2
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ Ringing will only occur if the transmission line is not terminated on both sides. Let me rephrase that for clarity. Ringing will not occur if at least one side of the transmission line is terminated in its characteristic impedance. \$\endgroup\$ – mkeith Feb 5 at 4:50
  • \$\begingroup\$ @mkeith can you please elaborate on the mechanism when the transmission line is opened at both ends. This is what I am trying to figure out. Why is ringing observed in this case. \$\endgroup\$ – Trilok Girish Kamagond Feb 5 at 5:22
  • \$\begingroup\$ @mkeith can you also reason why termination at one end at least causes no ringing? \$\endgroup\$ – Trilok Girish Kamagond Feb 5 at 5:31
  • \$\begingroup\$ Do you know what Fourier series is? A bunch of sine-waves of the right phase, frequency, and amplitude combined together make any possible wave. It is not the square wave that is causing the ringing, but the higher frequency sinusoids that make up that square wave. It's just square waves have very vertical edges which means very fast transitions which requires high frequency components. If you sent a sinusoid with high enough frequency down the line, it will ring too. \$\endgroup\$ – DKNguyen Feb 5 at 5:38
  • \$\begingroup\$ This may help. When a signal appears on one side of a transmission line, the signal does not know what is on the far end. Open circuit? Short circuit? 50 Ohm resistor? It can only see the intrinsic impedance of the line. Only after the signal makes its way to the end and finds what is there, and reflects back to the source does the source become aware of what is at the end. \$\endgroup\$ – mkeith Feb 5 at 8:22
2
\$\begingroup\$

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:

enter image description here

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.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Thank you so much for explaining properly. I want a clarification for this - since Transmission line is modeled as a series of LC section for lossless case, is this what you mean by stray inductance and capacitance? \$\endgroup\$ – Trilok Girish Kamagond Feb 6 at 2:47
1
\$\begingroup\$

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.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ thank you for the answer. Can you please elaborate the case of open circuit termination. \$\endgroup\$ – Trilok Girish Kamagond Feb 5 at 5:20
1
\$\begingroup\$

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.

Ringing

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.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Thank you. I had gone through the above before asking. I hadn't gone through properly I guess. \$\endgroup\$ – Trilok Girish Kamagond Feb 6 at 2:39
  • \$\begingroup\$ Link-only answers are discouraged. Posting a relevant excerpt from the article would improve this answer. \$\endgroup\$ – Adam Haun Feb 6 at 16:49
0
\$\begingroup\$

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.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.