Here's an interesting exercise. The source is a 100W signal generator
equipped with a circulator and load that dissipates all reflected power
reaching the source so Pfwd1 is a constant 100W. The lossless twinlead is
an integral number of wavelengths long. rho at the load is 0.707 so 1/2
of the incident power is reflected. Pfwd1 and Pref1 are measured just to
the left of the impedance discontinuity at '+'. Pfwd2 and Pref2 are
measured just to the right of the impedance discontinuity at '+'.

Source---50 ohm coax---+---n*WL lossless 291.5 ohm twinlead---50 ohm load
Pfwd1-- Pfwd2--
--Pref1 --Pref2

t0 will be when the source 100W is first incident upon point '+'.
t1 will be when the first reflections from the load arrive at '+'.
t2 will be when the second set of reflections from the load arrive at '+'.

At t1, Pfwd1=100W, Pref1=50W, Pfwd2=50W, Pref2=25W

Question: Between t1 and t2, what is the equation and magnitude of Pfwd2?
--
73, Cecil, W5DXP

Beats the heck out of me.

Jack

On 21 Nov 2003 17:38:55 GMT, (JDer8745) wrote:
Beats the heck out of me.
A Rodney King exercise?

For instantaneous values,
Vf2 = (2*Z2*Vf1+(Z1-Z2)*Vr2)/(Z1+Z2)
Vr1 = ((Z2-Z1)*Vf1+2*Z1*Vr2)/(Z1+Z2)

You get to think about the phase of Vr2 with respect to Vf1, and the
answer should then be obvious.

Cheers,
Tom

(Cecil Moore) wrote in message . com...
Here's an interesting exercise. The source is a 100W signal generator
equipped with a circulator and load that dissipates all reflected power
reaching the source so Pfwd1 is a constant 100W. The lossless twinlead is
an integral number of wavelengths long. rho at the load is 0.707 so 1/2
of the incident power is reflected. Pfwd1 and Pref1 are measured just to
the left of the impedance discontinuity at '+'. Pfwd2 and Pref2 are
measured just to the right of the impedance discontinuity at '+'.

Source---50 ohm coax---+---n*WL lossless 291.5 ohm twinlead---50 ohm load
Pfwd1-- Pfwd2--
--Pref1 --Pref2

t0 will be when the source 100W is first incident upon point '+'.
t1 will be when the first reflections from the load arrive at '+'.
t2 will be when the second set of reflections from the load arrive at '+'.

At t1, Pfwd1=100W, Pref1=50W, Pfwd2=50W, Pref2=25W

Question: Between t1 and t2, what is the equation and magnitude of Pfwd2?

(Tom Bruhns) wrote in message om...
For instantaneous values,
Vf2 = (2*Z2*Vf1+(Z1-Z2)*Vr2)/(Z1+Z2)
Vr1 = ((Z2-Z1)*Vf1+2*Z1*Vr2)/(Z1+Z2)

My ISP doesn't seem to be making it lately so I am posting from Google.
Your equation represents Vf1 times the transmission coefficient plus
Vr2 times the reflection coefficient. This is perfectly logical but
IMO doesn't represent what is happening in reality. I hope I am wrong.
Re-arranging the terms:

Vf2 = 2*Z2*Vf1/(Z1+Z2) + Vr2(Z1-Z2)/(Z1+Z2)

For the example presented, I don't think there is enough interference
to support the above superposition. Again, I hope I am wrong but please
bear with me while you prove I am wrong. If I am wrong, it will resolve
something I have been wrestling with for three years.

Please calculate the two voltages, superpose them, and calculate the
resulting Pfwd2 forward power and then let's discuss the results.
--
73, Cecil, W5DXP

So what do YOU get? As I recall, it comes out to a voltage equivalent
to 112.5W for Pf2, and 12.5W for Pr1. Maybe I can head you off at the
pass, and give you the values for further rounds of the echo:
round Pf2 Pr1
1 112.5 12.5
2 153.125 3.125
3 175.78125 0.78125
4 187.69531... 0.19531...
5 193.79883... 0.04883...
6 196.88721... 0.01221...
7 198.44055... 0.00305...
8 199.21951... 0.00076...
9 199.60957... 0.00019...
10 199.80474... 0.00005...

Don't see anything worth discussing about it at the moment.


Cheers,
Tom

(Cecil Moore) wrote in message . com...
(Tom Bruhns) wrote in message om...
For instantaneous values,
Vf2 = (2*Z2*Vf1+(Z1-Z2)*Vr2)/(Z1+Z2)
Vr1 = ((Z2-Z1)*Vf1+2*Z1*Vr2)/(Z1+Z2)

My ISP doesn't seem to be making it lately so I am posting from Google.
Your equation represents Vf1 times the transmission coefficient plus
Vr2 times the reflection coefficient. This is perfectly logical but
IMO doesn't represent what is happening in reality. I hope I am wrong.
Re-arranging the terms:

Vf2 = 2*Z2*Vf1/(Z1+Z2) + Vr2(Z1-Z2)/(Z1+Z2)

For the example presented, I don't think there is enough interference
to support the above superposition. Again, I hope I am wrong but please
bear with me while you prove I am wrong. If I am wrong, it will resolve
something I have been wrestling with for three years.

Please calculate the two voltages, superpose them, and calculate the
resulting Pfwd2 forward power and then let's discuss the results.

(Tom Bruhns) wrote:
So what do YOU get? As I recall, it comes out to a voltage equivalent
to 112.5W for Pf2, and 12.5W for Pr1.

Yep, that's what I got with an s-parameter analysis. After all the negative
things said here about an s-parameter analysis, I was expecting someone to
post an answer different from the s-parameter answer.

In the s-parameter equation, b2 = s21*a1 + s22*a2, |b2|^2 is Pf2. If the
right side of the equation is squared, the (2*s12*a1*s22*a2) term is the
magnitude of the interference which is constructive in this case. On the
other side of the impedance discontinuity, b1 = s11*a1 + s12*a2. |b1|^2
equals Pr1. If the right side of that equation is squared, the
(2*s11*a1*s12*a2) is the magnitude of the interference which is destructive
in this case. Note that |(2*s11*a1*s12*a2)| = |(2*s21*a1*s22*a2)|, i.e.,
the constructive interference magnitude on one side of the impedance
discontinuity always equals the destructive interference magnitude on the
other side.
--
73, Cecil, W5DXP