[SI-LIST] Re: Transmission lines reflections again

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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