On Jul 1, 8:53*am, K1TTT wrote:
On Jul 1, 12:37*pm, Cecil Moore wrote:





On Jun 30, 11:29*am, Keith Dysart wrote:

Check the a0 coefficient in the Fourier transform. This represents
the DC component of the signal.

And the result is zero EM waves, either forward or reflected, and your
argument falls apart.

Without this, how would you deal with a signal such as
* V(t) = 10 + 2 cos(3t)

If the cosine term is zero, there are zero EM waves, either forward or
reflected, and your argument falls apart.

Incidentally, V(t) = 10, is a perfect way to prove that energy and the
time derivitive of energy are not the same thing and your argument
falls apart.

Alternatively, one can use the standard trick for dealing with
non-repetitive waveforms: choose an arbitrary period. 24 hours
would probably be suitable for these examples and transform from
there. Still, you will have zero frequency component to deal
with, but there will be some at higher frequencies (if you
choose your function to make it so).

Windowing doesn't generate EM waves where none exist in reality and
your argument falls apart.
--
73, Cecil, w5dxp.com

a better argument is that a constant voltage produces a constant
electric field everywhere, since the field is not varying in time or
space there is no time or space derivative to create a magnetic field
so there can be no propagating em wave. *you could do the same with
zero or constant current producing a constant magnetic field.

The same question for you...

With an infinitely long transmission line excited by a step function,
is there an EM wave propagating down the line?

If not, what is it that is propagating down the line? Especially at
the leading edge?

essentially the dc case IS unique in that you must wait forever for it
to reach sinusoidal steady state since the lowest frequency component
is 0hz

You have used similar phrases before. Are you suggesting that an
open circuited transmission line excited with a step function takes
infinitely long to read steady state?

....Keith

On Jul 5, 1:26*am, Keith Dysart wrote:
On Jul 1, 8:53*am, K1TTT wrote:



On Jul 1, 12:37*pm, Cecil Moore wrote:

On Jun 30, 11:29*am, Keith Dysart wrote:

Check the a0 coefficient in the Fourier transform. This represents
the DC component of the signal.

And the result is zero EM waves, either forward or reflected, and your
argument falls apart.

Without this, how would you deal with a signal such as
* V(t) = 10 + 2 cos(3t)

If the cosine term is zero, there are zero EM waves, either forward or
reflected, and your argument falls apart.

Incidentally, V(t) = 10, is a perfect way to prove that energy and the
time derivitive of energy are not the same thing and your argument
falls apart.

Alternatively, one can use the standard trick for dealing with
non-repetitive waveforms: choose an arbitrary period. 24 hours
would probably be suitable for these examples and transform from
there. Still, you will have zero frequency component to deal
with, but there will be some at higher frequencies (if you
choose your function to make it so).

Windowing doesn't generate EM waves where none exist in reality and
your argument falls apart.
--
73, Cecil, w5dxp.com

a better argument is that a constant voltage produces a constant
electric field everywhere, since the field is not varying in time or
space there is no time or space derivative to create a magnetic field
so there can be no propagating em wave. *you could do the same with
zero or constant current producing a constant magnetic field.

The same question for you...

With an infinitely long transmission line excited by a step function,
is there an EM wave propagating down the line?

If not, what is it that is propagating down the line? Especially at
the leading edge?

essentially the dc case IS unique in that you must wait forever for it
to reach sinusoidal steady state since the lowest frequency component
is 0hz

You have used similar phrases before. Are you suggesting that an
open circuited transmission line excited with a step function takes
infinitely long to read steady state?

...Keith

'it depends'... in the special case you have concocted where the
signal source has no reflections it only takes one round trip. this
case is very misleading if you try to extend it to cover other cases.
in general it takes infinitely long and you must account for the
infinite series of reflections. that is why the approximations used
to come up with the sinusoidal steady state solution is so useful, and
exactly why it can not be applied to steps and square waves and other
non sinusoidal constant sources.


and in your infinite line example it never reaches steady state so the
step wave propagates forever.

On Jul 5, 6:19*am, K1TTT wrote:
On Jul 5, 1:26*am, Keith Dysart wrote:





On Jul 1, 8:53*am, K1TTT wrote:

On Jul 1, 12:37*pm, Cecil Moore wrote:

On Jun 30, 11:29*am, Keith Dysart wrote:

Check the a0 coefficient in the Fourier transform. This represents
the DC component of the signal.

And the result is zero EM waves, either forward or reflected, and your
argument falls apart.

Without this, how would you deal with a signal such as
* V(t) = 10 + 2 cos(3t)

If the cosine term is zero, there are zero EM waves, either forward or
reflected, and your argument falls apart.

