Talk:Phase (waves)

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The confusion below speaks to the need to re-write the article in toto, IMO. -Dawn McGatney —Preceding unsigned comment added by McGatney (talkcontribs) 10:06, 21 January 2008 (UTC)


Do physicists really use φ(x) for the wave function and φ0 for the phase? — Omegatron 22:37, 31 October 2005 (UTC)

Looks right to me. Smtomak 01:16, 28 June 2007 (UTC)

Text references[edit]

The test refers to entities called A, B and C. Are these waves? Are they the waves in the diagrams? Some clarification would be nice for people who do not know this material.

I think they refer to older diagrams that were replaced without updating the text. — Omegatron 20:00, 28 November 2005 (UTC)


Shouldn't the equation be:

(Note the minus phi rather than the plus)

As is shown here (albeit with some other variables included).

Surely this way of writing it makes more sense as well as it would mean that a wave that is time shifted in front of another would have a positive phase difference rather than a negative one. It would also make the example make more sense which shows a minus of a phase difference for a wave shifted forward.

--Datr 22:08, 14 February 2006 (UTC)

When I first arrived here, people were calling the phase of the sine function, which contradicts other articles that define frequency as the time-derivative of phase. So I think it is very helpful to be able to refer to as the initial phase, to clearly distinguish it from the time-variant phase. If you change the sign, it just becomes more awkward. When we tell them that is the negative of the initial phase, they are bound to wonder why we didn't just use a plus sign. --Bob K 02:46, 15 February 2006 (UTC)
By the way, if signal A has an initial phase of 1° and signal B has an initial phase of 0°, then signal A has a head start. In your words, it is "time-shifted in front of signal B". But apparently you disagree. So "in front of" means different things to different people. --Bob K 03:05, 15 February 2006 (UTC)
Sorry, I'm only doing this at a high school level and so am having problems understanding. Just to be clear are you telling me that is the negative of the initial phase difference? As unless i'm confusing the math that is the only way the equation using + would give the correct answer.
I thought that the initial phase difference would be measured positively (i.e. wave B is in front of wave A so . But for that to be the case the equation would have to use -. --Datr 11:49, 17 February 2006 (UTC)
I've realised I may be overlooking something, I assumed phase difference was measured from the beginning of one wave to the beginning of the next. However, the equation using + would work if the phase difference was measured from the beginning of the second wave to the end of the first. Is this my mistake? --Datr 12:08, 17 February 2006 (UTC)
The initial phase is the phase at t=0. So as it is currently defined, that would be just . Your definition would make it: , or in other words would be the negative of the initial phase, which is a bit awkward for a definition.
The phase-difference between two waveforms is something you define either way (A-B or B-A). It doesn't matter, as long as your interpretation of the results is consistent with your definition. --Bob K 19:39, 18 February 2006 (UTC)
I'm not sure but I think the problem might be I'm using an incorect defininition. At my school, phi is used to describe how ofset one wave is relative to another (which might be different from the intial phase of a wave). For example a wave that begins pi/4 after another would be said to be pi/4 out of phase relative to the first wave. If you were to use the equation in this article to plot both waves then for it to show the second wave as being pi/4 in front of the first wave you would have to subtract the phase difference. I.e. A-B would have the waves the correct way round but B-A (the intuitive way) would not, do you mean this doesn't matter though and you just refer to the wave in front as the one you know should be in front? Surely this wouldn't work with calculations that depend on the waves points at an exact time though such as working out superposition, would it?
Would a wave with an intial phase difference of a quarter of a wave have the same initial displacement as the same wave with no intial phase difference but a quarter of the way through an oscillation? Or would a wave with an intial phase difference of a quarter of a wave have the same displacement as a wave with no intial displacement a quarter of an oscillation before? --Datr 21:51, 18 February 2006 (UTC)
I don't have time to address your questions point-by-point, but I think what you are getting at is that various perspectives are possible, depending on the problem you are trying to model. Indeed, phi can be used "to describe how offset one wave is relative to another", as you said. But that presumes the situation you are modelling has two waves. A definition that requires only one wave is also applicable to more complex situations (i.e. two waves). But a definition that requires two waves is useless for just one wave. --Bob K 22:45, 18 February 2006 (UTC)

