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why the radio based on frequency modulation has very small transmission area?

It's nothing to do with the FM modulation, it's to do with the higher frequency used for FM radio broadcasts, compared with AM radio broadcasts. AM radio is on a lower frequency, so it travels further. The lower longer-range AM radio channels were handed out first, so they use the older (and more frequency-efficient, but less interference-tolerant) modulation. -- The Anome 11:04 18 Jul 2003 (UTC)

Actually, FM is more frequency efficent than AM, except that commercial FM uses wide FM to get more audio bandwidth into the signal. If FM used the same audio bandwidth as AM, it would use the same or less RF bandwidth than AM. Commercial frequencies (i.e., public safety) and amateur frequencies that use FM use narrow FM. --ssd 03:59, 23 Jun 2004 (UTC)

Note: we should have a radio propagation to explain why the lower frequency radio waves travel further. -- The Anome

I've started one, but there may be lots of factual inaccuracies, I'm not sure, I haven't done radio stuff in ages :) Dysprosia 11:26 18 Jul 2003 (UTC)

I started Category:Radio frequency propagation with the intent of writing an article including as many of those terms as sane. Feel free to use it as a seed. --ssd 03:59, 23 Jun 2004 (UTC)

Could we please stop moving the disambiguation page and setting to FM. FM does stand for more than one thing. Thank you. --Numerousfalx 23:51, 15 Dec 2004 (UTC)

Can someone give an example for frequency modulation? let's say there is a test signal low tone, high tone, and a sweep. this composed audio signal is supposed to be put on FM radio - how would the modulated tone look like? how much bandwidth does FM radio need? When I detune my favorite radio station "92.2 MHz Rock" by, let's say +0.2 MHz, will the music and all transmitted acoustics be pitched to higher frequencies (so that radio moderators talk like Mickey Mouse) - or is FM resistant against those mistunings.

It would be great if someone could work my (answered) questions into the article. Thanks, --Abdull 21:41, 17 Jan 2005 (UTC)

I agree this sort of information should be covered by the article, preferably with diagrams. Answering these questions (for FM) is somewhat tricky, though, and involves a lot of math.

I can answer your last question here: no, tuning away from an FM station does not shift the frequency components of the output, so you won't hear "Mickey Mouse" voices like you would with SSB. The capture effect inherent to analog FM receivers means that slight mistunings are "automatically" compensated for. Tune far enough away and you hear static-y, distorted sound; tune further away and you hear only static.

Simon 18:20, 23 Jan 2005 (UTC)

Simon is correct about detuning an FM receiver that has no AFC (Automatic Frequency Control) or has AFC switched off. However most FM receivers do have AFC (because it is quite easy to implement) which truly compensates for mistuning. Try to tune away and the sound remains good, until you reach the limit of the AFC and suddenly lose the signal completely. Then the sound changes to the hiss heard between stations. Some receivers have an additional "squelch" function that mutes the hiss noise. Note: all the above observations are for home VHF FM radios, which the frequency Abdull mentions indicates.Cuddlyable3 18:50, 12 February 2007 (UTC)[reply]

In theory, detuning adds a DC offset to the output. There is always at least one capacitor in the path, that will block DC. Gah4 (talk) 17:56, 27 November 2023 (UTC)[reply]

The article reads as "FM as used in radio" with a little theory thrown in at the last minute. Shouldn't an article titled "Frequency Modulation" primarily cover the theory and link to separate articles on radio/data transmission/encoding (and FM synthesis)? The FM synth article has almost no theory at all.

That might be because the authors of the FM Synth article are unaware
of that FM-synthesis á la Yamaha is based on phase-modulation. /ja

The first sentence "...represents information as variations in the instantaneous frequency of a carrier wave" makes sense if you're reading about radio, but is hard to "get" if you're coming at this from the FM synth perspective.

Please have a look at the discussion page under "phase modulation" where I
present a few C like oneliners to make the point. /ja 

I suppose it makes sense to add a bit more on the theory, whether the article should have its current focus or not; I'd do it myself, except for the fact that I don't really understand FM all that well yet.

magetoo 14:16, 27 Apr 2005 (UTC)

But why not add a see also in both articles Theking2 (talk) 16:52, 27 November 2023 (UTC)[reply]

"The phrase frequency-modulated, an adjective, should have a hyphen when used attributively."

Is this significant to FM for some reason? - Omegatron 21:39, May 25, 2005 (UTC)

Could the author consider changing the font on the maths? I found it very difficult to see on a computer screen whether I was reading "t" for time or "f" for frequency. Otherwise found the article well balanced.

I agree that the font used makes the formulae very hard to read, but the author correctly used TeX to create them; there is nothing to fix on this page.

