# If RF signals require matched impedance to remove reflections, why do approaches to shunt noise need only a “good path to ground?”

From my understanding, for signals to propagate without reflection the impedance of the path must be constant and can be shown via the telegraph equation.

When it comes to noise on the line, for example the power to an IC, you're told for high frequency noise to have a solid, nearby path for the capacitors to shunt the noise.

I'm assuming the high frequencies in this case see a low impedance to ground through these capacitors.

How does this apply with the idea of matching impedance? If an IC or whatever is generating the noise, and it sees this low impedance path, why does it not just get reflected back?

• "Treating the power supply as an RF circuit and tuning its characteristics impedance to the transient impedance of the load (divide voltage noise margin by transient load current, on the scale of milliohms) to minimize noise" is rarely used due to practical difficulties, but this is an indeed legitimate theory in PCB power integrity engineering and can demonstrate excellent decoupling and minimum noise up to several gigahertz in selected demos. Jan 16 at 23:29

Noise and reflections are different phenomena requiring different treatment.

While reflections back and forth super-impose to become what could be considered "noise" by the receiver, it's generally not something you can "filter out" in the frequency domain (with a low pass filter, for instance). Reflected signals have the same frequency components as the original unadulterated signal of interest, so once they've mixed in with the original signal you can't then selectively remove "certain frequencies" in the hope of being left with the original, without killing the original too.

The only solution is to prevent reflections as much as possible, across the entire frequency spectrum, which requires proper impedance-matched line termination.

The frequency spectrum of random noise, on the other hand, is broad, and presumably stretches beyond the spectrum of the signal of interest, in both directions. Therefore this is an element of the signal that can be attenuated by removing frequency components that fall outside the spectrum of components of the original signal.

So, to diminish the effects of random noise, a band-pass, or low-pass filter can be employed.

As to your question about (I presume) power-supply bypass (or "decoupling") capacitors, usually an explanation is with reference to the lumped element model of things, but I'm going to attempt a crude wave-based explanation, in terms of travelling potential waves and reflections.

An IC that produces sudden transients in its demand for supply current, and whose supply nodes are not bypassed with a capacitance, will cause large voltage fluctuations between those supply nodes. Those fluctuations will travel along the supply lines just as any signal would along a transmission line. There will be reflections back and forth, too, with superposition of potential waves occurring at all points. These reflections are a nightmare for anything else connected to those lines. So, we bypass the supply, near the IC, with a capacitor.

At high frequency there is now a dead short-circuit across the IC's supply terminals, which effectively removes that IC from the rest of the circuit. (By short-circuit, I am referring to the extremely low impedance of the capacitor, not the physical length of the transmission line between capacitor and IC). That short-circuiting capacitor absolutely does reflect any incident voltage fluctuations, but being a short-circuit (at frequency), the reflected signal is 180° out of phase with the incident signal.

In other words, any signal emerging from the IC at its power terminals, and which reaches the bypass capacitor, is immediately reflected by that short-circuiting capacitance, 180° out of phase with the incident wave.

The reflected waves travel back towards and inside the IC. They interfere destructively with waves coming the other way, resulting in a net superposition that cancels all voltage fluctuations.

For example, if the IC tries to instantly lower the voltage across its power supply from 5V to 4V (by suddenly drawing a lot of current), the new 1V lower potential difference travels towards the bypass capacitor, and is reflected 180° out of phase, becoming a 1V higher potential difference, which when superimposed onto the outgoing wave, restores the potential difference to 5V.

Of course, this only works if the distance between bypass capacitor and IC is small enough that the reflected signal arrives back at the IC practically instantly. If that distance is too great, then the reflected wave will take time to come back, and there will be enough time for the IC's power supply potential difference to fall to 4V and stay there for a while.

As I said, that was a crude explanation. In reality, the transition from 5V to 4V won't be instant, and the waves I talk of will have components at many frequencies and amplitudes. The superposition won't be a simple $$\(+5V) + (-1V) + (+1V) = +5V\$$, it will be the sum of goodness knows how many waves.

The result is the same though. A capacitor across a transmission line will tend to keep the potential difference constant, by reflecting incoming waves 180° out of phase, which interfere destructively on their way back.

