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Signal reflections in open and closed circuits

  1. Jan 3, 2009 #1
    I'll try to make this as concise as possible. I'm trying to get my head around how signals behave in coaxial cables. If you set up a circuit like this:

    [​IMG]

    Then the earthed end will force a node at that end, right? So you'll end up with your reflections having a phase shift of pi and looking like this:

    [​IMG]

    And you find that a standing wave arises with a node at the earthed end. I also understand that if that end were not earthed but left open (infinite resistance) then the signal would reflect without any phase shift, and a similar standing wave would arise but with an antinode at the open end of the circuit. But what if one were to construct a circuit like the one below?

    [​IMG]

    With the switch open, it should be just the same as an open circuit: a standing wave should arise with an antinode at the open end.
    With the switch closed, the sine waves should hit one another in phase and cause an identical standing wave to occur.

    However, I have measured the electric field strength of the parallel wires in the circuit above with both an open and closed switch, but the standing waves produced are not only of different amplitudes, but also out of phase:

    [​IMG]

    [​IMG]

    Could somebody explain to me why this happens? I feel I'm missing something blindingly obvious.

    Many thanks in advance for any help! :smile:
     
  2. jcsd
  3. Jan 3, 2009 #2

    berkeman

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    Staff: Mentor

    Thread moved to Homework Help. Welcome to the PF!

    Why do you say this?

    You seem to have gotten much of it right up until you say that you still have an antinode when you close the right-hand switch. The v(t) condition changes from an anti-node to a node when you close the switch.
     
  4. Jan 3, 2009 #3
    Ah, sorry, I wondered if I had posted in the wrong place.

    Well the reason is that I don't see how closing the switch will force a node. I can understand than an earthed end will force zero volts and thus a node (like wiggling a string which is nailed to a wall), and an open circuit will create an antinode (like wiggling a free string), but I don't understand what happens when the switch is closed.

    Maybe I should get a long piece of string and a friend to help me wiggle it from both ends at the same frequency.
     
  5. Jan 3, 2009 #4

    berkeman

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    Staff: Mentor

    By closing the switch, you create a short at the far end, which envorces v(t)=0.

    Don't think of the bottom wire as being "ground" -- as you get away from the signal generator, it is just part of the transmission line. The TL can be unbalanced (like with coax), or balanced (like with twisted pair).

    The string analogy works pretty well. To force a node, tie the string to a wall at the far end. To force an anti-node, you need a pretty low-friction ring on a rod, so that it can move freely up and down (or side-to-side, if you excite it that way).
     
  6. Jan 3, 2009 #5
    When the switch closed, the end shorted to ground and there cannot be any voltage swing therefore the end becomes a note.

    I still don't see why your scope traces have different amplitude between open and closed. Did you match the transmission line impedance with the generator source impedance? I guess you just roll two wires parallel onto a spool, no matching, no shielding to get the strange looking wave!!.

    If you match the Tx line Z to source without rolling into a roll, it should be a sine wave and the amplitude should be the same except they are out of phase between open and close of the switch if the generator drive a sine wave.
     
  7. Jan 4, 2009 #6
    But how can there be an earth when the switch is closed when there isn't an earth anywhere in the circuit?

    And I don't know why the two waves have different amplitudes; they were held taut and parallel (not on a spool). The equipment looked like this:

    [​IMG]
     
  8. Jan 4, 2009 #7
    From the wave form, the wave length is about 40cm. Assume speed of signal travel at light speed ( which should be close). The freq is 3EE8m/0.4m = 7.5 GHz!!! Is this true? That is too high a freq to run in the setup. Be careful of the setup. At RF, setup will very likely cause more error on the result than the circuit itself.

    Your scope probe ground lead hook up to the wire will change the circuit. Without a proper ground plane, the scope probe ground literally change the length of your wire between switch open and close. I can't even start to name all the parasitic effect also. The only way to do this is either Fab a long micro strip with ground along side but not close to the line with many VIAs stitch to the ground plane below for the prob return......Or coax cut at certain segment for measuring and need to be very short at the cut...... Hard!!!

    BTW I asked a stupid question. Of cause the amplitude should be different. That is the whole thing about STANDING WAVE. The amplitude is the sum of the forward and reflected wave. At different point you have different amplitude. The amplitude also different at the same point if the switch is open or close. I wasn't thinking when I asked, sorry.
     
    Last edited: Jan 5, 2009
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