Standing-Wave Current vs Traveling-Wave Current
 

"Yuri Blanarovich" wrote in message
...

"Dave" wrote in message
news:cwPbj.1073$ML6.117@trndny04...


you can do it when it makes physical sense. it does not make sense in
standing waves for all the obvious reasons that i have pointed out. it
does make sense in the individual traveling waves. just accept what your
little swr meter tells you, it shows the forward power and reflected
power, that is all you need and the only powers that make sense.


Little SWR meter shows forward AND reflected power in one direction, and
reflected power only in reverse direction. Why is the Bird wattmeter
calibrated in Watts, measuring power (forward and reverse) and has chart
to calculate SWR, when there are no standing waves and no power in them?
Laying waves or sitting waves???

Seems to me that the PROBLEM is that some consider standing wave to be
some imaginary, stopped, frozen wave, no good, while some of us consider
standing wave to be the result of superposition of forward and reverse
waves, that can be (their components) measured, current heats when flowing
through resistance, voltage "burns" when poor dielectric.
Like there is standing wave current, but no standing wave, huh????
Or are we forgetting that we are dealing with electromagnetic waves?
Can someone sort out the terminology and definitions?

Yuri, K3BU

last time, real simple. there ARE standing current waves. there ARE
standing voltage waves. There ARE NOT standing power waves.

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 24, 11:50*am, "Yuri Blanarovich" wrote:
Why is the Bird wattmeter
calibrated in Watts, measuring power (forward and reverse) and has chart to
calculate SWR, when there are no standing waves and no power in them?

Why indeed? The decision of Bird Electronic to build
an instrument that measured actual line voltage and
current and then compute forward or reverse voltage
but display the result in watts has lead to enormous
confusion about the nature of forward and reverse
waves.

If only they had decided to display forward or
reverse volts, life would be much better. People
would not have internalized "forward and reverse
power" to such a degree.

On the other hand, it would have then required
more arithmetic to compute actual power.

But they did it, and it can not be undone.

Do you have an unambiguous definition of "standing
wave power" that can be used?

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Except that V(x,t) and I(x,t) are not, in general, related by Z0.

From "Fields and Waves ..." by Ramo & Whinnery, 2nd edition:

V(x,t) = V*e^j(wt-kx) + V'*e^j(wt+kx)

I(x,t) = [V*e^j(wt-kx) - V'*e^j(wt+kx)]/Z0

No. The two wave view is merely an alternate set of expressions
which, when summed (i.e. using superposition), provide the actual
voltage and current on the line. These alternate expressions are
obtained by algebraic maniupulation of the more fundamental
descriptive equations.

Methinks you are confusing cause and effect. The standing
wave is not the cause of the two traveling waves.

In my analysis, P(x,t) = V(x,t) * I(x,t) is the equation that means
the power at any point and time can by obtained by measuring the
actual voltage and current on the line at the point and time of interest.

Make that the *NET* power and you will have it nailed.

Are you sure you want to throw away this ability? Are you sure you
want to claim that instantaneous power can NOT be obtained by
multiplying the instaneous measured voltage by the instanteous
measured current?

When the instantaneous voltage is the sum of two more elementary
voltages (same for current) then you are reporting the *NET*
results, not the underlying component results. The *NET* results
do not dictate reality.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Dave wrote:
try to look at it this way. when you look at the forward and reflected
waves separately it is intuitively obvious how the power calculation shows
the flow along the line with each wave.

This is a very old equation:

Pnet = PLoad = Pfor - Pref = Vfor*Ifor - Vref*Iref

For a pure standing wave it reduces to:

Pnet = Pfor - Pref = 0 or |Pfor| = |Pref|

There is zero net power transfered in a standing wave
so Vtot*Itot*cos(A) = ZERO *at every point* along a
lossless line containing pure standing waves, not just
at the Vtot=0 and Itot=0 points, but *everywhere*.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
With an open circuited line, I agree that there is no
net energy transfer at any point on the line. At most
points on the line, there is energy sloshing back and
forth, but netting to zero.

It is not "sloshing" back and forth unless you consider
the speed of light to be "sloshing". That's EM photonic
energy. If it's not traveling at the speed of light, it
doesn't exist. I would describe the movement of EM wave
energy as "racing" back and forth.

My statement about those 90 degree points where the
voltage or current is always 0 is much stronger: NO
energy transfer.