• +1 Excellently written explanation. Jan 16 at 13:22
• Absolutley answered exactly what i didnt know. The time taken to answer that has gone a long way thankyou. I would assume then there could well be some of the frequency that is matched (the capacitor and the noise) for it to litterally travel to ground, without the reflections to cancel? Jan 16 at 16:21
• @Georgekirby The impedance of the capacitive "short" is frequency dependent, so I suppose that there could be a specific frequency at which reflections go away. However, I'm hesitant to say there definitely is, because phase is equally dependent upon frequency; it's possible there's no frequency at which both real and imaginary parts of that impedance "align" with the transmission line's own characteristic impedance, to provide the "perfect match". I don't know for sure. Jan 16 at 16:39
• @Georgekirby SimonFitch is using the term "short" as in "short circuit", i.e. a shunt between two conductors. An object (such as a transmission line) which has a small length, is also said to be "short", but that is a different meaning of the word. A "short", i.e. shunt, in a transmission line causes reflections, and prevents transmission beyond that short/shunt. A "short" (i.e. not long) transmission line appears to be a direct connection. Jan 16 at 22:50
• @Georgekirby Soryy for the ambiguity of the meaning of "short", that was quite an oversight. MathKeepsMeBusy has cleared that up, I hope, but I'll do some editing to improve things. Jan 17 at 2:50

If rf signals required matched impedance to remove reflections, why do approaches to shunt noise need only a “good path to ground”

The purpose of decoupling capacitors is to reduce voltage disturbances at the IC and in the supply lines that would be caused by rapid changes in load at the IC.

Refections are the mechanism by which a component at one end of transmission line "sees" an impedance at the other end of the transmission line as a just that impedance, instead of "seeing" the characteristic impedance of the transmission line. If the transmission line is long, it may take a while before these reflections cause a distant impedance to make itself "felt" ("seen"). If the transmission line is short, reflections make the impedance at one end, felt quickly at the other end. It is by means of these rapid reflections that we can neglect the fact that there is a transmission line intervening between two components.

Good practice says to make the distance between IC and the decoupling capacitor small. Thus, the IC sees the capacitor as a capacitor with very little delay.

Since the impedance of a capacitor is mostly reactive at the frequencies of interest, while the impedance of a transmission line is mostly resistive, a shunt capacitor, by itself, will virtually never match the characteristic impedance of a transmission line. A capacitor (or any "good path to ground") will cause strong reflections when placed as a shunt across a transmission line. Since electrical disturbances are either reflected, transmitted, or dissipated, if most of the signal is reflected, little of it is transmitted.

Thus, the disturbance in current that is present on the IC side of the decoupling capacitor, is seen as a much smaller disturbance on the power supply side of the decoupling capacitor. It is the reflection of the disturbance back toward the IC that largely blocks that disturbance from being transmitted into the power supply line (and on to other components).

• Many thanks for the help, hopefully it dosent come across as me misreading (im new in learning most of this). But, now im a little confused when you say the reflections ocur back to the IC, and that the short distance meen the rapid reflections means a neglet of the transmission lines, is this the same reason why for example if two IC's that have a high speed between them at a certain impedance, if there short enough together, the transmission line matters less? Thanks Jan 16 at 16:31
• If I understand you correctly, yes. Short distance implies quick reflections, which implies transmission line transients disappear quickly, hence we can treat the transmission line as a simple direct connection. Jan 16 at 17:24
• Thankyou, but wouldnt that remove the message/data, if the data is modulated in high freqency? Jan 16 at 20:42
• If you have a transmission line, such as a coaxial cable, and there is a (sufficiently large) capacitor between the outer and inner conductors somewhere in the cable, then a high frequency signal that enters one end, will (largely) be reflected back to the end where it started, and little of it will be transmitted to the far end. Is that what you mean? Then yes. Jan 16 at 21:08
• Reflections don't cause a signal to destruct. By transmission line transients, I meant the differences between what is sent, and what is received. When those transients disappear, the signal has successfully "gotten through" the transmission line. Jan 16 at 22:46

External noise would be reflected. That's fine; the goal is to keep it out of the IC. The bypass capacitors you put next to ICs are there to provide a clean DC supply voltage. Impedance matching would allow high frequency noise to enter the IC through its supply pins.

Decoupling capacitors are deliberately placed with extremely short loops, often just a few millimeters. This corresponds to tens of GHz wavelength. However, decoupling capacitors are only effective to a few tens of MHz beyond which their impedance increases due to inductance.

