Henry Kolesnik wrote:

I know that any power not dissipated by an antenna is reflected back to the
transmitter. Then the transmitter "reflects" this reflection back to
antenna, ad nauseum until its all gone. I also know that a short or an open
is required to reflect power and I'm searching for which it is, an open or a
short. I'm inclined to think it's a virtual open but I'm at a loss to
understand that and I wonder if someone has a good explanation or analogy
and some math wouldn't hurt.
tnx
Hank WD5JFR

Hi Hank,

Here's a link which talks about (and illustrates) the physics of waves
at a boundary.

http://www.physicsclassroom.com/Class/waves/U10L3a.html

It shows what happens at the two extremes for transverse waves along a
string - not unlike electromagnetic waves in a transmission line in some
ways. In one instance, the string is fastened directly to the boundary
- a rod in this case. This is analogous to a short across a
transmission line. It will be seen that a reflection occurs, and that
the wave becomes inverted upon reflecting.

In the other case, the string is fastened to a ring which can slide
freely up and down the rod. This case is analogous to an unterminated,
or open transmission line. It can be seen that this too causes a
reflection, only this time the wave is reflected back without a phase
reversal. The amplitude of the reflected wave in both these cases
equals the amplitude of the incident wave.

Now imagine that some friction between the sliding ring and the rod can
be added in varying amounts. This friction would be proportional to the
conductance in an electrical circuit, and would be infinite at one
extreme and zero at the other. The greater the friction, the greater
the conductance (and the lower the electrical resistance). As we begin
to increase the friction (conductance) from zero, the amplitude of the
inverted, reflected wave begins to decrease. The decrease in amplitude
continues with increasing friction until the amplitude of the reflected
wave becomes zero. It could be said that this value is equal to the Z0
of the transmission line.

As the amount of friction is increased still further, a small reflection
once again begins to appear. Only now the phase is opposite from what
it was before. Further increases in friction produce further increases
in reflection amplitude until the amplitude of the reflected wave once
again equals the amplitude of the incident wave.

What we noticed in the exercise is that there is some value of friction
(conductance or resistance) for which no reflection occurs. The exact
value depends on the medium through which the wave is propagating. All
other values produced a reflection.

There is no perfect explanation, and this certainly isn't a perfect
analogy but I hope that it will at least help give you a little more of
a feel for the idea.

73, Jim AC6XG

Jim Kelley wrote:
It shows what happens at the two extremes for transverse waves along a
string - not unlike electromagnetic waves in a transmission line in some
ways.

Where are the E and H fields in the string response?
--
73, Cecil http://www.qsl.net/w5dxp



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Thanks, I knew this but what I don't know is why a final doesn't dissipate
the reflected wave but just reflects 100% I assume. Transistor and tube
finals dissipate a bunch producing the RF but what is the mirror, check
valve or diode that keeps it reflecting

--
73
Hank WD5JFR
"Jim Kelley" wrote in message
...
Henry Kolesnik wrote:

I know that any power not dissipated by an antenna is reflected back to
the
transmitter. Then the transmitter "reflects" this reflection back to
antenna, ad nauseum until its all gone. I also know that a short or an
open
is required to reflect power and I'm searching for which it is, an open
or a
short. I'm inclined to think it's a virtual open but I'm at a loss to
understand that and I wonder if someone has a good explanation or
analogy
and some math wouldn't hurt.
tnx
Hank WD5JFR

Hi Hank,

Here's a link which talks about (and illustrates) the physics of waves
at a boundary.

http://www.physicsclassroom.com/Class/waves/U10L3a.html

It shows what happens at the two extremes for transverse waves along a
string - not unlike electromagnetic waves in a transmission line in some
ways. In one instance, the string is fastened directly to the boundary
- a rod in this case. This is analogous to a short across a
transmission line. It will be seen that a reflection occurs, and that
the wave becomes inverted upon reflecting.

In the other case, the string is fastened to a ring which can slide
freely up and down the rod. This case is analogous to an unterminated,
or open transmission line. It can be seen that this too causes a
reflection, only this time the wave is reflected back without a phase
reversal. The amplitude of the reflected wave in both these cases
equals the amplitude of the incident wave.

Now imagine that some friction between the sliding ring and the rod can
be added in varying amounts. This friction would be proportional to the
conductance in an electrical circuit, and would be infinite at one
extreme and zero at the other. The greater the friction, the greater
the conductance (and the lower the electrical resistance). As we begin
to increase the friction (conductance) from zero, the amplitude of the
inverted, reflected wave begins to decrease. The decrease in amplitude
continues with increasing friction until the amplitude of the reflected
wave becomes zero. It could be said that this value is equal to the Z0
of the transmission line.

As the amount of friction is increased still further, a small reflection
once again begins to appear. Only now the phase is opposite from what
it was before. Further increases in friction produce further increases
in reflection amplitude until the amplitude of the reflected wave once
again equals the amplitude of the incident wave.

What we noticed in the exercise is that there is some value of friction
(conductance or resistance) for which no reflection occurs. The exact
value depends on the medium through which the wave is propagating. All
other values produced a reflection.

There is no perfect explanation, and this certainly isn't a perfect
analogy but I hope that it will at least help give you a little more of
a feel for the idea.

73, Jim AC6XG

Henry Kolesnik wrote:
Thanks, I knew this but what I don't know is why a final doesn't dissipate
the reflected wave but just reflects 100% I assume. Transistor and tube
finals dissipate a bunch producing the RF but what is the mirror, check
valve or diode that keeps it reflecting

There is none! The definition is the problem. It is simply a copout.
Sources can dissipate reflected energy. The amount is unknown.
--
73, Cecil http://www.qsl.net/w5dxp



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