Hello,

Consider a source impedance of Zo=50+j200 and Zl=0-j200.

Since these are both series equivalent impedances, Zo
is like a 50 ohm resistor with a series inductor, and Zl
is like a series capacitor.

At ONE test frequency, the inductive and capacitive
reactances will cancel out (series resonance). When this happens,
is will be equivalent to Zo=50 and Zl=0, which is a short.

If you incorrectly use the "normal" equation for rho (when
Zo is complex), you will get:

Rho = (Zload-Zo)/(Zload+Zo) = (-50-j400)/50
= 403.1 /_ -97 degrees

So some silly people on this NG think that a
short will reflect a voltage 403.1 times the incident voltage,
which is absolutely insane.


Now try the correct "conjugate" equation (for complex
Zo):

Rho = (Zload-Zo*)/(Zload+Zo) = -50/50= -1

Which is exactly what you should get for a short,
a full reflection with a phase shift of 180 degrees, but the ratio
can never be more than one for a passive network.




And consequently, the ratio of the reflected to incident
powers can also never be more than 1 for a passive network:

If you use ratios, it doesn't matter whether you use
peak or RMS voltages.

([Vpeak.incident/Vpeak.reflected])=
([Vrms.incident/Vrms.reflected])=sqrt(Pi/Pr)=[rho]

This is because the sqrt(2) is in the numerator and denominator.
And the 2 (after squaring) is also factored out in the Power RC!

And...the Zo is also factored out for Power RC!

The Zo is not needed for the Power RC, because the
impedance of the source is identical to the load for the
reflected power! Sure, you use the Zo in relation to Zl to
get rho, but once you get rho, you have the power RC.

Hah! the plot thickens a bit....


Slick

On 6 Sep 2003 04:37:43 -0700, (Dr. Slick) wrote:

Hello,

Consider a source impedance of Zo=50+j200 and Zl=0-j200.

Since these are both series equivalent impedances, Zo
is like a 50 ohm resistor with a series inductor, and Zl
is like a series capacitor.

At ONE test frequency, the inductive and capacitive
reactances will cancel out (series resonance). When this happens,
is will be equivalent to Zo=50 and Zl=0, which is a short.

If you incorrectly use the "normal" equation for rho (when
Zo is complex), you will get:

Rho = (Zload-Zo)/(Zload+Zo) = (-50-j400)/50
= 403.1 /_ -97 degrees

So some silly people on this NG think that a
short will reflect a voltage 403.1 times the incident voltage,
which is absolutely insane.


Now try the correct "conjugate" equation (for complex
Zo):

Rho = (Zload-Zo*)/(Zload+Zo) = -50/50= -1

Which is exactly what you should get for a short,
a full reflection with a phase shift of 180 degrees, but the ratio
can never be more than one for a passive network.




And consequently, the ratio of the reflected to incident
powers can also never be more than 1 for a passive network:

If you use ratios, it doesn't matter whether you use
peak or RMS voltages.

([Vpeak.incident/Vpeak.reflected])=
([Vrms.incident/Vrms.reflected])=sqrt(Pi/Pr)=[rho]

This is because the sqrt(2) is in the numerator and denominator.
And the 2 (after squaring) is also factored out in the Power RC!

And...the Zo is also factored out for Power RC!

The Zo is not needed for the Power RC, because the
impedance of the source is identical to the load for the
reflected power! Sure, you use the Zo in relation to Zl to
get rho, but once you get rho, you have the power RC.

Hah! the plot thickens a bit....


Hmm.... so you manually typed off page 13 of some
transmission line school book, so
what was the purpose of your exercise?
Typing skill increase?

Or did I miss something?

w.
--
On the Internet nobody knows that I am a dog.

In article ,
(Dr. Slick) wrote:

Hello,

Consider a source impedance of Zo=50+j200 and Zl=0-j200.
******

(1) A *source* impedance of Z_0 = 50 + 200j is easily arranged. A
*transmission line surge impedance* of Z_0 = 50 + 200j is impossible;
surge impedances of transmission lines must have angles between - Pi/4
radians and + Pi/4 radians.