Take one waveform, copy it and paste it a quarter of an oscillation away. I'm sure you can see that they are not the same. Figure this one out, and you'll also figure why it's + and not -. --Freanz5 03:27, 29 August 2006 (UTC)

I don't agree that it is that simple. For instance, does "away" mean to the left or the right?  Or what if is 3π/2 ? --Bob K 16:36, 29 August 2006 (UTC)

Removed obsolete material[edit]

I removed the following, which refers to a diagram that no longer exists:

--Heron 18:21, 18 June 2006 (UTC)

Why does the diagram no longer exist? — Omegatron 00:56, 19 June 2006 (UTC)
Because its been deleted? 8-)--Light current 01:11, 19 June 2006 (UTC)
Is it worth reproducing, though? — Omegatron 19:02, 23 June 2006 (UTC)
I dont know. I have not seen it 8-|--Light current 19:15, 23 June 2006 (UTC)


  • Agree with all proposed merges. 8-)--Light current 19:45, 23 June 2006 (UTC)
  • Ditto. --Bob K 14:21, 24 June 2006 (UTC)

NO! you guys are wrong! —Preceding unsigned comment added by (talk) 02:38, 3 October 2007 (UTC)

Misleading image??[edit]

The major problem with the image is that the phase difference between the two time-shifted waves should be measured vertically, not horizontally. Horizontal measurement only measures time. Phase difference would be measured as the difference of arccosines of unit-magnitude sinewaves. -swstadel 01 Oct 2010 —Preceding unsigned comment added by (talk) 20:51, 1 October 2010 (UTC)

Huh?? Please explain the difference between time shifting a sine wave and phase shifting a sine wave. — Omegatron 20:47, 23 June 2006 (UTC)

Ones got a horizontal axis of time, and the other of angle. Normally diagrams like this have the horiz axis as time. So the horiz axis needs labelling in degrees to remove confusion.. That would do it. Also, consider the prime directive: [1] 8-)--Light current 20:55, 23 June 2006 (UTC)
How can you justify this image in view of the prime directive? 8-? --Light current 23:46, 23 June 2006 (UTC)
  • The what? Please read the rest of that thread. It's mostly about biographies on living people, several others disagreed with Jimbo's sentiments, and he explicitly stated they are not policy. If you want to call it your Prime Directive (which would seem appropriate in light of your deletionist tendencies), you can refer to it as such, but having your own declared modus operandi is not sufficient justification for removing useful information from articles. Vandals have directives, too, but that doesn't make their edits valid.
  • Now, please explain the difference between a time-shifted sine wave and a phase-shifted sine wave. Perhaps you could show us some equations? You might want to read the article first, too. — Omegatron 00:20, 24 June 2006 (UTC)
I quote from Jimbo's email:
This is true of ALL information, but it is particularly true of negative information about living persons.

my bolding and caps

I have already explained the difference between a time-shifted sine wave and a phase-shifted sine wave diagram. THe diagram is misleading as it stands as it does not have the axes labelled. 8-|

--Light current 00:39, 24 June 2006 (UTC)

