Reflections at a short-circuit

  1. Hey

    I wounder why an approaching electrical impulse is completely reflected if you short-circuit the conductor. I have read some explanations suggesting it is because Kirchhoffs law must be satisfied, but that argument falls badly when the conductor is long and the pulse short, such that there will be no constructive interference and doubling of amplitude. According to the formula zc=(z1-z2)/(z1+z2) there should be no reflection, since the characteristic impedance of the two parts of the conductor have the same characteristics.

    If the above z1 instead is the impedance/resistance (instead of the characteristic impedance) of the first part of the conductor, this would mean that a resistor inserted between the two conductors that earlier composed the short-circuit would cause reflections even though its impedance matches the characteristic impedance of the conductor.

    Thanks
     
  2. jcsd
  3. nsaspook

    nsaspook 1,164
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  4. sophiecentaur

    sophiecentaur 13,913
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    The reflection coefficient is given by
    ρ = (zterm-z0)/(zterm+z0)
    (Your formula is not dimensionally correct.)

    If zterm =0
    the reflection coefficient becomes -1, which is what you get, isn't ti?
     
  5. Oh, that was a typo. I meant to write coefficient of reflection.

    Thanks you for your answers. :)
     
  6. The formula has, everywhere I have seen it, been given by your termination impedance exchanged for the characteristic impedance of the leader which the signal leaves at the joint. According to your formula all you have to do is match the termination impedance to the characteristic impedance of the second conductor, which is not the case.
     
  7. sophiecentaur

    sophiecentaur 13,913
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    I don't understand that comment. If you terminate the line with Z0 you will get a ρ value of zero. If you put a shunt of Z0 across the line you will be effectively terminating with Z0/2, which will be a mis-match with ρ of -1/3. Are we confusing the action of terminating with the action of adding a parallel element?
    When you short the line out, of course, it makes no difference whether the short is at the end of a line or at some point on the line; the ρ is -1, in that case.
     
  8. I interpreted your reply as if the characteristic impedance of the termination between two conductors (not necessarily with the same characteristic impedance) was denoted by Zterm. If you by Zterm mean the extra circuitry at the joint And the second conductor, then the formulas are identical.
     
  9. sophiecentaur

    sophiecentaur 13,913
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    Hopefull my subsequent post will have cleared that up. I used the word 'terminate' in the literal sense - i.e. what is connected to the 'end' of the line in question. This could be a resistor, another line or a dead sheep (antenna engineer's term for any load you care to think of but not an entirely fictional idea, I believe).
    But there seemed to be some doubt about the formula producing the right answer - it does.
     
  10. I really appreciate your help I must add before taking this further. :)

    Why is zterm zero at the short-circuit? Would not the stuff connected to the end of the conductor have a characteristic impedance that is not zero?


    Talking about all these reflections - if you have an open circuit which is subjected to alternating voltage, wouldn't this mean the circuitry before the point of breach(?) is constantly causing energy to be lost/dissipated to the surroundings?
     
  11. sophiecentaur

    sophiecentaur 13,913
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    A short circuit in parallel with another value of impedance (the downstream bit of line) is still a short circuit. No signal will propagate past a short, will it? It will all be reflected. Nothing can be dissipated in a short - that's why all the energy is reflected. This all assumes that your transmission line is not radiating (ideal). Sort out those basic issues before talking about a leaky line.
     
  12. Yes, I am aware of that. Still, at the short circuit there are at least two ends - one through which the incident signal propagates and one through which the signal could possably return, assuming that we don't already know that 100 % of the signal is reflected. Hypothetically removing the second one and measuring its characteristic impedance will leave us with a nonzero result. Reconnecting it, this will cause the characteristic impedance of the termination to be nonzero as well. Right?

    That an ideal transmission line does not cause any loss of energy is indeed intuitive, but I was looking for confirmation that this is not the case for a real line and implicitly wondering to what extent this loss is a consern, i.e., the magnitude of the loss (which of course is different for different constructions). Voltage differences in a line --> current. Current + resistance --> power...
     
  13. sophiecentaur

    sophiecentaur 13,913
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    I am completely confused now. What arrangement are you actually trying to describe? You quoted a formula, earlier on, and I assumed that this was a simple transmission line question (two parallel wires or a co-axial arrangement) and that you were familiar with it. A Google search will show you many hits which will relate that formula to a a practical transmission line. See this link
    How are you actually relating this theory to your practical setup?
    Can you give us a diagram, please?
     
