, not active. Consider
the following system:

XMTR---1/4WL 300 ohm feedline---1800 ohm load

What does it really mean when we say that the XMTR "sees" 50 ohms?

It means that an interference pattern is established at the XMTR
that re-reflects (re-directs) the reflected energy back toward the
load. The transmitter has absolutely nothing to do with that reflection.

We can reconfigure the system as follows:

XMTR---1WL 50 ohm feedline--+--1/4WL 300 ohm feedline---1800 ohm load

Now we can see that the match is passive and the XMTR has nothing to
do with the match.

Destructive interference on the XMTR side of '+' eliminates the reflections.
The resulting equal magnitude of constructive interference on the load side
of '+' re-reflects (re-directs) the reflected energy back toward the load.

Optics engineers understand interference. Seems RF engineers might not
fully understand interference. Interference is the cause of the
elimination of reflections in optic systems. Interference is the cause
of elimination of reflections in RF systems, including at antenna tuners.
--
73, Cecil http://www.qsl.net/w5dxp


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Cecil, W5DXP posed a problem in which a transmitter seeking a 50-ohm
load (I suppose) is attached to 1 WL of 50-ohm cable which is attached
to 1/4-wave of 300-ohm feedline which is attached to an 1800-ohm load.

Cecil wrote:
"Destructive interference on the XMTR side of "+" (where 50-ohm cable
meets the 300-ohm Q-matching section) eliminates the reflections."

It seems to me, the product of 50 (the cable Zo) and 1800 (the load Z)
is 90,000. The sq.rt. of 90,000 is 300 (the Q-section impedance). The
numbers are right, so in my opinion, the Q-section converts the
1800-ohms to 50-ohms. This is the 50-ohms likely prescribed for the
transmitter`s load. The 50-ohms presented by the Q-section to the cable
should result in a match and thus there should be no reflected energy in
the 50-ohm cable to cause interference. The reflections should all be in
the 300-ohm Q-section.

Best regards, Richard Harrison, KB5WZI

On Sun, 05 Sep 2004 18:06:07 GMT, Richard Clark
wrote:

|On Sun, 5 Sep 2004 12:47:45 -0500, (Richard
|Harrison) wrote:
|
|The tank circuit is mostly a harmonic filter providing a very high
|impedance to the fundamental frequency and shorting out the harmonics.
|
|Hi Richard,
|
|Even here, the Goatman offered in his notes that his finals tank
|(actually a series resonant Z match) offered a loaded Q of 2! (If I
|read his scribblings correctly.)

Yeah and he also calculates

(http://www.techatl.com/wrek/docs/gnm69_25.htm)

the required plate load resistance as:


Eb
Rl ~ -------
Idc

Which for class C is off by about a factor of 2, but with Eb = 2500
and Idc = .25, he does the division and comes up with Rl = 1000. Hey
what the heck, what's a factor of 10 among friends.

If the calculation is done more accurately:

Eb - Eg2
Rl = ---------
K * Idc

Where K = 2 for Class C and
Eg2 = 300 (screen voltage)

Then Rl ~ 4 Kohm

Since the minimum output capacitance (Cp) of a 4CX300 is 4.5 pF, the
parallel equivalent of Rl and Cp is Rp ~ 3920, Xp ~ -388.

Thus the minimum possible Q ~ 10, which to someone who has built a few
VHF amplifiers, sounds much more plausible.

For example here's one I designed and built not much later that the
WREK(ed) transmitter.

http://www.qsl.net/n7ws/K7CVT_Amp.html

But it gets worse. Try as I might with the component values he
specifies, I cannot develop a plate load Z anywhere close to what is
necessary. He has a lot more inductance that he thinks, so maybe that
helps and I suspect his output lowpass filter (seen in the photos but
not on the schematic) is part of the matching network.

I'm really surprised that with the construction and documentation
presented he could get FCC type acceptance.

Richard Harrison wrote:
Cecil, W5DXP posed a problem in which a transmitter seeking a 50-ohm
load (I suppose) is attached to 1 WL of 50-ohm cable which is attached
to 1/4-wave of 300-ohm feedline which is attached to an 1800-ohm load.

Cecil wrote:
"Destructive interference on the XMTR side of "+" (where 50-ohm cable
meets the 300-ohm Q-matching section) eliminates the reflections."

It seems to me, the product of 50 (the cable Zo) and 1800 (the load Z)
is 90,000. The sq.rt. of 90,000 is 300 (the Q-section impedance). The
numbers are right, so in my opinion, the Q-section converts the
1800-ohms to 50-ohms. This is the 50-ohms likely prescribed for the
transmitter`s load. The 50-ohms presented by the Q-section to the cable
should result in a match and thus there should be no reflected energy in
the 50-ohm cable to cause interference. The reflections should all be in
the 300-ohm Q-section.

Exactly, thanks to destructive interference. In S-parameter terms:

b1 = a1*s11 + a2*s12

where b1 is the normalized reflected voltage back toward the source
and is equal to zero because complete destructive interference between
a1*s11 and a2*s12 causes their phasor sum to equal zero.

Those two voltage components cancel to zero toward the source. That
destructive interference is necessary and sufficient for a matched
system. The result is constructive interference toward the load.
--
73, Cecil http://www.qsl.net/w5dxp


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Cecil Moore wrote:

Walter Maxwell wrote:

Thus, no transmission line is necessary. For example, the device can be
connected directly to the antenna terminals, or any other device you
desire to
determine the mismatch, and power it directly from the signal source--no
transmission line is needed on either port for the device to indicate
the degree
of mismatch.


Assume a 100+j100 ohm load and a 100-j100 ohm transmitter
directly connected with no transmission line. The system
is matched. Are there any reflections? Now install a
transmission line. Will an SWR meter read the same thing
in both cases?

What's the Zo of the transmission line? Is it sufficiently long to act
as a transmission line?

Cecil, methinks you posed an incomplete question :-)