on 5/25/01 11:09 AM, Billy Ward at billydeanward@hotmail.com wrote:
>> JOHN:
>> At the 3 dB points, half the power that is transmitted at that frequency by
>> the source or whatever, gets passed through to the load or whatever is on
>> the other side of the filter.
>>
>> So where does the other 3 dB go? Is it dissipated? Hardly. To understand
>> where it goes, one must look at the magnitude of S11 and S22 of the filter.
>> At the 3 dB point, the values of S11 and S22 are getting large and all of
>> that power is being REFLECTED back towards the source.
>
> BILLY:
> I need to study this further and think on it but I have a problem thinking
> that power can just be reflected back to the source and done away with. It
> has to go somewhere. If we have a conjugate match at a given frequency--and
> we can only have a CM at one frequency--then the output will be attenuated
> at any other frequency. Just as we cannot create power, we cannot destroy
> it either. We must dissipate it in a manner that does a job for us such as
> transmitting an rf wave or dissipate it as a loss.
Billy,
Take a look at a filter plot of both S21 and S11. I can send some to you if
you need any since I sell them.
At frequencies outside the passband of the filter, S11 is high. A filter is
a REFLECTIVE not an absorptive device. Attenuated power at the output is
NOT absorbed by the filter. How can it be? A filter has no lossy elements.
All elements are lossless or nearly so.
So yes, the power outside of our passband (and it SHOULD be small by the
way), gets reflected back to the source where if there's an impedance
mismatch some gets reflected and some absorbed, etc. The reflected energy
travels back down the line towards the filter, and back and forth. Just
like if we put a high VSWR load on a 50 Ohm transmission line.
Networks with pure Ls and Cs are lossless other than the intrinsic real
world loss that we all wish we didn't have to deal with.
In some applications, particularly in the commercial world, this energy
reflected back toward the source can cause some instability problems. It is
common practice commerically to put isolators or attenuating pads at source
outputs. An isolator since it has a resistive load on it, WILL absorb and
dissipate power. A resistive attenuator is obvious.
> Back to the Tuner instead of the filter:
>
> I have never had the need of thinking this out before but since we are
> discussing the Conjugate Match (rather than a band-pass filter), if there
> is a loss of 3db which, of course, is at a frequency that is close to but
> removed from the CM point, a portion of but not all of that power will be
> re-reflected toward the load.
Correct. However, the 3 dB point in most amateur matching networks will be
quite a ways away from the frequency of interest. The Q of our networks
isn't that large. So the energy transmitted by the source at the 3 dB point
should be rather low.
> It would seem that some of it must get by the
> re-reflection point of the tuner to the source and a portion of it should be
> re-reflected to the load. Does this just continue to go on and on until it
> has made so many trips up and down the feed line that it has dissipated in
> the resistive losses of the line?
Yes. The same thing occurs with an antenna with a bad SWR. That is why it
is called SWR or Standing Wave Ratio. The reflections back and forth on the
line create a "standing" wave. On each subsequent reflection the
transmission line absorbs a little more of the signal until it is completely
attenuated due to the Ohmic, real-world losses in the line.
Every hear anyone say that a high SWR increases the loss in your line?
Well, it doesn't affect the physical loss characteristics, but you do end up
losing power.
> And what about the part that will get past
> the re-reflection point to the source. Does it not dissipate in the
> lossiness of the source output impedance. My main point is that it has to
> go somewhere! I wish I had more time to spend here but I will take the file
> with me.
Yes, it will. I am sure your reasoning behind that is to try to get back to
the idea of the source having a lossy resistance that dissipated half the
total source power at a conjugate match. However, keep in mind that the
concept of a voltage source and a resistance is a MODEL. You cannot go
inside that model using the components of the model. Your HF transmitter
may have a source or output impedance of 50 Ohms. But you will not find a
50 Ohm resistor in series with the output line, not unless it's part of an
attenuator or some other circuit function.
So once we get back to the source, our model is no longer sufficient to
describe the source. It only works on defining things outside of it.
Remember, if the efficiency calculations are to work, they must work
properly for all combinations of source and load impedance and they must
work in BOTH the Thevenin and the Norton case since both are (and can be
shown) to be equivalent.
73,
Jon
NA9D
-------------------------------------
Jon Ogden
NA9D (ex: KE9NA)
Member: ARRL, AMSAT, DXCC, NRA
http://www.qsl.net/ke9na
"A life lived in fear is a life half lived."
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