In my opinion there is a problem with the HTML generation of formulae (it was just as unreadable in IE as Firefox). I will look around and see if there is somewhere I can raise this issue. In the meantime, I recommend changing your wikipedia math preferences to use png for all formulae (the default is to use HTML for 'simple' formulae and png for everything else). Happyharris 18:40, 18 August 2005 (UTC)[reply]

After reading some more doc, it seems that individual formula can be forced to PNG. That seems appropriate here so I did it (although I still think the default use of HTML is a wikepedia-wide fault) Happyharris 18:44, 30 August 2005 (UTC)[reply]

Happyharris, i really agree and have had some of my PNG converted to HTML like in Plank units at the bottom. but i think that consistency in notation and appearance is important. ironically, i was recently editing and was putting in in-line HTML math instead of the usual PNG. so maybe i got infected with WP orthodoxy. r b-j 05:15, 20 October 2005 (UTC)[reply]

The main theory discussion has a very common mistake in deriving frequency modulation. I picked up on it because I make the same mistake every time I start to review FM, and have to fix it. The formula x(t) = A*cos(2*pi*f(t)*t) is incorrect. It arises because A*cos(2*pi*f*t) is correct when f is a constant - f is the frequency. However, frequency is the rate of change of the cos() argument. If f(t) is time varying, then the rate of change of f(t)*t is not f(t), it is some complex derivative that is no fun at all.

A better derivation is to start with x(t) = A*cos(phi(t)). Frequency is defined as the rate of change of phi(t). To get sinusoidal FM, we need some phi(t) such that the derivative of phi(t) is 2*pi*deviation*cos(2*pi*rate*t). That isn't so hard - we know the derivative of sin(kt) is k*cos(kt), so we postulate phi(t) = deviation/rate *sin(2*pi*rate*t). It is inconvenient to carry the ratio of deviation/rate, so it is often called Beta. The final formula, then, is x(t)=A*cos(Beta * sin(2*pi*rate*t)) where Beta is deviation/rate, and both deviation and rate are in Hz.

This leads nicely to the Bessel functions, which are a formalized way of working out equations like cos(m sin(x)).

References: "Analog and Digital Communications" by Hwei Hsu (Schaum's Outline) and "Fundamentals of Electronics" by Aldo Vieira da Rosa. (above written by Doug olney)

In response to the above, I don't think anybody said that x(t) = A*cos(2*pi*f(t)*t) is even used for teaching FM, unless f(t) is defined as fo + m(t). I would expect that the formula should be something like x(t) = A*cos[ {wc + k.m(t)}.t ], where the message k.m(t) somehow alters the frequency inside the brackets { }. The issue here is: if we apply the formula to obtain the instantaneous frequency of x(t) = A*cos( {wc + km(t)}.t ), then it turns out we will get the instantaneous frequency to be something like: wc + k.{m(t) + t.dm(t)/dt}, which is nothing like what we 'hoped' to get for the instantaneous frequency, such as: wc + k.m(t). This sort of thing is absolutely confusing to people that are learning about frequency modulation. For example, why do they not begin with this particular expression x(t) = A*cos[ {wc + k.m(t)}t ] ? And why do they side-step it by beginning with phase modulation? Maybe because they are conveniently able to get the instantaneous frequency to become what they want to see when beginning with a phase modulation expression, ie. they are able to get the instantaneous frequency to be wc + k.m(t) when they start with phase modulation expressions. It really is something unusual. This really needs to be addressed in the Frequency Modulation Wikipedia page. It needs to be told straight-up to people. That is, why do they not begin with x(t) = A*cos( {wc + km(t)}.t )? The answer is of course, the equation does not model instantaneous frequency being proportional to message amplitude, even though one might incorrectly 'intuitively' or 'instinctively' assume this is the right model. You are right, it is not the right model. KorgBoy (talk) 00:48, 23 March 2017 (UTC)[reply]

Without thinking about it in too much detail, it seems to me that the wrong form should be close enough when the modulation frequencies are much smaller than the carrier frequency. That is, we can ignore the higher terms. That is, do a first order approximation. In theory, FM has infinite sidebands, which would preclude multiple stations, but in the practical case the sidebands fall off fast enough. Even when the modulating frequency isn't so small, such as the luminance signal in video tape, it will be filtered after the modulator to remove excess sidebands, which likely makes more difference than getting it right in the first place. Gah4 (talk) 06:12, 20 November 2019 (UTC)[reply]

I think that the angle modulation page does it right. It defines angle-modulated signals in terms of A cos(wt + phi(t)). Then it defines instantaneous phase and instantaneous frequency. It then shows how in FM the instantaneous frequency is linearly related to the modulating signal, and how in PM the instantaneous phase is linearly related to the modulating signal. That page is much less confusing than this one. Maybe this page should reference the angle modulation page, or adopt a similar approach. DigitalCommsEngineer (talk) 16:45, 11 July 2024 (UTC)[reply]

"FM is also used at intermediate frequencies by most analog VCR systems, including VHS, to record the luminance (black and white) portion of the video signal."