I agree that the diagram could/should be better labelled. But it is not wrong, and frankly it is thought-provoking in a good way. It gets your attention. IMO it is better than no diagram. So if you want to delete it, please replace it with a better one.
--Bob K 05:01, 24 June 2006 (UTC)
I didnt say it was wrong. I said it was misleading because it is. Misleading info should be removed according to Jimbos directive.> He doesnt say anything about waiting until a new version exists before deleting - does he?--Light current 17:19, 24 June 2006 (UTC)
  1. It is not misleading.
  2. Just because an article contains a misleading statement doesn't mean you delete the whole article. You fix it so that it isn't misleading anymore. This should be common sense to a Wikipedian with your experience. If you don't have time or are too lazy to fix it, leave it for someone else to fix. You could even tag it as needing attention so others will notice it needs work. That's why we have a multitude of cleanup and attention tags.
  3. Misleading information is better than no information? I disagree. Misleading information is better than no information. No information is better than incorrect information, but misleading ≠ incorrect. While we're at it, incomplete information is better than no information, and correct and complete information is better than all.
  4. Destroying information is bad. Wasting time with pointless discussions is bad. Especially over trivial things like this. This image has been there for three months with no complaints. If you have a minor complaint like this about something in the article, bring it up on the talk page. Don't go around deleting useful things without discussion. — Omegatron 17:56, 24 June 2006 (UTC)
And he didn't say anything about judging one element of an article totally out-of-context, which is what you are doing. You haven't said anything about the supporting text, most notably:   "Thus, a shift in time is equivalent to a change in the initial phase. Conversely, a change in the initial phase is tantamount to a shift in time." If we delete every figure in Wikipedia that is ambiguous without explanation, that would probably be a majority. Your opinion matters, but it is not the only one that matters. Mine is that in the context of the article, the image is not misleading. And I am asking you to please leave it alone, unless you can make a positive improvement. --Bob K 17:43, 24 June 2006 (UTC)
I think the diagram needs to be improved forthwith. If I could find out how to do it myself I would. 8-|--Light current 17:47, 24 June 2006 (UTC)
Thank you! Maybe these sources are useful to you:
I suspect Omegatron would be happy to do it himself, if you can agree on what needs to be done. It's worth a shot.
--Bob K 18:14, 24 June 2006 (UTC)
How would you improve it? — Omegatron 17:57, 24 June 2006 (UTC)

Well, IMO, all it needs really is the x axis labelling as Phase (deg)-- 0 to 360 say, or Phase (rad) 0 to 2pi.--Light current 18:32, 24 June 2006 (UTC)

Ive changed the caption now. Thats all I can do ATM to make it less misleading --Light current 18:46, 24 June 2006 (UTC)

And why would an x axis of time be incorrect? — Omegatron 19:39, 24 June 2006 (UTC)

It would be incorrect as it stands because the diag shows the distance between the two points shown as being a phase diff. Phase is not the same thing as time 8-|--Light current 20:45, 24 June 2006 (UTC)

So if the red waveform was , what would the equation for a phase-shifted version be, and what would the curve look like? — Omegatron 22:04, 24 June 2006 (UTC)

I assume you are asking these questions in good faith so I will answer likewise. The phase shifted version would of course be a sine wave displaced along the horizontal axis by degrees from the original waveform.

The equation of the phase shifted waveform is: NB phi may be +ve or -ve --Light current 22:31, 24 June 2006 (UTC)

Extract from WP:V[edit]

Jimmy Wales has said of this: "I can NOT emphasize this enough. There seems to be a terrible bias among some editors that some sort of random speculative 'I heard it somewhere' pseudo information is to be tagged with a 'needs a cite' tag. Wrong. It should be removed, aggressively, unless it can be sourced. This is true of all information, but it is particularly true of negative information about living persons." [1][2]

If its not policy, why is it on the policy page?--Light current 00:46, 24 June 2006 (UTC)

And this has what to do with the image? — Omegatron 03:51, 24 June 2006 (UTC)
Nothing. The image is not incorrect. It is not pseudo information. It does not require a source. Could it be even better?... yes. Is it better than nothing?... yes. Is this debate a waste of time?... yes. --Bob K 05:01, 24 June 2006 (UTC)

You are being obtuse 'O'. REread my above comments. Bob, tHe image is 'misleading'. The directive is to remove misleading info.--Light current 09:51, 24 June 2006 (UTC)

No. I am expressing a different 'O'pinion than yours. I was not misled by the image, but I did have to stop and think about it. Also note that the image does not stand alone in a vacuum. There is explanatory text, notably this:   "Thus, a shift in time is equivalent to a change in the initial phase. Conversely, a change in the initial phase is tantamount to a shift in time." --Bob K 13:08, 24 June 2006 (UTC)

Only for a fixed frequency!--Light current 20:47, 24 June 2006 (UTC)