Incidentally, V(t) = 10, is a perfect way to prove that energy and the
time derivitive of energy are not the same thing and your argument
falls apart.

Alternatively, one can use the standard trick for dealing with
non-repetitive waveforms: choose an arbitrary period. 24 hours
would probably be suitable for these examples and transform from
there. Still, you will have zero frequency component to deal
with, but there will be some at higher frequencies (if you
choose your function to make it so).

Windowing doesn't generate EM waves where none exist in reality and
your argument falls apart.
--
73, Cecil, w5dxp.com

a better argument is that a constant voltage produces a constant
electric field everywhere, since the field is not varying in time or
space there is no time or space derivative to create a magnetic field
so there can be no propagating em wave. *you could do the same with
zero or constant current producing a constant magnetic field.

The same question for you...

With an infinitely long transmission line excited by a step function,
is there an EM wave propagating down the line?

If not, what is it that is propagating down the line? Especially at
the leading edge?

essentially the dc case IS unique in that you must wait forever for it
to reach sinusoidal steady state since the lowest frequency component
is 0hz

You have used similar phrases before. Are you suggesting that an
open circuited transmission line excited with a step function takes
infinitely long to read steady state?

...Keith

'it depends'... in the special case you have concocted where the

'Concocted has such perjorative ring to it. Much better would be
'appropriately selected to illustrate a point'!

signal source has no reflections it only takes one round trip. *

Excellent. Some agreement.

this
case is very misleading if you try to extend it to cover other cases.
in general it takes infinitely long and you must account for the
infinite series of reflections. *

Of course. But this illustrates one of the benefits of "appropriately
selecting" examples. One can choose examples that do not take forever
to settle and therefore can be analyzed in finite time.

that is why the approximations

To which approximations do you refer?

used
to come up with the sinusoidal steady state solution is so useful, and
exactly why it can not be applied to steps and square waves and other
non sinusoidal constant sources.

Are you suggesting that it is inappropriate to use the reflection
coefficient computed at an impedance discontinuity to predict the
behaviour of a transmission line excited with a 'step, square wave or
other non sinusoidal constant sources"?

and in your infinite line example it never reaches steady state so the
step wave propagates forever

So is this 'step wave' an EM wave, according to your definition of an
EM wave? If not, what would you call it?

....Keith

On Jul 6, 12:17*am, Keith Dysart wrote:
On Jul 5, 6:19*am, K1TTT wrote:



On Jul 5, 1:26*am, Keith Dysart wrote:

On Jul 1, 8:53*am, K1TTT wrote:

On Jul 1, 12:37*pm, Cecil Moore wrote:

On Jun 30, 11:29*am, Keith Dysart wrote:

Check the a0 coefficient in the Fourier transform. This represents
the DC component of the signal.

And the result is zero EM waves, either forward or reflected, and your
argument falls apart.

Without this, how would you deal with a signal such as
* V(t) = 10 + 2 cos(3t)

If the cosine term is zero, there are zero EM waves, either forward or
reflected, and your argument falls apart.

Incidentally, V(t) = 10, is a perfect way to prove that energy and the
time derivitive of energy are not the same thing and your argument
falls apart.

Alternatively, one can use the standard trick for dealing with
non-repetitive waveforms: choose an arbitrary period. 24 hours
would probably be suitable for these examples and transform from
there. Still, you will have zero frequency component to deal
with, but there will be some at higher frequencies (if you
choose your function to make it so).

Windowing doesn't generate EM waves where none exist in reality and
your argument falls apart.
--
73, Cecil, w5dxp.com

a better argument is that a constant voltage produces a constant
electric field everywhere, since the field is not varying in time or
space there is no time or space derivative to create a magnetic field
so there can be no propagating em wave. *you could do the same with
zero or constant current producing a constant magnetic field.

The same question for you...

With an infinitely long transmission line excited by a step function,
is there an EM wave propagating down the line?

If not, what is it that is propagating down the line? Especially at
the leading edge?

essentially the dc case IS unique in that you must wait forever for it
to reach sinusoidal steady state since the lowest frequency component
is 0hz

You have used similar phrases before. Are you suggesting that an
open circuited transmission line excited with a step function takes
infinitely long to read steady state?

...Keith

'it depends'... in the special case you have concocted where the

'Concocted has such perjorative ring to it. Much better would be
'appropriately selected to illustrate a point'!

signal source has no reflections it only takes one round trip. *

Excellent. Some agreement.

this
case is very misleading if you try to extend it to cover other cases.
in general it takes infinitely long and you must account for the
infinite series of reflections. *

Of course. But this illustrates one of the benefits of "appropriately
selecting" examples. One can choose examples that do not take forever
to settle and therefore can be analyzed in finite time.

that is why the approximations

To which approximations do you refer?

used
to come up with the sinusoidal steady state solution is so useful, and
exactly why it can not be applied to steps and square waves and other
non sinusoidal constant sources.

