Talk:Q factor

Physics B‑class Low‑importance

	Physics portal This article is within the scope of WikiProject Physics, a collaborative effort to improve the coverage of Physics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.PhysicsWikipedia:WikiProject PhysicsTemplate:WikiProject Physicsphysics articles
B	This article has been rated as B-class on Wikipedia's content assessment scale.
Low	This article has been rated as Low-importance on the project's importance scale.

FWHM vs 3dB

The article states:

the bandwidth is defined as the 3 dB change in level besides the center frequency.

The definition of the bandwidth BW as the "full width at half maximum" or FWHM is wrong.

Unless the definition depends on the slight difference between -3dB (1/1.995) and half maximum (1/2), this would seem to be the same thing. Anyone know for sure, here? -- DrBob 17:57, 27 Sep 2004 (UTC)

I might be able to help here... The reason we use FWHM and not 3dB for the Q factor of mechanical and optical resonators is that the angular frequencies for which a driven oscillator will store half the peak power is $\omega =\omega _{0}\left(1\mp {\frac {1}{2Q}}\right)$ so that conviniently: $\Delta \omega =\omega _{0}\left(1+{\frac {1}{2Q}}\right)-\omega _{0}\left(1-{\frac {1}{2Q}}\right)=\omega _{0}\cdot {\frac {1}{Q}}$ . I would never have labelled that diagram as having bandwidth, it would always be FWHM (they are completely different and used in different situations). --152.78.72.172 (talk) 11:21, 7 August 2009 (UTC)[reply]

Strictly, the FWHM is full width at half-max, that is half power. In dB this is -3.01 dB, close enough to 3 dB down to not worry about it.

The definition "In optics..." appears strange to me. In mechanics the Q factor can be shown to be equal to 2 * Pi times the energy stored in the oscillator divided by the energy dissipated per cycle. I suspect this is the correct definition, applicable to optics, mechanics, and any other oscillating system.

24.245.15.183

I suspect you're absolutely correct about the definition applying to all oscillating systems. "Q-switching", though, is a very real phenomenon and used to great advantage in pulsed laser systems.

By the way, you can easily sign your "talk" posts by appending four tildes (~~~~) to the posting. When you "Save changes", this will be replaced by your username in a handy linked form and a timestamp of your edit.

Atlant 14:27, 10 August 2005 (UTC)[reply]

Q the cycles for energy to go to zero?

Alison Chaiken 00:00, 23 September 2005 (UTC): I've always thought of the quality factor as the number of cycles that it takes for energy to be dissipated from the system. Thus a Q of 1000 means that the excited oscillation will ring down to zero in 1000 cycles. I would think this article should mention this insight. I would add it except that I can remember if Q is the number of cycles to ring down to 1/2 the original energy or what exactly.[reply]

No, that's wrong. It never decays to zero. The article states correctly that it's the number of cycles required to decay to 1/535 of its original energy.--24.52.254.62 01:25, 21 October 2006 (UTC)[reply]

Alternative to 1/535 definition

I'd like to suggest a change to the definition of Q as the number of cycles for the response to decay to 1/535 of the original amplitude. Although this is certainly true, and used in the Crowell book, I've never seen it used elsewhere. The main problem is that as a definition for actually measuring Q (which is fun and instructive for students to do with a pendulum), it is practically useless. If you actually try to count the cycles, by the time the response decays to 1/535 it is so flat that the accuracy is lost. It also doesn't offer much insight into the mathematics.

The usual way of defining Q in engineering texts is that it is $2\pi \,$ times the number of cycles for the response to decay to $1/e=0.368\,$ which is one time constant of the exponential decay envelope. Although at first sight this looks more complicated, it shows that the Q is just the ratio of the exponential and sinusoidal time constants of the response: if the response is: $Ae^{\alpha t}cos{(\omega t+\phi )}\,$ , then $Q=2\pi f/\alpha =\omega /\alpha \,$ . And it is also a practical definition for measuring Q, since at 0.368 the response is still sloped enough to determine accurately when the amplitude falls below it.

I suggest changing the definition of Q to $2\pi \,$ times the number of cycles for the response to decay to $1/e=0.368\,$ of the original amplitude. --Chetvorno 21:02, 21 August 2007 (UTC)[reply]

I calculated

Q=f_{0}/B\,

, the definition of

Q\,

over the bandwidth and your formula

Q=2\pi f/\alpha \,

for a signal like

Ae^{\alpha t}cos{(\omega t+\phi )}\,

and find out that there is always missing a factor of 2 between these formula. It has to be

Q={\frac {2\pi f}{2\alpha }}\,={\frac {\omega }{2\alpha }}

like formula E.7 on this webpage http://ccrma.stanford.edu/~jos/filters/Quality_Factor_Q.html 77.10.132.145 (talk) 11:01, 5 December 2007 (UTC)[reply]

I noticed that Q is defined as number of cycles for the stored energy to decay to 1/535 of the original. I believe this is correct, but please note that stored energy is amplitude squared. So the amplitude will only decay to 1/23. Amplitude and energy seem to be confused in this discussion.Ken Wilsher 22:55, 5 February 2009 (UTC) —Preceding unsigned comment added by Kwilsher (talk • contribs)

Worth adding to this discussion that this https://ccrma.stanford.edu/~jos/fp/Decay_Time_Q_Periods.html page says this definition is an only approximation. In fact, 2pi(Energy stored)/(Energy dissipated in one cycle) is also an approximation https://ccrma.stanford.edu/~jos/fp/Q_Energy_Stored_over.html, though the main page gives it as the definition, so the "true" definition of Q isn't particularly clear (at least not to me!). It would be great if someone could give it as a function of an LTI system's matrix or eigenvalues or something like that. 174.252.36.13 (talk) 01:11, 9 December 2010 (UTC)[reply]

Reference for the Q-value equation in mechanical systems?

