Keith Dysart wrote:
On Jan 2, 1:41 am, Roger wrote:
Keith Dysart wrote:
On Jan 1, 6:00 pm, Roger wrote:
Roy Lewallen wrote:
Roger wrote:
Could you better describe how you determine that the source has a Z0
equal to the line Z0? I can guess that you use a Thévenin equivalent
circuit and set the series resistor to Z0.
Probably the simplest way is to put the entire source circuitry into a
black box. Measure the terminal voltage with the box terminals open
circuited, and the current with the terminals short circuited. The ratio
of these is the source impedance. If you replace the box with a Thevenin
or Norton equivalent, this will be the value of the equivalent circuit's
impedance component (a resistor for most of our examples).
If the driving circuitry consists of a perfect voltage source in series
with a resistance, the source Z will be the resistance; if it consists
of a perfect current source in parallel with a resistance, the source Z
will be the resistance. You can readily see that the open circuit V
divided by the short circuit I of these two simple circuits equals the
value of the resistance.
The power output of the Thévenin equivalent circuit follows the load.
Sorry, I don't understand this. Can you express it as an equation?
There seems to be some confusion as to the terms "Thévenin equivalent
circuit", "ideal voltage source", and how impedance follows these
sources.
Two sources we all have access to are these links:
Voltage source:
http://en.wikipedia.org/wiki/Voltage_source
Thévenin equivalent circuit:http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem
I don't disagree with anything I read there.
But you may not quite have the concept of impedance
correct.
The impedance of the Thevenin/Norton equivalent source
is not V/I but rather the slope of the line representing
the relationship of the voltage to the current.
When there is a source present, this
line does not pass through the origin. Only for
passive components does this line pass through
the origin in which case it becomes V/I.
Because the voltage to current plot for an ideal
voltage source is horizontal, the slope is 0 and hence
so is the impedance. For an ideal current source
the slope is vertical and the impedance is infinite.
Hoping this helps clarify....
...Keith
Yes, I can understand that there is no change in voltage no matter what
the current load is, so there can be no resistance or reactive component
in the source. The ideal voltage link also said that the ideal source
could maintain voltage no matter what current was applied. Presumably a
negative current through the source would result in the same voltage as
a positive current, which is logical if the source has zero impedance.

This is true.

During the transition from negative current passing through the source
to positive current passing through the source, the current at some time
must be zero.

This would occur, for example, when the output terminals
are open circuited. Or connected to something that had
the same open-circuit voltage as your source.

How is the impedance of the perfect source defined at
this zero current point?

This is not a different question than: How is the
resistance of a resistor defined when it has no
current flowing in it?

It continues to satisfy the relation V = I * R,
though one can not compute R from the measured
V and I.

How is the impedance of the attached system
defined?

It is unknown, but has a voltage equal to the
voltage of the source.

The resistor in the Thevenin/Norton equivalent source is selected with
some criteria in mind. What I would like to do is to design a Thevenin
source to provide 1v across a 50 ohm transmission line INFINATELY
long, ignoring ohmic resistance. I would like the source resistor to
absorb as much power as the line, so that if power ever returns under
reflected wave conditions, it can all be absorbed by the resistor. I
think the size of such a resistor will be 50 ohms.

That is the correct value to not have any reflections
at the source.

But do not expect the power dissipated in the resistor
to increase by the same amount as the "reflected power".
In general, it will not. This is what calls into question
whether the reflected wave actually contains energy.

Do some simple examples with step functions. The math
is simpler than with sinusoids and the results do not
depend on the phase of the returning wave, but simply
on when the reflected step arrives bach at the source.

Examine the system with the following terminations on
the line: open, shorted, impedance greater than Z0,
and impedance less than Z0.

Because excitation with a step function settles to
the DC values, the final steady state condition is
easy to compute. Just ignore the transmission line
and assume the termination is connected directly
to the Thevenin generator. When the line is present,
it takes longer to settle, but the final state will
be the same with the line having a constant voltage
equal to the voltage output of the generator which
will be the same as the voltage applied to the load.

Then do the same again, but use a Norton source. You
will find that conditions which increase the dissipation
in the resistor of the Thevenin equivalent circuit
reduce the dissipation in the resistor of the Norton
equivalent circuit and vice versa.

This again calls into question the concept of power
in a reflected wave, since there is no accounting
for where that "power" goes.

...Keith

Very interesting! It makes sense to me. I must be gone most of the day
today so will be QRT for a while.

I am gaining an appreciation for the time boundaries here. To properly
account for the power, we would need to integrate the source power over
the entire time from switch on to switch off plus the time required for
the reflected wave to completely return and die out.

If we consider the extended time period, within that period the power
delivered by the source should equal the power dissipated by the
resistor. The forward and reflected waves should only serve to be
storage for the power during portions of the extended time period.

73, Roger, W7WKB