Physics behind signal reflections and series termination

Question

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?

Answer 1

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.)

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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).

changes.