Paul Christensen wrote:
> > We have to remember that when a transmission line is involved and there are
> > mismatches, there is impedance transformation going
> on. Any
> > time the load is not matched to the line the line no longer presents the
> > same impedance at the watt meter or the transmitter that
> it
> > did before.
> > What I am saying is that I don't think the bird meter will read true power
> > unless it sees a 50 ohm resistive load. So what is
> seen on
> > the meter is not necessarily what is going on in the transmission line.
>
> Provided that the Bird directional coupler matches the characteristic
> impedance of the line, the metered results should be
> reasonably accurate. Interpreting the displayed results is another story.
It doesn't matter what the characteristic impedance of the line is. What
matters is what the meter sees. I can take a 1/2 wavelength
line that has a characteristic impedance of 100 ohms or 400 ohms or 900 ohms.
If I put a 50 ohm load on the end of that line and the
other end connected to the watt meter the meter will see 50 ohms with no
reflected power. Impedance transformation.
>
>
> > If a coax line is unterminated the bird meter reads 100 watts forward and
> > 100 watts reflected as you saw but is there really 100
> watts
> > there?
>
> I'm not too sure it matters. What does matter is that whatever power really
> is generated, all of it (less line loss power) is being
> self-absorbed in the transmitter's solid-state PA. (For those folks just
> jumping in, we're discussing transmitters with a fixed
> 50-ohm source impedance.)
There is an impedance matching network in a solid state transmitter too. It
just happens to be broad banded.
>
>
> If there is zero current and some finite voltage then is there really any
> power actually present? The transmission line is no
> > longer 50 ohms at the watt meter because of the impedance transformation.
> > (sorry about the additional questions but I don't have
> a
> > definitive answer)
>
> Perhaps not at that precise instant in time. Recall that power is defined as
> energy divided by time (e.g., 1 Joule per second = 1
> watt). The characteristic impedance of the line never changes as a function
> of impedance or SWR. It is held constant by the
> conductor sizing and spacing.
What instant in time does not matter. Power is averaged over time that we read
on the meter.
>
>
> > What I was saying about an open quarter wave line is that an open circuit
> at the end of a quarter wave line will be transformed
> into a
> > short at the other end due to impedance transformation of the line. If this
> > is hooked to the watt meter then the watt meter would
> be
> > looking into a short. Would it still read 100 watts? The transmitter
> > probably could not put out an infinite amount of current at
> zero
> > voltage to get a true 100 watts at that point.
>
> That point is an instant in time. If we are looking at voltage or current at
> any given instant snapshot in time, then by
> definition, there is no power. I do not know if the directional coupler will
> indicate any differently just because we made the open
> termination at the end of a quart-wave line. Any takers?
Pick any amount of time you want. The relationship between voltage and current
is not going to change. We have a certain amount of
current across a certain resistance, That is how we measure power. In this case
the resistance is zero. What is the power.
>
>
> > If the reflected power were to get absorbed by the transmitter finals then
> > an open wire transmission line would never work but in
> a few
> > instances where it was matched to the antenna.
>
> An open-wire line or a relatively loss-less coaxial line behave similarly.
> The reason that an open-wire balanced line works in most
> of our applications is because a *conjugate* network is placed somewhere on
> the transmission line (typically at the transmitter with
> an auto-tuner or transmatch), or a *matching* network is placed at the
> antenna terminals. In your example above, the same rules
> apply.
An open wire line can be and is often used for a single antenna to operate on
multiple bands. On one band the antenna may match the
characteristic impedance of the line and there is no reflected power. On the
next band that same antenna may present a very drastic
mismatch to the line. The line may be 450 ohms and the antenna may present a
2000 ohm load. There is no matching network used at the
antenna. There is reflected power from the antenna mismatched to the line. If
we cut the line the right length or we use the proper
lengths of another impedance line we can get it to present 50 ohms or any
impedance we wish at the transmit end. There will be high
reflected power on the line. But we do this with no tuner, only transmission
line. This will work fine and the final will not know any
difference than if we had a matched line and antenna hooked to it.
>
>
> Consider an RF amplifier (gotta' stay somewhat on topic here...heh) with a
> fixed, balanced 450-ohm source impedance terminated into
> 100-feet of 450-ohm balanced open line. If the opposite end of the open-wire
> line is unterminated, all generated power is returned
> back to, and absorbed by, the transmitter. Again, I ask the question: In
> this example where else can the power go?
Again, is there really any power generated when there is no load. Or is there
only voltage.
>
>
> In reality the reflected power on that type of line is often very high if you
> were to
> > measure it. The reason that it does work is that most all of the reflected
> > power gets re- reflected back to the antenna and is
> absorbed
> > by the antenna.
>
> Agreed, but for the right reasons.
>
> -Paul, W9AC
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