Where does the equation $Q={\frac {\sqrt {MK}}{R}}$ on mechanical systems derive from? I can't find a source and it's not cited. 130.233.189.53 14:07, 26 October 2007 (UTC)[reply]

Somehow it remained without attention until now, that the article suggests "In a parallel RLC circuit, Q is equal to the reciprocal of the above expression.", referring to the correct formula for Q in a RLC circuit as "the above expression". Which means that a parallel RLC circuit would have a Q-factor that grows with increasing R. Which, in turn, is an obvious nonsense. —Preceding unsigned comment added by 83.79.25.246 (talk) 19:58, 20 November 2007 (UTC)[reply]

Ooops, sorry. disregard the above. Of course, a parallel RLC circuit would prefer a higher R, ideally - no R at all (means open circuit istead of R), for an endless oscillation. —Preceding unsigned comment added by 83.78.43.62 (talk) 20:59, 20 November 2007 (UTC)[reply]

I think that in the Usefulness of 'Q' section, the Q factor for second order filters (Butterworth and Bessel) are derived from some equations and doesn't need to be cited. Probably someone with knowledge in these filters can show the derivations in this article. Hytar (talk) 15:08, 14 August 2008 (UTC)[reply]

Relationship to damping ratio

I'm not sure whether the equations relating $Q$ to $\zeta$ are marked "citation needed" because there is some doubt about their accuracy; or whether it's just there because it's good practice to reference these. When I first looked at them I was doubtful about them, but I think that was just because they were in a slightly unfamiliar form. I can't supply a citation as none of the texts I have to hand use the damping ratio as defined here. However, I can show that it follows straightforwardly from the definitions of $Q$ and $\zeta$ given in Wikipedia which hopefully is sufficient.

This article says $Q=1/2\zeta$ which we would like to verify. The damping ratio defines $\zeta$ in terms of the differential equation

{\frac {d^{2}x}{dt^{2}}}+2\zeta \omega _{0}{\frac {dx}{dt}}+\omega _{0}^{2}x=0.

which has solutions of the form

x=Ae^{-\zeta \omega _{0}t}\sin(\omega _{0}t+\delta ).

This means the energy stored is

E={\tfrac {1}{2}}m\omega _{0}^{2}A^{2}e^{-2\zeta \omega _{0}t},

where m is the mass being oscillated (or, if you prefer, an arbitrary constant to make it dimensionally consistent — it cancels out later); and the power loss is

-{\frac {dE}{dt}}=m\zeta \omega _{0}^{3}A^{2}e^{-2\zeta \omega _{0}t}.

Using the definition of Q, this gives

Q=\omega _{0}E/\left(-{\frac {dE}{dt}}\right)={\frac {1}{2\zeta }}

as required. A bit more manipulation gives the more familar (to me, anyway) form

\,Q=\pi f\tau

where $f=\omega _{0}/2\pi$ is the frequency, and $\tau$ , the time to decay to $1/e$ . (I.e. $\tau =1/\zeta \omega _{0}$ .)

Hope that helps. — ras52 (talk) 14:42, 30 December 2007 (UTC)[reply]

The equation for the Q-factor of a spring (in the Mechanical systems section) is similarly marked with a "citation needed". This one is even easier to see. The differential equation defining $\zeta$ was

{\frac {d^{2}x}{dt^{2}}}+2\zeta \omega _{0}{\frac {dx}{dt}}+\omega _{0}^{2}x=0.

Multiplying through by $M$ gives the familar "F = ma" equation, and from the definition of $R$ in the article, obviously $R=2\zeta \omega _{0}M$ . Then

Q={\frac {1}{2\zeta }}={\frac {\omega _{0}M}{R}}={\sqrt {\frac {K}{M}}}{\frac {M}{R}}={\frac {\sqrt {KM}}{R}}.

I think you're unlikely to find many references for these things as, at least in my experience, Q factors are not greatly used in mechanics. Looking through the indexes of three or four degree-level texts that I have to hand that mention this sort of mechanical oscillations, I can't find a single mention of Q factors.

