Tom Donaly wrote:
50 ohm Shorted line 8.5655 meters long. Frequency = 7 Mhz. 2 volt 50 ohm
generator. Current at input = 12.361 milliamps. Current at short =
40 milliamps. Divide the current at the imput of the line by the current
at the short and take the arc sine (in radian mode) of the result. This
is 1.2566. Now take Beta = .14671 and multiply it by the length 8.5655.
This also equals 1.2566, which is the angular length of the shorted
line. Will someone explain how this works to Cecil?

You won't understand what I am talking about until you perform
the stub experiments that I previously posted.

---600 ohm line---+---10 deg, 100 ohm line---open-circuit

How many degrees of 600 ohm line does it take to resonate
that stub to an electrical 1/4WL?

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

How many degrees of 600 ohm line does it take to resonate
that stub to an electrical 1/4WL?
--
73, Cecil http://www.w5dxp.com

On Dec 4, 11:13 pm, Cecil Moore wrote:
Tom Donaly wrote:
50 ohm Shorted line 8.5655 meters long. Frequency = 7 Mhz. 2 volt 50 ohm
generator. Current at input = 12.361 milliamps. Current at short =
40 milliamps. Divide the current at the imput of the line by the current
at the short and take the arc sine (in radian mode) of the result. This
is 1.2566. Now take Beta = .14671 and multiply it by the length 8.5655.
This also equals 1.2566, which is the angular length of the shorted
line. Will someone explain how this works to Cecil?

You won't understand what I am talking about until you perform
the stub experiments that I previously posted.

---600 ohm line---+---10 deg, 100 ohm line---open-circuit

How many degrees of 600 ohm line does it take to resonate
that stub to an electrical 1/4WL?

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

How many degrees of 600 ohm line does it take to resonate
that stub to an electrical 1/4WL?
--
73, Cecil http://www.w5dxp.com

You have said multiple times that the electrical length of a
quarter wave stub must be 90 electrical degress, so the
computation is too easy...

1) x + 10 = 90
x = 80 degrees for the 600 Ohm line
2) 5 + x + 5 = 90
x = 80 degrees for the 600 Ohm line

although I suspect others will disagree with your solution.

Keith Dysart wrote:
On Dec 4, 11:13 pm, Cecil Moore wrote:
Tom Donaly wrote:
50 ohm Shorted line 8.5655 meters long. Frequency = 7 Mhz. 2 volt 50 ohm
generator. Current at input = 12.361 milliamps. Current at short =
40 milliamps. Divide the current at the imput of the line by the current
at the short and take the arc sine (in radian mode) of the result. This
is 1.2566. Now take Beta = .14671 and multiply it by the length 8.5655.
This also equals 1.2566, which is the angular length of the shorted
line. Will someone explain how this works to Cecil?
You won't understand what I am talking about until you perform
the stub experiments that I previously posted.

---600 ohm line---+---10 deg, 100 ohm line---open-circuit

How many degrees of 600 ohm line does it take to resonate
that stub to an electrical 1/4WL?

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

How many degrees of 600 ohm line does it take to resonate
that stub to an electrical 1/4WL?

You have said multiple times that the electrical length of a
quarter wave stub must be 90 electrical degress, so the
computation is too easy...

1) x + 10 = 90
x = 80 degrees for the 600 Ohm line
2) 5 + x + 5 = 90
x = 80 degrees for the 600 Ohm line

although I suspect others will disagree with your solution.

I have not yet provided a solution. Your's is *wrong*.

The 90 degree physical solution is *wrong* because it results
in more than 90 electrical degrees. Please try again.
--
73, Cecil http://www.w5dxp.com

On Dec 5, 9:14 am, Cecil Moore wrote:
Keith Dysart wrote:
On Dec 4, 11:13 pm, Cecil Moore wrote:
You won't understand what I am talking about until you perform
the stub experiments that I previously posted.

---600 ohm line---+---10 deg, 100 ohm line---open-circuit

How many degrees of 600 ohm line does it take to resonate
that stub to an electrical 1/4WL?

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

How many degrees of 600 ohm line does it take to resonate
that stub to an electrical 1/4WL?

