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

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

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

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

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

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