Keith Dysart wrote:
It would be much more convincing if you were to compute the
magnitude of the ghost rather than handwaving an answer.

I made the rough computation of -25 dB before I responded.
It is below the ability of the human eye to detect TV
ghosting.
--
73, Cecil http://www.w5dxp.com

Dave wrote:
forget power, forget energy, they are products of other calculations. all
that is needed is impedance and voltage OR current. once you know voltage
OR current and the impedance you can calculate all the rest.

What would you do if you were dealing with EM light waves
instead of EM RF waves? Optical physicists don't have the
luxury of dealing with "impedance and voltage OR current".
They are forced to deal with the essence of EM waves which
is photonic energy. They have done it for over a century
and have learned a lot about the nature of EM waves. They
did not "forget power, forget energy".
--
73, Cecil http://www.w5dxp.com

On Mar 24, 8:45 am, Cecil Moore wrote:
Keith Dysart wrote:
It would be much more convincing if you were to compute the
magnitude of the ghost rather than handwaving an answer.

I made the rough computation of -25 dB before I responded.
It is below the ability of the human eye to detect TV
ghosting.

Would you care to reveal the details of the calculations
so that we can locate the error(s) that lead to a non-zero
answer?

....Keith

Keith Dysart wrote:
On Mar 24, 8:45 am, Cecil Moore wrote:
I made the rough computation of -25 dB before I responded.
It is below the ability of the human eye to detect TV
ghosting.

Would you care to reveal the details of the calculations
so that we can locate the error(s) that lead to a non-zero
answer?

As I said, it was a rough computation based on your 450
ohm series padding resistor and a lot of simplified
assumptions about the configuration of the system.

In order to avoid ambiguity, would you please give us
better specifications for the source? Ramo and Whinnery
warn us not to draw any conclusions about what happens
inside a Thevenin equivalent source so your specification
must be an achievable well-defined real-world source.
How about a class-A linear amplifier with no filters,
no impedance transformation, and no protection circuitry?
Vrms/Irms = 75 ohms Vrms*Irms*cos(0) = 1 watt

The amount of interference cannot be calculated without
magnitude and phase. Shall we assume that the 450 ohm
line is an integer number of wavelengths and lossless?
i.e. Z0-matched to 75 ohms?

If you require detailed calculations, more specified
information is needed.
--
73, Cecil http://www.w5dxp.com

On Mar 24, 10:02 am, Cecil Moore wrote:
Keith Dysart wrote:
On Mar 24, 8:45 am, Cecil Moore wrote:
I made the rough computation of -25 dB before I responded.
It is below the ability of the human eye to detect TV
ghosting.

Would you care to reveal the details of the calculations
so that we can locate the error(s) that lead to a non-zero
answer?

As I said, it was a rough computation based on your 450
ohm series padding resistor and a lot of simplified
assumptions about the configuration of the system.

In order to avoid ambiguity, would you please give us
better specifications for the source? Ramo and Whinnery
warn us not to draw any conclusions about what happens
inside a Thevenin equivalent source so your specification
must be an achievable well-defined real-world source.
How about a class-A linear amplifier with no filters,
no impedance transformation, and no protection circuitry?
Vrms/Irms = 75 ohms Vrms*Irms*cos(0) = 1 watt

The amount of interference cannot be calculated without
magnitude and phase. Shall we assume that the 450 ohm
line is an integer number of wavelengths and lossless?
i.e. Z0-matched to 75 ohms?

If you require detailed calculations, more specified
information is needed.

Standard analysis can tell us quite a bit about the situation
without the need for further information. The information
we have:
- Generator with 450 Ohm output impedance
- connected directly to a 1000 foot line with 450 Ohm
characteristic impedance
- connected to a load with 75 Ohm input impedance.

Remembering that reflection coefficient is RC = (Z2-Z1)/(Z2+Z1) let
use compute the RCs at all the connection points:
RCgen-line = (450-450)/(450+450) - 0
RCline-load = (75-450)/(75+450) - -0.714
RCline-gen = (450-450)/(450+450) - 0

Let the output of the generator into a 450 Ohm load be Vout.

Forward voltage supplied by the generator into the line:
Vfwd = Vout
This is also the Vfwd that reaches the load (for the purposes of
example we are dealing with lossless lines).