I also notice that whilst these mechanical relations are marked with "citation needed", equally simple electrical relations are not. For example, the Q factor of RLC circuit is as obvious as the mechanical ones, but that is not marked "citation needed". Clearly citations would be desirable for all of these, but it is seems we're requiring a higher standard of referencing for mechanical examples than electrical examples. — ras52 (talk) 16:04, 30 December 2007 (UTC)[reply]

A possible reference for $Q=1/(2\zeta )$ would be W.T.Thomson, Theory of vibration with applications, and in the 3rd edition (London, 1988) it is on p75 equation (3.10-4). Fathead99 (talk) 16:38, 12 June 2008 (UTC)[reply]

Qualitiy factors of common systems

The three examples for the filters are marked with "citation needed".

A reference that points in the right way is Active Filter Design Techniqes by Texas Instruments. One may find tables with factors for the polynoms of the respective filter types' transfer functions and the definition of the Q factor based on these factors:

Q=sqrt(b_i) / a_i

Here's an example for a 2nd order butterworth filter:

The component values (R, L, C) of a practical circuit are chosen so that a = sqrt(2) and b = 1. Thus, the generalizes transfer function

A(s) = A₀ / (1+a*s+b*s²)

becomes

A(s) = A₀ / (1+sqrt(2)*s+1*s²)

Using the definition of Q for low-pass and high-pass filters, we obtain:

Q = sqrt(b) / a

Q = sqrt(1) / sqrt(2)

Q = 1 / sqrt(2)

Q = 0.707

Similar calculations will yield the Q factors for the other filter types based on the factors that may be found in the reference's tables.

I am aware that this is a hands-on approach based on some tables and definitions and not a really thorough derivation towards the requested citations, but it's a start showing that the numbers in the article are o.k. at least for 2nd order types of the filters named in the article. If anyone would like to give further information based on filter theory, it would certainly help the article. Also, further information for filters of a higher order than 2nd would be appreciated. --Zb-de (talk) 13:03, 5 January 2010 (UTC)[reply]

External links reverted as spam

The following weblinks have really to to with the Q factor and are no spam, like Dicklyon means.

Q factor to/from 'Bandwidth per octave' converter:

http://www.sengpielaudio.com/calculator-bandwidth.htm

Q factor and center frequency - Find the cutoff frequencies of the bandwidth:

http://www.sengpielaudio.com/calculator-cutoffFrequencies.htm

--Robert 19:42, 08 Oct 2008 (UTC)
Dick Lyon says: If an editor's only contribution is multiple ext. links to one site, I call it spam.

They may be perfectly good links, but if an anon user does nothing but post multiple links to a site, then that editor is a de-facto spammer, in my estimation. If someone who is not a spammer wants to review them, and decides to add them as external links, then that's more likely to be acceptable. Dicklyon (talk) 18:57, 8 October 2008 (UTC)[reply]

"Energy of steady-state vibrations" figure

The figure under "Physical interpretation of Q" shows that the bandwidth is determined by HALF of the PEAK ENERGY. As discussed earlier in this Talk page, the bandwidth is determined by half of the PEAK POWER (the square of energy). So the figure should be changed to be $E_{\text{max}}/{\sqrt {2}}$ or each $E_{\text{max}}$ should be changed to $P_{\text{max}}$ . —TedPavlic | (talk) 14:51, 16 December 2008 (UTC)[reply]

You are incorrect here. Power is proportional to energy in a damped harmonic oscillation. You're thinking of amplitude, which is proportional to the square root of power or energy. Dicklyon (talk) 16:49, 16 December 2008 (UTC)[reply]

That's true. My mistake. P=E/t. I'm not sure that this will be clear to the average reader; "Half power" language is prolific in systems literature. —TedPavlic | (talk) 18:11, 16 December 2008 (UTC)[reply]

In this case, since the figure is about the steady state energy of a driven harmonic oscillator, the half-energy point is relevant. The power going into the system is not proportional to energy as the frequency changes. For a given frequency, the time at which the power being dissipated decreases to half is the same as the time at which the stored energy decreases to half, in a non-driven system. Dicklyon (talk) 01:28, 17 December 2008 (UTC)[reply]

Ok. So you're point is that the figure is not meant to depict the steady-state driven magnitude response, and so energy is the correct inner-product to use. Because the context of the section is a non-driven system that decays over time, the energy in the signal is finite, and so we use an energy signal. If it was driven, we'd have to use a power signal for the correct time normalization. I'll buy that explanation. Thanks. —TedPavlic | (talk) 14:08, 17 December 2008 (UTC)[reply]

It still appears like the section (and the figure, actually) are handling the driven case. The section talks about filtering and such. Perhaps the figure caption needs a change of wording... —TedPavlic | (talk) 14:10, 17 December 2008 (UTC)[reply]

You misunderstood me, and yes it depicts that driven case. I'm not sure what conflict you see in that. Dicklyon (talk) 14:12, 17 December 2008 (UTC)[reply]

The energy in a periodic signal is infinite. The output will be periodic if the input is periodic, and so the output will have infinite energy. However, both the input and output will have finite power. So (I think?) the figure only makes sense in terms of power. —TedPavlic | (talk) 15:25, 17 December 2008 (UTC)[reply]

The energy referred to is the energy in the system, the sum of its potential and kinetic energies, not the integrated power. Dicklyon (talk) 16:13, 17 December 2008 (UTC)[reply]

Error with attenuation constant?

