How do reflections of a square wave cause ringing in transmission lines?

Question

When a square wave is applied to a transmission line which is not terminated with its characteristic impedance, reflection and ringing occur.

I don't understand how sinusoidal ringing occurs when I apply a square wave.

I really don't understand the mechanism of reflection in case of square wave.

When a square wave is applied, is the square wave reflected or are its harmonics reflected?

If harmonics are reflected, why?

Edit: The present answer and comments clarifies regarding the above questions. I mainly wanted to know the case why ringing is observed in the open circuit termination at both ends. What is the mechanism behind it?

Please consider answering the questions in edit.

Answer 1

We'll start with what a square wave is.

A square wave is a bunch of sine waves added together. You have a primary, and a bunch of harmonics added to it. The harmonics are in a particular phase relationship to the primary. The sum of all the waves is a square wave. If you leave any of the sine waves out (or change their phase,) then the sum looks less like an ideal sine wave.

This image from the Wikipedia page illustrates it clearly:

Now, on to the reflections.

If your transmission line were terminated in a perfect resistor that doesn't match the line impedance (and there were no effects from capacitance or inductance) then the reflection would be a nice, sharp copy of your original square wave.

In any real circuit, you will have capacitances and inductances to deal with.

The inductance, capacitance, and resistance of your transmission line will interact with the inductance, capacitance, and resistance of the termination. The result is an impedance that varies with frequency.

A square wave is a collection of sine waves of different frequencies, so different parts of it are more or less strongly reflected. Some frequencies will be slightly delayed, and some won't.

The result is often that some part is reflected as a somewhat proper looking square wave while other parts are reflected differently enough that you can see the sine wave itself.

When neither end is properly terminated, then each sine in the square wave is reflected depending where exactly it "hits the end."

Say you have a transmission line 1 meter long, and a wave one meter long. In this case, there will be no reflection - the wave fits exactly in the transmission line.

Now keep the 1 meter transmission line, but change to a wave that is 67 centimeters long. The wave doesn't fit exactly in the transmission line anymore. Part of it will be reflected.

Put the one meter wave and the 67 centimeter wave into the same transmission line at the same time, and you will only see reflections from the 67 centimeter wave.

Now consider your square wave (made of waves of different lengths.) Some parts will reflect more strongly than other parts. If most of the sine waves reflect together, then you might get something that looks like a square wave. You might also get a situation where only a few (or one) frequency reflects strongly - in which case you get something that looks like a sine wave as a reflection. That's ringing.

Answer 2

pure resistive ( no L nor C) mis-terminations will produce perfect (square wave) reflections.

if there is any dampening, it must also be broadband.

If your setup has these, you will not see sin waves unless your input wave is also sin.

If open circuit termination, the parasitic open-end capacitance will prevent true broadband reflection.

If "short circuit" termination, the non-zero inductance (approximately 1 nanoHenry per millimeter) of the shorting wire will prevent true broadband reflection.

Answer 3

Dr. Eric Bogatin, one of the leading signal integrity (SI) experts in the industry, does a good job of explaning this in his Rule-of-Thumb-26:

https://www.edn.com/what-is-the-ringing-period-on-an-unterminated-line-rule-of-thumb-26/

EDIT 1:

One thing to note from that article is that the period of the ringing is determined by the round-trip (out and back) time of the signal, which is proportional to the length of the interconnect, and not so much by the harmonic content of the square (or pseudo-square) wave.

EDIT 2: Excerpt from the article, per Adam Haun's suggestion

The poster child for signal integrity problems is ringing at the RX end due to reflections on an unterminated transmission line. A great example is shown in the figure below. In yellow is the signal at the RX with no termination, and in red, the same, with a terminated line.

The root cause of the ringing is reflections from the mismatch at the RX (receiver, or destination) between the impedance of the line and the high input impedance of the RX, and the mismatch when the reflected signal makes its way back to the low output impedance of the TX (transmitter, or source).

The reflection coefficient is close to 1 at the RX (open circuit), but is a negative value at the TX (low output impedance). Let’s trace the signal path on its journey. When it first reaches the RX, it reflects. The initial voltage at the RX is high. The reflected signal makes its way back to the TX, where it sees a high to low impedance change and reflects. But the reflected voltage is negative. This negative reflected signal makes its way back to the RX, where it again reflects, but since it is negative, pulls the signal at the RX down. When it reflects from the RX, it is still negative, makes its way back to the TX, changes sign when it reflects, and travels back to the RX as a positive signal. From the initial positive signal at the RX, the signal takes one TD (time delay) to make it back to the TX, another TD to reflect back to the RX (that’s 2 × TD from a peak to a valley at the RX), another TD to get to the TX again, and another TD for the reflected signal to make it to the RX as a positive signal.

This is a rather surprising result: four time delays between successive peaks due to reflections in an unterminated line.

Answer 4

It's caused by the damping effect. For example, let's say your signal is 5V and your first overshoot voltage is 5.5V. When the signal overshoots to 5.5V, it's not a step response because there is damping in the signal line, so the square wave signal reaches to 5.5V over a period of time. This is your first sinusoidal curve on the graph, and the same principle applies to the ups and downs.

Answer 5

The ringing period is determined by the frequency of resonance of the total circuitry including the transmission line and the load (which not necessarily coincides with the length of a standing wave in the transmission line alone, although it approximates in cases when the most of the impedance is caused by the transmission line and the impedance of the load is too small... so technically a short circuit).

If both, line and load, have exactly the same impedance then the half of the length of the ringing will coincide exactly with the length of the transmission line and you wont see it cause coincides with the first harmonic of your signal and you will see a quasi perfect replica of your input signal (Except for the very HF harmonics that dissipates very fast) ... the size (or voltage) of the first ringing is in proportion of the energy that could not be absorbed by the load (Reflection Coefficient) which also determines the phase of the ringing.

The attenuation of the ringing will depend on the dissipation of power of the line and the load due their ohmic resistance or radiation. If there were no dissipation along the system you would see an infinite sequence of sinusoidal lobules mounted on the top of your square signal, increasing the voltage... in theory infinitely... that's why is not good idea connect high power transmitters without any load in the other side of the line or the proper circulators.