Which is what the whole article is about. I guess you just need to have the last word, so go ahead. Unless it contains some actual substance, I will leave it alone next time. --Bob K 22:33, 24 June 2006 (UTC)

Yes thats OK. The diagram(s) though, should reflect what the text is saying. Only when its the correct last word! 8-)--Light current 22:35, 24 June 2006 (UTC)

Good examples of phase difference: [2] The lh phasor diag shows phase diff. The rh waveform shows subsequent time delay between the waveforms.--Light current 23:24, 24 June 2006 (UTC)

Very nice. I suggest we add a link to that site. --Bob K 00:16, 25 June 2006 (UTC)

What's with the [single brackets] everywhere? Looks like those should be (parentheses)? — Omegatron 18:45, 26 June 2006 (UTC)

Doh! You are absolutely right, but I had to check google to find out. It is apparently a bad habit of mine that I have been doing for so long that it looks perfectly normal to me. And this is actually the first time anyone has asked about it. Feel free to fix them, if so inclined. I will try to do better in the future. --Bob K 07:21, 27 June 2006 (UTC)

Phase vs polarity[edit]

Is "phase shifting a wave by 180°" the same thing as "inverting the polarity of a wave"? Does it even make sense to say "phase shifting a wave" when the wave is not a sinusoid? See Talk:Active_noise_control#Phase_vs._polarity. — Omegatron 18:31, 10 August 2006 (UTC)

Consider:   .   Unless and are constants, this modulated signal is not a sinusoid. Yet a change in can be construed as a phase shift. Since   ,   a 180° shift is equivalent to sign inversion. --Bob K 20:17, 10 August 2006 (UTC)
When someone "phase shifts a square wave by 180°", the outcome is equivalent no matter how you think of it, but if they "phase shift a square wave by 35°", you could think of it as shifting the entire square wave by a fraction of its period, or of shifting each individual frequency component by 35°, which would make it not a square anymore. Does it make sense to talk about phase shifting arbitrary signals, or only sinusoids?
Also, a phase shift in real life implies a time delay, no? So phase shifting an entire signal by 180° would require a delay of at least half a period of the lowest frequency present? So a device that instantaneously shifts the phase of every frequency component of a signal by 33° is not physically realizable, similar to how non-causal ideal filters aren't physically realizable, and therefore a 180° device of the same description would also be impossible, so 180° phase shift in real life implies a time delay, and polarity inversion does not?
  • "If you think about it, it's obvious that there is no such thing as "out-of-phase" for a sound that has more than one frequency. As the figures above illustrate, any given delay time will yield an infinite number of phase shifts at an equally infinite number of different frequencies. So how can a sound be 180° out of phase? Answer: it can't, unless it's a sine wave." [3]
  • "Usage Note: polarity vs. phase shift: polarity refers to a signal's reference NOT to its phase shift. Being 180° out-of-phase and having inverse polarity are DIFFERENT things. We wrongly say something is out-of-phase when we mean it is inverted. One takes time; the other does not." [4]Omegatron 20:53, 10 August 2006 (UTC)

So going back to my example, are you saying that a sound wave can never have the form:
Also note that:
which represents destructive interference, as occurs when two speakers are wired in opposite phase.
--Bob K 22:34, 10 August 2006 (UTC)

I took a peek at [5]. I believe the confusion is that his conception of phase shift is limited to the special case of a time-delay that is constant for all frequency components. The corresponding phase shift is a linear function of frequency. So there is only one frequency that can have a 180° shift. His perspective makes sense for the case of sound propagation in air, because air is a non-dispersive medium [6]. But his perspective is not the general case of phase-shift possibilities. --Bob K 22:56, 11 August 2006 (UTC)


Just moving the template here where it belongs. -- Your friendly, neighborhood housekeeper, Flex 14:41, 6 September 2006 (UTC)

