Standing Wave Phase
 

David Ryeburn wrote:
...
This makes me wonder.

David, ex-W8EZE


And, you complete the circle--isn't that what we came to the hobby
for--to wonder? And, just when you think you have "wondered"
everything--it happens again ...

Regards,
JS

Standing Wave Phase
 

David Ryeburn wrote:
However, loaded mobile antennas presumably radiate, at least a little, and
my analysis (and W5DXP's discussion of angle lengths of transmission lines
and "phase shift" at their junction) is for *LOSSLESS* transmission lines.
This makes me wonder.

The current conditions for a mobile antenna can be simulated
by a lossy transmission line made out of resistance wire.
The equations simply include the losses.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Cecil Moore wrote:
David Ryeburn wrote:
However, loaded mobile antennas presumably radiate, at least a little,
and my analysis (and W5DXP's discussion of angle lengths of
transmission lines and "phase shift" at their junction) is for
*LOSSLESS* transmission lines. This makes me wonder.

The current conditions for a mobile antenna can be simulated
by a lossy transmission line made out of resistance wire.
The equations simply include the losses.

The energy content of a 1/2WL dipole is only about 20%
radiated which means that 80% of the energy doesn't
radiate, quite like a lossy transmission line.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

On Dec 13, 5:27 pm, Cecil Moore wrote:
Cecil Moore wrote:
David Ryeburn wrote:
However, loaded mobile antennas presumably radiate, at least a little,
and my analysis (and W5DXP's discussion of angle lengths of
transmission lines and "phase shift" at their junction) is for
*LOSSLESS* transmission lines. This makes me wonder.

The current conditions for a mobile antenna can be simulated
by a lossy transmission line made out of resistance wire.
The equations simply include the losses.

The energy content of a 1/2WL dipole is only about 20%
radiated which means that 80% of the energy doesn't
radiate, quite like a lossy transmission line.

Can you expand on the 20%, 80% above.

For convenience assuming the dipole is lossless, it seems
to me that after the transmitter is turned on, some of the
energy is stored in the antenna, but once the antenna
is charged, all the energy entering the antenna is radiated
until the transmitter is turned off, after which the energy
stored in the antenna is radiated until the antenna stores
no energy.

So everything that goes in to the antenna is radiated.

....Keith

Standing Wave Phase in Loaded Mobile Antennas
 

Cecil Moore wrote:
John Smith wrote:
There is no "built in" cotan function on my ti-86, ti-83, etc.

Poor guy - why can't you do cotangent functions in
your head? :-)

Help me out Cecil, anyone?

How about:

cot(x) = tan(90-x)

cot(10) = tan(80) = 5.67

Geesh, I had it, but sure didn't look right ... a walk around the block
and letting things go put all back to right ...

CoT(10) = (1/tan(10))

damn ... moments like these are humbling ...

Sometimes my progress forward is "inversely proportional" to the energy
I apply ... ;-)

Regards,
JS

Standing Wave Phase
 

Keith Dysart wrote:
For convenience assuming the dipole is lossless, it seems
to me that after the transmitter is turned on, some of the
energy is stored in the antenna, but once the antenna
is charged, all the energy entering the antenna is radiated
until the transmitter is turned off, after which the energy
stored in the antenna is radiated until the antenna stores
no energy.

So everything that goes in to the antenna is radiated.

Before steady-state is reached, a considerable amount of
energy is stored in the standing waves following key-down
(your "charging" time). That energy stored in the standing
waves will not radiate until the source is disconnected,
e.g. after key-up.

A 1/2WL dipole is a standing-wave antenna. The energy
radiated from such an antenna is considerably less than
the energy stored in the standing-waves. The SWR on the
antenna is probably around 20:1.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

On Dec 13, 11:16 pm, Cecil Moore wrote:
Keith Dysart wrote:
For convenience assuming the dipole is lossless, it seems
to me that after the transmitter is turned on, some of the
energy is stored in the antenna, but once the antenna
is charged, all the energy entering the antenna is radiated
until the transmitter is turned off, after which the energy
stored in the antenna is radiated until the antenna stores
no energy.

