On Sat, 03 Apr 2010 12:21:22 -0500, donroden@hiwaay.net wrote:
>Or would the "zero-ohm" load absorb most of the harmonic ?
Yes, the stub burns most of the harmonic power that hits it. While we talk
about a "zero ohm load," it's really some fraction of an ohm depending on
the loss in the coax used to make the stub, so the harmonic attenuation the
stub provides is approximately equal to the ratio of the short circuit
impedance of the stub to the impedance of the line at the point of
connection (and at the frequency of the harmonic). The "approximately" comes
from fact that the calculation is more complex than that, thanks to the
reactive component of that line impedance.
73,
Jim K9YC
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