I have been looking for cause of signal reflections in transmission lines. Everywhere it is concluded that the reason is impedance mismatch. I can understand if the impedance changes in the path of the signal travels, then reflection will occur, but what I could not understand is the physics behind it.

How does including a series resistor of the characteristic impedance suppress the reflections, since the direction of reflected voltage is opposite? Does it mean the point at which the impedance changes becomes high in potential or anything else? Can anyone give directions to understand the physics behind this?

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    \$\begingroup\$ You can play with this: helloworld922.blogspot.com/2013/04/… \$\endgroup\$
    – DKNguyen
    Commented Mar 3, 2020 at 14:31
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    \$\begingroup\$ Do you understand the analogy of injecting water into a pipe that changes diameter somewhere along its length? The resistor helps dampen reflections from reaching the transmitter and keeps the segment between transmitter and nearest discontinuity really short so reflections that do reach the transmitter die out quickly. It doesn't actually have to be related to the characteristic impedance, but it helps. If you do try to match the resistance you are basically adjusting the driver impedance to match the line. \$\endgroup\$
    – DKNguyen
    Commented Mar 3, 2020 at 14:38
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    \$\begingroup\$ You can study maxwell's equations for the lowest level understanding possible today. Another concept is that an open circuit at the end of a line enforces a certain rule (current = 0) and a short circuit enforces a different rule (voltage = 0). But when you apply a signal to a a transmission line, the signal does not know what is at the end. So power is delivered to the transmission line and energy travels down the line. Only when it gets to the end of the line does the energy learn whether it will be accepted or rejected. \$\endgroup\$
    – user57037
    Commented Mar 3, 2020 at 18:42
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    \$\begingroup\$ Very good mechanical visualisation: youtube.com/watch?v=DovunOxlY1k \$\endgroup\$
    – jusaca
    Commented Mar 4, 2020 at 9:30
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    \$\begingroup\$ I found this explanation very good: youtube.com/watch?v=Il_eju4D_TM \$\endgroup\$
    – AndreKR
    Commented Mar 4, 2020 at 9:35

5 Answers 5


With electrical transmission lines, it all has to do with the speed of light being finite, thus so is the speed of EM propagation in a wire. You can think of a wire as a long series of infinitesimal capacitors (connected by infinitesimal inductors). If you start charging the capacitors at one end, you have to keep pumping charge into the wire to charge more and more of the infinitesimal little distributed capacitors down the wire (at the speed of EM propagation).

Then a problem occurs when you get to the end of the wire.

If shorted, the last capacitor can't charge, and it in turn then discharges the capacitor one back. This discharging of capacitors then travels back to the source (at the speed of EM propagation) as a negative voltage reverse wave that cancels out the forward positive wave.

If open, remember there is still charge being pumped into the line. It has to go somewhere, so the "momentum" of the current (due to distributed inductance of the wire) charges up the last capacitor to double the voltage of the rest of the line capacitance. But this last capacitor is connected to all the ones before it. So this wave of charging to double the voltage then propagates backwards on the wire as a positive reflection wave.

Somewhere in between a short and an open at the end is a nice medium value of impedance which can absorb the wave of current hitting the end without either under or overcharging that last bit of capacitance at the end. Thus leaving no change in voltage to propagate back along the transmission line. That terminating impedance just happens to be the characteristic impedance of the transmission line, which is determined by the distributed capacitance and inductance of the wire (and its surroundings: permeability, ground plane or return geometry, coax shield, et.al.)

Pretty much the same thing occurs with air pressure waves in a tube or pipe (organ), or when whipping one end of a rope sideways. Different rope waveforms occur depending on whether the rope is tied or loose at the opposite end. etc.

(Added: For transmission lines where there is a discontinuity or mismatch in impedance somewhere in the middle, you can think of it as a superposition of two transmission lines, one line with no discontinuity over the total length, plus one shorter line with an open or short at the discontinuity. The reflection will relate to the ratio of the two superpositions.)

  • \$\begingroup\$ your answer has cleared and justified my understanding, i also heard the term return path as you mentioned in the comment in my understanding it refers to the other plate of the virtual capacitor formed by the transmission line, am i right ?? \$\endgroup\$
    – Vignesh C
    Commented Mar 4, 2020 at 5:49
  • \$\begingroup\$ @hotpaw2, may i know for an open-ended line how the voltage becomes double at the open end \$\endgroup\$
    – Confused
    Commented Jun 17, 2021 at 15:00
  • \$\begingroup\$ Consider an open ended line to be the folded superposition of a line twice as long with an electrical charge or current wave sent towards the middle from both ends simultaneously. What happens to the 2 voltage waves when they eventually meet in the middle? (Think linear systems.) \$\endgroup\$
    – hotpaw2
    Commented Jun 17, 2021 at 16:19

The impedance of a transmission line, in ohms, is the ratio of voltage wave and current wave that travels down the line. For a 100 ohm line for instance, a 1 volt wave will always be accompanied by a 10mA wave. Intuitively, the current wave delivers charge to the parts of the line that have to 'charge up' to the voltage of the voltage wave.

If the 100 ohm line is now connected to a 50 ohm line, the ratio of voltage and current waveforms in that line will be different. The junction between the lines cannot physically support two voltage and current waves with different ratios. A third wave is therefore generated that 'takes up the slack' between the mismatched waves, and is reflected back along the source line. Note that the reverse travelling current wave will subtract at the junction, whereas the voltage waves will add. It's therefore always possible to find an amplitude and phase of a reverse wave that will match any two lines.

