> 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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