Owen Duffy wrote:
Cecil Moore wrote:
Reg, I'm curious how you would solve this stub problem
without a Smith Chart.

| 45 deg | 45 deg |
Source====Z01=========Z02====open

Stub sections are lossless. Z01 = 600 ohms and is 45
degrees long. Z02 = 50 ohms and is 45 degrees long.
What is the impedance looking into the stub from the
source?

I missed the significance of this problem Cecil.

Is it principally a theoretical (being lossless) problem that a Smith
chart can solve, or does it have some other significance?

It's just a mental exercise with a hidden significance. This
is the type of problem that I would solve with a Smith Chart.

How about a solution? What impedance does the source see? The
physical length of the stub is 90 degrees. What is the electrical
length of the stub in degrees?
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:



It's just a mental exercise with a hidden significance. This
is the type of problem that I would solve with a Smith Chart.

How about a solution? What impedance does the source see? The
physical length of the stub is 90 degrees. What is the electrical
length of the stub in degrees?

KISS

But if you did that you couldn't get this thread to last forever.

tom
K0TAR

Tom Ring wrote:
KISS

Keep It Simple Stupid?

But if you did that you couldn't get this thread to last forever.

I have kept it as simple as possible. Wonder why nobody
has ventured an answer?
--
73, Cecil http://www.qsl.net/w5dxp

"Cecil Moore" wrote in message
. ..
Owen Duffy wrote:
Cecil Moore wrote:
Reg, I'm curious how you would solve this stub problem
without a Smith Chart.

| 45 deg | 45 deg |
Source====Z01=========Z02====open

Stub sections are lossless. Z01 = 600 ohms and is 45
degrees long. Z02 = 50 ohms and is 45 degrees long.
What is the impedance looking into the stub from the
source?

I missed the significance of this problem Cecil.

Is it principally a theoretical (being lossless) problem that a Smith
chart can solve, or does it have some other significance?

It's just a mental exercise with a hidden significance. This
is the type of problem that I would solve with a Smith Chart.

How about a solution? What impedance does the source see? The
physical length of the stub is 90 degrees. What is the electrical
length of the stub in degrees?
--
73, Cecil http://www.qsl.net/w5dxp

Hi Cecil

Is it posible that the length of the "stubs" change? I'd have thought the
length of the stub is always the same. 45 degrees should always be 45
degrees, shouldnt it?? An open circuit, 45 degrees back along a 50 ohm line
looks like 50 ohms capacitive. That 50 ohms looks like something like 500
ohms inductive as viewed 45 degrees back along a 600 ohm line.
I'd guess your point is that 500 ohms of pure inductive reactance is never
seen 90 degrees back from an open, no matter what the Zo of the line

Jerry

Jerry Martes wrote:
Is it posible that the length of the "stubs" change? I'd have thought the
length of the stub is always the same. 45 degrees should always be 45
degrees, shouldnt it?? An open circuit, 45 degrees back along a 50 ohm line
looks like 50 ohms capacitive. That 50 ohms looks like something like 500
ohms inductive as viewed 45 degrees back along a 600 ohm line.
I'd guess your point is that 500 ohms of pure inductive reactance is never
seen 90 degrees back from an open, no matter what the Zo of the line

The point I'm eventually going to make is about loading coils
in mobile antennas but let's stick with the above stub example.

| 45 deg | 45 deg |
Source====Z01=========Z02====open

Z01 = 600 ohms, Z02 = 50 ohms

If the Z0 were constant and the stub was 90 degrees long,
the source would see zero ohms. Yet in our above example
the stub is physically 90 degrees long and the source sees
+j500 ohms. The above stub is electrically 130 degrees long.
There is a 45 degree delay through the Z01 section of stub.
There is a 45 degree delay through the Z02 section of stub.
There is a 40 degree phase shift at the Z01 to Z02 junction.

If we want to turn the above stub into a functional 1/4WL
open stub such that the source sees zero ohms, we can
remove 40 degrees from the Z02 section. If we make the
Z02 section 5 degrees long, the entire stub will be
electrically 90 degrees long, and 1/4WL resonant.
There will be a 45 degree delay through the Z01 section
There will be a 5 degree delay through the Z02 section
There will be a 40 degree phase shift at the Z01 to Z02
junction.
--
73, Cecil http://www.qsl.net/w5dxp

"Cecil Moore" wrote in message
m...
Jerry Martes wrote:
Is it posible that the length of the "stubs" change? I'd have thought
the length of the stub is always the same. 45 degrees should always be
45 degrees, shouldnt it?? An open circuit, 45 degrees back along a 50
ohm line looks like 50 ohms capacitive. That 50 ohms looks like
something like 500 ohms inductive as viewed 45 degrees back along a 600
ohm line.
I'd guess your point is that 500 ohms of pure inductive reactance is
never seen 90 degrees back from an open, no matter what the Zo of the
line

The point I'm eventually going to make is about loading coils
in mobile antennas but let's stick with the above stub example.


Hi Cecil I sure dont want to get involved with any mobil antenna loading
coil discussions, I admit that I'm not qualified.



| 45 deg | 45 deg |
Source====Z01=========Z02====open

Z01 = 600 ohms, Z02 = 50 ohms

If the Z0 were constant and the stub was 90 degrees long,
the source would see zero ohms.

Yeah, but is isnt a line with a constant Zo


Yet in our above example
the stub is physically 90 degrees long and the source sees
+j500 ohms. The above stub is electrically 130 degrees long.

