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Re: Topband: 50 ohm direct burial coax cable‏

To: "Donald Chester" <k4kyv@hotmail.com>, <topband@contesting.com>
Subject: Re: Topband: 50 ohm direct burial coax cable‏
From: "Hardy Landskov" <n7rt@cox.net>
Date: Sat, 14 Jun 2014 18:53:39 -0700
List-post: <topband@contesting.com">mailto:topband@contesting.com>
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@hotmail.com>
To: <topband@contesting.com>
Sent: Saturday, June 14, 2014 8:19 AM
Subject: Re: Topband: 50 ohm direct burial coax cable‏





From: n7rt@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@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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