Brainteaser
 

On Fri, 26 May 2006 03:16:38 GMT, Cecil Moore
wrote:



If the source power is 100 watts, the forward power is 200
watts, the reflected power is 100 watts, and the load power
is 100 watts, all the joules per second needed to support
that forward and reflected power was supplied by the source
before steady-state was reached. Anything else would violate
the conservation of energy principle.

This is easily proven using a one second long lossless
transmission line as a conceptual training aid for the
uninitiated.

I gathered that from an earlier post, but I think the confusion comes
in the second second, doesn't it? 100 watts of power is generated
from the generator for two seconds. the first second, 100 watts
travels to the antenna and only 100 watts is in the transmission line.
Then for the second second, 100 forward watts is moving towards the
antenna and 50 reflected watts are returning to the transmitter for a
total of 150 watts in the transmission line? Of course, since the
transmitter isn't matched to the antenna, the reflected power is
reflected again for 25 watts being reflected back to the antenna
bringing the power from 100 forward watts to 125 watts (the reason
power meters go up in wattage when SWR rises) thus the antenna
reflects 62.5 watts which adds to the 125 for a total of 187.5 watts
in the transmission line. Sooner or later one will have a meltdown
that will make Chernoble look like a firecracker compared to a
fireworks display. lol


--
73 for now
Buck
N4PGW

Brainteaser
 

is that line 1 second long, or 1/2 second long?? a 1 second long line would
take 2 seconds worth of energy from the generator before the reflection
returned to let it know the line is terminated. this is of course very
important if you are measuring lines with tdr's where you can really see
those returned waves, of course the time it takes them to get back to the
tdr is double the one-way travel time... mess it up and you are looking for
faults twice as far down the line as you calculate.


"Cecil Moore" wrote in message
. com...
Jim Kelley wrote:
One should also carefully consider the more interesting variation of the
problem: an open transmission line. In the steady state we have 100
watts forward, 100 watts reflected, 200 Joules in the line, and 0 watts
being sourced by the generator. :-)

Expanding on my earlier response - For the first two seconds,
the source doesn't know it is looking into an open transmission
line so a 100 watt source would faithfully output 200 joules
into a one second long open circuit transmission line. That
200 joules cannot be destroyed. Is it mere coincidence that
the forward and reflected waves are 100 joules/sec*(one second),
exactly equal to the 200 joules supplied by the source?
--
73, Cecil http://www.qsl.net/w5dxp

Brainteaser
 

Buck wrote:
I gathered that from an earlier post, but I think the confusion comes
in the second second, doesn't it? 100 watts of power is generated
from the generator for two seconds. the first second, 100 watts
travels to the antenna and only 100 watts is in the transmission line.
Then for the second second, 100 forward watts is moving towards the
antenna and 50 reflected watts are returning to the transmitter for a
total of 150 watts in the transmission line? Of course, since the
transmitter isn't matched to the antenna, the reflected power is
reflected again for 25 watts being reflected back to the antenna
bringing the power from 100 forward watts to 125 watts (the reason
power meters go up in wattage when SWR rises) thus the antenna
reflects 62.5 watts which adds to the 125 for a total of 187.5 watts
in the transmission line. Sooner or later one will have a meltdown
that will make Chernoble look like a firecracker compared to a
fireworks display. lol

Actually, the forward power levels off at 200 watts during steady-
state. The reflected power levels off at 100 watts. 300 joules of
energy exist that have been sourced but not dissipated in the load.
300 joules are exactly what the forward wave and reflected wave
need to support their existence.

Here's the brainteaser configuration:

100W-SGAT---one second long lossless 50 ohm feedline---291.42 ohm load

The 'SGAT' is a signal generator equipped with a super fast auto-tuner
that re-reflects all the reflected power back toward the load so there
are no losses at the source end. The only losses in the entire system
are in the 291.42 ohm load which, to make the math easier, was chosen
for a power reflection coefficient of 0.5, i.e. half the incident
power is reflected at the load.

I will generate an EXCEL spreadsheet today that will give a second by
second summary of the powers and total energy.

In the meantime, here is a more simplified version of the brainteaser.

100W-SGCL---one second long lossless 50 ohm feedline---291.42 ohm load

The 'SGCL' is a signal generator equipped with a circulator and 50 ohm
resistor. During steady-state, half the generated power is dissipated
in the load and half in the circulator resistor. There are no re-
reflections at the source.

During the 1st second, the source supplies 100 joules into the feedline.
There is no dissipation either in the load or circulator resistor.

During the 2nd second, the source supplies another 100 joules into the
feedline, 50 joules of which are converted to heat in the load and 50
joules of which are reflected from the load as 50 watts of reflected
power. At the end of the 2nd second, the feedline contains 150 joules
of energy.

During the 3rd second, steady-state is reached. Of the 100 watts sourced,
50 watts are dissipated in the load and 50 watts are dissipated in the
circulator resistor. At the end of the 3rd second, the source has
generated 300 joules, 100 of which have made it to the load, and 50 of
which have made it to the circulator resistor.

300-100-50 = 150 joules stored in the feedline.

