[ExI] And Meta You Know

hkhenson hkhenson at rogers.com
Thu Jan 24 16:01:21 UTC 2008


At 10:07 PM 1/23/2008, you wrote:

>--- hkhenson <hkhenson at rogers.com> escreveu:
> >
> > Anyone need an explanation?
> >
>I kinda lost at the "Stand and Wave Ratio."
>But i would guess its a noise measure.

It's a play on "standing wave ratio."  High standing wave ratio is a 
result of impedance mismatch.

The best way to explain this is probably using a rope (the analogy is 
exact in math terms).  As a kid you or your friends may have tied a 
rope to something solid, pulled it out and induced a wave to go down 
the rope by giving it a good shake.  If you never did this as a kid, 
it's worth trying now.

Now if the rope was infinitely long, the wave would just go out and 
never come back.  But since you probably had 10 or 20 meters of rope, 
the wave would hit the place it was tied and come back to 
you.  That's a total reflection, like light bouncing off a mirror, 
and is the result of a complete mismatch.  In this case it's like a 
shorted electrical transmission line.

If you tied the rope to a "matched" damper the wave would be 
completely absorbed.

If you tied a heavy rope to a lighter one and tried the same trick, 
you would get a reflection back from the place the two were 
connected.  A really heavy rope would act like the lighter rope being 
tied to a building, the closer the ropes were in mass per unit of 
length, the smaller the reflection.

The same thing happens in electrical signal transmission lines, if 
there is a mismatch the signal bounces back from the connection.  The 
worse the mismatch, the more reflection back to the sender.

The wikipedia entry for standing wave ratio is accurate, but not much 
use if you don't have a lot of background.

Best wishes,

Keith 




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