Error correction
 

Reg Edwards wrote:
A correction - insert dt instead of dz.

Another correction is in order. The resistive term should be i(z) * R
and the conductance term should be v(z) * G.


The fundamental partial differential equations of transmission lines
are -

- dv/dz = R + L*di/dt

- di/dz = G + C*dv/dt

where volts v and current i are incremental functions of distance and
time, and z is incremental distance along line.

bart

Mismatched Zo Connectors
 

I've come around to that conservation of energy stuff ;-)

I understand that your argument involves the energy that enters the
line before it knows anything about the load, the energy that enters in
an initial transient, but unless you can show that nothing happens
during the initial transient to deliver some or all of that initial
energy to the load, your argument has a hole.

You're presupposing that there is some energy that enters the line
during an initial transient that cannot leave until you shut the source
off, so you get the 100J related to the 100W net power flow and 100J
that went into the line before the source knew about the load.. and
then there's another 100J that enters somehow? I guess to set up the
reflected wave?

The argument is circular. The initial transient supplies 200J of
stored energy to the line so there must be 300J in a one second line if
there's 100J in the steady-state fields associated with power flow.
Since there's 300J in the line, the initial transient must have
supplied 200J in stored energy. It's just not working for me.


Dan

Mismatched Zo Connectors
 

wrote:
I've come around to that conservation of energy stuff ;-)

I'm glad - most folks here ignore it. :-)

I understand that your argument involves the energy that enters the
line before it knows anything about the load, the energy that enters in
an initial transient, but unless you can show that nothing happens
during the initial transient to deliver some or all of that initial
energy to the load, your argument has a hole.

Let's return to the one second long lossless transmission line.
From a 100 watt transmitter, at the end of second number one,
the line will contain 100 joules and the load will have accepted
zero joules. Since the load is rejecting 1/2 of the incident energy,
at the end of the 2nd second, the source will have supplied 200 joules,
there will be 150 joules of energy in the line, and 50 joules will have
been accepted by the load. If the source is equipped with a
circulator+load, this is steady-state with 150 joules of energy stored
in the transmission line.

At t=0:
zero joules zero joules
100w--------one-second long feedline------load rho^2=0.5
Pfor=0-- --Pref=0 Pload=0

At t=1:
100 joules zero joules
100w--------one-second long feedline------load
Pfor=100w-- --Pref=0 Pload=0

At t=2:
150 joules 50 joules
100w--------one-second long feedline------load
Pfor=100w-- --Pref=50w Pload=50w

You're presupposing that there is some energy that enters the line
during an initial transient that cannot leave until you shut the source
off, so you get the 100J related to the 100W net power flow and 100J
that went into the line before the source knew about the load.. and
then there's another 100J that enters somehow? I guess to set up the
reflected wave?

Yes, at the end of the 2nd second, the source has supplied 200 joules
and the load has accepted 50 joules. That leaves 150 joules left over
that cannot be any place except in the line according to the
conservation of energy principle. In a circulator+load system, we
have reached steady state with 150 joules in the transmission line
that will not reach the load until after the source is powered down.

The argument is circular.

Proving that confusion exists. It's actually not circular. It's
based on cause, effect, and the conservation of energy principle.
I apologize if I have not explained it in a way that is easy to
understand. Please bear with me.

It's all linear cause and effect. With an ideal auto-tuner
at the source, none of the reflected energy is accepted back by
the source. Half the energy incident upon the load is rejected. There
is no other place for the extra energy to be except inside the
transmission line. I have an EXCEL spreadsheet that might help you
sort all of this out. A copy of its output is available at:

http://www.qsl.net/w5dxp/1secsgat.gif
--
73, Cecil http://www.qsl.net/w5dxp

Mismatched Zo Connectors
 

Cecil Moore wrote:
I have an EXCEL spreadsheet that might help you
sort all of this out. A copy of its output is available at:

http://www.qsl.net/w5dxp/1secsgat.gif

The EXCEL spreadsheet corresponding to the above can be
downloaded from: http://www.qsl.net/w5dxp/1secTline.xls
It includes a graph of forward power, reflected power,
and joules stored in the transmission line.
--
73, Cecil http://www.qsl.net/w5dxp

Mismatched Zo Connectors
 

Cecil Moore wrote:
Cecil Moore wrote:
I have an EXCEL spreadsheet that might help you
sort all of this out. A copy of its output is available at:

http://www.qsl.net/w5dxp/1secsgat.gif

The EXCEL spreadsheet corresponding to the above can be
downloaded from: http://www.qsl.net/w5dxp/1secTline.xls
It includes a graph of forward power, reflected power,
and joules stored in the transmission line.

I have enhanced that spreadsheet such that the resistive
load on the one second long lossless 50 ohm feedline is
a variable entered by the user. Please note the graph
of forward power, reflected power, and joules stored in
the feedline (chart1). The enhanced file is available at:

http://www.qsl.net/w5dxp/1secline.xls
--
73, Cecil http://www.qsl.net/w5dxp

Mismatched Zo Connectors
 

Cecil,

Have you included the fact that the *source* is properly terminating
the line on its end?
The source can accept power from the reflected wave, right?

Dan

Mismatched Zo Connectors
 

wrote:
Have you included the fact that the *source* is properly terminating
the line on its end?

The source is relying on an ideal autotuner to
terminate the line with a 50 ohm Z0-match.

The source can accept power from the reflected wave, right?

No, there is an ideal autotuner on the output of the
source. *Zero reflected energy reaches the source*. The
SWR between the Source and the Autotuner is 1:1 just
as it is in any properly tuned antenna system.
--
73, Cecil http://www.qsl.net/w5dxp

Mismatched Zo Connectors
 

So how are you taking into account the stored energy in the ideal
autotuner? Even if it's perfect, there's energy stored in the fields
of the impedance transforming device. Are there 0J in the tuner? If
not, how does the impedance transformation take place?

Dan

Mismatched Zo Connectors
 

Sorry should have been "If so, how does the impedance transformation
take place" not "If not"

Dan

wrote:
Are there 0J in the tuner? If
not, how does the impedance transformation take place?

Dan

Error correction
 

Bart Rowlett wrote:
Reg Edwards wrote:

A correction - insert dt instead of dz.


Another correction is in order. The resistive term should be i(z) * R
and the conductance term should be v(z) * G.


The fundamental partial differential equations of transmission lines
are -

- dv/dz = R + L*di/dt

- di/dz = G + C*dv/dt

where volts v and current i are incremental functions of distance and
time, and z is incremental distance along line.

bart

Good to know you're still lurking around here, Bart! Glad somebody
still keeps track of units. ;-)

73, ac6xg