100 joules are stored in the 100W forward wave. 100 joules/sec = 100W

50 joules are stored in the 50W reflected wave. 50 joules/sec = 50W

This conservation of energy works for any length feedline. If the feedline
were one microsecond long instead of one second long, there would be 150
microjoules stored in the feedline, 100 microjoules in the forward wave
and 50 microjoules in the reflected wave.
--
73, Cecil http://www.qsl.net/w5dxp

Brainteaser
 

Dave wrote:
is that line 1 second long, or 1/2 second long?? a 1 second long line would
take 2 seconds worth of energy from the generator before the reflection
returned to let it know the line is terminated. this is of course very
important if you are measuring lines with tdr's where you can really see
those returned waves, of course the time it takes them to get back to the
tdr is double the one-way travel time... mess it up and you are looking for
faults twice as far down the line as you calculate.

The line is 1 second long. The source is generating 100 watts. So the
number of joules generated starting at t=0 is 100N where N is the
total number of seconds. At the end of 30 seconds, the generator will
have sourced 3000 joules which must be conserved.

I will generate an EXCEL spreadsheet at work today and post it to
my web page when I get home. It will cover the first 30 seconds
in 1 second increments. It will show that of the 3000 joules sourced
by the generator during that first 30 seconds, only 2700 joules have
reached the load. The other 300 joules are contained in the 200W
forward wave and 100W reflected wave.

(200 watts)(one second) = 200 joules in the forward wave

(100 watts)(one second) = 100 joules in the reflected wave

200 joules + 100 joules = 300 joules not delivered to the load
--
73, Cecil http://www.qsl.net/w5dxp

Brainteaser
 

On Fri, 26 May 2006 12:41:45 GMT, Cecil Moore
wrote:

Buck wrote:
All this technical detail is interesting, but in the last thirty years
of hamming, I have taken note to know the following:


My radio is most happy when matched with an SWR of less than 1.8:1
If my antenna exceeds that, either adjust it or use a tuner
The higher I raise my antenna, the better.
A yagi works better than a dipole given the same height and direction.
A dipole works better to the west than a beam pointed north.
i would rather have a high gain antenna than a high power amp.


Most importantly:

If band conditions permit, I can talk across the ocean on an antenna
with a 30 db loss, but with certain band conditions, even a 30 db gain
antenna won't make the contact!

I can't tell you how many joules are in my coax, but, if you ask, I
can tell you how many jewels are in my logbook. :)

--
73 for now
Buck
N4PGW

Brainteaser
 

"Buck" wrote:
I can't tell you how many joules are in my coax, but, if you ask, I
can tell you how many jewels are in my logbook. :)

This is not aimed at people who don't care about such things.
This is aimed at people who do care about such things especially
those who are disseminating false technical information about
such things.

Some say that since there's 100 watts in and 100 watts out
during steady-state, there's nothing left over for the reflected
waves. They apparently forgot about the time immediately
following power up when there was 100 watts in and less
than 100 watts out.
--
73, Cecil, W5DXP

Brainteaser
 

I've lost the original question.

But there is a lossless transmission line 1 second long.
With a velocity factor of 1.0 it is 30,000 kilometres long.

Let Zo = 100 ohms.
Let applied volts = 100 VDC.
Let terminating resistance be 100 ohms.

Line Inductance = 10,000 Henrys.
Line Capacitance = 1 Farad.

Under steady state conditions -
Line current = 1 amp.
Energy stored in inductance = Sqr(I)*L/2 = 5000 Joules.

Volts across capacitance = 100 volts.
Energy stored in capacitance = SqrV)*C/2 = 5000 Joules.

Total energy stored = 10, 000 Joules = 10,000 watt.seconds.

Which has nothing whatsoever to do with all this nonsense about
reflections. Note that energy in inductance equals energy in
capacitance.

(I trust I have not made a mistake with the arithmetic.)
----
Reg.

"Cecil Moore" wrote in message
. net...
"Buck" wrote:
I can't tell you how many joules are in my coax, but, if you ask,
I
can tell you how many jewels are in my logbook. :)

This is not aimed at people who don't care about such things.
This is aimed at people who do care about such things especially
those who are disseminating false technical information about
such things.

Some say that since there's 100 watts in and 100 watts out
during steady-state, there's nothing left over for the reflected
waves. They apparently forgot about the time immediately
following power up when there was 100 watts in and less
than 100 watts out.
--
73, Cecil, W5DXP

Brainteaser
 

Reg Edwards wrote:
Let Zo = 100 ohms.

For what range of frequencies is the above true?
--
73, Cecil http://www.qsl.net/w5dxp

Brainteaser
 

For what range of frequencies is the above true?
--
73, Cecil http://www.qsl.net/w5dxp

====================================
If you ask silly questions you can expect silly answers.

Are you looking for a reason to introduce your irrelevant reflections
again.
---
Reg.

Brainteaser
 

Reg Edwards wrote:
I've lost the original question.

But there is a lossless transmission line 1 second long.
With a velocity factor of 1.0 it is 30,000 kilometres long.

Let Zo = 100 ohms.
Let applied volts = 100 VDC.
Let terminating resistance be 100 ohms.

Line Inductance = 10,000 Henrys.
Line Capacitance = 1 Farad.

Under steady state conditions -
Line current = 1 amp.
Energy stored in inductance = Sqr(I)*L/2 = 5000 Joules.

Volts across capacitance = 100 volts.
Energy stored in capacitance = SqrV)*C/2 = 5000 Joules.

Total energy stored = 10, 000 Joules = 10,000 watt.seconds.

Which has nothing whatsoever to do with all this nonsense about
reflections. Note that energy in inductance equals energy in
capacitance.

Yep, you don't have to mess with those pesky reflections when the line
is terminated in a load equal to its characteristic impedance. Then the
problem reduces to the trivial one you've posed. Simple problem, simple
solution.

Roy Lewallen, W7EL