Since these are both series equivalent impedances, Zo
is like a 50 ohm resistor with a series inductor, and Zl
is like a series capacitor.

At ONE test frequency, the inductive and capacitive
reactances will cancel out (series resonance). When this happens,
is will be equivalent to Zo=50 and Zl=0, which is a short.
**********

(2) Not equivalent in any reasonable sense. 50 and 50 + 200j aren't equal,
nor are - 200j and 0 equal.

If you incorrectly use the "normal" equation for rho (when
Zo is complex), you will get:

Rho = (Zload-Zo)/(Zload+Zo) = (-50-j400)/50
= 403.1 /_ -97 degrees

(3) You forgot the factor of 50 in the denominator. The quantity you are
calculating above is approximately a magnitude of 8.062257748 at an angle
of about - 97.12501636 degrees. Of course this is silly for a value of rho
(but not as silly as 403.1 at an angle of - 97 degrees). However see my
comment (1) above.

So some silly people on this NG think that a
short will reflect a voltage 403.1 times the incident voltage,
which is absolutely insane.

(4) I hope most readers believe the way to calculate rho when Z_L = 0 is
rho = (Z_L - Z_0)/(Z_L + Z_0) = (0 - Z_0)/(0 + Z_0) = - 1.

Hah! the plot thickens a bit....

Yes, it does.

David, ex-W8EZE

--
David or Jo Anne Ryeburn

To send e-mail, remove the letter "z" from this address.

(David or Jo Anne Ryeburn) wrote in message .. .
In article ,
(Dr. Slick) wrote:

Hello,

Consider a source impedance of Zo=50+j200 and Zl=0-j200.
******

(1) A *source* impedance of Z_0 = 50 + 200j is easily arranged. A
*transmission line surge impedance* of Z_0 = 50 + 200j is impossible;
surge impedances of transmission lines must have angles between - Pi/4
radians and + Pi/4 radians.


Ok, a source impedance then.

I don't fully understand why your last statement needs to be so.


Since these are both series equivalent impedances, Zo
is like a 50 ohm resistor with a series inductor, and Zl
is like a series capacitor.

At ONE test frequency, the inductive and capacitive
reactances will cancel out (series resonance). When this happens,
is will be equivalent to Zo=50 and Zl=0, which is a short.
**********

(2) Not equivalent in any reasonable sense. 50 and 50 + 200j aren't equal,
nor are - 200j and 0 equal.



I understand your point, but the reactances WILL cancel. And if
you are feeding from a lossless 50 ohm transmission line, the circuit
won't know the difference.



If you incorrectly use the "normal" equation for rho (when
Zo is complex), you will get:

Rho = (Zload-Zo)/(Zload+Zo) = (-50-j400)/50
= 403.1 /_ -97 degrees

(3) You forgot the factor of 50 in the denominator. The quantity you are
calculating above is approximately a magnitude of 8.062257748 at an angle
of about - 97.12501636 degrees. Of course this is silly for a value of rho
(but not as silly as 403.1 at an angle of - 97 degrees). However see my
comment (1) above.



My mistake. Wrote too quickly. A gain of about 8 is STILL insane
for a passive network!




(4) I hope most readers believe the way to calculate rho when Z_L = 0 is
rho = (Z_L - Z_0)/(Z_L + Z_0) = (0 - Z_0)/(0 + Z_0) = - 1.


rho = (Z_L - Z_0*)/(Z_L + Z_0)

I agree with you. But the incident voltage in this case will be
coming
out of a series inductor of +j200 reactance at the test frequency.

It will be charging up a capacitor, but the reflected voltage will
not be
8 times the incident.

Again, the reactances will cancel at the series resonance, so in
effect, if you are feeding a lossless 50 ohm tranmission line, you
will not be able to tell the difference. It will appear exactly like
a 50 ohm line shorted at the end.

