Reg Edwards wrote:

"Cecil Moore" wrote

On my system, there's a 50 ohm cable from the transceiver to
the input of the SWR meter and another 50 ohm cable from the
output of the SWR meter to the balun. Each of these cables
forces the ratio of the voltage to current in each of the
traveling waves to a value of 50 ohms. I have an in-line
Autek WM-1 and no tuner.
--
====================================

Cec,
You can't measure SWR on a line which is less than 1/4-wave long.
Preferably it should be as long as 1/2-wavelength to ensure the max
and min voltage points both occur on it.

Total and absolute nonsense.

Where do you come up with this stuff?

snip arm waving speech

--
Jim Pennino

Remove .spam.sux to reply.

Owen Duffy wrote:
On Tue, 27 Sep 2005 01:36:15 +0000 (UTC),
wrote:

Owen Duffy wrote:
On Mon, 26 Sep 2005 21:39:27 +0000 (UTC),
wrote:


Jim, that seems inconsistent with your earlier statemetn "No, the SWR
being measured is on the load side of the meter."

The load side is the side with the load, i.e. the antenna, on it.

In the example you quoted with a 100 ohm load on a 100 ohm line, were
the line loss low, and the line long enough to be sure to sample a
fully developed voltage maximum and voltage minimum it would be found
that the VSWR was 1:1.

Not for a 50 Ohm system, i.e. a transmitter expecting 50 Ohms and a
meter calibrated for a 50 Ohm system.

I am sorry Jim, the VSWR is a property of the transmission line and
its termination, and the VSWR on that 100 ohm line with a 100 ohm
termination is 1:1. The VSWR could be *MEASURED* on that line by
sampling the magnitude of the voltage at different points on the line
and it would be found that the magnitude of the voltage was constant,
which means VSWR=1:1.

No, the measured SWR is relative to the design impedance of the SWR
meter which is normally 50 Ohms.

The SWR on the line depends on the characteristic impedance of the
line and the impedance of the termination of the line. 50 ohms does
not come into it.

We are not talking about SWR on the line, we are talking about SWR
at the input END of the line; big difference.

The SWR on your proposed 100 ohm line with a 100 ohm termination is
1:1. If your measurement indicates anything else, then you need to
consider your measurement as invalid.

Once again, we are not talking about SWR *ON* the line, we are talking about
SWR as seen at the input *END* of the line; big difference.

Furthermore, the SWR *ANYWHERE* on the line is *NOT* 1:1 for a 50 Ohm
reference.

Try hooking a length of 93 Ohm line (which is easier to get than 100 Ohm
line) terminated with a 93 Ohm resistor to any 50 Ohm SWR meter of any type.

Then hook just the 93 Ohme resistor to the meter and tell me what the
difference is in the readings.

Owen
--

If anything is misnamed it is the term SWR.

SWR is nothing more than a dimensionless impedance ratio.

You do NOT need a transmission line to have SWR in spite of the W in
SWR standing for 'wave'.

--
Jim Pennino

Remove .spam.sux to reply.

On Tue, 27 Sep 2005 02:25:11 +0000 (UTC),
wrote:


SWR is nothing more than a dimensionless impedance ratio.

Is that so... Owen
--

Owen Duffy wrote:
On Tue, 27 Sep 2005 02:25:11 +0000 (UTC),
wrote:


SWR is nothing more than a dimensionless impedance ratio.

Is that so... Owen
--

Yes, it is so and it is equal to:

SWR = (A + B)/(A - B)

Whe

A = sqrt ( (R + Z)^2 + X^2 )
B = sqrt ( (R - Z)^2 + X^2 )

R = resistive component of load impedance in Ohms.
X = reactive component of load impedance in Ohms.
Z = reference impedance (purely resistive) in Ohms

If you don't believe it, get some resistors, capacitors and a half
way decent SWR meter and do some experiments; no transmission line
required.

--
Jim Pennino

Remove .spam.sux to reply.

On Tue, 27 Sep 2005 02:54:31 +0000 (UTC),
wrote:

Owen Duffy wrote:
On Tue, 27 Sep 2005 02:25:11 +0000 (UTC),
wrote:


SWR is nothing more than a dimensionless impedance ratio.

