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Author Topic: Transmission Line Question  (Read 9131 times)
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ashart
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« on: December 11, 2017, 07:18:26 PM »

I'm perplexed!

Walt Maxwell, in his book Reflections II, extensively discusses transmission-line reflections.  In Appendix 7-1, discussing the transmission line conjugately driven by a matched source, [ eg: Z(source) = R + jX and Z(line) = R- jX ], but terminated by a mismatch, [ Z(load) and Z(line) not equal ], Maxwell states:

   “The reflection at the mismatch gives rise to a second wave that travels back to the input of the line with the magnitude determined by multiplying the magnitude of the initial wave by the magnitude of [the reflection coefficient] ρ.  On arriving at the line input it sees total reflection ρ = 1 at the matching point in the network that achieved the conjugate match.”

Now I’ve read this idea enough times, in enough places, by enough authors, in enough different wording, that it would seem foolhardy to disbelieve it.  But!  Back on page 3-1, Maxwell states in Eq.  3-1 that:
      
         ρ =  [ Z(load) - Z(line ] / [ Z(load) + Z(line) ]

I’ve also seen that equation in lots of places, and in a conjugate-matched system, this equation, I believe, ends up with a zero numerator, and thus equates to zero.

And if so, how do I correlate p = 0 here with Maxwell’s later statement that the conjugate match at a line input gives ρ = 1?  Was he lying in Chapter 3 or is he lying in Appendix 7?

Ok, ok, so Walt didn’t lie.  Where am I missing the boat?  How does a first reflection from a mismatched load, arriving at the conjugately-matched line input, get fully reflected?

-al hart,
al@w8vr.org
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KA2DZT
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« Reply #1 on: December 11, 2017, 07:39:38 PM »

Without having read any of Maxwell's work, it seems p is a number between zero and one.  I concluded this from just what you say his equations are.

I think I see that with a perfect match p would be zero.

Probably not much help,

Fred

Just re-read your last question,  I'm not getting how the reflected wave reaching the matched input gets fully reflected (p=1).  I always thought it gets absorbed by the matched input.

Maybe I'm not understanding something.

I think I also see where p would be the absolute value because I think I see where p could be a negative number when Z(line) is greater than Z(load).
Been a while since I did any math with complex numbers.
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R. Fry SWL
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« Reply #2 on: December 12, 2017, 08:57:51 AM »

RE:“... The reflection at the mismatch gives rise to a second wave that travels back to the input of the line with the magnitude determined by multiplying the magnitude of the initial wave by the magnitude of [the reflection coefficient] ρ.  On arriving at the line input it sees total reflection ρ = 1 at the matching point in the network that achieved the conjugate match.” ...

If this concept was universally true, there would be no need for the load SWR protection circuitry included in many transmitters.

Such reflected power can (and has) caused the destruction of r-f output stage components of a transmitter -- particularly where high incident power is involved, such as in the broadcast industry.
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ashart
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« Reply #3 on: December 12, 2017, 09:29:51 AM »

Mr. Fry:

Thank you for your response.  I read Maxwell to opine that with a mismatched load, injury to a final amplifier caused by high line SWR is not caused by reflected power being dissipated by that amplifier per se, as so many do erroneously believe, but instead, is caused by excessive voltage or current at the antenna terminals resulting from a change in feedline impedance.

Supporting Maxwell, it seems reasonable to assume that phase-angle considerations (P = V I cos theta) might allow destructive voltages or currents to prevail even without the dissipation of any power.

73 de al hart
al@w8vr.org
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wa1sth
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« Reply #4 on: December 12, 2017, 09:55:48 AM »

Reading this made my head hurt....
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R. Fry SWL
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« Reply #5 on: December 12, 2017, 09:59:46 AM »

RE: ... excessive voltage or current at the antenna terminals resulting from a change in feedline impedance. ...

But a mismatch between the Zo of the termination at the far end of a transmission line to the Zo of that line does not really change the Zo of the transmission line.  That depends on the physical construction of the line, itself.

