NetNews Usenet Archive 1992 #31/NN_1992_31.iso/spool/rec/radio/cb/1584

Text File | 1992-12-28 | 5.3 KB | 98 lines

Newsgroups: rec.radio.cb
Path: sparky!uunet!paladin.american.edu!gatech!swrinde!emory!wa4mei!ke4zv!gary
From: [e-mail] (Gary Coffman)
Subject: Re: SWR Tuning and PLL
Message-ID: <[e-mail]>
Reply-To: [e-mail] (Gary Coffman)
Organization: Destructive Testing Systems
References: <[e-mail]> <[e-mail]> <[e-mail]> <[e-mail]>
Date: Mon, 28 Dec 1992 15:39:13 GMT
Lines: 86

In article <[e-mail]> [e-mail] (Yang Ming) writes:
>In article <[e-mail]> [e-mail] (Gary Coffman) writes:
>>In article <[e-mail]> [e-mail] (bill nelson) writes:
>>>[e-mail] (RUFUS) writes:
>>>:
>>>:  Jason,
>>>:   An SWR meter measures the Standing Wave Ratio hence SWR. What SWR is is the
>>>: amount of reflexed power recieved by your rig from a mistuned antenna. A
>>>
>>>Close. It is the ratio between the power reaching the antenna and the power
>>>reflected by the antenna.
>>
>>The SWR, properly the VSWR, is the ratio of the forward voltage on the
>>line to the reverse voltage on the line at any point along the line.
>
>This is conceptually incorrect. VSWR = Vmax/Vmin, not the ratio of
>the 'forward' to the 'reverse' voltage. Your definition is the inverse
>of the reflection coefficient.

Well to be precise, VSWR = (1+sqrt(Vr^2/Vf^2))/(1-sqrt(Vr^2/Vf^2)).
It's a ratio of the two voltages, Note that to find Vmax and Vmin you
need a slotted line and a movable RF probe because the two values will
not occur at the same point on the line. You also have to subtract out
the voltage that would be present in the matched case since it serves
as a fixed offset to the standing wave voltage. VSWR meters work by a
different principle. They directly measure the forward and reverse waves
at a particular point on the line using a pair of directional couplers.
And, using a specially marked scale, they use the indicating meter
as an analog computer to calculate the actual VSWR of the line from
those measured values. This works at any single point along the line.

>>The "standing wave" is the vector sum of these two voltages. All the
>
>Correct.
>
>>power injected into the transmission line by the transmitter is dissipated
>>in either the antenna or the line losses, none is absorbed by the radio. A
>
>The power reflected from the transmission line and the antenna due to
>mismatch CAN be absorbed by the radio, depending on the impedance
>when you look into the radio from the transmission line. If this impedance
>is equal to the characteristic impedance of the line, all reflected power
>will be absorbed by the radio. Thus, power injected into transmission line
>can be reflected and absorbed by the radio. You have to look at the
>match conditions on the two ends of the line.

This is where it gets interesting. A transmitter is not a load. An
impedance is made up of a combination of resistance and reactance,
either of which may be zero. Now a properly tuned antenna will have
the reactance cancelled and only the radiation resistance will be
present. It is a pure dissipative load. If uncancelled reactance is
present, it won't be a pure dissipative load, but will still accept
power. The reactance can't dissipate power, and after a bit of oscillating
back and forth, the power will be absorbed by the resistive part of the
impedance and dissipated. Only a real dissipative resistance can absorb
power.

But when you look into the output port of an operating transmitter, you
don't see a dissipative load. Instead you see the transformed dynamic
load line of the output device. This product of dynamic operating voltage
and current can be mathematically transformed to appear the same as a real
resistance, but it isn't a dissipative load. The output matching network,
a purely reactive transformer, will transform the load line "resistance"
to the characteristic impedance of the transmission line where the power
will flow to a dissipative load and be absorbed. Power flowing the other
way will be transformed up to the load line impedance by the output
network, now acting as an input network. These voltages and currents will
now superpose on the dynamic voltages and currents of the output device
and add vectorially with them. This shifts the value of the load line
such that the incoming reverse wave is not matched anymore. Remember this
isn't a real resistor, it's a dynamic mathematical fiction. Since the
reverse wave cannot be larger than the forward wave by definition, power
continues to flow outward from the transmitter toward the load, now
including the re-reflected reverse wave. This is called a conjugate
mirror. The only effect of the reverse wave on the transmitter is to
shift the operating point, which under certain conditions can reduce
the *efficiency* with which the stage turns DC power to RF power.
*That* can cause the output stage to get hot, but it has *absorbed*
none of the reverse power. It is not a load.

Gary
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