So everything that goes in to the antenna is radiated.

Before steady-state is reached, a considerable amount of
energy is stored in the standing waves following key-down
(your "charging" time). That energy stored in the standing
waves will not radiate until the source is disconnected,
e.g. after key-up.

I follow the principle, but I am not convinced that it is a
"considerable" amount of energy.

A 1/2WL dipole is a standing-wave antenna. The energy
radiated from such an antenna is considerably less than
the energy stored in the standing-waves. The SWR on the
antenna is probably around 20:1.

Consider a quarter wave-length of open circuited line connected
to a 100 W source. After 1/2 cycle it is fully charged. At 4 MHz
it stores 0.125E-6 J of energy. I can not see a mechanism
where a 1/2 wavelength antenna would store more than this
(SWR on the line would be inifinity, much worse than 20:1),
and it will be radiating 100 J/s. 0.125E-6 J is not much in a
system that is moving 100 J/s.

....Keith

Standing Wave Phase
 

Keith Dysart wrote:
Consider a quarter wave-length of open circuited line connected
to a 100 W source. After 1/2 cycle it is fully charged. At 4 MHz
it stores 0.125E-6 J of energy.

Only if the 100 W source is turned off after the first
1/2 cycle. If it is a constant power source and is not
turned off, the forward and reflected power in that stub
would increase without bounds except for I^2*R losses
and dielectric losses. Only when the power output of
the source equals the losses in the stub will steady-
state be reached and that could be when the forward
power is 1000 watts or more.

I can not see a mechanism
where a 1/2 wavelength antenna would store more than this

As I said above, if you have a constant power source,
the energy in that stub would increase without bounds.
The forward power is certainly not limited to the source
power. The source is only having to supply the losses in
the stub. The stub could be storing magnitudes more energy
than your calculation.

(SWR on the line would be inifinity, much worse than 20:1),
and it will be radiating 100 J/s. 0.125E-6 J is not much in a
system that is moving 100 J/s.

As pointed out above, your logic is flawed. I estimate
that with 100 watts being fed into the dipole and 100
watts being radiated from the dipole, the forward power
is about 500 watts and the reflected power is about
400 watts at the feedpoint of the 1/2WL dipole.

There is also another flaw in your logic. You are equating
the length of time it takes to charge the stub to one
second so you are off by almost 10 magnitudes.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Cecil Moore wrote:
Keith Dysart wrote:
Consider a quarter wave-length of open circuited line connected
to a 100 W source. After 1/2 cycle it is fully charged. At 4 MHz
it stores 0.125E-6 J of energy.

Only if the 100 W source is turned off after the first
1/2 cycle. If it is a constant power source and is not
turned off, the forward and reflected power in that stub
would increase without bounds except for I^2*R losses
and dielectric losses. Only when the power output of
the source equals the losses in the stub will steady-
state be reached and that could be when the forward
power is 1000 watts or more.

This is, of course, assuming that the constant power
source is seeing its designed-for impedance as it
would through a lossless matching network.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Keith Dysart wrote:
I follow the principle, but I am not convinced that it is a
"considerable" amount of energy.

This is a follow up to my other reply. An example
closer to the 1/2WL dipole would be a Z0=600 ohm
1/4WL open stub made out of resistance wire such
that the input impedance is 50+j0 ohms. The resistance
wire simulates the radiation "loss" in a 1/2WL dipole.

100w---600 ohm 1/4WL stub---open

The 100w source sees a 50 ohm load because the stub
is made out of resistance wire.

What is the forward power on the stub at the feedpoint? _______

What is the reflected power on the stub at the feedpoint? _______

Stubs like this one are easily modeled with EZNEC.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Cecil Moore wrote:
Stubs like this one are easily modeled with EZNEC.