There seems to be an error. The article lists one definition of the Q factor as the angular frequency divided by double the attenuation coefficient. That would give Q dimensions of length per time when in fact Q is dimensionless. The attenuation coefficient is a spatial decay measure not a temporal decay measure. I believe what was meant is that Q = angular frequency over decay rate. I will change it. Please explain to me how I am in error before changing it back. —Preceding unsigned comment added by 129.63.129.170 (talk) 20:11, 13 January 2009 (UTC)[reply]

Perhaps you are wrong in assuming the units you did for alpha? What source are you relying on for alpha and lambda? For now, let's leave it as it was; we should be able to quickly come to an agreement here. Dicklyon (talk) 05:52, 14 January 2009 (UTC)[reply]

According to the attenuation article, as used in ultrasound,

\alpha

is typically measured in dB/(MHz·cm) — i.e. its dimensions are [time · length^-1], dB being dimensionless. That would give

Q=\omega _{0}/2\alpha

units of [length · time^-2], not [length · time^-1] as the anonymous editor states. However, it strikes me as far more likely that a different definition of the attenuation coefficient,

\alpha

, is being used. It's hardly novel to find the same term being used in different branches of physics with slightly different meanings. Really, we should see Siebert's Circuits, Signals, and Systems says as that is the reference given for the equation. I rather expect that we'll find that he defines

\alpha

so that the damped harmonic equation reads

{\frac {d^{2}x}{dt^{2}}}+2\alpha {\frac {dx}{dt}}+\omega _{0}^{2}x=0.

That would a very common form of the equation, right down to using

2\alpha

as the coefficient of the first derivative, although I don't recall having seen that

\alpha

referred to as the attenuation. (I can't easily look it up as all my books are packed up in boxes while I have a new floor laid.) — ras52 (talk) 10:51, 14 January 2009 (UTC)[reply]

Note that the Attenuation coefficient article (which is linked from the ultrasound paragraph in the attenuation page, which is more of a disambig page anyway) gives

\alpha

in essentially the same units as are given here (in the Q factor article). However, it refers to the "linear attenuation coefficient." The point here is that different contexts are going to have different domains of interest, but in all cases there's some sort of decay (that may be in one or more dimensions) that behaves in roughly the same way. When dealing with Q factor, we are implicitly using the linear case. —TedPavlic (talk) 14:46, 14 January 2009 (UTC)[reply]

The

\alpha

given here is the real part of a complex pole of a second-order system. That is, it is the coefficient of oscillatory damping. The second-order complex poles will produce oscillations like...

e^{-\alpha t}\sin(\omega t)\,

Here,

\alpha

is a temporal rate of decay only because the domain of interest is time. In other contexts where the systems are functions of space,

\alpha

will represent the exponential decay rate of spatial oscillations. So I'm not quite sure what the problem is here. Certainly on the Q factor page time is going to be the most relevant domain of interest (at least for example sake), and so it's reasonable that

\alpha

be described as a temporal rate. Typically it is only used as a spatial rate of decay when space is the domain of interest.—TedPavlic (talk) 14:40, 14 January 2009 (UTC)[reply]

Also note that

\alpha

is frequently given in units of Nepers because it represents an exponential decay. —TedPavlic (talk) 14:40, 14 January 2009 (UTC)[reply]

Problem in "Usefulness of the Q factor"

Quote: "Likewise, a high-quality bell rings with a single pure tone endlessly after being struck."

If it rings (out in the air), then it dissipates power, and thus will not ring forever. —Preceding unsigned comment added by 129.240.84.144 (talk) 09:30, 1 April 2009 (UTC)[reply]

Who said anything about the bell ringing in the air? For one, it's a limiting example. However, technically speaking, "ringing" need not involve the transmission of sound through a medium. Power need not be dissipated for a bell to "ring." A pendulum need not swing through the air in order to swing. Finally, it's obvious that "endlessly" makes no sense for any passive system (i.e., when thinking of the universe as closed, then ANY system). The statement is an rhetorical ideal used to communicate a point. —TedPavlic (talk) 03:12, 29 May 2009 (UTC)[reply]

Can This Be Less Esoteric?

While it is true that Q describes how underdamped an osc is, that is far from the meaning of Q. I think the meaning of Q has more to do with not loosing energy. Can we make the lead article assessable by (and meaningful to) laymen? John (talk) 01:26, 29 May 2009 (UTC)[reply]

The opening paragraph, as it is, seems to already match your request.

In physics and engineering the quality factor or Q factor describes how under-damped an oscillator is. Higher Q indicates a lower rate of energy loss relative to the stored energy of the oscillator; the oscillations die out more slowly. A pendulum suspended from a high-quality bearing, oscillating in air, has a high Q, while a pendulum immersed in oil has a low one. Oscillators with high quality factors have low damping so that they ring longer.

