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-   -   Length of Coax Affecting Incident Power to Meter? (https://www.radiobanter.com/antenna/203-length-coax-affecting-incident-power-meter.html)

Roy Lewallen August 14th 03 12:09 AM
Dr. Slick wrote:

On second thought, i believe we are all wrong to equate S11 with
SWR!

Input S11 of a system will certainly never change. But the SWR
is absolutely dependant on the source impedance.

I give up. People will believe what they want to believe, no matter what
-- it's like arguing religion. But I hope some of the lurkers have
learned that SWR is independent of source impedance, even if some of the
active posters just can't seem to.

. . .

Roy Lewallen, W7EL

Dr. Slick August 14th 03 12:55 AM
"Ian White, G3SEK" wrote in message ...

Input S11 of a system will certainly never change. But the SWR
is absolutely dependant on the source impedance.

No! SWR, S11, return loss, rho, Y-parameters, Z-parameters, etc, etc are
all different derived functions of the same two variables: an arbitrary
complex impedance, and the system reference impedance Z0 (a constant
which may or may not be defined as complex).


Correct, but it doesn't have to be 50 Ohms.


Only those two variables are involved, so all of these functions are
locked together. If one variable changes, all the derived functions
change too. Either all change, or none change; nothing else is logically
possible.


How about a 50 Ohm resistor, which is always 50 Ohms (impedance
doesn't change), fed with 20 ohms? Or 75?

Cecil is correct in saying that the SWR meter would then have to
be designed for 20 or 75 ohms, but that is beside the point.

SWR doesn't have to be strictly 50 ohms, and will involve TWO
impedances. If your source doesn't match your reference impedance
(normalized center of Smith Chart), then you won't be measuring the
reflected power coming right off the source.

And because most PA are not 50 ohms output, and most SWR meters
are 50 Ohms, there is problem.


As Roy says, the equations relating any one of these parameters to any
other are all well known. NONE of them ever involves source impedance.




Assuming the source impedance is 50 ohms, which it usually isn't
with most PAs.



If you had a network
analyzer calibrated for 20 Ohms, you would certainly have reflected
power and high VSWR going into 50 Ohms, and a 1:1 SWR going into 20
Ohms.

This would be the same as re-normalizing the Smith Chart for 20
Ohms in the center. You certainly can do this in MIMP.

I don't blame anyone for believing it's a 50-Ohm-only world!

No argument about any of that... but it's a totally separate point that
has no relevance whatever to your earlier statements about source
impedance.


it's very relevant if you consider the port on a network analyzer
to be 50 ohms or not... It should be, but your PA may be quite far
off.


Slick

Richard Clark August 14th 03 01:04 AM
On Wed, 13 Aug 2003 16:13:13 -0700, Roy Lewallen
wrote:

Believe me, Ian, I know how frustrating it can get. But remember all the
lurkers out there who benefit from your insightful postings. Please keep
it up -- it is worth while. For their sake.

Roy Lewallen, W7EL

Hi Roy,

It is nice of you to commend Ian to continue the good effort. Why
didn't you do it directly instead of posting him through me? Are the
lurkers to take some lesson by this breach of netiquette? :-)

73's
Richard Clark, KB7QHC

Dr. Slick August 14th 03 01:12 AM
(Tom Bruhns) wrote in message om...
 (Dr. Slick) wrote in message . com...

impedance, you can determine the SWR on that line. But your "SWR"
meter isn't really an S11 meter; at best it's a |S11| meter.


Of course. We have no phase information. Only the magnitude of
the voltage reflection coefficient.


This also brings up another point: do YOU define S11 to be the same
as reflection coefficient?

Cheers,
Tom


I define the S11 as the complex impedance, which never changes,
but S11 can also refer to the magnitude of the reflection coefficient
at a particular phase, AT A PARTICULAR NORMALIZED REFERENCE IMPEDANCE.
If you define the reflection coefficient as having phase
information, then yes, they are the same, but only at a particular
reference impedance.

