Standing Wave Phase
 

AI4QJ wrote:
So far, I find this very interesting. Not all -j567 impedors are
equal when
it comes to transmission lines.

An impedor has an impedance but all impedances are not
impedors. The IEEE Dictionary gives three different
definitions for "impedance".
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

On Dec 11, 11:24 pm, "AI4QJ" wrote:
"Keith Dysart" wrote in message

...





On Dec 10, 11:43 pm, "AI4QJ" wrote:
I may have misspoken. Once you get to -j567 at the discontinuity
(travelling
10 degrees along the 100 ohm line), now you interface with the 600 ohm
line.
At that point you have to normalize the -j567 ohms to -j(567/600)
= -j0.945
on the smith chart (you normalize to Zo for it to calculate properly).
This
abrupt switch increases the angle from 10 degrees to arctan (0.945)) or
43
degrees. I think the effect to look for is that the abrupt impedance
change
when Zo changes. A lumped component is not enough to make the model
correct.
Comments welcome

I follow the arithmetic, and it still has a certain attractiveness
but how can it make such a difference how the -j567 is produced.

What if you were offerred 3 black boxes, each labelled -j567?
Would it make much difference what was in them?
How does one compute the phase shift at the terminals?

I use the smith chart in my response.

If you have 3 black boxes each labeled "input impedance = -j567", they could
contain a number of different things but since I am using the smith chart,
they will contain 3 different lengths of open transmission lines.

Box 1 contains 10 degrees length at 4MHz of 100 ohm transmission line. Based
on VF=1, this line would be 3/4*E10E2*10/360 = 2.08m long.

On the smith chart, plot from circle 10 degrees (transmission coefficient)
and read -j5.67. Normalize to 1 = 100. The impedance at the input of the
line is -j567. Label this box "input impedance = -j567".

Box 2 contains 19.2 degrees length at 4MHz of 200 ohm transmission line.
Based on VF=1, this line would be 3/4*10E2*19.2/360 = 4.0m long

On the smith chart, plot from infinte impedance circle 19.2 degrees
(transmission coefficient) and read -j2.84. Normalize to 1 = 200. The
impedance at the input of the line is -j(200*2.84) = -j567. Label this box
"input impedance = -j567".

Box 3 contains 27.2 degrees at 4MHz of 300 ohm transmission line. Based on
VF=1, this line would be 3/4*10E2*27.2/360 = 5.67m long.

On the smith chart, plot from infinte impedance circle 27.2 degrees
(transmission coefficient) and read -j1.89. Normalize to 1 = 300. The
impedance at the input of the line is -j(300*1.89) = -j567. Label this box
"input impedance = -j567".

So, do all boxres labeled "input impedance =-j567 ohm" transmission lines
behave the same when connected to the 600 ohm transmission line?

No.

For each of these, impedance at the discontinuity will be -j567. However,
each has different electrical lengths, thus the 600 ohm line connecting to
it will have to be cut to different electrical lengths, for all of the
degrees to add to 90 total.

You can only say that all -j567 of a given Zo will affect phase shifts in
connected 600 ohms lines in the same way.

So far, I find this very interesting. Not all -j567 impedors are equal when
it comes to transmission lines.- Hide quoted text -

To find the length of the 600 ohm line, do I not just plot -j567/600
and work from there? It seems to me that it yields the same
answer regardless of the content of the black box.

....Keith

Standing Wave Phase
 

Keith Dysart wrote:
To find the length of the 600 ohm line, do I not just plot -j567/600
and work from there? It seems to me that it yields the same
answer regardless of the content of the black box.

Double the frequency and see what happens.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

On Dec 12, 8:21 am, Cecil Moore wrote:
Keith Dysart wrote:
To find the length of the 600 ohm line, do I not just plot -j567/600
and work from there? It seems to me that it yields the same
answer regardless of the content of the black box.

Double the frequency and see what happens.

There are certainly many experiments one can conduct
to deduce an equivalent circuit for the interior of the black
box.

But for this sub-thread, which started with 90 "electical degrees",
a single frequency seems to me to be strongly implied.

....Keith

Standing Wave Phase
 

On Dec 11, 10:48 pm, Cecil Moore wrote:
Keith Dysart wrote:
Open circuit 46.6 degrees of 600 ohm line attached
to 80 degrees of 100 ohm line. The drive point impedance
will be 0 ohms. And the impedance at the junction will
be the same -j567. Using your arithmetic, there is
90-46.6-80 - -36.6 degrees of phase shift at the
junction. The physical length is 126.6 degrees
while the system is 90 "electrical degrees".

Yes

So this means I could take an antenna element
which is longer than 90 physical degrees, i.e.
greater than 1/4WL, and with the appropriate
matching network make a system that was
90 "electrical degrees".

I detect progress on the definition.

....Keith

Standing Wave Phase
 

On Wed, 12 Dec 2007 07:21:21 -0600, Cecil Moore
wrote:

Double the frequency and see what happens.

Hi Dan,

The Lottery Commission is asking to re-examine your Ticket. Not to
worry, winners WILL be notified at a later date. All that is asked is
that you be patient as some have been waiting longer than you.

73's
Richard Clark, KB7QHC

Standing Wave Phase
 

Keith Dysart wrote:
But for this sub-thread, which started with 90 "electical degrees",
a single frequency seems to me to be strongly implied.

Of course, but handicapping oneself to a steady-state
single sinusoidal frequency is like target shooting
while blindfolded and spinning on a Lazy Susan. Why
enforce a silly arbitrary handicap?
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Keith Dysart wrote:
So this means I could take an antenna element
which is longer than 90 physical degrees, i.e.
greater than 1/4WL, and with the appropriate
matching network make a system that was
90 "electrical degrees".

Of course, if you reverse the position of the
600 ohm line and 100 ohm line in the previous
open-stub example, you will *lose* 8.3 degrees
of phase shift at the impedance discontinuity.
The transmission line will have to be physically
98.3 degrees long to get an electrical 90 degree
phase shift.

In like manner, if you have a straight (Hustler)
base rod and make the rest of the antenna a
helical coil with no stinger, you will wind up
with a resonant mobile antenna that is more than
90 degrees long physically.

If the Z0 of the base rod is 400 ohms, the
Z0 of the loading coil is 4000 ohms, and the
length of the base rod is 10 degrees, you will
lose 9 degrees at the base rod-to-coil impedance
discontinuity. The coil will need to be 89 degrees
long, i.e. almost self-resonant. The antenna will
be 99 degrees long physically.
--
73, Cecil http://www.w5dxp.com

Standing Wave Phase
 

Keith Dysart wrote:

To find the length of the 600 ohm line, do I not just plot -j567/600
and work from there? It seems to me that it yields the same
answer regardless of the content of the black box.

...Keith

It darn well better.

Roy Lewallen, W7EL

Standing Wave Phase
 

Keith Dysart wrote:
I detect progress on the definition.

Are you ready to find out why center-loading takes
more coil in a mobile antenna than does base loading?

Take the 10 degrees of 100 ohm line and move 5
degrees of it to the other end of the stub.

From this:

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

To this:

--5 deg 100 ohm line--+--600 ohm line--+--5 deg 100 ohm line--open

Now how many physical degrees of 600 ohm line is needed
to make the stub electrically 1/4 wavelength?
--
73, Cecil http://www.w5dxp.com