Has this method ever been tried for audio recording, aside from the HiFi modes on some analog VCR systems such as VHS and Betamax?144.139.87.8

On magnetic tape, the common method of adding a supersonic frequency to the audio signal reduces distortion very effectively, and exploits the dynamic range of the tape to the full. An FM signal with its sidebands would demand a faster tape speed, hence allow less recording time.

FM of an audio-frequency carrier has been useful as a means of recording low-rate digital data on audio tape (cassette) recorders. Such audio equipment could not handle long strings of repeated 0's or 1's directly because of limited low-frequency response.Cuddlyable3 19:01, 12 February 2007 (UTC)[reply]

"FM is the only feasible method of recording video to and retrieving video from magnetic tape without extreme distortion, as video signals have a very large range of frequency components — from a few hertz to several megahertz, too wide for equalisers to work with due to electronic noise below -60 dB. This sentence tries to say too many things and needs attention. The megahertz-wide frequency range of video does not force any particular choice of modulation method, nor is it difficult to "equalise" over the video frequency range. I think the subject of FM in video recording belongs in a new section, where the significant issues treated are linearity, pre- and post-emphasis, tape speed, head gaps and noise distribution.Cuddlyable3 18:53, 9 February 2007 (UTC)[reply]

It is that the frequency goes down to almost zero that is what makes it hard to do other than FM. I suppose another possibility would be to mix it up to a much higher frequency, but higher frequencies are also hard to record. Gah4 (talk) 22:29, 16 July 2018 (UTC)[reply]

The statement of "Commonly, the chrominance component is recorded as a conventional AM signal, using the higher-frequency FM signal as bias." of appears to be referring to the VHS system and this isn't true. The chrominance signal is recorded using the "color under" system. The 3.58 MHz subcarrier is simply downconverted to a lower frequency and recorded along with the luminance signal. Maybe this whole section should be scrapped and links given to FM video recording systems such as VHS and data recording such as the Kansas City Standard. (Even though the latter is rarely used these days.) — Preceding unsigned comment added by 96.240.175.232 (talk) 01:31, 20 December 2013 (UTC)[reply]

After mixing down, it is then recorded as conventional recording. I suppose you could even include the mixing done by an AM modulator in the comparison. Gah4 (talk) 22:29, 16 July 2018 (UTC)[reply]

The animation is nice. However the AM example is jerky, gives an impression that carrier cycles are individually modulated, and they don't look sinusoidal. There is also redundant text "modulatie", not even in English (the M's stand for modulation already). I think we have to look extra critically at animated drawings because of the relatively high data overhead they represent.Cuddlyable3 15:19, 15 February 2007 (UTC)[reply]

I have fixed the above with a new animation. Besides being accurate and non-language specific, it is a much smaller file, and we all like pages to load quickly, don't we?Cuddlyable3 01:04, 18 February 2007 (UTC)[reply]

I may be wrong, since I'm still studying elementary physics, but does the first graph show a frequency over time (signal) superimposed on an amplitude over time (carrier)? To remain consistent, I would think that the two graphs should have the same two variables. --Pyg 01:09, 11 March 2006 (UTC)[reply]

Your question confuses me but the graphs do not. They all have the variables voltage (vertical) and time (horizontal).Cuddlyable3 18:18, 9 February 2007 (UTC)[reply]

The red curve shows the signal as advertised, which is a varying voltage input to the modulator. Only after the modulation does the carrier frequency over time take on the same curve (or it could be inverted since there is no fixed convention about the direction of frequency deviation).Cuddlyable3 21:22, 11 March 2007 (UTC)[reply]

I think the first image (AM, green waveform) is incorrect - it should be showing a varying signal amplitude (vertical scale) at a fixed frequency. Nogami 00:47, 12 August 2006 (UTC)[reply]

It is correct. The carrier drawn in green has neither AM nor FM.Cuddlyable3 18:18, 9 February 2007 (UTC)[reply]

There seems to be something wrong with the formula. It is different from the one shown in one of the external links (http://www.fas.org/man/dod-101/navy/docs/es310/FM.htm). In this page the formula does not involve an integral at all. It seems to me that the formula as it is now requires the time integral of the signal Xm(t) from 0 to t to be restricted to [-1,1] for all t. This is not necessarily true even if |Xm(t)| < 1. Also it would be interesting to know how to recover the transmitted signal from the modulated carrier wave.

