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Re: Bang the rocks together harder lumpophiles
Hi, Antonio,
In the lasts paragraph as was attempting explain that our different views
were probably due to the fact that I view it all from a propagating wave
perspective where as you appear not to make this distinction.
For example you suggest that there is no average power can flow in a driven
open circuit line and that the current is out of phase with the voltage.
Correct but I would put a directional power meter on the line and it would
show that there was average power flowing one way (in phase V and I) and
average power (in phase V and I) flowing the other way. The power sum would
be zero with the actually line V and I at 90deg.
We appear to agree an most thing. I did not understand that you where
referring to the impedance looking in to the line and I guess you did not
understand the significance of the capitalised word "propagating" after my
comment about all lines are resistive. I was referring to the wave
impedance of the line.
In these types of discussion its easy to misunderstand things.
Regards Bob
-----Original Message-----
From: Tesla List <tesla-at-pupman-dot-com>
To: tesla-at-pupman-dot-com <tesla-at-pupman-dot-com>
Date: 24 April 2000 21:33
Subject: Re: Bang the rocks together harder lumpophiles
>Original Poster: acmq-at-compuland-dot-com.br
>
>>Original Poster: "Robert Jones" <alwynj48-at-earthlink-dot-net>
>
>>>A lossless transmission line, or a lossless coil, is a purely reactive
>>>element, and so, in sinusoidal steady state conditions (CW), all
>>>voltages
>>>are at +/-90 degrees with all currents. In the mechanical analogs,
>>>this would be the relation between positions and speeds.
>>
>>I am sorry to be blunt but this is not correct the impedance of a
>>losseless transmission line is resistive, hence for a PROPAGATING
>>wave the voltage is ALWAYS in phase with the current.
>
>This is wrong. The line is open at the other end. It would look resistive
in
>
>STEADY STATE only if terminated on its characteristic resistance. It's
true
>that
>a long line behaves as a resistive element, but only for SHORT signals,
that
>
>last for less time that the required for travel along the entire line and
back
>
>to the input side. And even so, with the other end open, there is a
reflection,
>
>and current is drawn from the input after the signal returns.
>Think at what happens at DC, for example. There is no return path for the
>input current, and the open-ended line can't be considered a resistor. It's
>
>just a capacitor. It behaves as a capacitor until the first resonance. It
>alternates between inductive and capacitive phase input characteristics at
>each resonance.
>
>>Your mechanical analogue appears odd, V x I = power, position x speed=?
>
>>V x dV/dt=? I would expect force x velocity = power
>
>The equations and the behavior are similar, not the units. You can consider
>
>the tension in the cord (a force proportional to the displacement from
rest)
>
>and the speed to get power.
>
>>I dont think I understood the above paragraphs. If the line is not
>>resonating
>>the in to out phase could be anything it depends on the length of line.
At
>
>>resonance what is being measured is the phase of a standing wave with
>>respect
>>to the input. A standing wave is just that it does not move. Its the sum
of
>
>>a forward and backward travelling. It has constant phase it only
>>changes in amplitude. Its phase is pinned to the reflection at the end of
>>the line
>>that created it. If you could remove the standing wave you would be able
to
>
>>observe the progressive phase change of the input signal.
>
>With the output open, in sinusoidal steady state, in a lossless line, no
>average
>power
>can enter or leave the line, or even move along the line. At any point
where
>
>you cut the line you will observe 90 degrees phase relationship between
>voltage and current, at any frequency. Otherwise there would be energy
>moving along the line.
>
>>At resonance the input current is in phase with the input voltage thats
one
>
>>definition of resonance. ( assuming resistive drive). If the were no
losses
>
>> the input voltage would be zero ie it would look like a short.
>
>Correct, but with a resistive drive you are effectively measuring the
relation
>
>between the input current (the input voltage is just a small voltage drop
in
>
>the resistive losses of the line) and the output voltage. 90 degrees
>relationship.
>
>
>>I suspect you are talking about the relative phase of the current and
>>voltage in a standing wave which is 90deg.
>>I think there has been a lot of confusion about standing wave and
>>propagating
>>waves. This is what produced the "no phase shift so its not a
transmission
>
>>line" view.
>>In effect the two types of waves are spatial and temporal.
>>One moves in space and one moves in time.
>>The difficulty is that if you put a scope probe on either one you get the
>>same sinewave hence confusion. At first it fooled me too.
>>
>>It may be possible to have a internally consistent theory of how the coil
>>works
>>using standing waves with particular properties. I prefer the main stream
or
>
>>traditional theory of forward and backward travelling waves.
>>I missed the posting of your model I will look for it.
>
>I don't like to think about travelling waves. For short signals the concept
>is
>useful, but things become very confuse as reflected signals appear back at
>the input.
>
>Antonio Carlos M. de Queiroz
>http://www-dot-compuland-dot-com.br
>
>
>