The first sentence of this article could really bear clarification - maybe even breaking into different sentences. "fraction of a complete cycle corresponding to an offset in the displacement from a specified reference point.." - what?? offset, displacement, and reference point are all so closely conceptually related I wonder if they aren't the same thing explained three different ways.. andf how does this fraction correspond to all that whatever? -- (talk) 16:49, 30 June 2008 (UTC)
Moreover the diagram does apparently nothing (for a novice's eyes anyway) to show where the phase is identified in the waveform. Shouldn't it say "phase" somewhere? And is "period" in the diagram another way of saying "cycle" (in the sentence)? -- (talk) 16:52, 30 June 2008 (UTC)

I concur. I have no idea what the first sentence means. (talk) 04:10, 17 September 2010 (UTC)


It is important to mention that the phase is zero when the wave is maximal. It is mentioned in the definition of "Instantaneous phase", but it should be mentioned here too.

Please sign your comments. First of all, sin(0) = 0, and max( sin ) = 1. Secondly, t=0 is kind of a "starting point", and it seems more intuitive (to me) for a wave to start at level 0 and rise to its apex, like a mountain, rather than for the apex to just "appear" out of nowhere. So I prefer to keep the sin() function. Third, why would it be "important" to mention any of this? It seems rather minor.
--Bob K 15:15, 27 June 2007 (UTC)

Merge with Antiphase[edit]

Yes please MarkAnthonyBoyle 08:48, 27 September 2007 (UTC)

Merge completed, since the little serious replies seen are positive. Fuzzygenius 06:40, 15 October 2007 (UTC)

Question worth mentioning[edit]

Just one question: When you spend 1 watt of sending a wave to meet another 1 watt wave in opposite phase, you end up with zero energy because they cancel each other. According to the energy laws; where have those 2 watts in total gone? —Preceding unsigned comment added by (talk) 21:46, 8 October 2008 (UTC)

To paraphrase a famous quote: "You can cancel at a point in space for all time, and at a moment in time for all space. But you can't cancel all points at all times." (assuming the waves are actually "meeting" from opposite directions) Compensating for the point/moment cancellations are other points and moments of constructive interference.
--Bob K (talk) 01:56, 9 October 2008 (UTC)
I'm guessing the follow-up question would be... what if they're traveling in the same direction? (e.g., joined at a summing junction somewhere) I assume the answer is that the power that is not transmitted gets eaten at each source. — TedPavlic | (talk) 13:18, 9 October 2008 (UTC)
I assume we would need to get more specific about the nature of the sources and the "summing junction". Theoretically, if 2 electrical sources (e.g. batteries) are placed in series and each connected to a resistor, their voltages add. If the sources are equal and opposite, no current flows. No power is expended. Each source effectively sees an infinite resistance, like an open circuit.
--Bob K (talk) 11:22, 10 October 2008 (UTC)
Consider the electrical generation of a virtual ground. Two equal and opposite sources with R output impedance are jointed together. The junction between the sources sits steady at 0V, while at each point in time each output impedance dissipates V2/R power. Now remove one of the sources and add an R between the old junction and a true ground (representing power transmission). Now V2/(4R) is dissipated in the source and V2/(4R) is transmitted. If the other source is added again, the transmission will cease, and the total 2 V2/R will be dissipated over the sum of the two output impedances. So, again, the destructive interference that prevents transmission results in constructive interference at the source. That's what I was saying. I think you'll find a similar effect in any type of junction scenario — reflections will end up generating constructive interference at the source. —TedPavlic | (talk) 14:06, 10 October 2008 (UTC)
I'm not curious enough to follow that line of reasoning, but it appears to corroborate what I said about "getting more specific about the nature of the sources and the summing junction." Hopefully we have at least answered the original question.
--Bob K (talk) 15:51, 10 October 2008 (UTC)

Instantaneous phase[edit]

"Instantaneous phase has a formal definition that is applicable to more general functions and unambiguously defines a function's initial phase at t=0." This is definitely wrong, compare with the article about instantaneous phase. Pk72 (talk) 08:54, 7 July 2011 (UTC)

θ or φ?[edit]

