[SI-LIST] Re: Transmission lines reflections again

Skin effect, if you will.

QED?

Bill Hargin

> -----Original Message-----
> From: si-list-bounce@xxxxxxxxxxxxx=20
> [mailto:si-list-bounce@xxxxxxxxxxxxx] On Behalf Of Todd=20
> Westerhoff (twesterh)
> Sent: Thursday, October 20, 2005 10:56 AM
> To: si-list@xxxxxxxxxxxxx
> Subject: [SI-LIST] Re: Transmission lines reflections again
>=20
> You see?
>=20
> Good signal integrity =3D3D=3D3D good skin tones.
>=20
> Q.E.D.
>=20
> Todd ;-)
>=20
>=20
> Todd Westerhoff
> High Speed Design Group Manager
> Cisco Systems
> 1414 Massachusetts Ave - Boxboro, MA - 01719 email:twesterh@xxxxxxxxx
> ph: 978-936-2149
> =
=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D=
3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3
D=3D3D=3D3D=3D3D=3D3D=3D
> =
=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D=
3D=3D3D=3D3D=3D3D=3D3D
> =3D20
> "Always do right.
>  This will gratify some people and astonish the rest."
> =3D20
> - Mark Twain
>=20
>=20
> -----Original Message-----
> From: si-list-bounce@xxxxxxxxxxxxx=20
> [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
>=20
> Interesting thing is, lots of Olympic pools are designed to=20
> 'impedence match' at the sides (with water-level gutters) to=20
> reduce reflected waves. Also, the racing lanes have floats=20
> that are designed to act as resistances to reduce lane-to-lane waves.
>=20
>     -tom
>=20
> -----Original Message-----
> From: si-list-bounce@xxxxxxxxxxxxx=20
> [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
>=20
>=20
> Leonard,
>=20
> This is not a direct answer to your specific question, but I=20
> hope it helps...
>=20
> If you are a visual person, you can explain it to yourself by=20
> imagining what happens when you pour water into one end of a=20
> long narrow channel, like a gutter under your roof.
>=20
> The water will flow to the other end, and when it hits the=20
> closing wall, it splashes up.  Why?
> There is nowhere to go.  Imagine what happens when the end=20
> wall is missing?  It flows straight out with the levels=20
> lowering down to its bottom.
> What if the impedance is matched, i.e. you have an identical=20
> channel filled with the same amount of water in it?  The=20
> water wave in your first channel will propagate nicely into=20
> the second channel without splashes...
>=20
> Arpad Muranyi
> Intel Corporation
> =
=3D3D3D3D=3D3D3D3D=3D3D3D3D=3D3D3D3D=3D3D3D3D=3D3D3D3D=3D3D3D3D=3D3D3D3D=3D=
3D3D3
D=3D3D3D3D=3D3D=3D
> 3D3D=3D3D3D3D=3D3D3D3D=3D3D3D3D=3D3D3
> D=3D3D
> =
3D=3D3D3D3D=3D3D3D3D=3D3D3D3D=3D3D3D3D=3D3D3D3D=3D3D3D3D=3D3D3D3D=3D3D3D3=
D=3D3D3D3D
> =3D3D3D3D=3D3D3D
> =
=3D3D3D3D=3D3D3D3D=3D3D3D3D=3D3D3D3D=3D3D3D3D=3D3D3D3D=3D3D3D3D=3D3D3D3D=3D=
3D3D3
D=3D3D3D3D=3D3D=3D
> 3D3D=3D3D3D3D=3D3D3D3D=3D3D3D3D=3D3D3
> D=3D3D
> =
3D=3D3D3D3D=3D3D3D3D=3D3D3D3D=3D3D3D3D=3D3D3D3D=3D3D3D3D=3D3D3D3D=3D3D3D3=
D=3D3D3D3D
>=20
> =3D3D3D20
>=20
> -----Original Message-----
> From: si-list-bounce@xxxxxxxxxxxxx=20
> [mailto:si-list-bounce@xxxxxxxxxxxxx]
> =3D3D3D On Behalf Of Leonard Alexman
> Sent: Wednesday, October 19, 2005 10:54 AM
> To: si-list@xxxxxxxxxxxxx
> Subject: [SI-LIST] Transmission lines reflections again
>=20
>=20
> Hi ,
>=20
> I am still trying to figure out how in a simple open=20
> transmission line =3D =3D3D =3D3D3D the voltage gets doubled at the=20
> end of the line. I have seen the formulas =3D3D3D and rope=20
> drawings but only found one article that kind of goes into=20
> what I =3D3D3D what.
>=20
> The article I read had a battery connected to a serries50 ohm=20
> resistor =3D =3D3D =3D3D3D and a 50 ohm transmission line. The=20
> equivelant circuit of the transmission linsmission line is a=20
> series inductor with a capacitor to the return =3D3D3D path to=20
> the battery  When the last capacitor in the =3D line =3D3D is=20
> charged, there is no voltage =3D3D3D across the last inductor=20
> and current =3D3D flow through the last inductor stops. With=20
> =3D3D3D no current =3D flow to maintain =3D3D it, the magnetic=20
> field in the last inductor collapses =3D and forces current to=20
> continue to flow in the same direction =3D3D3D into the last =
capacitor.
> Because the direction of current has not changed, =3D3D3D the=20
> capacitor charges in the same direction, thereby increasing=20
> the charge =3D3D3D in =3D the capacitor. Since the energy in the=20
> magnetic field equals the energy =3D =3D3D3D =3D3D in the=20
> capacitor, the energy transfer to the capacitor doubles the=20
> voltage across the capacitor. The last capacitor is now=20
> charged to the battery voltage and the current in the last=20
> inductor drops to zero.
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> My question is=3D3D3D20
>=20
> 1. Since the second to the last cap is charged to 1/2 the=20
> battery =3D3D3D voltage where does the current flow from the=20
> left end of the last inductor to =3D3D3D the bottom of the last=20
> cap in order to double the =3D3D voltage on the last cap ?
>=20
> Can anyone point me to an article that explains the above in detail ?
>=20
> Leonard Alexman
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