Can you suggest how it might be changed? It already discusses energy loss/storage and "ringiness." —TedPavlic (talk) 03:08, 29 May 2009 (UTC)[reply]

Suggesting an improvement is somewhat harder than complaining, I am still in the complaining stage. Ill keep thinking... I am having trouble convincing myself Q describes an oscillator vice a resonator. Describing it in terms of how underdamped it is is not too satisfying either, few laymen will relate to that. John (talk) 20:46, 1 June 2009 (UTC)[reply]

Can you explain your objections a little bit more? Perhaps give an example. A resonator is just a specialized oscillator. Perhaps you're having trouble picturing how a Q factor could be used when the device is, say, active because the energy in the system is not simply the energy from the input. Is that the problem? A couple of comments...

In an active device, the energy being added to the system is being used to control the dissipation of the input energy. That is, it can work to dissipate the input energy quickly, or it can work to prevent the loss of that input energy. Input energy that is lost might get "re-supplied" by the external source, which means that you'll have a higher energy "stored" to energy "lost" ratio. We're just using an active device to supply extra power rather than polishing surfaces to remove the friction that causes the loss; however, it's basically the same idea.
Stepping back to the math, any system that can be described by a second-order linear ordinary differential equation (ODE) has a Q factor. If it helps, think of Q as a mathematical construct that has a special physical meaning in certain contexts. However, I don't think you have to stretch the physical interpretation much to apply it to any general second-order linear ODE. If you perturb the ODE off of its equilibrium, it follows a certain trajectory back to it. That trajectory might be slow and oscillatory, or it might be fast and monotonic. Either way, without the initial perturbation, the movement wouldn't occur. The Q factor describes how quickly the input influence is damped out. It's possible that it will be sustained forever. In a perfect world, that would occur with lossless components. Because it's not a perfect world, we can use an active device to give us the behavior of lossless components. Sure, energy is being added from another source, but the net effect is that the energy from the input is "sustained" forever.
From a black box perspective, an active device that implements the same differential equation as a passive device may as well be the passive device. All we really care about is input–output relationship. If it helps to think of it as a passive device (some resonance well or RLC circuit or mass–spring system or whatever), then that's fine, but it's not necessary.
Finally, consider:
- A bell that has some fancy inertial units (e.g., motors) mounted on its sides that can induce additional oscillations in the bell. If we like, we can set the motors up to make sure induced oscillations from a strike of the bell never die out. The extra energy will be coming from remote (external power supplies powering the actuators), but the net effect will be that the bell rings forever. Is that not an infinite Q factor?
- A cart with a motor-driven wheels and a control system that ensures that the motor back-EMF stays constant. Assuming that a quick shove on the cart can overpower the control, the cart will roll forever despite friction trying to slow it down. We've implemented ice with an external power supply (motor batteries). Sure, the system isn't truly frictionless, but what do we care? It behaves frictionless, and that's all that really matters. We're SUSTAINING the input kinetic energy by RE-supplying it when it gets lost to friction. That makes it have a high Q.

So maybe the best definition of Q factor would refer to "input energy sustained" rather than lost and stored. Does that make sense? —TedPavlic (talk) 13:02, 2 June 2009 (UTC)[reply]

Seems to me energy and its loss, and sustain doesn't capture enough of the notion of Q. There is also a selectivity feature that is not apparent in the doorbell-with-motor view. Your notes sure help focus the question. I am not sure this captures the essence of Q. Still thinking (kinda slow, huh?) John (talk) 20:22, 2 June 2009 (UTC)[reply]

Remember that energy, when discussing Q factor, is meant in a signal processing sense. That is, it's a more general property than the "energy" you see in physical systems. It's referring to the energy in the signal (say, at the output of a bell). Hence, you need not worry about physical energy lost and gained when batteries are included. Instead, you think about how much energy is tied up in an output (especially after an impulse). —TedPavlic (talk) 18:18, 3 June 2009 (UTC)[reply]

Ted, Not sure its relevant, but I think energy is just metadata for signal processing resonance. A passive, real inductive & capacitive electrical resonant circuit ringing with finite Q looses real energy to real heat and its amplitude decays over time. An appropriately connected amplifier (active element) can replace that energy and make up for the heat loss and keep it ringing. In a software emulation of that, nothing real gets really warmer. But that’s not the thread I thought I was pulling on. I was thinking about addressing some kind of selectivity notion: that a resonant system with high enough Q does not ring as much as a lower Q system if the excitation is too far off frequency. John (talk) 20:25, 6 June 2009 (UTC)[reply]

The energy referred to in Q factor is the energy discussed on Energy (signal processing). Using that mathematical definition of the energy of a time signal, "Q factor" applies to all systems – passive, active, and abstract. If you would like to give some qualitative comparison of Q factor that more obviously applies to a wide range of systems (which may or may not be physical), then that's probably a good thing. —TedPavlic (talk) 14:11, 7 June 2009 (UTC)[reply]

I strongly agree that this article, especially its introduction, should be much less esoteric and more accessible to laypersons. What would be wrong with leading off the article with something like: "The Q Factor is a measure of the relationship between stored energy and rate of energy dissipation in certain electrical components, devices, etc, thus indicating their efficiency. It is a measure of the quality of an electric circuit; the ratio of the reactance to the resistance. For a capacitor, inductor or tuned circuit, the Q factor, or Q, is a figure of merit. The higher the Q, the lower the loss and the more efficient the component." --Westwind273 (talk) 16:57, 8 March 2011 (UTC)[reply]