  14. f95toli

    f95toli 2,418
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    I wonder if the source of confusion is that the OP is thinking that characheristic impedance is a property of a SINGLE conductor; whereas in fact it is always a property of a PAIR of conductors. This is easier to understand if you realize that we are (nearly) always assuming that you have a TEM wave propagating along the line, and that much of the energy is actually travelling in-between the conductors (which is why the speed of light in a coaxial line is set by the dielectric).

    Hence, it does not make sense to talk about the impedance of a short circuit, this would always be a single conductor that connects the signal conductor to ground; in the case of a coaxial cable you could make short by simply bending the inner conductor and soldering it to the outer conductor (and if you wanted to terminate a coaxial line you could solder a 50 ohm resistor between the centre conductor and ground).

    Also, when talking about ideal short circuits it is implied that you are talking about something much shorter than the wavelength with zero resistance and reactance. Now, this could obviously never happen in real life, but it is not difficult to make a short with almost perfect reflectivity.
     
    Last edited: Feb 12, 2013
  15. sophiecentaur

    sophiecentaur 13,913
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    You could be right there. Let's wait for a reaction from ff.
     
  16. Okay, this should clarify my thoughts:

    I have attached a picture of a simple circuit. In the original circuit there is no blue line, but then there is a short circuit introduced, represented by the blue line. The length of the blue line has, of course, nothing to do with the "length of the short circuit", but rather we are talking about an ideal short circuit. If the equation for the reflection coefficient ρ is applied at the point which is contradictory represented by the blue line, then Zterm=Z3 and Z0=Z1.

    On the other hand, this would mean that if Z3=Z1 there would be no reflection at all, which is not the case. This reasoning would imply that the reflection is caused by the point of connection alone. Then the term Zterm would somewhat loosely be the characteristic impedance of only the point of connection, and the termination is completely independent of what the return path, which for that matter could be more than just a wire, has in store.
     

    Attached Files:

  17. sophiecentaur

    sophiecentaur 13,913
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    You clearly haven't read f95toli's last post. He was right about your fundamental misconception about transmission lines and what is meant by 'characteristic impedance'. Rather than try to give you the 'right' version, I will suggest you look at this link and many others, in which the term is defined and used correctly.
    It amazes me that you haven't already looked up the term. You clearly were having difficulty yet you seem to be using PF as your sole source of information. We have all (you included) been talking at cross purposes and wasting our and your time. What would you have done in the past, when all that was available was a good text book :rolleyes:?
     
  18. nsaspook

    nsaspook 1,164
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    I just remembered this classic video about wave reflections.
     
    Last edited by a moderator: Sep 25, 2014
  19. davenn

    davenn 3,884
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    WOW, what a classic video ... pun intended. Produced in the year I was born haha

    Tho I have learn and know those principles, for my work in RF transmission lines, I have never seen them demonstrated in such a clear way

    thanks nsaspook :)

    Dave
     
    Last edited by a moderator: Sep 25, 2014
  20. sophiecentaur

    sophiecentaur 13,913
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    Should be required viewing for everyone, before they think of asking questions about waves 'n' things on PF.
    So much for Computer Graphics.
    Brilliant.
     
  21. Actually - I never totally got my brain around this until I used a Time Domain Reflectometer (TDR) - the wiki is http://en.wikipedia.org/wiki/Time-domain_reflectometer

    Basically is it a RADAR for a transmission line - when set to the right line type, it will tell you distance to AND type of fault ( Short, Open, impedance mis-match - etc).

    I started using a thumper ( http://www.hvinc.com/downloads/cds_brochure.pdf ) which evolved into a TDR-Thumper ( http://green.rapidsw.com/images/pdfs/primary-fault-detectors/XF_Series_TDR_andThumper.pdf ) The thumper name comes from the sound, used on Medium V ( 1000-69KV) buried cable, when you shoot a 1000J down the cable - and it hits the fault - it makes a thump - of course when you dig up the cable and still can not find the fault you do it again - then it is a BANGer!
     
    Last edited: Mar 2, 2013
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