Where do you stand David?


Slick

In article ,
(Dr. Slick) wrote:

(David or Jo Anne Ryeburn) wrote in message
.. .
In article ,
(Dr. Slick) wrote:

Hello,

Consider a source impedance of Zo=50+j200 and Zl=0-j200.
******

(1) A *source* impedance of Z_0 = 50 + 200j is easily arranged. A
*transmission line surge impedance* of Z_0 = 50 + 200j is impossible;
surge impedances of transmission lines must have angles between - Pi/4
radians and + Pi/4 radians.


Ok, a source impedance then.

In that case you shouldn't be using a formula intended to apply to the
surge impedance of a transmission line.

I don't fully understand why your last statement needs to be so.

I assume that by "last statement" you mean

"A *transmission line surge impedance* of Z_0 = 50 + 200j is impossible;
surge impedances of transmission lines must have angles between - Pi/4
radians and + Pi/4 radians."

This follows immediately from the formula Z_0 = sqrt((R + jwL)/(G + jwC)),
the facts that none of w, R, L, G, or C are negative, the way angles work
when one divides complex numbers and takes square roots, and the fact that
the real part of Z_0 can't be negative (which decides which of the two
square roots should be used).

Where do you stand David?

I believe that algebra speaks for itself. I believe that whether a model
accurately depicts reality has to be tested by experiment. And I believe
that when many such experiments have been previously carried out, all
confirming the accuracy of the depiction, any claim that the model is
inaccurate and that another one is accurate has to be supported with
extraordinarily strong empirical evidence.

David, ex-W8EZE

--
David or Jo Anne Ryeburn

To send e-mail, remove the letter "z" from this address.

(David or Jo Anne Ryeburn) wrote in message .. .
In article ,
(Dr. Slick) wrote:

(David or Jo Anne Ryeburn) wrote in message
.. .
In article ,
(Dr. Slick) wrote:

Hello,

Consider a source impedance of Zo=50+j200 and Zl=0-j200.
******

(1) A *source* impedance of Z_0 = 50 + 200j is easily arranged. A
*transmission line surge impedance* of Z_0 = 50 + 200j is impossible;
surge impedances of transmission lines must have angles between - Pi/4
radians and + Pi/4 radians.


Ok, a source impedance then.

In that case you shouldn't be using a formula intended to apply to the
surge impedance of a transmission line.


You mean i can't use Zo=50 + j200 with

Rho = (Zload-Zo*)/(Zload+Zo), for complex Zo?

Only up to Zo=50 + j50?

Ok, well, the conjugate formula still makes more sense to
me.




Where do you stand David?

I believe that algebra speaks for itself. I believe that whether a model
accurately depicts reality has to be tested by experiment. And I believe
that when many such experiments have been previously carried out, all
confirming the accuracy of the depiction, any claim that the model is
inaccurate and that another one is accurate has to be supported with
extraordinarily strong empirical evidence.

David, ex-W8EZE


If the algebra speaks for itself, what does it say to you? Is
Besser and Kurokawa and the ARRL incorrect? If you're not too sure
and you don't wanna say, i wouldn't blame you.


Slick

"Dr. Slick" wrote:

Hello,

Consider a source impedance of Zo=50+j200 and Zl=0-j200.

An excellent example.

Since these are both series equivalent impedances, Zo
is like a 50 ohm resistor with a series inductor, and Zl
is like a series capacitor.

At ONE test frequency, the inductive and capacitive
reactances will cancel out (series resonance). When this happens,
is will be equivalent to Zo=50 and Zl=0, which is a short.

If you incorrectly use the "normal" equation for rho (when
Zo is complex), you will get:

Rho = (Zload-Zo)/(Zload+Zo) = (-50-j400)/50
= 403.1 /_ -97 degrees

Corrected arithmetic error - -1-j8 = 8.062/_ -97.125

So some silly people on this NG think that a
short will reflect a voltage 403.1 times the incident voltage,
which is absolutely insane.

