Resistance & Impedance

[k7buc]

Hi,
Reactances have a real "resistance" portion.
This is non-dissapative, the energy remains in the system.
There is also dissapative resistance, as in a resistor where electrical energy is converted to heat.
I'm "assuming" when we look at the resistance trace on the AIM4170 we're seeing the overall resistance, which would include the sum of the actual resistance (dissapative) and the real portion of the impedance?
Am I correct?
-Del

[Bob]

k7buc said:

Hi,
Reactances have a real "resistance" portion.
This is non-dissapative, the energy remains in the system.
There is also dissapative resistance, as in a resistor where electrical energy is converted to heat.
I'm "assuming" when we look at the resistance trace on the AIM4170 we're seeing the overall resistance, which would include the sum of the actual resistance (dissapative) and the real portion of the impedance?
Am I correct?
-Del

Yes, the resistance (Rs or Rp) is the real part of the impedance, Z. Xs or Xp is the imaginary part of Z.

--73/Bob

[n9fyx]

I don't think you could say the real resistance portion
of a reactance is non-dissapative. It would be just as
dissapative as a pure resistor with no reactance. It is true
that the voltage times current for a reactive load isn't
going to be the power but there will still be power dissapated
depending on the phase angle (magnitude of the real
resistance projection). Am I wrong?

[k7buc]

I think W2DU has a paper published on this subject. Also the IEEE.
They show 2 definitions. Its unfortunate the industry has used the same word "resistance" for two different meanings.
According to W2DU's info, the real part of the impedance is non-dissipative. Included in the resistance is the specific case where there is actually energy converted to heat.
What I'm saying, is that the R= E/I resistance as far as I know is just a ratio, and does not mean that energy is lost to heat unless some physical material is involved.
I'm not sure that one can determine the real power loss without a bolometer to measure temp rise. If I'm wrong, I'd sure like to know.
I don't have the instrumentation to experiment and prove it either way.
-Del

[n9fyx]

I can see if you had a ferrite core balun that a portion of the resistive component would result in heating, where with an iron powder core, more of the resistive component might be a reflection of the antenna resistance. In both cases there would be power dissapated but with antennas some goes out as EM radiation and not heat. I'm skeptical about the real portion of impedance having two different meanings, although I haven't read those papers.
-Steve

[sp2wdw]

The real part of antenna impedance is a sum of two components - radiation resistance and loss resistance. If you measure a hypotethical antenna made of superconductor located in free-space with no ground and other near-field losses, the real part of impedance will be totaly "non-dissipative" - all energy will be radiated in the form of RF field. In practical cases a fraction of this value will be dissipative (in the form of heat), too.
You may want to look up the definition of "antenna (radiation) effciency".

[n9fyx]

OK, so I found the article where w2du quotes a IEEE dictionary definition talking about 2 types of the real part of the impedance, dissipative and non-dissipative. They claimed that coax and impedance matching networks frequently had "non-dissipative"
impedances and transmitter outputs would be "non-dissipative".
I find the use of this type of terminology is unfortunate, since the problem isn't that the industry is using resistance for
2 different meanings. The problem is they concocted this definition that is misleading since there is only one type of real resistance.
They should just point out that with coax and matching networks the impedances just refer to the expected loads they are supposed to match, and not any real measured impedance. The output impedance of a transmitter is the load into which it it has been designed to deliver its rated power, and has nothing to do with any impedance at the transmitter. A 50 ohm coax doesn't refer to 50 ohms resistive measurement inside the coax. It is just a matching network where if you put 50 ohms on one end you will see 50 ohms on the other. Any real part of the impedance you measure with an AIM won't be any mystical "non-dissipative" energy that stays in the system, it will dissipate energy and be true to physics.