The fundamental definition of SWR flows from the behaviour and
properties of RF transmission lines.

When a transmission line is terminated in an impedance other than its
characteristic impedance, there will be both a forward wave and a
reflected wave of such magnitude to resolve the conditions that must
apply at the termination.

The forward wave and the reflected wave sum at all points along the
line having regard for their magnitudes and relative phase to produce
a "standing wave". The Standing Wave Ratio (SWR or VSWR) is defined to
mean the ratio of the maximum to the minimum of the magnitude of the
standing wave voltage pattern along the line.

The SWR on a lossless line can be calculated knowing the complex
characteristic impedance of the line and the complex load impedance.

The SWR on the line does not depend in any way on some unrelated
independent reference resistance as you suggest in your formula.

You seem to be suggesting that your redefined SWR is a really good
(obscure) way to talk about an impedance (independently of a
transmission line) in terms of some standardised reference value, and
you can throw away the fundamental meaning of SWR to support your
SWR(50) concept. In your terms (independently of a transmission line),
for instance, a Z of 60+j10 would be SWR(50)=1.299, and so would an
infinite number of other Zs have SWR(50)=1.299... how is that of
value. To know Z is 60+j10 is to know more than to know SWR(50)=1.299.

Owen
--

Owen,

SWR meters with a sampling line.

The only experience I've had has been I once made one for HF. It was
of the type where a second wire is drawn alongside the inner conductor
of a short length of coaxial line of impedance in the same street as
the system it is to work with.

Operating frequencies covered the whole of the HF band. That is a very
wide band. Which indicates that line length plays no part in
measuring accuracy once calibrated.

To explain how the thing works it is necessary to return to what it
really is. It is a resistance bridge. All so-called SWR meters,
whatever the circuit or form of construction, are resistance bridges.

The bridge has 3 internal ratio arms. The 4th arm is the variable
transmitter load. If all 4 arms are of same resistance we have a very
sensitive arrangement suitable for QRP transmitters. However, 3/4 of
the TX power is dissipated in the 3 internal bridge arms.

For higher power transmitters it is necessary to use high ratios for
the ratio arms. In the case of meters which use a little ferrite ring
as a current transformer, a resistor of the order of 30 to 100 ohms
can be shunted across the current transformer secondary winding while
the primary winding has an input resistance of the order of 0.1 ohms
which forms the value of the ratio arm in series with the external
load. This 0.1-ohm arm is capable of carrying the load current of
several amps with only a small power loss.

The other two ratio arms can be a pair of high value resistors in the
same ratio as occurs via the current transformer. If the input
resistance of the current transformer is 0.1 ohms then the bridge
ratio is 50 / 0.1 = 500:1 where 50 ohms is the usual value of the
load resistance when the bridge is balanced and SWR = 1:1

The two high impedance arms can be capacitors in the same ratio of
500:1 which have zero power dissipation but have a minor effect on
accuracy. They introduce a small phase angle into the load as seen by
the transmitter through the meter. The error increases with
increasing frequency.

It will be seen that the take-off point is effectively the same for
both current and voltage.

Returning to the so-called sampling line.

There is a bridge configuration which is not quite so obvious. But
instead of a current transformer the current is picked off by means of
a short length of wire in parallel with the coaxial inner conductor by
virtue of their mutual inductance. The line is too short for
propagation effects to play a significant part.

Voltage is picked off at the same point by virtue of the capacitance
between the wire and coaxial inner conductor. The phase relationship
between volts and amps can be reversed just by reversing the direction
of propagation through the meter.

The bridge ratio is set partially by the ratio of the impedances Zo of
the additional wire and inner coax conductor. The length of coaxial
line affects only the bridge sensitivity and power dissipated in the
meter. As you must be aware, sensitivity falls of fast with decreasing
frequency and 160 meters was my favourite band. So the home-brewed
meter was soon discarded and I returned to ferrite rings.

I was left with the impression it was very easy to make and that
almost anything would work.

Hope you can understand the foregoing.
----
Reg.

Jim,

Perhaps there's some misunderstanding about location of the meter and
what it is measuring. Let's try to clear it up.

Would you please do us both a favour by answering the following simple
question?

There is a 50 ohm line feeding a 100 ohm antenna.

