On Mon, 2008-07-07 at 15:39 +0800, Marinus Loewensteijn wrote:
> When Coax is terminated at the transmitter end and the far end with an
> impedance that is equal to its defined impedance we will have a SWR of 1:1.
> Only when the currents on inner conductor and on shield are equal and the
> current / voltage nodes coincide at the same locations then they'll cancel
> If we have anything else that creates a SWR that is not 1:1 then waves will
> be reflected from the ends. Hence no longer will the current and voltage
> nodes be coinciding.
So long as the shield is intact, it matters not whether the nodes match,
there is current only on the inside of the shield and that can't
> When nodes are no longer coinciding then we will have radiation.
> What is wrong about this, what am I missing here?
You are forgetting that the coax is shielded and that limits radiation.
> Thanks, 73, marinus, ZL2ML
73, Jerry, K0CQ
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