Are you suggesting that it is inappropriate to use the reflection
coefficient computed at an impedance discontinuity to predict the
behaviour of a transmission line excited with a 'step, square wave or
other non sinusoidal constant sources"?

and in your infinite line example it never reaches steady state so the
step wave propagates forever

So is this 'step wave' an EM wave, according to your definition of an
EM wave? If not, what would you call it?

...Keith

correct, the 'step wave' is not AN EM wave, it is an infinite
summation of EM waves.

On Jul 6, 6:01*pm, K1TTT wrote:
On Jul 6, 12:17*am, Keith Dysart wrote:
On Jul 5, 6:19*am, K1TTT wrote:

On Jul 5, 1:26*am, Keith Dysart wrote:

On Jul 1, 8:53*am, K1TTT wrote:

On Jul 1, 12:37*pm, Cecil Moore wrote:

On Jun 30, 11:29*am, Keith Dysart wrote:

Check the a0 coefficient in the Fourier transform. This represents
the DC component of the signal.

And the result is zero EM waves, either forward or reflected, and your
argument falls apart.

Without this, how would you deal with a signal such as
* V(t) = 10 + 2 cos(3t)

If the cosine term is zero, there are zero EM waves, either forward or
reflected, and your argument falls apart.

Incidentally, V(t) = 10, is a perfect way to prove that energy and the
time derivitive of energy are not the same thing and your argument
falls apart.

Alternatively, one can use the standard trick for dealing with
non-repetitive waveforms: choose an arbitrary period. 24 hours
would probably be suitable for these examples and transform from
there. Still, you will have zero frequency component to deal
with, but there will be some at higher frequencies (if you
choose your function to make it so).

Windowing doesn't generate EM waves where none exist in reality and
your argument falls apart.
--
73, Cecil, w5dxp.com

a better argument is that a constant voltage produces a constant
electric field everywhere, since the field is not varying in time or
space there is no time or space derivative to create a magnetic field
so there can be no propagating em wave. *you could do the same with
zero or constant current producing a constant magnetic field.

The same question for you...

With an infinitely long transmission line excited by a step function,
is there an EM wave propagating down the line?

If not, what is it that is propagating down the line? Especially at
the leading edge?

essentially the dc case IS unique in that you must wait forever for it
to reach sinusoidal steady state since the lowest frequency component
is 0hz

You have used similar phrases before. Are you suggesting that an
open circuited transmission line excited with a step function takes
infinitely long to read steady state?

...Keith

'it depends'... in the special case you have concocted where the

'Concocted has such perjorative ring to it. Much better would be
'appropriately selected to illustrate a point'!

signal source has no reflections it only takes one round trip. *

Excellent. Some agreement.

this
case is very misleading if you try to extend it to cover other cases.
in general it takes infinitely long and you must account for the
infinite series of reflections. *

Of course. But this illustrates one of the benefits of "appropriately
selecting" examples. One can choose examples that do not take forever
to settle and therefore can be analyzed in finite time.

that is why the approximations

To which approximations do you refer?

used
to come up with the sinusoidal steady state solution is so useful, and
exactly why it can not be applied to steps and square waves and other
non sinusoidal constant sources.

Are you suggesting that it is inappropriate to use the reflection
coefficient computed at an impedance discontinuity to predict the
behaviour of a transmission line excited with a 'step, square wave or
other non sinusoidal constant sources"?

and in your infinite line example it never reaches steady state so the
step wave propagates forever

So is this 'step wave' an EM wave, according to your definition of an
EM wave? If not, what would you call it?

...Keith

correct, the 'step wave' is not AN EM wave, it is an infinite
summation of EM waves

Well at least there is no attempt at diversion here.

What is the shape of these EM waves of which there is an infinite
number which sum to a step?

....Keith

On Jul 4, 8:26*pm, Keith Dysart wrote:
Are you suggesting that an
open circuited transmission line excited with a step function takes
infinitely long to read steady state?

Your infinitely long open-circuited transmission line example
certainly takes infinitely long to reach steady-state so the leading-
edge EM wave continues forever with zero reflected EM waves and your
argument involving reflected waves falls apart.