There is an SWR meter located at the line-antenna junction.

The meter has a reading. Does the reading apply to SWR of the
antenna, or does it apply to the SWR along the feedline?

Antenna or Feedline?
----
Reg.

"Reg Edwards" wrote:
There is a 50 ohm line feeding a 100 ohm antenna.
There is an SWR meter located at the line-antenna junction.
The meter has a reading. Does the reading apply to SWR of the
antenna, or does it apply to the SWR along the feedline?
______________

It applies to the match of the RF network that follows the SWR meter to the
impedance for which the SWR meter was calibrated.

And if in your example the SWR meter has been calibrated for 50 ohms, and is
moved to the input end of that line+antenna RF network, it will also have a
reading -- which will be the same as when it was inserted at the
antenna-line junction, less the round-trip RF attenuation of the
transmission line (assuming that the transmission line is 50 +/- j0 ohms
throughout its length).

In fact it is a common practice to optimise the transmission line/antenna
match of commercial FM and TV broadcast antenna systems by use of a variable
transformer inserted at the antenna input, whose adjustment is made by
reference to the far-end reflection seen at the sending end of the
transmission line, using a high-directivity reflectometer, or SWR meter.

The same physics applies to ham antenna systems and methods/means of
measurement.

RF

Visit http://rfry.org for FM transmission system papers.

Reg Edwards wrote:
There is a bridge configuration which is not quite so obvious. But
instead of a current transformer the current is picked off by means of
a short length of wire in parallel with the coaxial inner conductor by
virtue of their mutual inductance. The line is too short for
propagation effects to play a significant part.

The pickup lines in my Heathkit HM-15 are terminated on one
end with a 50 ohm resistor. One pickup line thus attenuates
the reflected traveling wave and allows the forward traveling
wave to be rectified. The other pickup line attenuates the
forward traveling wave and allows the reflected traveling wave
to be rectified. Knowing the peak values of both of these two
traveling waves allows a calibrated meter to indicate SWR.
--
73, Cecil http://www.qsl.net/w5dxp

Owen Duffy wrote:
On Tue, 27 Sep 2005 02:54:31 +0000 (UTC),
wrote:

Owen Duffy wrote:
On Tue, 27 Sep 2005 02:25:11 +0000 (UTC),
wrote:


SWR is nothing more than a dimensionless impedance ratio.

The fundamental definition of SWR flows from the behaviour and
properties of RF transmission lines.

And power=EI. And it also equals I^2*R and E^2/R.

SWR can be expressed in terms of power ratios, current ratios, and
impedance ratios.

When a transmission line is terminated in an impedance other than its
characteristic impedance, there will be both a forward wave and a
reflected wave of such magnitude to resolve the conditions that must
apply at the termination.

Irrelevant.

The forward wave and the reflected wave sum at all points along the
line having regard for their magnitudes and relative phase to produce
a "standing wave". The Standing Wave Ratio (SWR or VSWR) is defined to
mean the ratio of the maximum to the minimum of the magnitude of the
standing wave voltage pattern along the line.

Is is also defined as a current ratio and an impedance ration.

The SWR on a lossless line can be calculated knowing the complex
characteristic impedance of the line and the complex load impedance.

What no waves, just impedences!! Now you are contidicting yourself.

The SWR on the line does not depend in any way on some unrelated
independent reference resistance as you suggest in your formula.

Read it again.

The R is the R of the thing at the end of the line.

The X is the X of the thing at the end of the line.

The X is the impedance of the line.

You seem to be suggesting that your redefined SWR is a really good
(obscure) way to talk about an impedance (independently of a
transmission line) in terms of some standardised reference value, and
you can throw away the fundamental meaning of SWR to support your
SWR(50) concept. In your terms (independently of a transmission line),
for instance, a Z of 60+j10 would be SWR(50)=1.299, and so would an
infinite number of other Zs have SWR(50)=1.299... how is that of
value. To know Z is 60+j10 is to know more than to know SWR(50)=1.299.

The equations given are general and can be derived from first priciples.

The Z in the equations is the Z of your reference, i.e. 50 for a 50
Ohm system.

SWR is *ALWAYS* relative to some reference impedance.

Owen
--

--
Jim Pennino

Remove .spam.sux to reply.