While I applaud your goal of making the article clearer to laypeople, I feel your suggested lead doesn't do that. Your proposed text only covers the use of Q in electronic components, and also doesn't mention that it has anything to do with resonance or oscillation. The concept of Q is broader than its use in electronics; it is used in connection with all types of resonators. I think introducing it in the context of mechanical vibration gives laypeople a clearer picture of what it's all about. --Chetvorno^TALK

Thank you for the critique of my suggestion. It has helped me understand the topic better in general. Although a purist would probably say that the mechanical definition is the most true basic explanation of Q, it seems to me that my electrical explanation is much closer to the practical knowledge that a layperson would be seeking. Note that English language dictionaries usually focus on Q in its electrical context. At the very least, it seems like the sentences I wrote should be somewhere in the introduction. I would argue that Q's meaning of "efficiency" is more central to its real world application, than Q's meaning of "oscillation". I would compare this Q factor article with the use of the Q factor term in the Wikipedia article on Equivalent Series Resistance. Put differently, let's say a layperson was looking over an inductor datasheet and saw "Q" for the first time. Would the current article be instructive to them in understanding this concept? I think not. It seems that there is a kind of battle for the soul of Wikipedia here: Is Wikipedia meant to be instructive to the population at large? Or is Wikipedia a place where academics come together to record the purest definition of terms in their field of specialty? --Westwind273 (talk) 11:14, 9 March 2011 (UTC)[reply]

Parallel RLC Q factor

Looking at the recent edit history, some editors do not fully understand Q factor in a parallel RLC circuit. It used to be correct here. The wikibooks circuit theory page has it correct, Q=R*sqrt(C/L): http://en.wikibooks.org/wiki/Circuit_Theory/RLC_Circuits . I am going to update the article page now. The intuitive justification (for editors who disagree) is that as R goes to infinity, Q must go to infinity as well because it because an ideal LC resonator with no resistor. As C increases, the peak voltage across the resistor for a given starting energy will decrease, so resistor power dissipation decreases, thus Q increases. As L increases, peak voltage increases, thus increasing resitor power dissipation, decreasing Q. Mattski (talk) 08:26, 24 February 2010 (UTC)[reply]

I'm glad someone is looking into this! I noticed the recent edits also and I was suspicious. I had this on my list to try to find reliable sources, but had not yet gotten around to it. Do you have any reference citations that could be used? Especially for equations and such, it is nice to have a place to double check things. Someone recently changed a constant in list of moments of inertia, and I had to sit there and spend 10 minutes re-work the integral from scratch, just to convince myself the old version was right and the new one was wrong! (I need to find references for that article too...) CosineKitty (talk) 18:33, 24 February 2010 (UTC)[reply]

Most books seem to just talk about the series case and skip the parallel, but I'll see if I can find a good source. Are posted lecture notes a good source, or should I go for academic paper/textbook? Mattski (talk) 23:13, 1 March 2010 (UTC)[reply]

X=?

In the last paragraph of the RLC section, the sentence "In this case the X and R are interchanged." occurs and yet X has had no previous mention. 62.49.27.35 (talk) —Preceding undated comment added 14:52, 20 April 2010 (UTC).[reply]

The guy who added that on Feb. 22 doesn't seem to be around, but you can probably figure out a fix from this version: [1]. Dicklyon (talk) 06:37, 22 April 2010 (UTC)[reply]

Removed the offending sentence. I don't see that it adds anything.Tunborough (talk) 13:23, 22 November 2011 (UTC)[reply]

Complex impedances

I don't see any justification for the equations in this section. The complex impedance is frequency dependent, but Q is not. In particular, the phase angle is zero at resonance, so tan $\phi$ would be zero at resonance, whereas Q would not be zero. I suggest this section should be removed, but I'm reluctant to do this much damage alone. Tunborough (talk) 13:05, 22 November 2011 (UTC)[reply]

I found a source for what I hope is the correct calculation: http://www.qsl.net/va3iul/Impedance_Matching/Impedance_Matching.pdf I will go about implementing it now.siNkarma86—Expert Sectioneer of Wikipedia
^{86 = 19+9+14 + karma = 19+9+14 + talk} 05:34, 26 November 2011 (UTC)[reply]

Here is another good document which can be used for later: http://www.edn.com/contents/images/159688.pdf siNkarma86—Expert Sectioneer of Wikipedia
^{86 = 19+9+14 + karma = 19+9+14 + talk} 03:44, 3 January 2012 (UTC)[reply]

The section referred to has been renamed to Q factor#Individual reactive components; I repaired an internal section link just now, and added a comment about why Q is frequency dependent. Dicklyon (talk) 04:10, 3 January 2012 (UTC)[reply]

Originated

"The concept of Q factor originated in electronic engineering, as a measure of the 'quality' desired in a good tuned circuit or other resonator. [edit]"

Please, please! Anyone have a reference for the above claim?? It's very important to me.

i.e. not originated WRT mechanical oscillators?

