I don't think that you can do much better than if you regard the output of
the pi network of an amp as a low pass filter. It is not resonant at the
harmonic frequency and acts just as a low pass filter with a low source
impedance at the output of the amp at the harmonic frequency.
Placing the stub a quarter wave away from the output of the amp you will
have it at a relatively high impedance point. If at that point (where the
stub is attached) the impedance of the line is modified by any reflections
from the antenna end of the line it will only help. If the reflection causes
a higher impedance then the stub will still provide the same amount of
attenuation. If there is a lower impedance caused then that lower impedance
will further attenuate the harmonic somewhat as if there was another stub
attached to the line.
73
Gary K4FMX
> -----Original Message-----
> From: towertalk-bounces@contesting.com [mailto:towertalk-
> bounces@contesting.com] On Behalf Of Jim Brown
> Sent: Tuesday, March 30, 2010 7:58 PM
> To: Tower and HF antenna construction topics.
> Subject: Re: [TowerTalk] coax stubs /bandpass filter
>
> On Tue, 30 Mar 2010 18:56:34 EDT, TexasRF@aol.com wrote:
>
> >Wonder how you go about determining where that is? Could you assume the
> >plate load capacitor is a point of minimum impedance and go from there?
>
> I see it as far more complex than that. The output of a tube stage is
> essentially the output Z of the tube transformed by the output network,
> where the output network is the plate tank and everything between it,
> the
> coax connector, and the transmission line. And, because it's a passive
> network, the TX end of the transmission line is part of the filter! And,
> of
> course, the transformation that the network provides is different at
> every
> frequency.
>
> >If so, the high impedance points would occur every odd quarter
> wavelength
> >(at the suppression frequency) along the cable, right?
>
> Well, they certainly vary with that spacing, but it's hard to know where
> "zero" is. :)
>
> Another point. Any filter that you build from discrete components
> presents
> equivalent problems with respect to analysis. It's convenient to design
> a
> filter assuming fixed values of source and load impedance, but that is a
> LONG way from the real world. You could easily measure 60 dB of
> attenuation
> from a 50 ohm generator into 50 ohms resistive, but if the source is 30
> -j10
> and the load is really 500 - j300 you'll get a LOT less attenuation!
>
> 73,
>
> Jim Brown K9YC
>
>
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>
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