So to answer your question, at wavelengths where impedance is low, you're orders of magnitude too small for transmission line effects to matter. At wavelengths where they do matter the impedance of the circuit is irrelevant.

• I agree with this answer. If you really wanted to lay out decoupling caps far away from the IC, you could do so with a transmission line that has impedance equal to the ESR of the capacitor. Though you would need pretty fancy technology to reach ceramic capacitor impedance of 0.01 ohms :)
– jpa
Jan 16 at 8:34

Firstly, impedance matching is designed for a specific frequency. With different circuits you can achieve higher or lower bandwidth of matching.

As you probably already know, noise is random. Ideal white noise covers the entire frequency spectrum with equal amplitude (unfimorm distribution). Other noise sources have different distributions but common is, that all distributions presents a distribution (a characterstics over a bandwidth). See chapter 4 of this book https://rftoolbox.dtu.dk/#book. What I am trying to state is, that noise is not a matter of impedance matching. Some parts of the noise spectrum may not be reflected where it is matched, but outside of matching, you will have reflections of noise as well. Instead, noise is minimized by limiting the bandwidth (filtering).

When impedance matching you may use simple circuits like an L-network. In this case you may create matching so that your shunt element is a capacitor. In that case you effectively create a low pass filter as your matching element.

In any case, a larger capacitor presents a lower reactance. If one needs a DC stable reference at the IC you may use larger caps. Even though you have done matching, picking a decoupling capacitor (10x, 100x, 1000x) larger than your matching capacitor, it shouldn't affect your matching circuit enough to be a problem.

• Thankyou, so why would the filtering minimise the noise? I understand the idea from that the capacitors impedance decreases with frequency, acting as a filter, but then i just cant seem to compare this with reflections when it comes to high freqencys Jan 16 at 20:49
• If you look into a datasheet for any opamp you will find a measure for noise, often called e_n (input-Referred Voltage Noise). You will see that it is a unit of V/sqrt(Hz). The sqrt(Hz) has always bothered me, instead squaring the unit, it is a power per Hertz. but the key take away is, that it is per Hz, meaning, that the wider bandwidth, the more noise you will end up integrating (summing over). Limiting the bandwidth decreases the "per Hertz" term and you will sum over a smaller area. Jan 31 at 7:36
• Regarding the reflections, I believe you are over complicating the thing. Reflections occur when the impedance changing along a wire. This is getting more and more important when the wavelength is so short, that a quarter of it can exist between two components. Ideally, at a single frequency, only, there may be a match. The further away from this frequency the more of the signal is reflected. If you add noise, only a single frequency is absorbed without any reflections and how often this one frequency occurs is random and can be described statistically by a distribution. Jan 31 at 7:44
• Every other part of the noise distribution will experience that some portion of that noise spectrum is being reflected. The amount of reflection depends on the matching characteristics. You may create a wide band of matching, then less of the noise is reflected. So, limiting the bandwidth makes it so, that less noise is traveling towards the impedance mismatch, still, some part is reflected, some part is absorbed, but minimizing the amount of noise being absorbed by the component, it could be an amplifier, then limits the amount of noise being amplified/mixed/modulated/etc in the next stage. Jan 31 at 7:51

From an electromagnetic point of view, what the bypass capacitor does is reflect RF energy. RF energy generated within the IC is reflected back to be absorbed by the IC. RF energy from outside is denied entry.

It is often useful to use decoupling resistors in the power distribution net to absorb unwanted energy. It's similar to using black material inside a camera to absorb unwanted light.

Your error: You imagine an IC outputs internally generated noise through its DC supply pins.

An IC does not transmit noise from its internal energy storage. The noise is born when IC's current consumption varies. If the DC power to the IC comes through a transmission line the suddenly starting current pulse to the IC is taken from the line capacitance and the line voltage drops at the end of the line. Effectively this is the same as a pulse is transmitted from the IC along the line to disturb others. But the noise doesn't come from the interior of the IC. The IC is only a varying load.

If there's a big capacitor in parallel with IC's DC supply pins the same current pulse makes the voltage to drop much slower - the started pulse at the end of the line has much lower voltage amplitude - maybe only millivolts instead of volts. Thus the caused noise is reduced. The harmful, say 1 volt pulse is never born.