[SI-LIST] Re: Transmsision lines
- From: ray jiang <jlray2000@xxxxxxxxx>
- To: lalexman@xxxxxxxxxxx
- Date: Sun, 16 Oct 2005 19:28:58 +0800
Hi, Leonard
First you have to believe in that voltage on a transmission line is
U(z) = U+(z) + U-(z), where U+(z) means input, U-(z) means
reflect.
given U+(z), with different loads, U-(z) is different,
in open load, U-(l)=U+(l), means all input is reflected back.
so at the load, U(l)=2U+
when load is shorted, U-(l)= - U+(l), all input is also reflected
back, but phase difference is 180 degree. This is to meet the
requiment that U(l)=0.
when load is matched, U-(l)=0, so U(l)=U+.
Hope this can help you.
--
Best Regards,
Ray Jiang
On 10/16/05, Leonard Alexman <lalexman@xxxxxxxxxxx> wrote:
>
> Hi,
>
>
> I am trying to figure out transmission lines and reflections and trying to
> understand why if the load is open or a high resistance the voltage that
> arrives at the load is doubled and the signal is them reflected back to
> the
> source. I understand there is an impedance mismatch but in all the
> articles
> I have found not explains in basic terms wht the voltage doubles and
> reflects back down the line. Can anyone point me to an article that might
> explain this in basic terms ?
>
>
>
> TIA
>
>
>
> Leonard Alexman
>
>
>
>
>
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