You have said multiple times that the electrical length of a
quarter wave stub must be 90 electrical degress, so the
computation is too easy...

1) x + 10 = 90
x = 80 degrees for the 600 Ohm line
2) 5 + x + 5 = 90
x = 80 degrees for the 600 Ohm line

although I suspect others will disagree with your solution.

I have not yet provided a solution. Your's is *wrong*.

The 90 degree physical solution is *wrong* because it results
in more than 90 electrical degrees. Please try again.

I thought that when you specified 5 and 10 degrees in your
problem statement, you meant electrical degrees. That is, the
phase shift encountered by the forward travelling wave.

Certainly, the answer was in terms of electrical degrees. That is,
the phase shift encountered by the forward travelling wave.

Or have I misunderstood the meaning of 'electrical degrees'?

Or perhaps 'electrical degrees' do not sum either?

Keith Dysart wrote:
I thought that when you specified 5 and 10 degrees in your
problem statement, you meant electrical degrees. That is, the
phase shift encountered by the forward travelling wave.

That specification is the same for physical and electrical
degrees because we are dealing with a single Z0 piece of
transmission line. The 100 ohm line is indeed 10 degrees
long both physically and electrically.

Certainly, the answer was in terms of electrical degrees. That is,
the phase shift encountered by the forward travelling wave.

You, and others, are going to be surprised to find out the
600 ohm section is only 43 degrees of physical length. How
can 43 degrees of 600 ohm line add to 10 degrees of 100 ohm
line to equal 90 electrical degrees of stub? Hint: Like I
told Roy and Tom years ago, there's a 37 degree phase shift
at the impedance discontinuity between the 600 ohm line and
the 100 ohm line. 43+37+10 = 90 electrical degrees.

Understand that simple stub example and you will understand
loaded mobile antennas. Most of the "experts" here are just
full of you-know-what.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Keith Dysart wrote:
On Dec 4, 11:13 pm, Cecil Moore wrote:
Tom Donaly wrote:
50 ohm Shorted line 8.5655 meters long. Frequency = 7 Mhz. 2 volt 50
ohm
generator. Current at input = 12.361 milliamps. Current at short =
40 milliamps. Divide the current at the imput of the line by the
current
at the short and take the arc sine (in radian mode) of the result. This
is 1.2566. Now take Beta = .14671 and multiply it by the length 8.5655.
This also equals 1.2566, which is the angular length of the shorted
line. Will someone explain how this works to Cecil?
You won't understand what I am talking about until you perform
the stub experiments that I previously posted.

---600 ohm line---+---10 deg, 100 ohm line---open-circuit

How many degrees of 600 ohm line does it take to resonate
that stub to an electrical 1/4WL?

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

How many degrees of 600 ohm line does it take to resonate
that stub to an electrical 1/4WL?

You have said multiple times that the electrical length of a
quarter wave stub must be 90 electrical degress, so the
computation is too easy...

1) x + 10 = 90
x = 80 degrees for the 600 Ohm line
2) 5 + x + 5 = 90
x = 80 degrees for the 600 Ohm line

although I suspect others will disagree with your solution.

I have not yet provided a solution. Your's is *wrong*.

The 90 degree physical solution is *wrong* because it results
in more than 90 electrical degrees. Please try again.

No, Cecil, it's your theory. You have to provide the method and
then everyone else will decide whether or not they agree with you.
You're not chicken are you?
73,
Tom Donaly, KA6RUH

Tom Donaly wrote:
No, Cecil, it's your theory. You have to provide the method and
then everyone else will decide whether or not they agree with you.
You're not chicken are you?

Actually, I wanted to see if anyone besides me could
solve the problem - no one else has.
--
73, Cecil http://www.w5dxp.com

Tom Donaly wrote:
No, Cecil, it's your theory. You have to provide the method and
then everyone else will decide whether or not they agree with you.

I appreciate your crediting me with a transmission
line theory, Tom, but after hundreds of years of
established theory, I seriously doubt that I have
discovered anything new. It is much more likely
that you slept through a few days of Fields and
Waves 301.
--
73, Cecil http://www.w5dxp.com