[SI-LIST] Re: Transmission lines reflections again

But whose signal integrity creme do you recommend?
At 01:55 PM 10/20/2005 -0400, Todd Westerhoff \(twesterh\) wrote:
>You see?
>
>Good signal integrity =3D=3D good skin tones.
>
>Q.E.D.
>
>Todd ;-)
>
>
>Todd Westerhoff
>High Speed Design Group Manager
>Cisco Systems
>1414 Massachusetts Ave - Boxboro, MA - 01719 email:twesterh@xxxxxxxxx
>ph: 978-936-2149
>=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
>"Always do right.
>  This will gratify some people and astonish the rest."
>=20
>- Mark Twain
>
>
>-----Original Message-----
>From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx]
>On Behalf Of Tom Biggs
>Sent: Wednesday, October 19, 2005 2:50 PM
>To: si-list@xxxxxxxxxxxxx
>Subject: [SI-LIST] Re: Transmission lines reflections again
>
>Interesting thing is, lots of Olympic pools are designed to 'impedence
>match' at the sides (with water-level gutters) to reduce reflected
>waves. Also, the racing lanes have floats that are designed to act as
>resistances to reduce lane-to-lane waves.
>
>     -tom
>
>-----Original Message-----
>From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx]
>On Behalf Of Muranyi, Arpad
>Sent: Wednesday, October 19, 2005 11:19 AM
>To: si-list@xxxxxxxxxxxxx
>Subject: [SI-LIST] Re: Transmission lines reflections again
>
>
>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
>=3D3D3D=3D3D3D=3D3D3D=3D3D3D=3D3D3D=3D3D3D=3D3D3D=3D3D3D=3D3D3D=3D3D3D=3D=
>3D3D=3D3D3D=3D3D3D=3D3D3D=3D3
>D=3D
>3D=3D3D3D=3D3D3D=3D3D3D=3D3D3D=3D3D3D=3D3D3D=3D3D3D=3D3D3D=3D3D3D
>=3D3D3D=3D3D
>=3D3D3D=3D3D3D=3D3D3D=3D3D3D=3D3D3D=3D3D3D=3D3D3D=3D3D3D=3D3D3D=3D3D3D=3D=
>3D3D=3D3D3D=3D3D3D=3D3D3D=3D3
>D=3D
>3D=3D3D3D=3D3D3D=3D3D3D=3D3D3D=3D3D3D=3D3D3D=3D3D3D=3D3D3D=3D3D3D
>
>=3D3D20
>
>-----Original Message-----
>From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx]
>=3D3D 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 =
>=3D
>=3D3D the voltage gets doubled at the end of the line. I have seen the
>formulas =3D3D and rope drawings but only found one article that kind of
>goes into what I =3D3D what.
>
>The article I read had a battery connected to a serries50 ohm resistor =
>=3D
>=3D3D and a 50 ohm transmission line. The equivelant circuit of the
>transmission linsmission line is a series inductor with a capacitor to
>the return =3D3D path to the battery  When the last capacitor in the =
>line
>=3D is charged, there is no voltage =3D3D across the last inductor and
>current =3D flow through the last inductor stops. With =3D3D no current =
>flow
>to maintain =3D it, the magnetic field in the last inductor collapses =
>and
>forces current to continue to flow in the same direction =3D3D into the
>last capacitor.
>Because the direction of current has not changed, =3D3D the capacitor
>charges in the same direction, thereby increasing the charge =3D3D in =
>the
>capacitor. Since the energy in the magnetic field equals the energy =
>=3D3D
>=3D 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=3D3D20
>
>1. Since the second to the last cap is charged to 1/2 the battery =3D3D
>voltage where does the current flow from the left end of the last
>inductor to =3D3D the bottom of the last cap in order to double the =3D
>voltage on the last cap ?
>
>Can anyone point me to an article that explains the above in detail ?
>
>Leonard Alexman
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