A recent mass change replaced all the θ's with φ's, with the explanation: "Try to impose a little bit of consistency in notation. Most of the time, the convention is θ for the entire angle or argument to the sin() function and it's φ for the phase offset."  I'm all for consistency, but is that statement a correct one?  In my experience, instantaneous phase is always written as φ(t), and the instantaneous phase of    is    So θ is a better choice (than φ) for x.   --Bob K (talk) 14:13, 16 February 2012 (UTC)

Lower case phi – φ – is the correct symbol for phase in terms of practical electronics. As a live sound engineer, I might point out that the Danish letter Ø has often been used to label a switch in polarity. I bring that up because a flip in polarity is often considered by sound guys to be the same as 180° of phase shift at all frequencies. Binksternet (talk) 19:33, 16 February 2012 (UTC)

Thanks. I assume you are referring to the "phase offset" (a constant), rather than φ(t). So perhaps the convention is that when talking about φ(t) (instantaneous phase), phase offsets are represented by θ to avoid writing    And otherwise (such as here), a constant phase offset is most often represented by φ. Another case where that works is the narrowband signal model (Phase_(waves)#In-phase and quadrature (I&Q) components), where the carrier and the phase modulation are considered separately, rather than as a whole instantaneous phase. --Bob K (talk) 20:20, 17 February 2012 (UTC)

Phase term in "magnitude-phase" representation of signals[edit]

I am wondering if it may be worth to make a difference of the phase term discussed here with the phase term in the complex domain representation of signals. For example, a sinusoidal signal (real) can be represented in magnitude-phase, and its phase (which in this case can be understood as the angle with respect to the positive real axis) takes only 0 or pi radians. We even can add a "phase shift" to the aforementioned signal but the signal phase will still take 0 or pi radians. For example, cos(wt) and cos(wt + phi) can be thought of a real vector in the complex domain varying its magnitude from 1 to -1 periodically. What do you think? — Preceding unsigned comment added by EthanZur (talkcontribs) 11:42, 3 May 2012 (UTC)

Clarification requested on the Phase Compensation section[edit]

The explanation is vague and it isn't tied in with the figure. What part of this circuit represents phase compensation and what part represents the phase to be compensated? What amplifier stage is switched into the signal path? Is the amplifier in the figure part of a larger system? Constant314 (talk) 00:54, 4 September 2012 (UTC)

Upon further reflection, phase is a subject that transends electronics. I'm not sure that an electronic circuit should even be part of this subject.Constant314 (talk) 13:55, 6 September 2012 (UTC)

Phase compensation, figure is dubious[edit]

The figure purports to illustrate a phase compensation circuit. But, the circuit has no dc feedback and could not work as a standalone circuit, although it could be an element of a larger system where there was overall dc feedback. But even then it’s not likely to produce useful phase compensation. At low frequency this circuit is an integrator with produces -90° degrees of phase shift. As frequency increases (assuming C1 >> C2) the circuit starts to resemble straight gain for frequency greater than ω₁ = 1/ (C1 x R2) and less than ω₂ = 1/ (C2 x R2). Above ω₂ it again resembles an integrator with phase shift of -90°. So, looking at the phase shift of this circuit, it starts out at -90°, then it increases toward 0°, but stays slightly negative and then decreases toward -90°. This circuit adds negative phase at most frequencies which is usually the opposite of what is needed to stabilize a loop. A lead-lag configuration would be a typical example of phase compensation of a larger loop. Constant314 (talk) 13:44, 6 September 2012 (UTC)

Removing Phase Compensation[edit]

The article is about phase which is a very general concept. Phase compensation is a technique used to change the phase shift of an electrical circuit or other linear system. Phase compensation is often used to stabilize circuits and systems that use feedback. A factually correct section about phase compensation might logically appear in an article about feedback, opamps or servos among others. It is out of place in this article.

The figure has no explanation of how the circuit works, what it does or how to use it. The particular circuit depicted is unlikely to be used for phase compensation because its phase shift is always negative.

Phase compensation is not a correction for phase error. It can be used to adjust the phase shift of a circuit.

