[TowerTalk] Probably dumb question

[jimlux]

On 8/14/18 12:00 PM, Jim Rhodes wrote:
> Notice the huge difference between the two sets of calculations. They 
> are "ratios" as long as all the measurements are in the same units, 
> measured on the same system you are good to go.
> 

Yeah, you could just take the square root of the power (ignoring the 
impedance, since it's the same everywhere)


So it's SWR  = (sqrt(fwd)+sqrt(refl))/(sqrt(fwd)-sqrt(refl))

or any algebraically equivalent form (there's 1+x/1-x forms, for 
instance, that you see when converting from return loss (refl/fwd = 
10^(RL/10), so

swr = (1+ sqrt(refl/fwd))/(1-sqrt(refl/fwd))




> 
> On Tue, Aug 14, 2018 at 1:08 PM jimlux <jimlux at earthlink.net 
> <mailto:jimlux at earthlink.net>> wrote:
> 
>     On 8/14/18 9:19 AM, k7lxc--- via TowerTalk wrote:
>      >      If there are 100 watts out and 10 watts reflected, what's
>     the swr?
>      >
>      >      If there are 200 watts out and 10 watts reflected, what's
>     the swr?
>      >
>      >      Tnx.
>      >
> 
>     Since SWR is the ratio between high and lowest  "voltage", you need to
>     calculate that.
> 
>     For a 50 ohm line, 100W is 70.7 V, the 10W reflected is 22.4V - so max
>     voltage is 70.7+22.4 =93.1 and min voltage is 70.7-22.4 = 48.3 so
>     VSWR =
>     93.1/48.3 or 1.93
> 
>     (the answer is impedance independent.. I just used 50 for an example)
> 
>     This is handy to remember a 2:1 is about 10dB return loss  (actually
>     it's 9.5 or something...)  1.5:1 is about 14 dB
> 
> 
> 
> 
>     For the 200W and 10W case the forward voltage is 100V, the reverse is
>     22.4 so it's 122.4/77.6  = 1.58
> 
>     (this is 13 dB return loss, so it's a bit worse than 1.5:1, which would
>     be 14 dB)
> 
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> -- 
> Jim K0XU
> jim at rhodesend.net <mailto:jim at rhodesend.net>