The integral is correct - it comes from the fact that FM is actually a special case of PM (phase modulation), and phase is the integral of frequency w.r.t. time. Also, I don't see why the time integral must be restricted to [-1,+1]. The only reason that the restriction ${\displaystyle |X_{m}(t)|<1}$ exists is so that the frequency is restricted to ${\displaystyle f_{c}\pm f_{\Delta }}$. Oli Filth 09:58, 22 January 2007 (UTC)[reply]

Could someone expand the theory section significantly? I think it could use more math/physics background and a sub-section on the engineering of actually accomplishing frequency modulation. Ryanluck 15:55, 6 November 2006 (UTC)[reply]

I disagree that the theory section needs more math/physics background. FM and radio is essentially the product of engineering and the mathematics behind it is nice to understand - but not fundamental. The explanation of FM given here is equivalent to asking someone the time and getting and answer on how his watch was built with a precise explanation of the location of every cog in the structure. It may in fact include the information being sought, however more than likely the questioner still won't know what time it is. A far more practical explanation of the theory of frequency modulation is given in the reference http://www.fas.org/man/dod-101/navy/docs/es310/FM.htm:

Frequency modulation uses the information signal, Vm(t) to vary the carrier frequency within some small range about its original value. Here are the three signals in mathematical form:

   * Information: Vm(t)
   * Carrier: Vc(t) = Vco sin ( 2 p fc t + f )
   * FM: VFM (t) = Vco sin (2 p [fc + (Df/Vmo) Vm (t) ] t + f) 

Df is the peak frequency deviation. In this form, you should be able to see that the carrier frequency term: fc + (Df/Vmo) Vm (t) now varies between the extremes of fc - Df and fc + Df. The interpretation of Df becomes clear: it is the farthest away from the original frequency that the FM signal can be. Sometimes it is referred to as the "swing" in the frequency.

We can also define a modulation index for FM, analogous to AM:

b = Df/fm , where fm is the maximum modulating frequency used.

The simplest interpretation of the modulation index, b, is as a measure of the peak frequency deviation, Df. In other words, b represents a way to express the peak deviation frequency as a multiple of the maximum modulating frequency, fm, i.e. Df = b fm.

Example: suppose in FM radio that the audio signal to be transmitted ranges from 20 to 15,000 Hz (it does). If the FM system used a maximum modulating index, b, of 5.0, then the frequency would "swing" by a maximum of 5 x 15 kHz = 75 kHz above and below the carrier frequency.

Here is a simple FM signal:

Here, the carrier is at 30 Hz, and the modulating frequency is 5 Hz. The modulation index is about 3, making the peak frequency deviation about 15 Hz. That means the frequency will vary somewhere between 15 and 45 Hz. How fast the cycle is completed is a function of the modulating frequency. —Preceding unsigned comment added by 67.142.130.42 (talk) 18:26, 5 October 2007 (UTC)[reply]

There is no explanation given of how the transmitted signal is recovered. That is cheating the reader after promising "The rest of this article...concentrates on the FM modulation [sic] and demodulation process." A solution could be instead to link to descriptions of the established FM detectors (Ratio, Foster-Seeley, Slope and PLL) in the article on Detector (radio).Cuddlyable3 17:54, 9 February 2007 (UTC)[reply]

Common methods used in FM receiver are slope detection and phase-locked loops.

The above is not true if by "common" we understand the majority of domestic radios.Cuddlyable3 19:06, 12 February 2007 (UTC)[reply]

That's what Carlson mentions. Please update the article if you have a more up-to-date reference. Alinja 21:16, 14 February 2007 (UTC)[reply]

Please help me by identifying what you mean by "Carlson". Until I check that, I like the solution I proposed i.e. a link to Detector (radio). BTW I have the impression that Slope detection of FM is too unwieldy for common use and is only kept mentioned in textbooks.Cuddlyable3 15:30, 15 February 2007 (UTC)[reply]

By Carlson I mean 'A. Bruce Carlson: "Communication systems, 2nd edition", McGraw-Hill, Inc, 1981'. A link to a more complete article about detectors is of course ok, but we want some kind of mention of how FM can be detected here anyway. Alinja 19:28, 15 February 2007 (UTC)[reply]

Thank you Alinja for that book reference. However FM signals having no amplitude variation are rather incompatible with AM detectors. The phenomenon called slope detection must have been an accidental discovery that AM radios nevertheless register some sound from FM transmissions. Slope detection is inferior in distortion, noise rejection and adjacent signal rejection compared to the established FM detectors and seems to be kept in textbooks only for its educational value.Cuddlyable3 00:48, 18 February 2007 (UTC)[reply]