Phase compensation is not required to obtain stability in an opamp but it can be used to improve the phase margin of an opamp and it can be used to stabilize a circuit that includes an opamp.

Phase compensation does not subtract out an amount of phase shift from a signal which is equal to the amount of phase shift added by switching one or more additional amplifier stages into the amplification signal path. Constant314 (talk) 22:37, 23 December 2012 (UTC)

Merge with Instantaneous Phase[edit]

I came here looking for the promised discussion about a merge with Instantaneous Phase. Nobody home. Well anyhow, I vote "no", because this article is about a constant, and the other article is about a time-variant function.

--Bob K (talk) 17:57, 18 October 2013 (UTC)

I vote no also. Constant314 (talk) 00:30, 19 October 2013 (UTC)

I vote no. Article instantaneous phase is about concepts related to but distinct from the topic of article phase (waves). (What's more, the "encyclopedic voices" adopted by the two articles are different. Merging would probably mean cacophony.) --νημινυλι (talk) 04:05, 5 March 2014 (UTC)

No. for all of the above reasons. But the Instantaneous phase article needs improvement. (talk) 03:50, 23 May 2014 (UTC)

Merge with phase angle[edit]

I vote no, since phase angle can be applied to any complex number or 2D vector, while the phase of this article is only applicable to waves. Ulflund (talk) 11:50, 12 January 2014 (UTC)

Weak support. I believe Ulflund's concerned could be addressed in the merged article.Constant314 (talk) 14:46, 12 January 2014 (UTC)

  • Oppose. While these articles are related and could no doubt be merged, I don't think that they could be merged under this title. Trying to shoehorn in too many things loses the focus of an article and all in all they are best left separate. While I don't agree with Ulfund that phase angle can be applied to any complex number (simply angle is more appropriate for that) there is a difficulty with things like electrical impedance. This certainly has a phase angle, but where is the wave? It has this property whether or not a signal is applied, and besides, as a lumped component there is no travelling wave even when it is "live". SpinningSpark 00:43, 6 October 2014 (UTC)
  • Oppose. Phase angle is related to complex numbers, not phase when related to waves. (talk) 21:52, 22 March 2016 (UTC)

I've removed the merge templates. In nearly three years even the person who added the templates, user:Mario Castelán Castro, hasn't bothered to come here with a rationale for merging. SpinningSpark 13:21, 23 March 2016 (UTC)

Linear combination[edit]

The following statement is incorrect, unnecessarily complicated, and not particularly relevant:

An equivalent formula is (see List of trigonometric identities#Linear combinations):

where and .

The incorrect part (fixed) is:

A simpler statement, but still not helpful would be:

--Bob K (talk) 13:52, 24 November 2014 (UTC)

Bob K I'd call the missing minus sign a typo, not a completely incorrect expression. Could you please to elaborate why you find it unhelpful? I find the expression helpful in practice because it shows how to extract phases out of FFT coefficients, for example. It also illustrates the more theoretical point that Cartesian and polar complex representations are equivalent. Thanks. Fgnievinski (talk) 18:46, 24 November 2014 (UTC)
(after ec) Actually, I think that that expression gives the co-angle of φ because φ is defined in the article as the phase shift from the cosine, not the sine. I think the point being made is that a phase shifted sinusoid is always the sum of a sine and a cosine wave (and vice versa). That is a worthwhile point to make, with or without formulae. You need to know this in Fourier analysis for instance, and if you are the designer of a spectrum analyzer. SpinningSpark 18:56, 24 November 2014 (UTC)

I didn't say the formula isn't important. I said it isn't relevant to the point of this particular article. But perhaps that was just a lazy way of saying that it just shows up without sufficient justification. And it did seem unnecessarily complicated to me. I'll give it some more thought too.
--Bob K (talk) 13:35, 25 November 2014 (UTC)

The point you would make here is also made at Phasor#Addition and at In-phase_and_quadrature_components#Narrowband_signal_model, which are both listed in the See Also section. But we could make it more obvious by adding a footnote here containing a simple statement that would include those links.
--Bob K (talk) 14:04, 25 November 2014 (UTC)