# The Dark Side of the Conjugate Match

Discussion in 'General Technical Questions and Answers' started by KL7AJ, Mar 12, 2010.

Not open for further replies. 1. You seem to be talking about visible light waves. When a physicist talks about light waves in general, he is talking about the entire spectrum of EM waves from the lowest frequency to the highest frequency. RF waves are part of the spectrum of invisible light. The energy in each RF photon is easy to calculate:

Photon Energy = h*f

where h is Planck's constant and f is the frequency.

h = 6.626x10^-34 joules/sec.

So the energy in an individual RF photon at 10.125 MHz would be:

10.125 MHz Photon Energy = 6.708825x10^-27 joules

At 100 watts (joules/second) on 10.125 MHz, I am radiating approximately
149,057,398,300,000,000,000,000,000 photons per second.

The only difference in 10.125^6 Hz RF and 10.125^15 Hz visible light is frequency.

2. You are doing it the hard way. If 60 watts out of every 100 watts is reflected at the load (antenna) then the power reflection coefficient is:

rho^2 = 60/100 = 0.6

The voltage reflection coefficient is:

rho = SQRT(0.6) = 0.7746

Power transmission coefficient = 1-rho^2 = 0.4

In order to get 100 watts (source power) into the load, steady-state forward power is:

Pfor = 100w/(1-rho^2) = 100w/0.4 = 250 watts

Pref = 250(0.6) = 150 watts

SWR = (1 + rho)/(1-rho) = 1.7746/0.2254 = 7.87:1

3. I know what "physicists think". What I am saying is that in my opinion, they are wrong. They have been wrong about things before, you know.

It has been proven that photons can not exist as a wave nor can a wave exist as a photon. Light is either a particle(photon) or a wave or even possibly a third form. But it cannot be both wave and particle at the same time.

When it is a photon, it can not exhibit wave characteristics. When it is a wave, it can not exhibit particle characteristics.

Photons can not travel down a wire. As a matter of fact, even electrons don't travel very far. It's the electric and magnetic fields that travel.

4. I was just keeping it simple and not at all concerned with calculating the SWR. It's just that according to the article, if there is a 60% power reflection at the antenna then there has to be a 60% power reflection at the tuner also but in opposite phase because of the complex conjugate.

But I did err. Since there is now a power coefficient at the tuner then:

100 watts is sent out by the xmitter. 60 watts (assume in phase with the original forward power) is reflected by the antenna. That 60 watts travels back to the tuner. It reflects 60% of that (because it has the same power reflection coeficient) and lets 24 watts back into the xmitter to be converted to heat. So 60% of 60 watts goes back to the antenna or 36 watts.

100 watts is sent out by the xmitter which now has an additional 36 watts out of phase. So there is 136 going to the antenna. The antenna reflects 60% of that so about 82 watts goes back. And 40 % of the 82 watts is passed on to the xmitter to be converted to heat and 60% or about 50 watts reflected.

Now we have 150 watts going to the antenna and, of that, about 60% or 90 watts is reflected.

According to that scenario eventually the power will increase until we will burn out the final amplifier.

5. Just remembered that I forgot something. Happens as we get older.

We know that photons are created when an electron drops from an atomic orbit of high energy to an orbit of low energy. The energy difference between the two orbits goes into the photon and that determines the frequency/wavelength/color of the photon.

We know that electromagnetic waves are created by the acceleration of electrons without the use of an atomic nucleus or it's orbits.

Photons and EM waves are created by two different mechanisms. Their energies are measured two different ways. Therefore photons and EM waves are two different things.

Can a photon transform into an EM wave and vice-versa? Evidently, that is so but are certain energy levels necessary? Are there certain requirements necessary for the transform?

Or, perhaps photons only mimic waves under certain conditions. There is the deBroglie effect to consider.

6. ### K9ASEXML SubscriberQRZ Page

Great article! where can I get a print copy to send to my brother an IT guy who has been asking me about radio. 7. IF there is a mis-match at the antenna which has a certain complex formula and a reflection coefficient, then the conjugate at the tuner must also have the same coefficient but with an opposite phase shift.

Let us say that the complex formula at the antenna is
47 - j34 indicating that the antenna is capacitive, thus too short.

Then the tuner should have 47 + j34 to compensate.

It is conceivable that another antenna has a complex formula of 47 + j34 indicating that the antenna is inductive, thus too long.

Then the tuner should have 46 - j34 to compensate.

If 47 + j34 causes part of the signal to reflect then 47 - j34 must also cause part of the signal to reflect with the same reflection coefficient.

If there is a reflection at the antenna because the antenna doesn't match the coax then there has to be a reflection at the tuner because the tuner doesn't match the coax.

As Eric said, ANY discontinuity, i.e. mismatch, no matter where it is, will cause a reflection. Even if it's at the tuner.

8. Ah yes. Lossless means only that there is no real resistance. It does not mean that there is no reactance.

A discontinuity can be lossless and still reflect power because of the reactance.

Remember the complex formula is REAL + j IMAGINARY.

A complex formula of 0 + j456 is entirely feasible and realizable and is a reflective discontinuity.

9. For a class C amplifier we want to efficiently transfer very short pulses from the tube to the tank -- but efficiently extract CW RF over a much longer time. Should be a duty-cycle factor for efficiency somewhere. Looks to me we're talking apples and pineapples.

Cortland
KA5S

10. A Class C amplifier does indeed produce very short pulses. However, in RF amplifiers, the cathode/collector/drain of the amplifier is connected to a parallel tank circuit. A parallel tank circuit operates somewhat like a tuning fork. When a tuning fork is struck, it produces a sinusoidal tone for quite some time but eventually dies off.

When a tank circuit is pulsed, it too produces a sinusoidal "tone" (at its resonant frequency) for quite some time that eventually dies off. However if the tank circuit is pulsed once per cycle, the "tone" continues.

Since the tank circuit is in series with the output device and the power supply, transferring the pulse to the tank is automatic and, really, child's play.

The output "tone" from the tank is then transferred to the output circuit (usually a pi-network) through a capacitor.