The reflected voltage at the load:
Vref = Vfwd * RCline-load = Vfwd * -0.714 = Vout * -0.714

The voltage applied to the load:
Vload = Vfwd + Vref = Vout * 0.286

The voltage reflected from the load that arrives back at the
generator:
Vref = Vout * -0.714

The amount of Vref reaching the generator that is reflected back
to the load:
Vrefref = RCline-gen * Vref = 0 * Vref = 0

There for no ghosts since there is no re-reflection.

A few other things of possible interest...
Power delivered to the load:
Pload = Vload * Vload / 75 = Vout * Vout * 0.0816 / 75
= Vout * Vout * .0011
This is the actual power that flows in the line.

Forwarwd power as indicated by a directional wattmeter in the line:
Pfwd = Vfwd * Vfwd / 450
Reverse power as indicated by a directional wattmeter in the line:
Prev = Vref * Vref / 450

Net power flowing in the line:
Pfwd - Prev = (Vfwd*Vfwd)/450 - (Vref*Vref)/450
= Vout*Vout/450 - Vout*RC*Vout*RC/450
= Vout*Vout*(1-(RC*RC))/450
= Vout * Vout * 0.0011
Which is the same as the load power; this being as expected
since directional wattmeters are useful for obtaining the
net power.

It is also useful to know what we can not compute from the
provided information....

The voltage at the generator output and its phase with respect
to the current since this is dependant on the velocity factor
of the line and the signal frequency. Though if the output is DC
(e.g. a step function) than this can be easily computing for
the situation after one round trip delay.

Generator dissipation since we would need to know the internal
construction of the generator and the actual voltage and current
phase at the output of the generator.

-----

Still, using conventional techniques we have derived quite a
bit about the system without knowing the internals of the generator.

Cecil, are you saying that your techniques can not produce any
derivations without knowing the internal construction of the
generator and the length of the line (in wavelengths)?

....Keith

"Cecil Moore" wrote in message
...
Dave wrote:
forget power, forget energy, they are products of other calculations.
all that is needed is impedance and voltage OR current. once you know
voltage OR current and the impedance you can calculate all the rest.

What would you do if you were dealing with EM light waves
instead of EM RF waves? Optical physicists don't have the
luxury of dealing with "impedance and voltage OR current".
They are forced to deal with the essence of EM waves which
is photonic energy. They have done it for over a century
and have learned a lot about the nature of EM waves. They
did not "forget power, forget energy".
--
yes they do. but photons are propagating em waves, they aren't currents and
voltages, they are E and B fields. you can do the exact same calculations
using either E or B and the refractive index or epsilon/mu. again, all you
need is 2 of them, and you can compute the third, and then you know it all.

Keith Dysart wrote:
Standard analysis can tell us quite a bit about the situation
without the need for further information. The information
we have:
- Generator with 450 Ohm output impedance

There are a number of techniques to ensure that the
reflected waves encounter an impedance of 450 ohms
but none of those techniques are implemented in
amateur radio transmitters. Your generator is
obviously equipped with a circulator load of 450
ohms, 450 ohm attenuation pads, or some active
feedback. Given that, your analysis is correct but
moot. Your point is already known and accepted by
every initiated person including me.

I freely admit that if the proper circulator load
exists or if 450 ohm attenuator pads exist, they
will dissipate the reflected power by heating
up the circulator load resistor or the pads.

But it's tough to argue that there is no energy in
the reflected waves while they are being used to boil
water.
--
73, Cecil http://www.w5dxp.com

Dave wrote:
yes they do. but photons are propagating em waves, they aren't currents and
voltages, they are E and B fields. you can do the exact same calculations
using either E or B and the refractive index or epsilon/mu. again, all you
need is 2 of them, and you can compute the third, and then you know it all.

True, but how do they measure two of them? Ever
tried to measure the E-field of visible photons?
Hint: It is an irradiance (power density) time-
averaged measurement proportion to E^2.
--
73, Cecil http://www.w5dxp.com

Richard Clark wrote:
Cecil Moore wrote:
your analysis is correct but moot

Is he stealing your style?

"Moot" is an interesting word, Richard. From
Webster's - "moot - 1. a: debatable, b: disputed"
--
73, Cecil http://www.w5dxp.com

On Sat, 24 Mar 2007 11:40:37 -0500, Cecil Moore
wrote:

your analysis is correct but moot

Is he stealing your style?