Now try the correct "conjugate" equation (for complex
Zo):

Rho = (Zload-Zo*)/(Zload+Zo) = -50/50= -1

Which is exactly what you should get for a short,
a full reflection with a phase shift of 180 degrees, but the ratio
can never be more than one for a passive network.

So for this example using the 'revised' rho Vr = -Vi
so the voltage across the capacitor would be Vi + Vr = 0 .

Let us do some circuit analysis.

As you say above, the equivalent circuit is 3 elements in
series: a 50 ohm resister, a +j200 ohm inductor and -j200
ohm capacitor.

Let us apply 1 volt to this circuit...
Total impedance
50+j200+0-j200 = 50 ohms
Total current (volts/impedance)
1/50 = .02 A
Voltage across resistor
.02 * 50 = 1 V
Voltage across inductor
.02 * (0+j200) = 4/_ 90 Volts
Voltage across capacitor (the load)
.02 * (0-j200) = 4/_ -90

Now for the check...
Vi = 0.5 V
With classic rho
Vr = 0.5 * 8.062/_ -97.125 = 4.031/_ -97.125
Vload = Vi+Vr = 0.5 + 4.031/_ -97.125 = 4/_ -90
The same as computed using circuit theory.

With revised rho
Vr = -.5
Vload = 0.5 -0.5 = 0

Which is more useful?
classic rho which properly predicts the voltage across the load
or
revised rho which just provides some number

I know which one I'd pick.

The thing to remember is that in circuits with inductors it is
very easy to achieve voltages greater than any of the supplies.

....Keith

wrote in message ...
"Dr. Slick" wrote:

Hello,

Consider a source impedance of Zo=50+j200 and Zl=0-j200.

An excellent example.

Since these are both series equivalent impedances, Zo
is like a 50 ohm resistor with a series inductor, and Zl
is like a series capacitor.

At ONE test frequency, the inductive and capacitive
reactances will cancel out (series resonance). When this happens,
is will be equivalent to Zo=50 and Zl=0, which is a short.

If you incorrectly use the "normal" equation for rho (when
Zo is complex), you will get:

Rho = (Zload-Zo)/(Zload+Zo) = (-50-j400)/50
= 403.1 /_ -97 degrees

Corrected arithmetic error - -1-j8 = 8.062/_ -97.125


i corrected this.


So some silly people on this NG think that a
short will reflect a voltage 403.1 times the incident voltage,
which is absolutely insane.

Now try the correct "conjugate" equation (for complex
Zo):

Rho = (Zload-Zo*)/(Zload+Zo) = -50/50= -1

Which is exactly what you should get for a short,
a full reflection with a phase shift of 180 degrees, but the ratio
can never be more than one for a passive network.

So for this example using the 'revised' rho Vr = -Vi
so the voltage across the capacitor would be Vi + Vr = 0 .

Let us do some circuit analysis.

As you say above, the equivalent circuit is 3 elements in
series: a 50 ohm resister, a +j200 ohm inductor and -j200
ohm capacitor.

Let us apply 1 volt to this circuit...
Total impedance
50+j200+0-j200 = 50 ohms
Total current (volts/impedance)
1/50 = .02 A
Voltage across resistor
.02 * 50 = 1 V
Voltage across inductor
.02 * (0+j200) = 4/_ 90 Volts
Voltage across capacitor (the load)
.02 * (0-j200) = 4/_ -90

Now for the check...
Vi = 0.5 V
With classic rho
Vr = 0.5 * 8.062/_ -97.125 = 4.031/_ -97.125
Vload = Vi+Vr = 0.5 + 4.031/_ -97.125 = 4/_ -90
The same as computed using circuit theory.



Ok, well this seems to make a bit of sense,
but where did the circuit theory predict the
Vr=4.031/_ -97.125?