Change the reference impedance, and you will have a new reflection
coefficient, but the complex impedance will still be measured to be
the same.

You can do this is Motorola's Impedance Matching Program (MIMP).

What this all comes down to is that your 50 Ohm SWR meters only
measure reflected powers after it, not before, so any mismatch from PA
to reference impedance (50 ohm coax from PA to meter) is not measured.


Slick

Tom Bruhns August 14th 03 01:13 AM
(Dr. Slick) wrote in message . com...
The SWR is based on the ratio of the
forward to the reflected power.

Exactly! And (in a linear, time-invariant system at steady-state,
with enough forward power to be interesting,) that ratio does not
depend on where the forward power comes from; it's determined by the
load attached to the line and the line itself. (I trust we ARE
talking about a situation with only one source, feeding only one end
of the line...)

Cheers,
Tom

Dr. Slick August 14th 03 01:19 AM
"Ian White, G3SEK" wrote in message ...

How would you explain what Cecil wrote?

Who else but Cecil would dare attempt that? :-)


I can understand your fear...




Do you think the series reactance a system offers a PA may
actually improve it's incident power?


For example, reducing the load impedance will usually make the output
device operate in a more linear way... but the efficiency drops and the
greater heat dissipation and current are likely to shorten the lifetime
of the device. Is that an improvement?


By improve, i mean increase the incident power. I think this is
possible, and i've actually measured it, if you read my original post.

If you can improve the incident power (tuning) of a PA with
varying the coax length, you might be able to adjust the SWR too, in
certain cases... though in my case, the SWR stayed about the same.


Slick

Roy Lewallen August 14th 03 02:10 AM
I apologize for the discourtesy. It did indeed set a bad example.

Roy Lewallen, W7EL

Richard Clark wrote:
On Wed, 13 Aug 2003 16:13:13 -0700, Roy Lewallen
wrote:


Believe me, Ian, I know how frustrating it can get. But remember all the
lurkers out there who benefit from your insightful postings. Please keep
it up -- it is worth while. For their sake.

Roy Lewallen, W7EL


Hi Roy,

It is nice of you to commend Ian to continue the good effort. Why
didn't you do it directly instead of posting him through me? Are the
lurkers to take some lesson by this breach of netiquette? :-)

73's
Richard Clark, KB7QHC

W5DXP August 14th 03 02:32 AM
Roy Lewallen wrote:

Dr. Slick wrote:
I disagree on this point. You are caught up in the 50 Ohm world,
which i admit is easy to do. The SWR is based on the ratio of the
forward to the reflected power.

That's not correct. The SWR (more correctly VSWR) is, by definition, the
ratio of the highest to lowest voltages which appear on a line long
enough to have both a maximum and minimum. It can be calculated from the
forward and reverse voltage waves. ISWR, the current standing wave
ratio, is numerically equal to the VSWR.

For Dr. Slick: Knowing the forward and reflected powers, one can use the
following equation to obtain SWR. 'Sqrt' means "square root of".

SWR = [Sqrt(Pfwd)+Sqrt(Pref)]/[Sqrt(Pfwd)-Sqrt(Pref)]
--
73, Cecil http://www.qsl.net/w5dxp



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W5DXP August 14th 03 02:54 AM
Dr. Slick wrote:
So we never really measure the reflected
power coming right out of the PA, even if we attach the meter directly
to it's output.

If the forward power out of a PA is 100 watts and the reflected power
into the PA is 50 watts, the PA is generating 50 watts, *by definition*.

Given that definition, the implication is clear. All PA's, by definition,
must re-reflect 100% of the incident reflected energy. Thus, everything
you are worried about has already been defined out of existence. :-)
--
73, Cecil http://www.qsl.net/w5dxp



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W5DXP August 14th 03 02:56 AM
Ralph Mowery wrote:
I have not seen a whole lot of equations on SWR but the few I have seen
never mention the source at all.

When someone defined the generated power as the forward power minus the
reflected power, the entire problem was defined out of existence. :-)
--
73, Cecil http://www.qsl.net/w5dxp



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