Good morning gentlemen!!! Well I got a good nights sleep, but all night I
was dreaming about where that reflected power goes, and if we assume we know
where it goes, can we proove it. If we make a statement as to where it goes,
we should be able to back-up our statement with proof. So I will try to
proove my theroy as to where the reflected power is consumed. Here we
go....Firstly, do we all agree that if we produce power from our amps and
convey X watts to an antenna system, we must consume X watts somewhere? Dont
think you will argue that...Right? Ok so... lets try to look at it
mathmatically, cause formulas dont lie!!!! Theoretically speaking now....If
our swr is 1:1 then our antenna is resonant and therefore purely resistive,
which means it is not reactive(be it inductively or capacitivly). Everybody
Agree? Ok moving on...We know that our signal (RF) is AC in nature, so the
AC resistance the antenna presents to our amp is really called impedence,
and if we look at our math formulas we know that impedence is effected by
inductance and capacitance Right?
No arguements yet? We also know that impedence in a specific antenna will
vary with frequency because of the Inductive reactance=Xl or
capacitance=Xc...Right?
Ok no arguements? lets continue...So we know that at 1:1 swr, the antenna is
said to be resonant and therefore is purely resistive (at that freq) and Xl
andXc =0 and the R value is 50 ohms. Great! thats what we want. Ok now for
an example...If we insert a tuner into a line with an antenna system at
resonace and we adjust the tuner to present a 1:1 swr to our linear,
essentially what will the settings be on our tuner (theoretically) and
mathmaticaly speaking our tuner would tune and it would report no inductance
setting and no capacitive setting, therefore the Xl and Xc of the tuner would
be 0. We know that Xl cancels Xc and vise-versa Right? So if there is no Xl
or Xc presented by the tuner, then there is no AC resistance introduced by
the tuner, there for with no AC resistance offered by the tuner it is like
not being there and so then there is NO voltage drop across the tuner as our
RF current passes through it. AGREE? Moving on...lets change frequency and
measure the swr....Ok we did...our swr is now 3:1...Ok now what? Well our
antenna is no longer 50 ohms nor is it any longer purely reistive. It is now
reactive either inductively or capacitively and depending on which it is,
the new impedence value of our antenna at this Freq. is either higher or
lower than 50 ohms. Ok everybody agree so far? So now we will all agree that
our antenna is no longer resonant so Xl or Xc does NOT equal 0, as it did
before...Right? Ok so we now use our tuner to match our linear to the
reactive antenna system, (no longer purely reistive) remember our swr is 3:1
now. So we tune our tuner, and we find, that we either added capacitance or
inductance to our antenna system, hence Xl or Xc now equaling some value.
Now remember Xl or Xc carry labels of OHMS....so...if we now have a value of
Xl or Xc presented in the tuner, and we pass our RF currrent through it, wont
we now have a voltage drop across the tuner? I think so, dont you? So if
this is true our voltage drop in the tuner would be calculated by E=IxZ and
therefore we will INDEED consume power in our tuner because if we do have a
voltage drop we will have power loss or consumption. So theoretically
speaking what will that power consumption be? I,ll bet is that reflected
value of power we spoke about before, that we agreed needed to be consumed
somewhere. Remeber that the antenna has a true R value which never changes
with frequency and if not resonant then also has a Xl or Xc componet which
will INDEED change with frequency. Ok now fellas, Do I get a metal of honor
or the firing squad?
Remember....e-mail amps@contesting.com with your ideas about this not
me.....Have a nice day all....73 LOU
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