Wikiecorrect (talk) 19:51, 4 March 2012 (UTC)[reply]

p.s. my name is cleyet. Search and answer off the talk page in addition to adding the reference on the Wiki page -- thanks!

I have given a reference to this - as requested. There is another article on the history, also by Bertha, Lady Jeffreys in either Wireless Word or Physics World (or Physics Bulletin) round about the same time. - Brian Cowan. — Preceding unsigned comment added by 80.229.155.170 (talk) 21:43, 5 June 2012 (UTC)[reply]

Apparent discrepancy between fundamental and reactive based definitions

In this section of the article I believe that the first equation is wrong, because the current used in the energy stored is the peak current, and the energy used in the power dissipated is the rms current whereas the text says they are both rms. If the same current is used for both the worrying factor of 1/2 disappears. The following discussion which explains it in terms of total stored energy is incorrect, because the energy stored in the capacitor and the energy stored in the inductor are the same energy, not separate energies that can be added together, the energy swaps place 4 times a cycle, from the capacitor to the inductor, back to the capacitor and then back to the inductor again, maintaining the same total energy all the time. When the voltage is highest on the capacitor there is no current in the inductor, and vice versa. The total energy can be calcuated at any point in the cycle and is half the capacitance times the peak voltage squared (or the capacitance times the the RMS voltage squared), or half the inductance times the peak current squared (or the inductance times the rms current squared), or proportions of each added together, but they always give the same total (at the resonant frequency). 192.93.164.28 (talk) 10:29, 7 March 2012 (UTC)[reply]

You are right, the equation is wrong for the reason you say. Constant314 (talk) 02:47, 8 June 2012 (UTC)[reply]

I think the section has a point. On average, the inductor stores 1/2L<I^2> energy (where <I^2> is the time average of the squared current, i.e. the RMS current squared), and -- on resonance -- resonators store one half their energy in the inductor and one half in the capacitor (when time averaged), so the total energy stored is 2(1/2L<I^2>)=L<I^2>. Likewise, the time averaged power dissipated in the resistor is R<I^2>, so the quality is w(energy stored)/(power dissipate)=wL<I^2>/R/<I^2> = wL/R = 1/R(L/C)^(1/2). However, when you just have the inductor, you're off by a factor of 2. Dlenmn (talk) 18:55, 8 April 2012 (UTC)[reply]

Parse errors?

There is suddenly a lot of red text about parse errors on the page. I think something is wrong there? CodeCat (talk) 15:52, 7 September 2012 (UTC)[reply]

It must be a temporary error on Wikipedia servers. It appears that the formulas work just fine under the edit preview. Go to (http://en.wikipedia.org/w/index.php?title=Q_factor&action=submit) and hit "Show preview" to see what I mean.siNkarma86—Expert Sectioneer of Wikipedia
^{86 = 19+9+14 + karma = 19+9+14 + talk} 19:56, 7 September 2012 (UTC)[reply]

Q factor for individual reactive components

First posting for Wikipedia, so I apologize in advance if I have not posted according to all guidelines. In studying the Q factor for individual reactive components, I have discovered an possible inconsistency with the definition

Q_{C}={\frac {X_{C}}{R_{C}}}={\frac {1}{\omega CR_{C}}}

,

because the capacitive reactance $X_{C}$ is defined (properly, I believe) in the provided link as:

X_{C}=-{\frac {1}{\omega C}}=-{\frac {1}{2\pi fC}}

This would imply a more suitable definition for the Q factor should include a negative sign (recognizing that $Q$ is non-negative):

Q_{C}=-{\frac {X_{C}}{R_{C}}}={\frac {1}{\omega CR_{C}}}

or perhaps a more generic equation that could be applied to the inductive case as well:

Q_{C}={\frac {|X_{C}|}{R_{C}}}={\frac {1}{\omega CR_{C}}}

.

The trouble is that many of the references I have studied and checked seem to agree with the current Q factor equation in Wikipedia:

Q_{C}={\frac {X_{C}}{R_{C}}}={\frac {1}{\omega CR_{C}}}

.

Since $X_{C}<0$ (agreed?), can anyone shed light on this possible inconsistency? Thanks.Stribs17 (talk) 19:23, 30 November 2012 (UTC)[reply]

Sign ambiguity of this sort is very common in many fields. In this case, there's a general inconsistency in how to treat the sign of capacitive reactance (see Talk:Electrical reactance for lots of discussion about it). It can be negative, so that it measures the same direction as inductive reactance, or positive, so it's measuring the opposite direction. When people write the ratio, they conveniently ignore whichever sign convention they might otherwise use. It would be good to fix it; I like your absolute value idea. Dicklyon (talk) 20:10, 30 November 2012 (UTC)[reply]

Depending on your definition,

X_{C}

and

X_{L}

can be seen as imaginary:

X_{C}=1/j\omega C=-j/\omega C

, so the absolute value is definitely in order. Q is inherently real and positive. Go for it.Tunborough (talk) 16:50, 1 December 2012 (UTC)[reply]

Attenuation Coefficient

In the section "Physical Interpretation of Q" the main article states: "The factors Q, Damping Ratio ζ, and Attenuation α are related such that..."
...and then proceeds with different formulas relating the variables Q, ζ and α .

