"Dr. Slick" wrote:
ok, Keith, i look forward with great interest on your
imaginary passive circuit which can reflect more power than
what you feed it (incident power).

I can't wait to hook it up to see more reflected power than
incident on my DAIWA meter, that would be very interesting.

I would never claim there is a passive circuit which violated
the rules of conservation of energy but if you define
A = Vf**2/Z0
and
B = Vr**2/Z0
then there are circuits for which B is greater than A and if you
accept the often held view that Pfwd = A and Prev = B then YOU
are also claiming the Pref is greater than Pfwd. I claim that A
and B don't, in reality, have much to do with power at all.

On the other hand Vr Vf has been demonstrated by several
examples of which Roy's August 20 post

http://groups.google.ca/groups?dq=&hl=en&lr=&ie=UTF-8
&threadm=vl88o381tks0cc%40corp.supernews.com
&prev=/groups%3Fdq%3D%26num%3D25%26hl%3Den%26lr%3D%26ie%3 DUTF-8
%26group%3Drec.radio.amateur.antenna%26start%3D25

is just one example.

This example had a line with Z0 of 68-j39 ohms connected to a load
with impedance 10+j50 ohms.

While this example demonstrated lossy lines, for this analysis
we can simplify the line to its Thevenin equivalent:
- an ideal voltage source
- producing a 10 kHz sinusoid
- at 52.68 V
- with a source impedance of 68-j39 ohms
connected to a load of 10+j50 ohms.

The incident voltage is 26.34 V.

Using polar notation...
- source voltage: 52.68/_ 0.0 [52.68+j0.0] V
- source impedance: 78.4/_ -29.8 [68-j39] ohms
- load impedance: 51.0/_ 8.7 [10+j50] ohms

Total circuit impedance is source + load impedance:
78.4/_ -29.8 + 51.0/_ 8.7 = 78.8/_ 8.03 [78+j11]
Circuit current (voltage/impedance)
52.68/_ 0.0 / 78.8/_ 8.03 = 0.669/_ -8.03 [0.662-j0.0934]
Voltage at load (current * impedance)
0.669/_ -8.03 * 51.0/_ 8.7 = 34.1/_ 70.7 [11.3+j32.2]
which agrees with Roy's.
Reflected voltage (load voltage - incident)
34.1/_ 70.7 - 26.34/_ 0.0 = 35.5/_ 115.1 [-15.0+j32.2]
which also agrees.

So reflected voltage is greater than incident voltage
which leads to rho being greater than unity.

Now about that Daiwa....

Directional wattemeters compute the Vf and Vr using
Vf = (V + I*Z0) /2
Vr = (V - I*Z0) /2

Your Daiwa is probably calibrated for Z0 = 50 ohms, but let's
assume we can recalibrate for Z0 = 78.4/_ -29.8 [68-j39] ohms.

Then it will obtain
Vf = (34.1/_ 70.7 + 0.669/_ -8.03 * 78.4/_ -29.8) /2
= 26.34/_ 0.0 [26.34+j0.0]
Vr = (34.1/_ 70.7 - 0.669/_ -8.03 * 78.4/_ -29.8) /2
= 35.5/_ 115.1 [-15.0+j32.2]
as expected. Please note that your Daiwa DOES think that
Vr is greater than Vf.

Assuming your Daiwa works like most directional wattmeters
it will feed these voltages (appropriately scaled) to a
meter which will move linearly in response to the voltage.

So if your Daiwa had a linear scale it would have no
difficulty showing Vf and Vr (except that it would need
to be adjusted for the different Z0) and it would show
a greater Vr than Vf (i.e. rho 1).

But displaying power (even though meaningless in this case)
is somewhat more difficult. Your Daiwa likely computes power
by having non-linear markings on the meter representing
V**2/Z0. This works fine for real Z0, but will not do for
complex Z0. For this, you need more sophosticated computation
than is possible with just a non-linear scale, so the
power indicated by your Daiwa will be quite incorrect.

But it does get Vf and Vr correct (assuming it is adjusted for
the different Z0).

....Keith