In fact, here it is:

http://www.w5dxp.com/stub_dip.EZ
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

On Dec 14, 11:17 am, Cecil Moore wrote:
Cecil Moore wrote:
Stubs like this one are easily modeled with EZNEC.

In fact, here it is:

http://www.w5dxp.com/stub_dip.EZ

So how much energy is stored in the stub?

Compared to the 100 J dissipated each second?

....Keith

Standing Wave Phase
 

Cecil Moore wrote:

---43.4 deg 600 ohm line---+---10 deg 100 ohm line---open

---43.4 deg 300 ohm line---+---10 deg 50 ohm line---open

So how long does the 600 ohm line have to be in the following
example for the stub to exhibit 1/4WL of electrical length?

---??? deg 600 ohm line---+---10 deg 50 ohm line---open

Strange, for the 600/100/10 degree I get:

Arctan(100\600)*(1/tan(10)) = 53.6635

for 300/50/10 degree:

Arctan(50\300)*(1/tan(10)) = 53.6635

and for 600/50/10 degree:

Arctan(50\600)*(1/tan(10)) = 27.0160

What am I missing here?

Arctan(50\600)*(tan(80)) still equ 27.0160

Don't tell me the calc is on the fritz!

Regards,
JS

Standing Wave Phase
 

Keith Dysart wrote:
On Dec 14, 11:17 am, Cecil Moore wrote:
Cecil Moore wrote:
Stubs like this one are easily modeled with EZNEC.
In fact, here it is:

http://www.w5dxp.com/stub_dip.EZ

So how much energy is stored in the stub?

Compared to the 100 J dissipated each second?

You are still comparing apples and oranges.

Why not compare it to the 6000 joules dissipated
each minute? Or the 360,000 joules dissipated each
hour? Or the 8+ megajoules dissipated each day?

The length of time that we need to use for a fair
comparison is the length of time it takes the forward
energy to propagate from one end of the stub to the
other. That time is about 62.63 ns for a 4 MHz 1/4WL
stub, or 6.263E-8 seconds.

100 joules/sec times 6.263E-8 seconds is
6.263E-6 joules or 6.263 microjoules lost to
radiation. That's 6.263 microjoules per 62.63 ns
so the power remains the same.

The forward power is about 31 microjoules per
62.63 ns. The reflected power is about 25
microjoules per 62.63 ns.

The forward energy is about five time the radiated
energy. The reflected energy is about four times
the radiated energy. That's why the standing-wave
current completely swamps the traveling-wave current
such that it is extremely difficult to use that
current for phase measurements.

Make the lossless stub one second long plus 1/4WL
and then recalculate the energy stored in the stub.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

John Smith wrote:
Cecil Moore wrote:

---43.4 deg 600 ohm line---+---10 deg 100 ohm line---open

---43.4 deg 300 ohm line---+---10 deg 50 ohm line---open

So how long does the 600 ohm line have to be in the following
example for the stub to exhibit 1/4WL of electrical length?

---??? deg 600 ohm line---+---10 deg 50 ohm line---open

Strange, for the 600/100/10 degree I get:

Arctan(100\600)*(1/tan(10)) = 53.6635

for 300/50/10 degree:

Arctan(50\300)*(1/tan(10)) = 53.6635

and for 600/50/10 degree:

Arctan(50\600)*(1/tan(10)) = 27.0160

What am I missing here?

Your calculator is not missing anything. Hard to
tell what you are missing. :-)

Arctan(50\600)*(tan(80)) still equ 27.0160

Don't tell me the calc is on the fritz!

Nope, but your brain seems to be. Everything
above is correct.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Cecil Moore wrote:

...
Nope, but your brain seems to be. Everything
above is correct.

Cecil:

Oh gesus, that ain't the half of it, I won't bore you with details!

Thanks for the double check ... between the smith chart and attempting
to apply Ryeburns' input and attempting to work out maths' to figure
loading coil impedances in a useful way(s) ... well, I just ain't a guru.

When you are rich and famous and have full fledged Guru status--remember
us little guys ... ;-)

Regards,
JS