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On Sun, 23 May 2004 02:07:13 GMT, "Henry Kolesnik"
wrote: I know that any power not dissipated by an antenna is reflected back to the transmitter. Then the transmitter "reflects" this reflection back to antenna, ad nauseum until its all gone. I also know that a short or an open is required to reflect power and I'm searching for which it is, an open or a short. I'm inclined to think it's a virtual open but I'm at a loss to understand that and I wonder if someone has a good explanation or analogy and some math wouldn't hurt. tnx Hank WD5JFR Hi Hank, Round Two. A short or an open certainly reflects. However, as you have observed through your question, so does a poorly matched antenna; thus you must agree that it presents neither a short nor an open. As such, reflection is not confined to these two conditions. We can display a condition where we have a 2:1 mismatch. This is fairly commonplace as a consideration, if not as a reality. Here, the reflected power is less than the total applied (some 12% if dead reckoning is correct). That is, if the antenna presents a 2:1 mismatch to the power applied to it, nearly 90% of that power will proceed to be radiated with a trivial portion returned to the source (or the tuner). If we boost that mismatch to 10:1, that increases the reflection substantially (let's call it 90% in this spirit of dead reckoning) and naturally less is radiated. The math is quite as simple as a balance ledger. 10:1 for a 50 Ohm source would mean either the load presents a Z of 500 Ohms, or 5 Ohms. From this you can see that we are approaching either an open (hi-Z) or a short (lo-Z) and either perform the same job of reflection - short of total reflection. As such, there is no distinction to the power whether it encounters either, the reflection ensues by virtue of the simple mismatch, not by the literal condition. The computation of mismatch defines how reflective the interface is. 73's Richard Clark, KB7QHC |
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