[SI-LIST] Re: Transmission lines reflections again
- From: "Tom Biggs" <tbiggs@xxxxxxxxxxxxxxxxxxxxx>
- To: <si-list@xxxxxxxxxxxxx>
- Date: Wed, 19 Oct 2005 11:50:07 -0700
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
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3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D
=3D20
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx]
=3D 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
the voltage gets doubled at the end of the line. I have seen the
formulas =3D and rope drawings but only found one article that kind of
goes into what I =3D what.
The article I read had a battery connected to a serries50 ohm resistor =
=3D
and a 50 ohm transmission line. The equivelant circuit of the
transmission linsmission line is a series inductor with a capacitor to
the return =3D path to the battery When the last capacitor in the line =
is
charged, there is no voltage =3D across the last inductor and current =
flow
through the last inductor stops. With =3D 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 =3D into the last capacitor.
Because the direction of current has not changed, =3D the capacitor
charges in the same direction, thereby increasing the charge =3D in the
capacitor. Since the energy in the magnetic field equals the energy =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=3D20
1. Since the second to the last cap is charged to 1/2 the battery =3D
voltage where does the current flow from the left end of the last
inductor to =3D 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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=20
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