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@earthlink.net
<mailto:jimlux@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@rhodesend.net <mailto:jim@rhodesend.net>
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