I have tidied up the references to FM detection, relegating the explanation of actual detector circuits to Detector (radio).84.210.139.189 18:26, 9 March 2007 (UTC)[reply]

The most common method used in commercial FM broadcast band radio receivers is the quadrature detector. See LM3189 datasheet for an IC that used to be the workhorse of FM detectors. Motorola AN189 gives some theory on how a quadrature detector works. The Foster-Seeley detector was common before ICs became common. — Preceding unsigned comment added by 96.240.175.232 (talk) 01:24, 20 December 2013 (UTC)[reply]

Could this article include the assigned frequencies for the different uses? Such as 118.00 – 136.975 for Aviation, 108.000-117.975 Navigation, 151.5125- 158.400 BRS. I think a side bar would be best served for this purpose. —The preceding unsigned comment was added by 70.41.64.103 (talk) 00:50, 15 February 2007 (UTC).[reply]

There should be a linking with Radio frequency and Frequency allocation pages to avoid duplication. Probably Wikipedia should only link to the most detailed frequency allocations with their national variations, found on the web. I would like it kept clear that FM is a modulation type, and not intrinsically any frequency range.Cuddlyable3 15:42, 15 February 2007 (UTC)[reply]

...although FM broadcasting is better allocated to higher frequencies than those historically allocated to AM broadcasting, because of issues of coverage and bandwidth.Cuddlyable3 21:29, 11 March 2007 (UTC)[reply]

Specifically, ${\displaystyle A\cos \left(2\pi \int _{0}^{t}\left[f_{c}+f_{\Delta }x_{m}(\tau )\right]\,d\tau \right)}$

makes it difficult to understand the underlying circuitry.

${\displaystyle \int _{0}^{t}\left[f_{c}+f_{\Delta }x_{m}(\tau )\right]\,d\tau =t\times f_{c}+\int _{0}^{t}f_{\Delta }x_{m}(\tau )\,d\tau }$

There's no electronic circuit that can indefinitely accumulate a non-zero average input as implied by this equation; while the equation may or may not be correct, it will inevitably be an indirect parallel to the physical model. Also, ${\displaystyle f_{\Delta }x_{m}\left(\tau \right)}$ is not clearly defined; the reader must infer it's nature from the integral, which may be incorrect.

${\displaystyle A\cos \left(2\pi \times f(t)\times t\right)}$ would be a simpler, more direct mathematical model. It has a constant amplitude with a varying frequency.

${\displaystyle f(t)=\left(f_{c}+k\times input(t)\right)}$ is a possible scenario; however, the article does not state (though it implies) that this the actual modulation.

The set of possible modulations is covered by ${\displaystyle f_{c}+m\left(input(t)\right)}$where${\displaystyle input\left(t\right)}$ is flat long-term.

note: I don't know how to get a proper multiplication dot; hence my multiplication looks like a cross-product; feel free to fix that if you know how.

It would be more helpful to state the exact modulation, or the exact circuit.

—Preceding unsigned comment added by 66.245.28.149 (talk • contribs) 09:06, 4 July 2007

The simpler mathematical model that you propose hides the continuous-phase nature of most frequency-modulation schemes. You're right that the existing model is not the easiest of equations to understand (in terms of intuition); however all the terms are clearly defined.

In most circumstances, it will be a direct representation of what's going on in the circuit. A VCO (or NCO) accumulates phase indefinitely; that's exactly what the equation shows. Oli Filth 11:33, 4 July 2007 (UTC)[reply]

Thank you for your help; I found one of the external links (www.fas.org) presented FM in a set of equations that cleared everything up for me personally. The confusion in the Wikipedia article is the result of the definition for ${\displaystyle f_{\Delta }}$ as "instantaneous maximum deviation": it's actually just a constant, but 'instantaneous implies that it is a variable in time.

The integral form might be mathematically correct, but it isn't physical. Its Fourier components go to infinity, which real radio transmitters can't do. For low modulation index, the f(t) form is close enough. The other terms get small really fast! Usual is an oscillator with a varicap diode in it, instead of a fixed capacitor. So simple to build! Gah4 (talk) 18:23, 27 November 2023 (UTC)[reply]

After Carson's rule, Bessel Functions and their role in calculating aspects of the sidebands created by FM needs to be added. I have added it, but it needs more explanation, and the table title needs repair as below.

