# Coaxial Cable loss

by nelectrode
Tags: cable, coaxial, coaxial cable, loss, optical fibers, transmission lines
PF Gold
P: 11,105
 Quote by Baluncore The instruction to pilots regarding crash landings, is that they must look at the gap between the trees, rather than the trees that are in the front of their mind. It is well known that, where they focus on the obstacle, they tend to collide with it. I want to navigate a safe course between Scylla and Charybdis. I do not doubt the MPTT. There are often engineering solutions that avoid the close encounter and come through it economically. The fascination with engineering a collision in order to prove the MPTT exists is quite unnecessary. It completely distracts attention away from the methods available to avoid poor efficiency.
More stories and still no Maths. All you need to do is to show that an amplifier that is 90% efficient when feeding into a 50Ω load on the end of a 50Ω feeder (so the transmitter has a source resistance of 6Ω), when operating to its spec., can, itself, be made to present a 50Ω termination for signals reflected from the load. The problem can't be reduced further so it should be easy for you to show that you are correct.

Is that simple bit of analysis not allowed in your world? Whatever you remember - or think you remember having been told or having measured in the past, the above has to be proved for your story to hold. Of course, it cannot.

Steer between the trees of nonsense and aim at the reliable landing strip of Mathematics. It will not fail you.
 P: 1,081 There is no reflected energy. QED.
P: 1,081
 Quote by sophiecentaur All you need to do is to show that an amplifier that is 90% efficient when feeding into a 50Ω load on the end of a 50Ω feeder (so the transmitter has a source resistance of 6Ω), when operating to its spec., can, itself, be made to present a 50Ω termination for signals reflected from the load. The problem can't be reduced further so it should be easy for you to show that you are correct. Is that simple bit of analysis not allowed in your world? Whatever you remember - or think you remember having been told or having measured in the past, the above has to be proved for your story to hold. Of course, it cannot.
If you want maximum power you should use a transformer to match the amplifier's Rs of 6 ohm to the 50 ohm line. Since the transformer is broadband and reciprocal it will correctly match the line to the amplifier's Rs for all reflected or received energy.

If the transmitter was operated near others in the same band, I would avoid a broadband transformer and use a hybrid cross connected transformer pair since the presence of the received signal from a nearby transmitter will cross modulate in the output stage of the transmitter and produce all sorts of out of band spurious signals. It is best to reflect the other signal or absorb it in a loaded hybrid than to allow it access to the final stage.

I am not an imbecile.
 Sci Advisor PF Gold P: 11,105 You really don't get this, do you? If you use a transformer to match the 6 Ω to a 50 Ω load, the transmitter will not be 90% efficient - will it? So it will not be operating as it was designed (to work into 50Ω). It will see 6Ω and not 50Ω. It will possibly melt, in fact because, instead of only 10kW dissipation in its anode, it will be dissipating around 100kW. What have all your specifics got to do with the basic principles? We are operating a single transmitter under ideal conditions. The style of transformer is totally irrelevant. It has just two ports and is a reciprocal network. It is not magic. If you cannot show the analysis to demonstrate how, in your world, the MPT does not apply then it does apply. I thought Engineering was a discipline. Professional Engineers can justify their designs and practice with calculations. Where are yours?
P: 1,081
 Quote by sophiecentaur You really don't get this, do you?
Oh yes I do.

 Quote by sophiecentaur All you need to do is to show that an amplifier that is 90% efficient when feeding into a 50Ω load on the end of a 50Ω feeder (so the transmitter has a source resistance of 6Ω), when operating to its spec., can, itself, be made to present a 50Ω termination for signals reflected from the load. The problem can't be reduced further so it should be easy for you to show that you are correct.
It is absolutely clear that there can be no reflected energy in this situation.
It does not take mathematics to show that 50 ohm = 50 ohm = perfect match.

Was this a trick question, or are you so buried in your worship of theorems and mathematics that you don't understand the real world ?

Now if you want to be even more malevolent and mismatch the load to reflect some energy back down the line, which would be a very poor engineering practice, then I can protect my valuable transmitter, while still maintaining the 90% efficiency, by using a hybrid cross connected transformer pair with a 1:1 ratio. If this was a microwave installation then instead of a hybrid I would use a circulator. That would divert all the reflected energy into a dummy load. It will also prevent cross-modulation problems from received signals because the reverse travelling energy will not reach the output stage.
P: 1,081
 Quote by sophiecentaur The style of transformer is totally irrelevant. It has just two ports and is a reciprocal network. It is not magic.
A hybrid cross connected transformer pair has four ports and is the low frequency equivalent of a circulator. A hybrid will work from the audio frequencies in a telephone all the way up to VHF. A hybrid is not the most efficient way to handle reflected energy but it is reliable and quite independent of line length variation, which really helps in switched networks and multicouplers.
PF Gold
P: 11,105
 It is absolutely clear that there can be no reflected energy in this situation. It does not take mathematics to show that 50 ohm = 50 ohm = perfect match.
So how do you claim 90% efficiency is possible for the transmitter?
You are trying to shift the goal posts now. Originally you were implying that everything was matched in practice (both ways) and yet a class C transmitter can still operate at high efficiency. What is your position now?
Why do you keep bringing up irrelevant things like hybrids and circulators? More smoke and mirrors rather than a short bit of Maths. HF transmitters use neither. Also, real HF antenna arrays do not present a perfect match over their band. VSWRs of worse than 2 are not uncommon. You will be aware that one HF array may need to operate at several different frequencies in one day. They cannot be tiffled every 15 minutes.
BTW, are there any Class C microwave amplifiers? Why do you need to introduce waveguides into this? More smoke and mirrors.

Can you just concentrate on the basic question here. Can you show that you can perfectly 'match' with a load and still get 90% efficiency? If reactances are necessary to prove your point then show how they work (quantitatively). None of the Maths involved need be too difficult and no magic is allowed. Also, no 'helpful' information about operating in the presence of other signals is needed. If there are some practicalities are involved, then include them in your calculations. (All this stuff is calculable, you realise) Failing that, give a good reference about standard transmitter practice with evidence of what you claim. The Marconi HF Transmitter spec that I read, recently, made no mention of output impedance - just the range of load impedances it works into.
That's all there is to it.
P: 1,081
 Quote by Baluncore Was this a trick question, or are you so buried in your worship of theorems and mathematics that you don't understand the real world ?
You have not answered my question.
You are an behaving like an incessant bully.