With revised rho
Vr = -.5
Vload = 0.5 -0.5 = 0

Which is more useful?
classic rho which properly predicts the voltage across the load
or
revised rho which just provides some number

I know which one I'd pick.

The thing to remember is that in circuits with inductors it is
very easy to achieve voltages greater than any of the supplies.

...Keith

wrote in message ...

Rho = (Zload-Zo)/(Zload+Zo) = (-50-j400)/50
= 403.1 /_ -97 degrees

Corrected arithmetic error - -1-j8 = 8.062/_ -97.125

So some silly people on this NG think that a
short will reflect a voltage 403.1 times the incident voltage,
which is absolutely insane.

Now try the correct "conjugate" equation (for complex
Zo):

Rho = (Zload-Zo*)/(Zload+Zo) = -50/50= -1

Which is exactly what you should get for a short,
a full reflection with a phase shift of 180 degrees, but the ratio
can never be more than one for a passive network.

So for this example using the 'revised' rho Vr = -Vi
so the voltage across the capacitor would be Vi + Vr = 0 .

Let us do some circuit analysis.

As you say above, the equivalent circuit is 3 elements in
series: a 50 ohm resister, a +j200 ohm inductor and -j200
ohm capacitor.

Let us apply 1 volt to this circuit...
Total impedance
50+j200+0-j200 = 50 ohms
Total current (volts/impedance)
1/50 = .02 A
Voltage across resistor
.02 * 50 = 1 V
Voltage across inductor
.02 * (0+j200) = 4/_ 90 Volts
Voltage across capacitor (the load)
.02 * (0-j200) = 4/_ -90



But you need to define Vi as the voltage
coming out of the series inductor. I don't believe
you do that here.



Now for the check...
Vi = 0.5 V
With classic rho
Vr = 0.5 * 8.062/_ -97.125 = 4.031/_ -97.125
Vload = Vi+Vr = 0.5 + 4.031/_ -97.125 = 4/_ -90
The same as computed using circuit theory.



Where does circuit theory predict
Vr = 0.5 * 8.062/_ -97.125 = 4.031/_ -97.125?





Vr = -.5
Vload = 0.5 -0.5 = 0

Which is more useful?
classic rho which properly predicts the voltage across the load
or
revised rho which just provides some number

I know which one I'd pick.

The thing to remember is that in circuits with inductors it is
very easy to achieve voltages greater than any of the supplies.

...Keith


i agree with your last statement, but do you think that the
power RC is around 64 for this circuit?

If you use ratios, it doesn't matter whether you use
peak or RMS voltages.

([Vpeak.incident/Vpeak.reflected])=
([Vrms.incident/Vrms.reflected])=sqrt(Pi/Pr)=[rho]

This is because the sqrt(2) is in the numerator and denominator.
And the 2 (after squaring) is also factored out in the Power RC!

And...the Zo is also factored out for Power RC!

The Zo is not needed for the Power RC, because the
impedance of the source is identical to the load for the
reflected power! Sure, you use the Zo in relation to Zl to
get rho, but once you get rho, you have the power RC.



Slick

Keith wrote:
"Stick to the incident and reflected voltage (or current) waves for
analysis. They work."

Bird Technical Series #1 may be enlightning:
"By proper combination of the two samples (derived from volts and amps)
we obtain an RF voltage proportional to the square root of main line
power---."

"---the scale which has been marked in watts corresponding to the power
being sampled from the main line."

With a little review of wave behavior on transmission lines (I like
Terman) in almost any good text, it is seen that voltage and current
continuously vary along a line containing a reflection. This results
from interference between the forward and reflected waves.

However, The power, forward or reflected, is free of the oscillations
produced in the volts and amps by interference between forward and
reflected waves. This steady power flow makes power the electrical
property to measure and this was the impetus for the Bird wattmeter.
It`s been around for about 50 years, and I`d wager it will be around for
another 50 years.

Best regards, Richard Harrison, KB5WZI