Phillipbeynon 04:24, 1 August 2007 (UTC)[reply]

I don't think this table is a good (so-called) illustration. A graph would show better that:

• Sidebands shown blank in the table are not non-existent, they are merely below an arbitrary quantising limit. The sidebands produced by sinusoidal FM are an infinite series.
• A graph shows that particular sideband pairs are zero at particular modulation indices, which gives a common method of calibrating frequency modulators using a spectrum analyzer. Please sign your posts. Cuddlyable3 10:09, 30 July 2007 (UTC)[reply]
  • Valid point about the undefined values. The harmonics produced by changing frequency continue ad infinitum; however, the arbitrary limit defining what are significant sidebands is 2% using Carson's rule and 1% for Bessel table values. I found it important to include the table for reference, as I myself was looking for it when I first read this page. A spectrograph would illustrate this infinite series better.--Phillipbeynon 04:24, 1 August 2007 (UTC)[reply]
    • I agree that a spectrograph would help explain the sideband behaviour. I notice that your table does identify the special values of modulation index where the carrier disappears, something which can surprise folks. I have another thought: since no FM radio channel can pass ALL the infinite sideband harmonics, every practical FM radio channel distorts a sine signal. I think Carson's rule is intended to be used to design the overall half-power bandwidth of a radio channel. Can we quote the maximum signal distortion that results? Cuddlyable3 13:49, 1 August 2007 (UTC)[reply]

I think this table is valid only when the modulating signal is a sinusoid - isn't it so? This should be mentioned. If the table has some meaning for non-sinusoidal modulating signals (as some kind of rule of thumb) it should be explicitly said. Matteosistisette (talk) 13:31, 17 February 2010 (UTC)[reply]

Why is it that most sources claim that an FM receiver only demodulates the strongest signal available, and yet one will sometimes hear two stations simultaneously? Does this indicate something is wrong with the receiver or is this 'normal' if the signals' strengths are the same order of magnitude? I have a few times been able to enjoy a talk and music at the same time using my car radio (KXBL 99.5 FM and KAKS 99.5 FM).

-User: Nightvid

That is normal and does not indicate anything wrong with your receiver. Unlike AM signals, only a small increase in the received power from the stronger station is needed for it to "overpower" (make inaudible) the weaker station. Descriptions of this capture effect are sometimes exaggerated.Cuddlyable3 23:12, 7 October 2007 (UTC)[reply]

FM and PM are really the same thing, separated only by the mathematical viewpoint. Perhaps they should be merged into one article and then have FM treated as a special case? HatlessAtless (talk) 21:48, 23 April 2008 (UTC)[reply]

One might equally argue that PM is a special case of FM. The naming of the Wikipedia articles is influenced by the fact that AM and FM are worldwide broadcasting standards whereas PM outside narrow technical fields is relatively little known or used. Cuddlyable3 (talk) 06:01, 24 April 2008 (UTC)[reply]

There actually are practical differences between the two. Noise effects arising from demodulation is one difference. A second is the modulation constant: phase modulation wraps after 2 pi radians, but frequency modulation does not, so the bandwidth spreading isn't limited in FM.

Serrano24 (talk) 15:37, 20 June 2008 (UTC)[reply]

There should be a more intuitive way to explain these phenomenon. The derivations of the SNR's are horribly complicated. Does anyone have an easy way to explain this? Maybe we can even have a whole section on calculating SNR? —Preceding unsigned comment added by Daviddoria (talk • contribs) 22:14, 15 September 2008 (UTC)[reply]

I have removed this section:

Wideband noise reduction

Since ${\displaystyle B_{T}}$ increases with the deviation ratio and the SNR is proportional to the deviation ratio, we can reduce signal power and keep the SNR constant as long as we increase the bandwidth.

The above is not correct. Thermal noise power at a receiver increases proportionally to its bandwidth. The implication that one can always maintain a SNR while reducing signal power is wrong. Cuddlyable3 (talk) 09:49, 26 November 2008 (UTC)[reply]

This page has attracted several versions of animated diagrams that all attempt to illustrate the same thing. Animation "C" (see right) has recently been inserted without explanation for the change. We should now compare the available diagrams and gain a consensus on which one to use. (Please see some earlier discussion above.) Below is my assessment.

• Animation A

• Oppose It is a jerky animation. The waves are not sinusoidal. The modulating signal is not shown but is different for the two carriers. The FM wave is not continuous.—Preceding unsigned comment added by Cuddlyable3 (talk • contribs)
• Oppose for the same reasons. Mange01 (talk) 15:00, 5 November 2008 (UTC)[reply]
• Oppose. No time, no source signal.--Berserkerus (talk) 19:13, 21 November 2008 (UTC)[reply]

• Animation B

• Support Two editors have worked on this animation which was nominated for FP.—Preceding unsigned comment added by Cuddlyable3 (talk • contribs)

• Oppose It presumes that the reader understands the oscilloscope horizontal sync mechanism. It is hard to understand for many readers why the that there are two time scales, one fine scale on the horizontal axis, and one course scale illustrated by the animated waveform change. Mange01 (talk) 15:00, 5 November 2008 (UTC)[reply]