It causes an energy reflection back toward the input end of that line which in some situations can be enough to cause its failure, along with possible failure of the transmitter output stage.
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KA2DZT
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« Reply #6 on: December 12, 2017, 10:30:35 AM »

It would seem that the phase angle between the transmitted and reflected waves at the transmitter would determine what excessive voltages or currents may or may not occur.  The phase angle is determined by the length of the transmission line between the transmitter and the mismatched load.
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ashart
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« Reply #7 on: December 12, 2017, 10:35:43 AM »

To WA1STF:

You're not alone, OM!

Years ago, I assigned a tech to graph some data I had taken, and to do so on "log-log" graph paper.

He eventually reported back without the graph, but informing me that the log-log paper had itself given him a headache!

Big smile!

-al hart
al@w8vr.org
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WD8BIL
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« Reply #8 on: December 12, 2017, 12:08:05 PM »

Quote
“The reflection at the mismatch gives rise to a second wave that travels back to the input of the line with the magnitude determined by multiplying the magnitude of the initial wave by the magnitude of [the reflection coefficient] ñ.  On arriving at the line input it sees total reflection ñ = 1 at the matching point in the network that achieved the conjugate match.”

OK, if I'm reading this right (phat chance) It states the reflection coefficient at the mismatch is some what less than 1. So the magnitude of the reflected wave is accordingly less than the incident. OK, I got that.

But you are correct. It still doesn't explain why, at the matched end, the reflection is perfect. I thought the whole idea of matched impedances was maximum power transfer? Huh If that is so, and a match occurs at source/line then why is the wave reflected and not transferred to the source?

We have to be missing something here! Huh
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KA2DZT
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« Reply #9 on: December 12, 2017, 12:59:25 PM »

Bud,

Unlikely that quoted statement is correct.  Hard to know exactly what Maxwell was thinking when he came up with this theory.

I think the reflected wave would add to the incident wave (depending on the phase relationship) at the transmitter and the sum of the two waves would travel back to the load.  Maybe this is what he is talking about.

My head is beginning to hurt.

Fred
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WA4WAX
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« Reply #10 on: December 12, 2017, 05:57:20 PM »

I can help with this one.

Assume you have a Johnson Matchbox feeding a 105 ohm loop with a length of 450 ohm ladder line.  The box is adjusted to give 50 ohms at the input.

When the reflected wave reaches the match point inside the box, 100% of it is reflected back towards the antenna.  That is where rho = 1 comes from.  When you look in the input, rho = 0, assuming a 50 ohm transmitter.

So, he is talking about rho for two different points in the circuit.
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ashart
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« Reply #11 on: December 13, 2017, 09:23:38 AM »

To:  WA4WAX

Well!  That certainly explains how rho can be 0 in Chapter 3 and 1 in Appendix 7!  I never thought of his referencing two different points!  Thank you!

Might you have time to address a follow-on question?

How can one define a “match point” within a hypothetical tuner, be it L or T, and absent a tuner, where within a PI tank or a link-coupled tank, would one find that match point?

Thank you again!

-al hart
al@w8vr.org
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WD8BIL
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« Reply #12 on: December 13, 2017, 09:33:18 AM »

It still doesn't add up.

According to ;
Quote
ñ =  [ Z(load) - Z(line ] / [ Z(load) + Z(line) ]
, if there is a match Z(load) - Z (line) = 0. Therefore p=0 and there is no reflection.


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Steve - K4HX
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« Reply #13 on: December 13, 2017, 09:45:16 AM »

Only true on the transmitter side of the tuner.
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WD8BIL
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« Reply #14 on: December 13, 2017, 10:44:44 AM »

What if there is no "tuner"? If the source and line are matched at the source end, and a mismatch is present at the load end the resulting reflected wave will see p=0 at source. The source will then have to absorb/dissipate the reflection, right?

Now I can see, with a tuner you actually have 2 systems working here. 1 source/line/load system from radio to tuner and one source/line/load system from tuner output to load. In the second case the tuner will then have to deal with reflection when tuned to a match and p again =0.

Confusing stuff.

