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Parallel coax
On 9/28/2015 7:52 PM, Jerry Stuckle wrote:
On 9/28/2015 5:17 PM, rickman wrote: On 9/28/2015 4:33 PM, Jerry Stuckle wrote: On 9/28/2015 4:15 PM, rickman wrote: On 9/28/2015 4:02 PM, Jerry Stuckle wrote: On 9/28/2015 3:02 PM, rickman wrote: On 9/28/2015 2:51 PM, Jerry Stuckle wrote: On 9/28/2015 1:42 PM, rickman wrote: On 9/28/2015 12:54 PM, Jerry Stuckle wrote: On 9/28/2015 12:47 PM, rickman wrote: On 9/28/2015 10:38 AM, Jerry Stuckle wrote: On 9/28/2015 12:03 AM, rickman wrote: On 9/27/2015 10:39 PM, Jerry Stuckle wrote: On 9/27/2015 9:46 PM, Wayne wrote: "John S" wrote in message ... On 9/27/2015 1:20 PM, Wayne wrote: "rickman" wrote in message ... On 9/27/2015 10:41 AM, kg7fu wrote: Matching the antenna won't make the Return Loss go away but it will make the transmitter happy. Can you explain this? I thought matching the antenna would *exactly* make the return loss go away because it would eliminate the mismatch. Not wanting to put words in his mouth.... I read that to mean that the high SWR between the ATU and the antenna would remain, but the transmitter would be happy with the SWR on the transmitter/ATU coax. # Rick is correct. If the antenna (load) is matched to the line, there is # no return loss, hence no SWR. The ATU will be adjusted (hopefully) to # make the transmitter operate properly with the impedance as seen at the # transmitter end of the line. # Yes, the SWR due to mismatch of the antenna (load) and line will remain. # Even if the real part of your load impedance is matched to the line, you # will still have a high SWR if the reactance remains. # Does this make sense? Yes. That's what I was trying to say using SWR instead of return loss. Return loss numbers get bigger with lower SWR. For example: SWR 1:1 = infinite return loss. Incorrect. Return loss increases with an increased SWR. An SWR of 1:1 has no return loss because there is no returned signal to lose. 100% of the signal is radiated. From LUNA web site regarding optical measurements which should be no different from RF... It "shouldn't be" - but optical measurements are handled differently than electrical measurements. Fiber Optics have their own way of measuring loss, reflection and refraction (which doesn't exist in feedlines). That's like applying electrician's color codes to electronics. They both have color codes - but don't hook the electrician's black wire to ground - or the transformer's green wires to safety ground. I thought you would claim optical was different. That's why I included the VSWR vs return loss table link. You didn't comment on that. I didn't because I thought it was obvious. But I guess not to you. Return loss is calculated with logs. Logs of values 1 are negative. And -10db is smaller than -5 db. As the SWR approaches 1:1, the reflected power approaches 0, and the returned loss approaches NEGATIVE infinity. Note that I said NEGATIVE infinity. At the same point, the returned power measured in watts is 0. I believe that is exactly what I said in the portions of my post which you trimmed. These values for RF return loss match exactly the equation which you are saying is not used in RF. So which is it, the return loss table is correct with negative values of return loss or the equation I posted is incorrect even though it gives the values in the table? You said return loss increases with lower SWR. It does not. Are you being pedantic that -1 dB is not lower than -10 dB? It is not numerically lower in value, but is lower in magnitude and it is a lower loss. I even referred to the magnitude in my post. But then that was the same part you snipped which I referred to earlier. "It shows a higher return loss (assuming you mean magnitude since the values are all negative) for lower VSWR." No, I am not being pedantic. -1 is greater than -10. The fact it is a negative number makes all the difference - as any engineer who knows what he's talking about will tell you. Except that I clearly stated I was referring to the magnitude. Magnitude without direction is meaningless. Lol. There is only one direction for return loss. Yes - the loss is a negative gain - hence the use of negative db values. You can't just give a magnitude. You need to specify the direction, also. You seem to have a very hard time understanding such a simple concept - one that was taught in freshman physics. Ok Jerry. You just said the value can only be negative, so if the magnitude is specified it is a complete specification. I'll toss in the towel as you will never stop arguing such a silly point. -- Rick |
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