A, B and C all purport to show what might be seen on an oscilloscope. Certainly one reader labours under a misunderstanding that B has both a fine and a course[sic] time scale. There is only one horizontal time scale common to all 3 waveforms. Cuddlyable3 (talk) 22:15, 6 November 2008 (UTC)[reply]

• Oppose. FM wave have abnormal phase speed directions. --Berserkerus (talk) 19:13, 21 November 2008 (UTC)[reply]

Abnormal? Not really. You see a similar effect to the wagon wheels in cowboy films that seem to turn backwards. The equations are:

s = ${\displaystyle pi}$ frame / 16

audio = - sin( x/15 + s )

am = 1.25 cos( x / 3 ) ( 1 + .7 sin ( x/15 + s ))

fm = 1.25 cos ( p / 2 ),

p = p + x ( 1 + .5 sin ( x/15 + s ))

Cuddlyable3 (talk) 21:09, 25 January 2009 (UTC)[reply]

• Animation C

• Weak support DoubtfulSee below. While mathematically correct, this animation has the signal and carrier phase locked together; this is not true in normal broadcasting. The movement and the use of colour does not add any information to what could be a simple diagram. The file size is large at over half a megabyte. I appreciate that the file size is now reduced. I withdraw my weak support and agree with Dicklyon that animation adds nothing to this diagram. Showing unrealistically synchronised waves in an animation encourages the sort of misunderstanding seen above. I would weakly support a non-animated version of this diagram. Cuddlyable3 (talk) 21:09, 25 January 2009 (UTC) (updated)[reply]

• 84 KB now--Berserkerus (talk) 19:13, 21 November 2008 (UTC)[reply]

• Support The "sliding window" makes it easy to understand. Of course it could be made more realistic if the two signals were out of sync, but that is not important. An unmodulated carrer could be added as a forth signal, to illustrate to concept of carrier waveform. A version that shows the difference between AM-UC and AM-SC (both DSB) would also be interesting, as well as a version that shows the difference between FM and PM. Mange01 (talk) 15:00, 5 November 2008 (UTC)[reply]

That is a nice wish list for new diagrams. Are you offering to provide some? Cuddlyable3 (talk) 22:15, 6 November 2008 (UTC)[reply]

Hmmm, no answer yet. Perhaps answering is another thing that Mange01 dismisses as not important. Actually providing realistic information is important and is what Wikipedia is supposed to do.Cuddlyable3 (talk) 17:01, 17 February 2010 (UTC)s[reply]

Hm. "Another thing"? If I was convinced my suggestion were supported, and if the original source code was available on the picture description page, I might make an attempt. Mange01 (talk) 22:45, 18 February 2010 (UTC)[reply]

• Support. I am author:)--Berserkerus (talk) 19:13, 21 November 2008 (UTC)[reply]
• Oppose. When the modulating signal is phase locked to the carrier the FM transmitter will be unable to handle a continuous analog waveform! Since there is no sign of this unrealism being corrected in this animation I withdraw my "support" and "doubtful" votes. I apologise for this turnaround! The animation in the article should not have been changed from B which is correctly realistic. Cuddlyable3 (talk) 10:25, 12 May 2010 (UTC)[reply]

What do other editors think?—Preceding unsigned comment added by Cuddlyable3 (talk • contribs)

• Oppose all – none of these adds any clarity to what you can get from a static diagram. Dicklyon (talk) 05:10, 6 November 2008 (UTC)[reply]

Specifically, the diagram FM Message and Modulated Signals.svg.

Which I clicked on the diagram, it says that f_delta = 0.15. I tried my best to work out and plot the equation of the green curve on my computer, but somehow it didn't work out. On closer examination, if f_c is 3 as stated on the file page, then f_delta of 0.15 seems too small for its effect to be noticeable on the graph. Are we sure f_delta is 0.15?

Also, I tried using other values of f_delta like 0.5, 1.5 etc and I couldn't get the green curve on the diagram.

Now let's say I simply made a mistake somewhere in my workings and that's why I have been unable to get the green curve, but there's something I've seen that doesn't seem quite right about the curve. If we examine equation (1) as referred to in the article, y(t) = A_c*cos(2*pi*f_c*t + 2*pi*f_delta*∫x_m(τ)dτ, the integral ∫x_m(τ)dτ represents the signed area of the region bounded within the data signal curve. Now, let's assume that this area between τ=0 and τ=t is represented by an average value of x_m_ave(t), so that ∫x_m(τ)dτ between τ=0 and τ=t equals x_m_ave(t)*t. Going back to Equation (1), it can be rewritten as

y(t) = A_c*cos[2*pi*(f_c + x_m_ave(t)*f_delta)*t]

What the above equation shows is that at time t, the instantaneous deviation between the frequency of the transmitted signal and f_c is dependent on the instantaneous amplitude of the data signal (i.e. x_m(t)). That is, at time t, the larger the value of x_m(t), the higher the instantaneous frequency of the transmitted signal; the smaller the value of x_m(t), the smaller the frequency.