I'd disagree with a conclusion that, just because the impedance seen by
the source is 500 ohms, the line connecting it to a load is 90 degrees
long.


There is a 45 degree delay through the Z01 section of stub.
There is a 45 degree delay through the Z02 section of stub.
There is a 40 degree phase shift at the Z01 to Z02 junction

If I disagree, do I have to get involved with some lengthy mathmatical
discussion? I'm not skilled enough to argue with you Cecil. I'm not even
smart. But, I sure dont see how anyone can conclude there is a phase shift
at the junction of two transmission lines. There is a shunt capacitive
reactance that results from that abrupt change in dimensions, but I wouldnt
have thought it would be enough to result in 40 degrees of phase shift.
..

If we want to turn the above stub into a functional 1/4WL
open stub such that the source sees zero ohms, we can
remove 40 degrees from the Z02 section. If we make the
Z02 section 5 degrees long, the entire stub will be
electrically 90 degrees long, and 1/4WL resonant.
There will be a 45 degree delay through the Z01 section
There will be a 5 degree delay through the Z02 section
There will be a 40 degree phase shift at the Z01 to Z02
junction.

I think I agree with the concept you are using. You are giving the
condition where a purely capacitive reactor of 50 ohm impedance is required
to be resonated by introducing a series inductor. A short length of higher
impedance transmission line will sure do that. The higher the line Zo,
the shorter it needs to be to resonate that 50 ohm capacitor.


--
73, Cecil http://www.qsl.net/w5dxp

Jerry

Jerry Martes wrote:
"Cecil Moore" wrote in message
Yet in our above example
the stub is physically 90 degrees long and the source sees
+j500 ohms. The above stub is electrically 130 degrees long.

I'd disagree with a conclusion that, just because the impedance seen by
the source is 500 ohms, the line connecting it to a load is 90 degrees
long.

Well, it is physically 90 degrees long because the two
physical pieces are physically 45 degrees each. That's a
given. However, the +j500 result tells us that it is
electrically 130 degrees removed from the open circuit
at the far end.

There is a 45 degree delay through the Z01 section of stub.
There is a 45 degree delay through the Z02 section of stub.
There is a 40 degree phase shift at the Z01 to Z02 junction

If I disagree, do I have to get involved with some lengthy mathmatical
discussion? I'm not skilled enough to argue with you Cecil. I'm not even
smart. But, I sure dont see how anyone can conclude there is a phase shift
at the junction of two transmission lines.

There is an abrupt change in the Gamma angle of the reflection
coefficient at the impedance discontinuity. I can show you why
on a phasor graphic. Simplified, it goes something like this.

Itotal = 21.5*sin(25) = 10*sin(65)

where 21.5 is the phasor amplitude of the current in the 50
ohm section at the junction and 10 is the phasor amplitude
of the current in the 600 ohm section at the junction.

The values must be the same even though the magnitude of Z0,
which controls the amplitude of the current, has changed.
If those values must be equal and the amplitude changes because
the Z0 changed, the only other thing that can change is the
phase angle.
--
73, Cecil http://www.qsl.net/w5dxp

"Cecil Moore" wrote in message
. ..
Jerry Martes wrote:
"Cecil Moore" wrote in message
Yet in our above example
the stub is physically 90 degrees long and the source sees
+j500 ohms. The above stub is electrically 130 degrees long.

I'd disagree with a conclusion that, just because the impedance seen by
the source is 500 ohms, the line connecting it to a load is 90 degrees
long.

Well, it is physically 90 degrees long because the two
physical pieces are physically 45 degrees each. That's a
given. However, the +j500 result tells us that it is
electrically 130 degrees removed from the open circuit
at the far end.

There is a 45 degree delay through the Z01 section of stub.
There is a 45 degree delay through the Z02 section of stub.
There is a 40 degree phase shift at the Z01 to Z02 junction

If I disagree, do I have to get involved with some lengthy mathmatical
discussion? I'm not skilled enough to argue with you Cecil. I'm not
even smart. But, I sure dont see how anyone can conclude there is a
phase shift at the junction of two transmission lines.

There is an abrupt change in the Gamma angle of the reflection
coefficient at the impedance discontinuity. I can show you why
on a phasor graphic. Simplified, it goes something like this.

Itotal = 21.5*sin(25) = 10*sin(65)

where 21.5 is the phasor amplitude of the current in the 50
ohm section at the junction and 10 is the phasor amplitude
of the current in the 600 ohm section at the junction.

The values must be the same even though the magnitude of Z0,
which controls the amplitude of the current, has changed.
If those values must be equal and the amplitude changes because
the Z0 changed, the only other thing that can change is the
phase angle.
--
73, Cecil http://www.qsl.net/w5dxp

Hi Cecil

Thanks for pointing me toward learning about reflection coefficient. I
am really surprised that there is such a large amount of phase shift at that
junction.

Jerry

Jerry Martes wrote:
Thanks for pointing me toward learning about reflection coefficient. I
am really surprised that there is such a large amount of phase shift at that
junction.

W8JI has theorized an even larger phase shift there at the
junction of a loading coil and the stinger of a mobile antenna.
Unfortunately, he attributes 100% of the phase shift below
the stinger to that single junction point, while ignoring
the phase shift provided by the loading coil. This thread
is my attempt at setting the technical record straight and
correcting W8JI's new old wives' tale. This discussion
started years ago on QRZ.com where both sides were wrong
and nobody was 100% technically correct, including myself.
--
73, Cecil http://www.qsl.net/w5dxp