I have a problem with the link to the Attenuation article because this article does not explain what the Alpha α variable is and this article does not contain even one formula with this variable α.

Thus, the better article to link in its place is the Attenuation Coefficient article, because it better defines the variable Alpha α.
Alternatively, see the Attenuation Constant article, which also rigorously defines Alpha α in another exponential formula.

The way the article is written now, it is not easy to discover that the Attenuation Coefficient or Attenuation Constant Alpha α is related to the Exponential Decay Constant λ.

This relation is important to be able to answer "time of decay" questions. For example such as the one below:

Q: "It takes N cycles for the amplitude of a sine wave to decay to 50% of its original amplitude. What is its Q Factor ?"

In engineering texts, the Q Factor is usually defined as 2π times the number of cycles for the response to decay to 1/e = 0.368, which is 1 time constant of the exponential decay envelope.

The Attenuation Coefficient ( which is the same as the Exponential Decay Constant ) provides the clearest conceptual link to other exponential decay functions in such problems.

Verpies (talk) —Preceding undated comment added 13:32, 22 January 2014 (UTC)[reply]

"Attenuation rate" is a good name for it, since it's a rate constant (nepers per second). But the articles you linked, attenuation coefficient and attenuation constant both use spatial definitions (nepers per meter), not temporal. If you can find something more appropriate to link (like the lambda in Exponential decay), or create something, that would be better than those. Adding a formula with

\exp(-\alpha t)

is a good idea. Dicklyon (talk) 15:36, 22 January 2014 (UTC)[reply]

I will look for a temporal definition of Alpha.

Verpies (talk) 00:47, 23 January 2014 (UTC)[reply]

It's in Exponential decay. It doesn't really matter that they call it lambda there, as it's the same thing. I usually call it sigma, personally. Dicklyon (talk) 08:04, 23 January 2014 (UTC)[reply]

Actually, alpha is the decay rate for amplitude, not energy; it is in nepers per second, since nepers are defined as natural log of ratio of field quantities, not powers or energies. The energy decay rate is 2alpha per second, which is probably worth clarifying with a formula with

\exp(-2\alpha t)

. Dicklyon (talk) 16:19, 22 January 2014 (UTC)[reply]

Indeed the distinction between field decay rates and energy decay rates is an important one.

Verpies (talk) 00:47, 23 January 2014 (UTC)[reply]

A resonant system with Q = 5, plotting the oscillation and the decaying (from 1) amplitude and energy, with key times and factors marked. For any other Q value, the graph would be the same except for a different number of cycles of oscillation under the curve.

For your consideration: I made this figure. Use it if you think it's helpful. Dicklyon (talk) 00:46, 24 January 2014 (UTC)[reply]

Q factor definition in the context of individual reactive components

In the context of individual reactive passive components (e.g. inductors & capacitors), I'm familiar with the following definition:

Q(f)\ {\stackrel {\mathrm {def} }{=}}\ {\frac {X(f)}{R(f)}}.

The aforementioned definition is presented, in some form, in section Individual reactive components of the article, however, the section Definition of the quality factor has no mentioning of it. Instead, the following definition is presented:

Q(\omega )=\omega \times {\frac {\text{Maximum Energy Stored}}{\text{Power Loss}}},\,

a definition which is practically the same as the other energy-based definition in the section, and furthermore, includes a reference to James W. Nilsson's Electric Circuits 1989 edition. I've went through a recent (much newer) edition of the book, and found no energy-based definition of any kind for Q - whether for resonator circuits or individual components. I therefore tagged the reference as "Failed verification", and additionally, I wonder what is the point of this definition, which is practically the same as the previous one in the article. It seems to me that one of them is superfluous.

--Sagie (talk) 14:15, 6 February 2015 (UTC)[reply]

It seems that the only difference between that formula and the preceding one is the substitution of ω for 2πf which I would allow as trivial mathematical substitution. However, it also doesn't add any new information so I would not object to its removal. I do, however believe that the energy based definition should be the first definition because it applies to all situations and simple elements are never simple in reality. All inductors have capacitance.Constant314 (talk) 20:27, 7 February 2015 (UTC)[reply]

The substitution of ω for 2πf is a minor difference. The main, and major, difference between the two formulae, which is surprisingly subtle, is the explicit indication that the Q factor definition for the individual components is frequency-dependent. Another issue is that this definition isn't equivalent to the definition in section Individual reactive components of the article (

Q_{C}={\frac {X_{C}}{R_{C}}}={\frac {1}{\omega _{0}CR_{C}}}

).

I'm not sure what makes you say that "energy based definition ... applies to all situations". The frequency-to-bandwidth ratio definition is just as applicable for any system for which one would like to express a Q factor. Anyhow, the bottom line is that in my experience, the frequency-to-bandwidth ratio definition is the most prevalent definition in textbooks, as well as the one which is mostly used in technical documentation and specifications of filters and other systems, hence its placement as the first defintion.

--Sagie (talk) 21:59, 11 February 2015 (UTC)[reply]