Your finite open-circuited transmission line example reaches DC steady-
state where EM waves cease to exist so your argument involving forward
and reflected waves falls apart.
--
73, Cecil, w5dxp.com

On Jul 5, 1:16*pm, Cecil Moore wrote:
On Jul 4, 8:26*pm, Keith Dysart wrote:

Are you suggesting that an
open circuited transmission line excited with a step function takes
infinitely long to read steady state?

Your infinitely long open-circuited transmission line example
certainly takes infinitely long to reach steady-state so the leading-
edge EM wave continues forever with zero reflected EM waves and your
argument involving reflected waves falls apart.

Your finite open-circuited transmission line example reaches DC steady-
state where EM waves cease to exist so your argument involving forward
and reflected waves falls apart.
--
73, Cecil, w5dxp.com

'dc steady state' is an oxymoron...

On Jul 5, 8:38*am, K1TTT wrote:
'dc steady state' is an oxymoron...

Webster's says an "oxymoron" is self-contradictory. "DC transient
state" would be an oxymoron. "DC steady-state" is merely
redundant. :-)
--
73, Cecil, w5dxp.com

On 5 jul, 10:16, Cecil Moore wrote:
On Jul 4, 8:26*pm, Keith Dysart wrote:

Are you suggesting that an
open circuited transmission line excited with a step function takes
infinitely long to read steady state?

Your infinitely long open-circuited transmission line example
certainly takes infinitely long to reach steady-state so the leading-
edge EM wave continues forever with zero reflected EM waves and your
argument involving reflected waves falls apart.

Your finite open-circuited transmission line example reaches DC steady-
state where EM waves cease to exist so your argument involving forward
and reflected waves falls apart.
--
73, Cecil, w5dxp.com

Hi Richard, good day:

Again you give me another rethoric answer... Please, tell us how to
measure to distinguish Osc. A from Osc. B, having Osc. A 4*10^28
quanta and Osc. B 4*10^28 +1 quanta, having each 80 m quantum 2.3 *
10^-19 J.

Your answers are making me remember = "It was the only explicit
answer you will ever get" or "Superman's cataracts with his xray
vision. This is probably going to be your only direct
answer." (Please do not go upsetting, I am joking).

You dislike my examples, you dislike R & H & K classic and obviously
really good peer reviewed book reference (and examples), you dislike
university notes, you dislike analogies, you dislike Sagan... Today I
know all things you dislike, what I do not know is how measure A and B
oscillator to distinguish each other... :D
Remember, you are rebutting things stated in standard university
physics book, does not reverse the burden of proof.
Please be a good boy, be plain and do not resort to old tricks such as
posting esoteric rocket science hiper-specialized incomprehensible
answers :)

73 Miguel Ghezzi - LU6ETJ

On 5 jul, 14:20, lu6etj wrote:
On 5 jul, 10:16, Cecil Moore wrote:





On Jul 4, 8:26*pm, Keith Dysart wrote:

Are you suggesting that an
open circuited transmission line excited with a step function takes
infinitely long to read steady state?

Your infinitely long open-circuited transmission line example
certainly takes infinitely long to reach steady-state so the leading-
edge EM wave continues forever with zero reflected EM waves and your
argument involving reflected waves falls apart.

Your finite open-circuited transmission line example reaches DC steady-
state where EM waves cease to exist so your argument involving forward
and reflected waves falls apart.
--
73, Cecil, w5dxp.com

Hi Richard, good day:

Again you give me another rethoric answer... Please, tell us how to
measure to distinguish Osc. A *from *Osc. B, having Osc. A 4*10^28
quanta and Osc. B 4*10^28 +1 quanta, having each 80 m quantum 2.3 *
10^-19 J.

Your answers are making me remember = "It was the only explicit
answer you will ever get" or "Superman's cataracts with his xray
vision. *This is probably going to be your only direct
answer." (Please do not go upsetting, I am joking).

You dislike my examples, you dislike R & H & K classic and obviously
really good peer reviewed book reference (and examples), you dislike
university notes, you dislike analogies, you dislike Sagan... Today I
know all things you dislike, what I do not know is how measure A and B
oscillator to distinguish each other... :D
Remember, you are rebutting things stated in standard university
physics book, does not reverse the burden of proof.
Please be a good boy, be plain and do not resort to old tricks such as
posting esoteric rocket science hiper-specialized incomprehensible
answers :)

73 Miguel Ghezzi - LU6ETJ- Ocultar texto de la cita -

- Mostrar texto de la cita -

SRI, I ommited to say the example of the University of New Mexico link
it is similar to the one given in "Physics for scientists and
engineers" (Serway & Beichner, my copy is in spanish). They say the
same about it.

Humoroues note: Richard Feynman do not share your dislike for
analogies he compare corks in water with charged objects fields :)