[SI-LIST] Re: Transmission lines reflections again
- From: Yafei Bi <yafei_bi@xxxxxxxxx>
- To: arpad.muranyi@xxxxxxxxx, si-list@xxxxxxxxxxxxx
- Date: Wed, 19 Oct 2005 11:30:21 -0700 (PDT)
Nicely explained.
I just hope all Professors in the school can do that,
rather than developping an intuitive understanding
after years of confusion by the students. :-)
The best lecturer I've experienced is Tom Lee of
Stanford EE...
best,
Yafei
--- "Muranyi, Arpad" <arpad.muranyi@xxxxxxxxx> wrote:
> Leonard,
>
> This is not a direct answer to your specific
> question, but I hope it helps...
>
> If you are a visual person, you can explain it
> to yourself by imagining what happens when you
> pour water into one end of a long narrow channel,
> like a gutter under your roof.
>
> The water will flow to the other end, and when it
> hits the closing wall, it splashes up. Why?
> There is nowhere to go. Imagine what happens
> when the end wall is missing? It flows straight
> out with the levels lowering down to its bottom.
> What if the impedance is matched, i.e. you have
> an identical channel filled with the same amount
> of water in it? The water wave in your first
> channel will propagate nicely into the second
> channel without splashes...
>
> Arpad Muranyi
> Intel Corporation
>
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
>
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
>
> =20
>
> -----Original Message-----
> From: si-list-bounce@xxxxxxxxxxxxx
> [mailto:si-list-bounce@xxxxxxxxxxxxx] =
> On Behalf Of Leonard Alexman
> Sent: Wednesday, October 19, 2005 10:54 AM
> To: si-list@xxxxxxxxxxxxx
> Subject: [SI-LIST] Transmission lines reflections
> again
>
>
> Hi ,
>
> I am still trying to figure out how in a simple open
> transmission line =
> the
> voltage gets doubled at the end of the line. I have
> seen the formulas =
> and
> rope drawings but only found one article that kind
> of goes into what I =
> what.
>
> The article I read had a battery connected to a
> serries50 ohm resistor =
> and a
> 50 ohm transmission line. The equivelant circuit of
> the transmission
> linsmission line is a series inductor with a
> capacitor to the return =
> path to
> the battery
> When the last capacitor in the line is charged,
> there is no voltage =
> across
> the last inductor and current flow through the last
> inductor stops. With =
> no
> current flow to maintain it, the magnetic field in
> the last inductor
> collapses and forces current to continue to flow in
> the same direction =
> into
> the last capacitor. Because the direction of current
> has not changed, =
> the
> capacitor charges in the same direction, thereby
> increasing the charge =
> in
> the capacitor. Since the energy in the magnetic
> field equals the energy =
> in
> the capacitor, the energy transfer to the capacitor
> doubles the voltage
> across the capacitor. The last capacitor is now
> charged to the battery
> voltage and the current in the last inductor drops
> to zero.
>
> My question is=20
>
> 1. Since the second to the last cap is charged to
> 1/2 the battery =
> voltage
> where does the current flow from the left end of the
> last inductor to =
> the
> bottom of the last cap in order to double the
> voltage on the last cap ?
>
> Can anyone point me to an article that explains the
> above in detail ?
>
> Leonard Alexman
>
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