But to Al's original observation:

If p =  [ Z(load) - Z(line ] / [ Z(load) + Z(line) ] is true, and Z (load) = Z (line) then the numerator will =0 and p cannot equal 1. Therefore the statement; "On arriving at the line input it sees total reflection ñ = 1 at the matching point in the network that achieved the conjugate match” cannot be true.



 
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KA2DZT
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« Reply #15 on: December 13, 2017, 11:33:13 AM »

Some folks completely agree with Maxwell's work.  There also is a lot of folks that don't agree.  I think the reflected wave is added to the incident wave and the sum of the two waves are sent back to the load.  Phase difference determines what the sum of the two waves ends up being.  So the reflected wave does return to the load.

Suppose a single cycle pulse was transmitted to the load.  The reflected wave would then return to a passive transmitter.  What happens to the reflected wave??  Assume a perfect match between the transmitter and transmission line.  Mismatch at the load.

Fred
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Steve - K4HX
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« Reply #16 on: December 13, 2017, 02:15:49 PM »

How could the source be matched and the load not be and no "tuner" be involved?

Quote
What if there is no "tuner"? If the source and line are matched at the source end, and a mismatch is present at the load end the resulting reflected wave will see p=0 at source.
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ashart
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« Reply #17 on: December 13, 2017, 02:49:08 PM »

Steve:

Would any source with a Thevenin internal impedance of Z = 50 + j0 not do the job or am I misunderstanding your question?

-al hart
al@w8vr.org
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WD8BIL
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« Reply #18 on: December 13, 2017, 04:36:30 PM »

Should have been more detailed Steve. Source, 50 ohms - Line, 50 ohms - Load, something other than 50.

This puts the mismatch at the load end of the line (which has an impedance of it's own) not at the source. Splitting hairs??? Perhaps!
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Steve - K4HX
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« Reply #19 on: December 13, 2017, 10:09:14 PM »

If the source is 50 Ohms and the line is also 50 Ohm but the load is not, there will be a mismatch at the source.  Example:

50 Ohm source driving 100 feet of RG213 with 200 Ohms connected at the far end - the Z seen by the source will be 105 - j80.
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ab3al
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« Reply #20 on: December 14, 2017, 01:41:42 AM »

But all of this only makes sense until you include the angle of the dangle in proportion to the droop of the hoop
   
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R. Fry SWL
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« Reply #21 on: December 14, 2017, 08:16:25 AM »

Just to note that, although a transmitter may be specified for the nominal r-f output power it can safely produce across a given load Z connected to its output terminals, this does not mean that the r-f source within the transmitter has the same impedance as that load.

If that was true, then 50% of the r-f power produced by the final amplifier would be dissipated in the amplifier circuitry itself, rather than being available at the output connector of the transmitter.

It is evident that this belief is not valid just by comparing the r-f power in the specified load Z at the transmitter output connector to the total d-c input power supplied to its final r-f amplifier.  The d-c to r-f conversion efficiency of a transmitter using a Class C or D final stage r-f amplifier can be 75%, or better.  Some AM broadcast transmitters using digital processes in the final r-f amplifier have d-c to r-f conversion efficiencies of better than 90%.

The final r-f amplifier circuits in many transmitter designs using FM, CW or high-level amplitude modulation, and specified for 50 Ω loads have an effective output source impedance of just several ohms.  The lower that source impedance, the greater the conversion efficiency of that amplifier.
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WD8BIL
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« Reply #22 on: December 14, 2017, 08:47:57 AM »

Quote
50 Ohm source driving 100 feet of RG213 with 200 Ohms connected at the far end - the Z seen by the source will be 105 - j80.

Uh duh.... thus the advent of the coaxial transformer! Thanks Steve! I forgot!  Undecided

Which brings up another point: Not working for a few months slows the brain down!!

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KD6VXI
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« Reply #23 on: December 14, 2017, 09:51:31 AM »

But all of this only makes sense until you include the angle of the dangle in proportion to the droop of the hoop
   

Now see, we where taught the angle of the dangle was inversely proportional to the square root of the beat of the meat???

I'm confused now.

--Shane
KD6VXI
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