No *NET* energy transfer because the energy being transfered
in either direction is equal to the other direction. This
would be true of incoherent EM waves and even applies to
any and every type of wave energy.

This follows inexorably from P(x,t) = V(x,t) * I(x,t).

Yes, it certainly does for *NET* power.
Pnet = Pfor - Pref = 0 for ideal standing waves.

If you disagree with the general applicability of this
equation, please indicate when it can and can not be
applied.

It can be correctly applied when the person doing the
applying understands what it really means in reality.

I suggest you take a look at HP's s-parameter AP 95-1.
If you square the normalized s-parameter voltage equations,
you get power directly. The s-parameter equations are
designed that way.

b1 = s11*a1 + s12*a2 standard s-parameter voltage equation

a1^2 = Forward Power incident upon the impedance discontinuity

b1^2 = Reflected Power = (s11*a1 + s12*a2)^2
Power reflected from the impedance discontinuity

When a1^2 - b1^2 = 0

what do you think is the physical meaning of Reflected
Power being equal to:

Pref = (s11*a1)^2 + 2(s11*a1)(s12*a2) + (s12*a2)^2 ???

Do you recognize the irradiance equation from optical
engineering?

Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(A)
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Gene Fuller wrote:
This photonic limitation
is something that exists only in your head.

Good Grief, Gene, I don't have time to teach you
quantum electrodynamics. Go read a book that tells
you about the nature of photons. It is also the
cornerstone of relativity.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
They are the real current and voltage. I can measure them
with voltmeters and ammeters.

They are the *NET* current and voltage. You can separate
them into their elementary components with directional
couplers.

"Forward and reflected power" are derived from the
real voltage and current.

From the *NET* voltage and current by someone who understands
the nature of forward and reflected EM waves.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Gene Fuller wrote:
With such profound statements as, "a pure standing wave is technically
NOT an EM wave", it you might either offer some sort of reference or
start planning your trip to Stockholm.

Gene, here is a true/false quiz for you. If you have a
reference that disagrees with the obvious answers, please
quote it.

1. Is EM wave energy photonic in nature? ________

2. Do photons move at the speed of light in a medium? ______

3. Do standing waves move at the speed of light? _______

If the answers are yes, yes, and no, then standing waves
have been eliminated from the set of EM waves.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
And for a challenging use case, please consider two
circuits connected together. The circuits are in black
boxes so you do not know their details, but the voltage
on the connection between the circuits is measured as
10 V RMS at 4 MHz. The current is measured as 0.

How much energy is being transferred between the
circuits?

Inside each black box is a 50 ohm signal generator equipped
with a circulator and a 50 ohm load resistor. The signal
generators are outputting identical 10 V RMS phase-locked
signals.

Signal generator #1 "sees" the 50 ohm resistor in
signal generator #2 as it's load and supplies 200
milliamp, thus heating up the load resistor.

Signal generator #2 "sees" the 50 ohm resistor in
signal generator #1 as it's load and supplies 200
milliamp, thus heating up the load resistor.

The net voltage is measured as 10 V RMS at 4 MHz.
The net current is measured as 0.

How much energy is being transferred between the
circuits?

Install a one wavelength 50 ohm lossless transmission
line between the two signal generators. Nothing changes
but the standing waves become very visible and measurable.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
When does P(x,t) not equal V(x,t) * I(x,t)?

Any time the angle between V(x,t) and I(x,t) is
not 0 or 180 degrees. The correct equation is:

P(x,t) = V(x,t) * I(x,t) * cos(A)

If you write the above equation as:

P(x,t) = V(x,t) * I(x,t)

then you have implied the *dot product* of those
terms. The *cross product* will give you the wrong
answer. The dot product and cross product are both
ways of multiplying so to be technically correct,
you must specify which method of multiplying that
you are talking about.

When is my bank account zero? When I have drawn out
all the money I deposited. Power is a scalar like
money. If you put in 10 watts and take out 10 watts,
the balance is zero.

If you try to put 10 watts into a load and the load
rejects it, that energy flows in the opposite direction.

P(x,t) = V*I*cos(A) = Pfor - Pref

If Pfor = 10 watts and Pref = 10 watts then
P(x,t), i.e. the *NET POWER* equals zero.
--
73, Cecil http://www.w5dxp.com