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Why doesn't a voltage source re-reflect the reflected wave in a transmission line

  1. May 20, 2010 #1
    scenario 1:- as we all know in a transmission line the forward travelling wave can't be completely absorbed by a load other than Z0 (where Z0 is characteristic impedance) . So part of the wave is reflected back by the mismatched load. My doubt is when this reflected wave reaches the energy source why doesn't the source reflect part of it back again. I guess this is related to some basics which I haven't understood. If the source were to re-reflect this wave , the wave which is in forward direction would undergo some change which would depend upon the phase difference. But this doesn't happen and my doubt is why the source doesn't re-reflect the wave.

    Scenario 2 :- I came across this scenario while studying the tutorials on Tx lines on IIT distance education website. Please see details of the scenario at this IIT-Bombay webpage - http://www.cdeep.iitb.ac.in/nptel/Electrical%20&%20Comm%20Engg/Transmission%20Lines%20and%20EM%20Waves/slides%202/13.4.html [Broken]

    [PLAIN]http://www.cdeep.iitb.ac.in/nptel/Electrical%20&%20Comm%20Engg/Transmission%20Lines%20and%20EM%20Waves/Figures/2.34.jpg [Broken]
    As we see the energy source is coupled somewhere along the line at a location x. The line is resonant length [tex]\lambda[/tex]/4 . The voltage source sends two waves in two different directions. As each wave sees the open circuit load it is reflected then it goes in the other direction sees the short circuit load and is again reflected. The wave keep on doing round-trips in the circuit. Now the situation is such that after each round trip phase change in the wave is 2*pi . and so the standing wave pattern keeps building up .

    But my question is how is this scenario different from scenario 1 , i.e what is different in the way the source is connected here that the waves are able to do round-trips. I understand that the standing waves build up because after each round trip the phase difference is 2*pi , but my doubts are in the way the source is connected. how is that different from the scenario 1 (which happens to be the more conventional scenario)
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 20, 2010 #2
    The source can reflect the wave. Ideally, the backwards reflection of the source is nearly 0 (incredibly low dB) so that the reflection is negligible. The reason is that the source's output impedance is matched to the transmission line.

    Have you studied scattering parameters yet? Typically, a transmission device has four very important parameters: forward gain, forward reflection, backward gain, and backward reflection. That's the more complete picture. Usually you're going to want everything to be 0 (low dB) except for the forward gain but real components are not going to be ideal.
  4. May 20, 2010 #3
    Thanks for the reply . :biggrin:You have somewhat resolved my doubt . No I haven't studied scattering parameters yet. Can you explain how the output impedance of the source is matched to the characteristic impedance? (I hope that doesn't need knowledge of scattering parameters) .Actually I am undergoing the course of transmission lines at undergrad level , so don't know whether that thing is included in it . But will try to cover that once the basics are done.

    Also the scenario 2 which I pointed out remains unadressed.
  5. May 20, 2010 #4
    You know how the reflection at the load can be zero, right? The reflection parameter is this:

    [tex]\Gamma = \frac{Z_l - Z_0}{Z_l + Z_0}[/tex]

    If the impedance of the transmission line and load are equal then the reflection is 0. The same thing happens for a backward traveling wave returning to the source. It's going to see the output impedance of the source just as if it was a load.

    The output impedance is what you see when you look into the output. It's the Thevanin equivalent resistance referenced from the output. It would be like shutting the source off and connecting an Ohmmeter across the output terminals. You can't really measure it directly like that in all circumstances but I hope you get the idea.

    In your second part, you have at least three transmission lines. There's the two that are described in the image going left and right from point X and a third one that delivers the signal from the source to point X. We could arrange it so that when we look back towards the source down the delivery line from point X all we see is an open circuit. So, a reflected wave comes back towards point X and sees a nearly infinite impedance to go back towards the source or a finite impedance to travel to the other end. It's going to go to the other end. At least that's what I think is going on. I could be wrong.
  6. May 20, 2010 #5


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    It's worth while pointing out that the output impedance of a power amplifier may be far from an ideal match because it is an active and, possibly, a non linear device. There will be other considerations when you want as much power out as you can get. But it is easier (but not trivial) to match well to the load and that will reduce the effect of reflections.
  7. May 20, 2010 #6


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    why doesn't a voltage source re-reflect the reflected wave in a transmission line

    There can only be one voltage at a time at any point on the transmission line (or anywhere else).

    So the returning wave gets back to the signal source and, just by being there, it affects the input impedance of the line as seen by the signal source. If the returning voltage is the same polarity as the incoming wave, the input impedance is increased and it is decreased if the returning wave is out of phase with the incoming wave.

    This is identical to the effects you get with batteries in Kirchoff Law problems. You can have batteries in a circuit but no current flowing if the voltages exactly oppose each other.

    So, it is not really a reflection, but it certainly affects the behaviour of the transmission line.

    In practice, the returning voltage is never as big as the outgoing one, so you can't get complete nulling of the input current.

    To answer your next question, if you turned off the signal generator before the wave returned, yes, you would get another reflection back along the line if there was a mismatch.
  8. May 20, 2010 #7
    Thanks people , 90% of my doubts are resolved. I can quite appreciate how if load is matched to Z0 most of our issues are taken care of.

    @Okefenokee I just have one doubt about the output impedance.
    see while calculating [tex]\Gamma[/tex] at load end we consider ZL as seen at the load end and not the transformed load impedance as seen from source end.
    So while calculating [tex]\Gamma[/tex] at the source end ,should we consider :- a)the source impedance as seen from source end or b) the transformed source impedance as seen from load end.

    Also is this source output impedance, simply the source impedance Zs i.e the internal resistance of our energy source ?
    If yes, does it mean we have to take care that it is matched with Z0 or as some people pointed out that as the source is an active load we cant do much about it.

    I have the above doubt because in the course that I am studying , it seems quite implicit that no reflection takes place at the source. is the reflection very negligible in pratice?
  9. May 21, 2010 #8
    The answer is a. if you want to know the backward reflection at the point where the source connects to the TL then all you need to know is the output impedance of the source and the characteristic impedance of the TL.

    If you want to know how to calculate the total backward reflection of a number of components in a chain then you really need to learn about scattering parameters and matrices. You should know that an ideal TL does not reflect anything by itself. It has to have a load on either end that is not matched to 50Ohm (or maybe 75Ohm depending on the system).

    If you analyzed the source at it's output for the Thevenin equivalent circuit you would get a truly ideal voltage source and some output impedance. Zs is that output impedance. It acts as the load for a backward traveling wave. Think of it this way. You would basically have a voltage divider between the output impedance of the source and any load that you connect directly to it. If you took a number of accurately measured loads with different impedances and directly connected them to the source one at a time then measured the voltage, you could get a graph that reveals the output impedance of the source.

    Do we have to take care that it is matched? Test equipment is usually matched and/or calibrated closely to 50 Ohms. In the practice that I've seen, which was only lab work at college, you don't have to worry about it. If it's a problem in some system then they likely do impedance matching with some network of capacitors and inductors to reduce the reflection.
    Last edited: May 21, 2010
  10. May 21, 2010 #9


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    We certainly do!
    If you don't have a good match on a long feeder (Up a TV mast, for instance - and the transmitter will not be a good match) your echos can either give you a nasty frequency tilt on your signal or edge distortions / ghosts. That really crucifies a high quality TV signal and can also seriously degrade a digital signal. Remember, transmitters aren't often used for CW transmissions.
  11. May 21, 2010 #10
    @sophiecentaur , @Okefenokee , @vk6kro

    Thanks people for helping me clear my doubt and that right quickly. o:)
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