Tom Donaly wrote:
Cecil Moore wrote:
I already did it on another thread, Tom. Adding 43 degrees
of Z0=600 ohm feedline to the 10 degrees of Z0=100 ohm
feedline will turn the stub into an electrical 1/4
wavelength (90 degree) open stub. And that's exactly
how base-loaded mobile antennas work.

It will, will it? I'm waiting for you to prove it. Do you
really expect it to be resonant at the right frequency?

I have proved it in a reply to Dan and verified it
with MicroSmith. I don't know what else you are
asking for. Yes, it will resonate at the design
frequency. Are you incapable of those simple
calculations? Note that everything is rounded
off to the nearest degree.
--
73, Cecil http://www.w5dxp.com

John Smith wrote:
You mean at the frequency where the 600 line length is 53 degrees and
the 100 line 10 degrees length ... well, I guess that already answers
your own question, doesn't it?--but then, you should have already knew
that ...

Make that 43 degrees instead of 53 degrees.
--
73, Cecil http://www.w5dxp.com

Tom Donaly wrote:
O.K., Cecil, I finally figured out what you want to do. You want
a zero ohm input impedance, just like a 1/4 wave open stub. In that
case, you're absolutely right, the 600 ohm line should be 43.387
degrees long. If you call the 100 ohm line, line 1, and the 600
ohm line, line 2, then the criterion for what you want is:
tan(Bl1)*tan(Bl2)= Z01/Z02. This behaves sort of like a backwards,
transmission-line, Helmholtz resonator. I still don't know where you
come up with the 90 degree stuff.

For an open stub to exhibit a zero ohm input impedance,
it must be electrically 90 degrees long (or 270 ...).
That's where the 90 degrees comes from. The example
stub is electrically 90 degrees long while being 53
degrees long physically.

Good for you, Tom, now you have it - "just like a 1/4WL
open stub" from 53 degrees of transmission line. Here's
another tidbit for you.

Using 600 ohm line and 100 ohms line, if you make the two
sections equal length, the dual-Z0 stub will be very close
to 1/2 the physical length of a single-Z0 stub, i.e. physically
45 degrees long for an electrical 1/4WL (90 deg) stub. On 75m,
that cuts the 1/4 stub physical length from ~66 feet to ~33
feet, a much more manageable length.

Here's a useful equation. For a 1/4WL stub with equal length
sections, the physical length in degrees of each section is:

ARCTAN[SQRT(Z0Low/Z0High)]
--
73, Cecil http://www.w5dxp.com

"Cecil Moore" wrote in message
t...
Tom Donaly wrote:
O.K., Cecil, I finally figured out what you want to do. You want
snip

Using 600 ohm line and 100 ohms line, if you make the two
sections equal length, the dual-Z0 stub will be very close
to 1/2 the physical length of a single-Z0 stub, i.e. physically
45 degrees long for an electrical 1/4WL (90 deg) stub. On 75m,
that cuts the 1/4 stub physical length from ~66 feet to ~33
feet, a much more manageable length.

Here's a useful equation. For a 1/4WL stub with equal length
sections, the physical length in degrees of each section is:

ARCTAN[SQRT(Z0Low/Z0High)]
--
73, Cecil http://www.w5dxp.com


lurking off

NEAT

lurking back on

Cecil Moore wrote:

...
Make that 43 degrees instead of 53 degrees.

Sorry, don't know if that is brain atrophy from age or the 20mg
hydrocodone the doc has me on for the fractured bone and arthritis in
the spine--laying down a bike at ~60mph is something better left for the
younger generation.

Can't seem to find my glasses after I lay 'em down--wife claims she has
already given me something, I claim no, then find 'em in my pocket.

By the way, one week in the hospital cost: Hospital $80,000, emergency
room $1,900, ambulance $1

John Smith wrote:
Cecil Moore wrote:

...
Make that 43 degrees instead of 53 degrees.

Sorry, don't know if that is brain atrophy from age or the 20mg
hydrocodone the doc has me on for the fractured bone and arthritis in
the spine--laying down a bike at ~60mph is something better left for the
younger generation.

Can't seem to find my glasses after I lay 'em down--wife claims she has
already given me something, I claim no, then find 'em in my pocket.

By the way, one week in the hospital cost: Hospital $80,000, emergency
room $1,900, ambulance $1,500, etc.

Oh yeah, and then there is constantly hitting the wrong key and sending
email early ...

Sorry guys/gals ... keep your medical PAID UP!

Regards,
JS

Cecil Moore wrote:
Tom Donaly wrote:
O.K., Cecil, I finally figured out what you want to do. You want
a zero ohm input impedance, just like a 1/4 wave open stub. In that
case, you're absolutely right, the 600 ohm line should be 43.387
degrees long. If you call the 100 ohm line, line 1, and the 600
ohm line, line 2, then the criterion for what you want is:
tan(Bl1)*tan(Bl2)= Z01/Z02. This behaves sort of like a backwards,
transmission-line, Helmholtz resonator. I still don't know where you
come up with the 90 degree stuff.

For an open stub to exhibit a zero ohm input impedance,
it must be electrically 90 degrees long (or 270 ...).
That's where the 90 degrees comes from. The example
stub is electrically 90 degrees long while being 53
degrees long physically.

(The rest deleted.)
O.k., Cecil, you said it, now prove it. There's no requirement
for a 90 degree phase shift when you do the math. Don't expect me
to do it for you this time. Since I did some math for you, you can
do some for me: Given the above formula, if you know l1, l2, and
Z01, and Z02, what's the formula for B? It should be easy, right?
73,
Tom Donaly, KA6RUH

Cecil Moore wrote:
John Smith wrote:
You mean at the frequency where the 600 line length is 53 degrees and
the 100 line 10 degrees length ... well, I guess that already answers
your own question, doesn't it?--but then, you should have already knew
that ...

Make that 43 degrees instead of 53 degrees.

Some, like me, might like to review some info like this, easily
digest-able ...:

http://courses.ece.uiuc.edu/ece450/N...sionLines2.pdf

Regards,
JS

John Smith wrote:
By the way, one week in the hospital cost: Hospital $80,000, emergency
room $1,900, ambulance $1

I'm sorry that happened. Get well quick.
--
73, Cecil http://www.w5dxp.com

Tom Donaly wrote:
O.k., Cecil, you said it, now prove it. There's no requirement
for a 90 degree phase shift when you do the math.

Are you using the equation for forward and reflected
current? If not, you need to do so. The phase shift
is not in the standing-wave current. Standing-wave
current phase is fixed with respect to the source.
Absolutely *nothing* happens to the standing-wave
current at the impedance discontinuity.

The reflected current is known to be in phase with
the forward current at the feedpoint. The forward
current is reflected at the tip of the antenna and
undergoes a 180 degree phase shift. Something must
account for the other 180 degrees or else the
feedpoint impedance would not be resistive. I am
working on a graphic that illustrates what happens
at the impedance discontinuity.

Please enlighten us on how the reflected current
gets back in phase with the forward current without
undergoing a phase shift of 180 degrees in its
round-trip path. It is my understanding that the
forward phasor rotates in one direction while the
reflected phasor rotates in the opposite direction.
The key concept there is that a phasor is always
rotating.

I have waded through the math before but I cannot
locate my notes after moving. If you can figure
out a reasonable answer to the above question, I
will certainly consider it.
--
73, Cecil http://www.w5dxp.com