If the line is loaded, or sourced, with a resistor equal to the impedance, then a voltage/current wave arriving at that junction finds that it can continue to propagate with just the right ratio of voltage to current, and so no reflected wave is generated.

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    \$\begingroup\$ i have edited the question because i can understand what is happening, what i can't get is why it is happening. \$\endgroup\$
    – Vignesh C
    Commented Mar 3, 2020 at 11:34
  • \$\begingroup\$ Small clarification for TL theory: The characteristic impedance always relates the voltage and current amplitudes in the same direction. E.g. the forward traveling components V+ and I+ will always be related by Z0, reflection or not. But the total voltage or current on the line, the sum of waves in both directions, will only be determined by real resistances in the line. The reflected and transmitted waves, individually, are characterized by Z0, but not their sum \$\endgroup\$ Commented Mar 12, 2020 at 14:20

out on a limb here ---- the (energy, as Tim Wescott writes) differential equations are used, with boundary conditions that require the sudden appearance of waves traveling in the reverse direction IF that boundary is not exactly Zo; the energy is preserved in the new mix of voltage/current values for each of the forward and (now) reverse waves.

regarding "series terminations" --- a series termination, best used with the resistor installed at the SOURCE end of the transmission line, exploits reflections in its operation. Initially the line voltage is only half the Source voltage because of the voltage division of the lumped resistor driving the Z of the line; any circuit monitoring the line will see only Vin/2 and that often is a FORBIDDEN VALUE for logic circuitry; however, at the far end, the receiving end, the math tells us that unterminated receiving end HAS A REFLECTION, and the math tells us the voltage doubles as part of preserving the energy. Thus ONLY at the far end, the receiving end, will a useful full amplitude waveform be created.

At all other points along the line, the voltage will be HALF for some time, and then the reflected energy doubles the voltage. This doubling occurs, eventually, at all points. In general, trying to extract data from this 50% then 100% waveform is a bad idea.

Only at the far/receiving end does a safe-to-use waveform exist.

On the other hand, the use of series-at-the-source termination will reduce overall power consumption.

  • \$\begingroup\$ This is the bottom line: the voltage and current distribution on the line follow a differential equation with boundary conditions at discontinuities. In the differential equations class, you learn to solve this with Fourier series. When you take your electromagnetic theory class, you will learn how the differential equation leads to waves and how the boundary conditions lead to reflected waves. The phase and amplitude of the reflected wave depends on the nature of the discontinuity: short, open, restance, capacitive reactance, inductive reactance, or a complex impedance. \$\endgroup\$ Commented Mar 9, 2020 at 20:42

If you take an electrical engineering course, then in your third year (at least in the US) you'll have a chance to take "Electricity and Magnetism" or "Engineering Electrodynamics" or some similarly title course. Physics courses will have an equivalent, I'm sure, although I don't know if they'd go into transmission lines to the depth of the EE course.

It'll teach you how light propagates in free space, and then it'll teach you how electromagnetic waves propagate in a transmission line.

It turns out that the easiest way to solve the equations is to express the energy in the transmission line as propagating at the speed of light for the line (it's slowed down by the material used for the dielectric, but that's another story). So the only way (in that view) that energy can travel in the line is by going at the speed of light, either forward or back. Moreover, the current a forward wave has to be equal to the forward wave's voltage divided by the line's characteristic impedance. Always. The equations don't allow for anything else, and they match the actual physics pretty darned well.

So when a forward wave hits a discontinuity in the line that forces the voltage/current relationship to be different than the line's characteristic impedance, a reflection must be generated. Otherwise, the physics don't work.

If the line is terminated by its characteristic impedance, then the voltage/current ratio at the termination is equal to the voltage/current ratio of the forward wave, and the physics do not allow a reflection to occur.


Here's an outline that should give you pointers to find more details yourself:

The physics behind reflections in electric transmission lines is a restriction of the general 2-dimensional case to a simplified case with only 1 dimension.

The general case, reflection and transmission of electromagnetic waves at a boundary, is completely described by the Fresnel equations.

(The 1-dimensional case is easier as you don't have to care about the incidence/transmission/refletion angle; it is always 0° or 180°, which can be expressed by a positive or negative sign instead of an angle).

The Fresnel equations in turn are based on very basic continuity conditions of \$E\$ and \$B\$ fields at the interface (boundary) between two media and Maxwell's equations.

A boundary between two media is

  • (in 2-D case) where the index of refraction \$n\$ (depending on \$\epsilon\$ and \$\mu\$)
    (e.g. at the interface of water and air)
  • (in 1-D case) where impedance \$Z\$ (depending on capacitance and inductance per length)
    (e.g. at the connection point of a RF-PA with 50Ω and a 75Ω coax cable).


  • \$\begingroup\$ I feel sorry for the student reading the above answers. Each answer points him in a different direction, and all of the answers are correct! \$\endgroup\$ Commented Mar 9, 2020 at 20:44
  • \$\begingroup\$ It's rather beneficial to have different aspects of a thing. Though I feel sorry about the fact that actually few of the answers really answer the question of the physics behind the phenomena. Most of them only describe or illustrate it more detailed but don't refere to more basic principles or laws of physics. \$\endgroup\$
    – Curd
    Commented Mar 10, 2020 at 7:07

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