W5DXP wrote:
A directional wattmeter reads 133.33W forward power.
100W + 33.33W = 133.33W. 100W/|1-rho|^2 = 133.33W
Forward power - reflected power = power delivered to the load.
The standing waves confirm that there is indeed 133.33W forward and
33.33W reflected.

So where does the 33.33 watts actually go? The 100 watt number seems to
account for all of the transfer of energy.

Let's look at the following system with two sources. Each source is a
signal generator equipped with a circulator and load. SGCL1 is equipped
with a 50 ohm circulator and load and SGCL2 is equipped with a 150 ohm
circulator and load. The two signal generators are phase locked but can
be turned on and off independently.

rho=0.5
100W SGCL1---50 ohm feedline---+---150 ohm feedline---33.33W SGCL2
Pfwd1-- Pfwd2--
--Pref1 --Pref2

Using the principle of superposition: With SGCL1 on and SGCL2 off,
Pfwd1=100W, Pref1=25W, Pfwd2=75W, and Pref2=0W. SGCL2 dissipates 75W
(Pfwd1)(|rho|^2) and SGCL1 dissipates 25W (Pfwd1)(1-|rho|^2).


With SGCL2 on and SGCL1 off, Pfwd1=0W, Pref1=25W, Pfwd2=8.33W,
Pref2=33.33W. SGCL2 dissipates 8.33W (Pref2)(|rho|^2) and SGCL1 dissipates
25W (Pref2)( 1-|rho|^2). This is how the s12 and s22 s-parameters are
measured.

Note that SGCL1 dissipates 25W in both of the above cases.

With SGCL1 and SGCL2 both on, Pfwd1=100W, Pref1=0W, Pfwd2=133.33W, and
Pref2=33.33W. This is similar to the earlier single source example.

Except for the amount of power input. Although you didn't mention it
explicitly, I assume SGCL2 inputs 33.33 watts. Obviously there's that
much more power being input to the network in this example that the
others we've worked on.

Note
that the two 25W waves obviously engage in wave cancellation and their
combined 50W of destructive interference joins the forward wave as constructive
interference.

I note that interference takes place.

Assume there is a set of feedline lengths that will accomplish
the given values. Vref2 must be 180 degrees out of phase with Vfwd1 at the
impedance discontinuity. I think 1WL of 50 ohm feedline and 1/2WL of 150 ohm
feedline will do that if the voltages from the two sources are max+ at the
same time.

Well, you won't get a phase reversal from reflection at the 150/50 ohm
boundary as you would with the 150/450 ohm boundary in our other
problem, so I think the 150 ohm feedline would need to be an odd number
of 1/4WL's in order to put Vref1 out of phase with Vfwd1 at the first
boundary. But since we're really interested in the relative phase
between Vref2 and Vref1, and Vref1 becomes inverted at the 50/150 ohm
boundary, you would want to maintain the phase of Vref2 by using an even
multiple of 1/4 wave lengths of the 150 ohm line.

b1 = s11*a1 + s12*a2 = 0V s11*a1 and s12*a2 engage in destructive interference.

Pref1 = Pfwd1(|rho|^2) + Pref2(1-|rho|^2) - destructive interference
0W = 25W + 25W - 50W

Pfwd2 = Pfwd1(1-|rho|^2) + Pref2(|rho|^2) + constructive interference
133.33W = 75W + 8.33W = 50W

I'll trade you a plus sign for your equal sign. ;-)

We also know that 133.33W = 100W + 33.33W.
So 100W = 100W + 33.33W - 33.33W. Which is equally revealing. ;-)

73, Jim AC6XG

Jim Kelley wrote:
So where does the 33.33 watts actually go? The 100 watt number seems to
account for all of the transfer of energy.

I told you, the 33.33W joins the 100W forward wave at the impedance
discontinuity and heads toward the load where a new 33.33W is rejected
by the mismatched load. Maybe that is the point of your confusion. The
33.33W that makes a round trip to the impedance discontinuity and back
to the load is not the same 33.33W that is reflected by the load. TV
ghosting proves that to be true.


Let's look at the following system with two sources. Each source is a
signal generator equipped with a circulator and load. SGCL1 is equipped
with a 50 ohm circulator and load and SGCL2 is equipped with a 150 ohm
circulator and load. The two signal generators are phase locked but can
be turned on and off independently.


rho=0.5
100W SGCL1---50 ohm feedline---+---150 ohm feedline---33.33W SGCL2
Pfwd1-- Pfwd2--
--Pref1 --Pref2

Except for the amount of power input. Although you didn't mention it
explicitly, I assume SGCL2 inputs 33.33 watts.

"33.33W SGCL2" seems pretty explicit to me.

Obviously there's that
much more power being input to the network in this example that the
others we've worked on.

We are also taking out more power than we previously were.

I note that interference takes place.

Please note that the 25W reflected when SGCL1 only is on is 35.36V at
zero deg, and 0.707A at 180 deg. When SGCL2 only is on, the 25W not
re-reflected is 35.36V at 180 deg, and 0.707A at zero deg. That's why
they cancel when both are on.


Assume there is a set of feedline lengths that will accomplish
the given values. Vref2 must be 180 degrees out of phase with Vfwd1 at the
impedance discontinuity. I think 1WL of 50 ohm feedline and 1/2WL of 150 ohm
feedline will do that if the voltages from the two sources are max+ at the
same time.

Well, you won't get a phase reversal from reflection at the 150/50 ohm
boundary as you would with the 150/450 ohm boundary in our other
problem, so I think the 150 ohm feedline would need to be an odd number
of 1/4WL's in order to put Vref1 out of phase with Vfwd1 at the first
boundary. But since we're really interested in the relative phase
between Vref2 and Vref1, and Vref1 becomes inverted at the 50/150 ohm
boundary, you would want to maintain the phase of Vref2 by using an even
multiple of 1/4 wave lengths of the 150 ohm line.

Nope, you are wrong about that but I'll let you figure our your own error.
The only voltage phase reversal in the given example is in the re-reflected
wave associated with the 8.33W. Vfwd1 travels 1WL while Vref2 travels 1/2WL.
That is enough to put them 180 degrees out of phase. Vfwd1(|rho|^2) undergoes
zero phase shift. Vref2(1-|rho|^2) undergoes zero phase shift.
--
73, Cecil http://www.qsl.net/w5dxp
"One thing I have learned in a long life: that all our science, measured
against reality, is primitive and childlike ..." Albert Einstein



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