Cecil Moore wrote:

Tam/WB2TT wrote:
You can get the same answer from knowing that the impedance looking
into a 1/4 wave section of shorted transmission line is infinite.

Ever measure the forward and reflected currents halfway into a
shorted 1/4WL stub? How can currents be flowing unimpeded into
and out of an infinite impedance?

The question is a little misleading because the direction of the flow of
current changes every half cycle and is transverse, or orthogonal to the
direction of wave propagation. In a transmission line, the current
flows through Z0, ostensibly, which is essentially the impedance from
one conductor to the other at every point along the transmission line.
Other than that, superposed forward and reflected waves behave just as
you described, naturally.

73, Jim AC6XG

"Jim Kelley" wrote in message
...
Cecil Moore wrote:

Tam/WB2TT wrote:
You can get the same answer from knowing that the impedance looking
into a 1/4 wave section of shorted transmission line is infinite.

Ever measure the forward and reflected currents halfway into a
shorted 1/4WL stub? How can currents be flowing unimpeded into
and out of an infinite impedance?

The question is a little misleading because the direction of the flow of
current changes every half cycle and is transverse, or orthogonal to the
direction of wave propagation. In a transmission line, the current
flows through Z0, ostensibly, which is essentially the impedance from
one conductor to the other at every point along the transmission line.
Other than that, superposed forward and reflected waves behave just as
you described, naturally.

73, Jim AC6XG
There is no current in the steady state. The steady state voltage is
independent of source impedance, which affect only how long it takes to
reach that. I ran a simulation on this, and you can see that as the voltage
builds up, the current decreases

Tam

Tam/WB2TT wrote:

"Jim Kelley" wrote in message
...
Cecil Moore wrote:

Tam/WB2TT wrote:
You can get the same answer from knowing that the impedance looking
into a 1/4 wave section of shorted transmission line is infinite.

Ever measure the forward and reflected currents halfway into a
shorted 1/4WL stub? How can currents be flowing unimpeded into
and out of an infinite impedance?

The question is a little misleading because the direction of the flow of
current changes every half cycle and is transverse, or orthogonal to the
direction of wave propagation. In a transmission line, the current
flows through Z0, ostensibly, which is essentially the impedance from
one conductor to the other at every point along the transmission line.
Other than that, superposed forward and reflected waves behave just as
you described, naturally.

73, Jim AC6XG
There is no current in the steady state. The steady state voltage is
independent of source impedance, which affect only how long it takes to
reach that. I ran a simulation on this, and you can see that as the voltage
builds up, the current decreases

Tam

Hi Tam,

The simulation would be fun to play with. What do you use?

73 de jk

SWCAD from Linear. I also built up a model of an ideal SWR meter.

Tam

Tam/WB2TT wrote:

SWCAD from Linear. I also built up a model of an ideal SWR meter.

Tam

I almost bought LLTC when they came out with that. Should have. Thanks
Tam.

jk

Tam/WB2TT wrote:
There is no current in the steady state.

There is no *net* current in the steady state at the input of a
shorted 1/4WL stub. What do you think the current at the short
is? How did that large amount of current get there without flowing?
--
73, Cecil http://www.qsl.net/w5dxp



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"Cecil Moore" wrote in message
...
Tam/WB2TT wrote:
There is no current in the steady state.

There is no *net* current in the steady state at the input of a
shorted 1/4WL stub. What do you think the current at the short
is? How did that large amount of current get there without flowing?
--
73, Cecil http://www.qsl.net/w5dxp


Cecil,
I think it is more instructive to turn on a DC voltage at t=0, rather than a
sine wave. In either case, the voltage builds up step wise, with smaller and
smaller steps until you don't see them.

Tam

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Tam/WB2TT wrote:

"Cecil Moore" wrote:
There is no *net* current in the steady state at the input of a
shorted 1/4WL stub. What do you think the current at the short
is? How did that large amount of current get there without flowing?
--
I think it is more instructive to turn on a DC voltage at t=0, rather than a
sine wave. In either case, the voltage builds up step wise, with smaller and
smaller steps until you don't see them.

DC doesn't tell us anything about a 1/4WL shorted stub which is a
network problem, not a circuit problem. The net current at the mouth
of the stub is close to zero while the voltage is at a maximum. At
the short at the other end of the stub, transmission line theory holds.
The voltage is close to zero while the current is at a maximum. Now
exactly how can maximum current be flowing through that short if
no current is flowing into and out of the stub. Hint: it can't!

The forward current and reflected current cancel at the mouth of
the shorted 1/4WL stub. However, they add in-phase 1/4WL away at
the short, maybe to many amps of RF current at the shorted end.
Since there is no physical impedance at the mouth of a stub, nothing
except superposition of forward and reflected waves happens there
and nothing except a virtual impedance exists there. All of the action
is at the shorted end of the stub where there exists 100% reflection.
--
73, Cecil http://www.qsl.net/w5dxp



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"Cecil Moore" wrote in message
...
The forward current and reflected current cancel at the mouth of
the shorted 1/4WL stub. However, they add in-phase 1/4WL away at
the short, maybe to many amps of RF current at the shorted end.
Since there is no physical impedance at the mouth of a stub, nothing
except superposition of forward and reflected waves happens there
and nothing except a virtual impedance exists there. All of the action
is at the shorted end of the stub where there exists 100% reflection.
--
73, Cecil http://www.qsl.net/w5dxp

I just did the simulation again, and you are right. For a 1V signal feeding
a 50 Ohm line through a 450 Ohm series resistor, there is 20 ma of
current flowing through the short. Don't understand why I didn't see it the
other time. There is 0 current through the 450 Ohm
resistor.

Tam/WB2TT

For Gawd sake, don't tell Cecil he's right. There'll be no stopping him!