Topband: 50 ohm direct burial coax cable‏

Sat Jun 14 21:53:39 EDT 2014

Don,
Tnx for your reply. I see exactly whar's happening. In fact I do the same 
thing with my Delta Loop on 160. It's about 110 to 120 ohms at the 
feedpoint, so I use a ¼ wavelength of 75 ohm Belden down to a ½ wavelength 
of RG213. And that's it. The confusing point is there are two impedance 
variations going on as you change frequency because of changing line Zo 
values:
1) The impedance of the antenna is changing as you go up/down in frequency, 
and
2) The impedance transformation of the ¼ wave matching section is changing 
because it is getting shorter/ longer in electrical length as you go up/down 
in frequency.

So now there are two variables that you can't really do anything about. 
Therefore, the 2:1 bandwidth narrows the more frequency dependent variables 
there are in the matching network. There is a paper which was written by 
Fano which comes to the conclusion that about 4 matching elements is the 
practical limit of wideband matching. Anyway please Google: "Fano"  and I am 
sure you will find a lot of info if you are interested.

This is not rocket science. I must apologise because I had never heard the 
terms "apparent", "virtual" before. Everything Don has said is fine. It was 
not clear until I drew it on the Smith Chart.

This Delta Loop has sure worked for me !!

73 Hardy N7RT

----- Original Message ----- 
From: "Donald Chester" <k4kyv at hotmail.com>
To: <topband at contesting.com>
Sent: Saturday, June 14, 2014 8:19 AM
Subject: Re: Topband: 50 ohm direct burial coax cable‏

>
>
>
>> From: n7rt at cox.net
>
>>
>> Please explain virtual SWR. I never heard that in any college classroom I
>> have been in.
>> Hardy N7RT
>>
>
> That's a phrase I coined in response to the situation Tom described; maybe 
> another term would be used in the textbooks. A quarter wavelength 75 ohm 
> coax working into a 50 ohm load, transforms the 50 ohm load to 112.5 ohms, 
> non-reactive, as it appears at the end of the coax next to the 
> transmitter, as previously discussed. If we place a 50-ohm SWR meter at 
> the near end of the coax, between it and the rf source (the transmitter), 
> the meter will "see" 112.5 ohms, not 50 ohms nor 75 ohms. It will read 
> 2.25:1 SWR. But the actual SWR on the coax line remains 1.5:1. If we add 
> another quarter-wave of coax to the line, making its total length a half 
> wavelength, the meter will now "see" a 50 ohm non-reactive load, and read 
> 1:1 SWR. The SWR READING has changed from 1.25:1 to 1:1 as the length was 
> changed. But the actual SWR along the transmission line has not changed, 
> because Zo and Zr are still the same, and adjusting the length of the line 
> doesn't affect the standing wave generated by the 1.5:1 mismatch at the 
> far end. That meter reading is what I called 'virtual SWR'. Maybe a better 
> term would have been APPARENT SWR.
>
> Don k4kyv
>
>
>
> ----------------------------------------
>
>>> From: w8ji at w8ji.com
>>
>>> The worse case SWR of a 50 ohm system with 75 ohm cable isn't 1.5:1 when
>>> normalized to 50 ohms. It is 2.25:1. 1.5*1.5 = 2.25
>>>
>>> A 50 ohm load with 1/4 wave of 75 ohm is 112.5 ohms, and that is 2.25:1.
>>> This is why the cable needs to be 1/2 wave long, so impedance is back
>>> around
>>> 50 ohms. If you are unlucky and pick an odd 1/4 wave, and the load is 
>>> 50,
>>> the input SWR is 2.25 in the lossless cable case at the radio.
>>
>> The SWR on the line is still 1.5:1. SWR= Zr/Zo or Zo/Zr, whichever case
>> gives a ratio greater than one. Zr is the load at the far end of the
>> transmission line, and Zo is the characteristic impedance of the line. In
>> the above case, Zr=50 ohms and Zo=75 ohms. Thus, SWR=75/50=1.5
>>
>> The quarter-wave line (or odd multiple thereof) is a special case, in 
>> which
>> the line acts as a transformer. The impedance looking into the line, Zs =
>> (Zo)^2/Zr
>>
>> In the above case, Zs=(75)^2/50 = 5626/50 = 112.5 IOW, the transmitter
>> "sees" a 112.5 ohm load looking into the line instead of a 50 ohm load,
>> because the quarter-wave line has "transformed" the impedance. 
>> Consequently,
>> the tuning network at the output of the transmitter would have to be 
>> tweaked
>> in order for the final amplifier to be properly matched to the load. A
>> 50-ohm SWR meter inserted between the transmitter and the transmission 
>> line
>> would indeed read 2.25:1 - but this is only a virtual reading. The actual
>> SWR along the feedline, which by definition is the ratio of the maximum 
>> line
>> voltage to the minimum line voltage, would be 1.5:1
>>
>> The longtime confusion between real and virtual SWR as read on a meter 
>> has
>> led to a popular misconception in the amateur community that trimming the
>> length of coax can reduce or eliminate standing waves.
>
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