After having said all that, if we look back at the green curve in question, it is at odd with the above conclusion. Can someone please comment on this? 222.152.22.151 (talk) 14:34, 13 April 2010 (UTC)[reply]

This shows the equations and you can contact the artist here.Cuddlyable3 (talk) 16:27, 30 May 2010 (UTC)[reply]

The "green curve" is wrong by inspection! Unless there is significant audio filtering, the instantaneous frequency of the modulated signal must be the carrier freq. plus the modulation input times a freq. deviation, as in the animated figure. This is clearly not the case in the figure. (The equations given along with the SVG file have an erroneous integration - and no references.) Furthermore the "red curve" is misleading and unhelpful. Superimposing the carrier and the modulating waveform this way suggests that they are co-present at some point in the system, and they are not. I suggest deleting both red and green figures. Even if they were fixed, they would not add anything compared with the animation. --Albany45 (talk) 14:52, 18 September 2010 (UTC)[reply]

I took it out. If someone wants to redo it correctly and put it back, that might be OK. Dicklyon (talk) 16:14, 18 September 2010 (UTC)[reply]

FM is currently a redirect to this article. That said, an non-exhaustive survey suggests that majority of incoming links are really referring to (the related) FM broadcasting article.

In this context, User:Georgia guy and I have been having a conversation over at Talk:FM about where the FM redirect should point.

If folks have a moment to read the conversation so far, we'd love some additional thoughts from other involved Wikipedians. Thanks in advance! —mako๛ 05:44, 27 September 2011 (UTC)[reply]

"The rest of this article concentrates on the FM modulation and demodulation process, which is identical in stereo and monaural processes." – No, it doesn't. Does anyone know where this sentence is supposed to go; it is now almost at the end of the article... -- megA (talk) 21:55, 6 November 2011 (UTC)[reply]

It was originally in an early "Stereo FM" section, when it was inserted in 2003, here. I fixed it. Dicklyon (talk) 22:41, 6 November 2011 (UTC)[reply]

This article says "In analog applications, the difference between the instantaneous and the base frequency of the carrier is not directly proportional to the instantaneous value of the input-signal amplitude but it is proportional to frequency." where as article on FM Broadcasting says "In analog applications, the instantaneous frequency of the carrier is directly proportional to the instantaneous value of the input signal. This form of modulation is commonly used in the FM broadcast band." They are contradicting. Which is right? wikieinstein (talk) 13:03, 17 January 2013 (UTC)[reply]

Good catch, I think. I can't figure out what the Wikipedia statement is getting at. --Bob K (talk) 14:23, 17 January 2013 (UTC)[reply]

In simple FM, the frequency is strictly proportional to modulating voltage. FM broadcasting uses pre-emphasis, however to boost SNR by increasing modulation at high frequencies. Maybe that's what our writer had in mind, but it's a mistake to generalize it to all of FM.--Albany45 (talk) 22:20, 3 February 2013 (UTC)[reply]

The article claims that Armstrong invented wideband FM. I'm not sure why "wideband". His early patents mention a modulation index somewhat less than 1 (delta phase at highest modulating frequency) which describes narrowband FM. In any case, the distinction between the two flavors is just a matter of degree, it is not fundamental. For broadcast normally wide band FM is used because of the signal quality benefits, while radio communication uses narrowband FM for spectrum efficiency, but both can be traced back to Armstrong.

Incidentally, Armstrong was not the original inventor of FM, though he arguably is the one whose work became known and led to what we use today. Hanso Idzerda in the Netherlands, the builder of broadcast station PCGG in 1919, invented and patented FM transmission and described its benefits for transmitter simplicity and efficiency. What he did not do (but Armstrong did) is describe receiver technology that takes advantage of the properties of FM, and the "slope detection" that Idzerda's listeners presumably used does not achieve those benefits. Paul Koning (talk) 18:55, 20 December 2017 (UTC)[reply]

As part of another discussion, I started to look at what Armstrong and Idzerda did. I suspect that Idzerda might have thought about wideband if the technology (high frequency vacuum tubes) existed. As well as I know, Armstrong understood the advantages in noise reduction, and also that the spectrum was available and usable. I don't know how to argue fundamentalness. There is a complicated relationship between bandwidth and S/N of the demodulated signal. Gah4 (talk) 22:39, 16 July 2018 (UTC)[reply]

The section titled "Sinusoidal baseband signal" begins, "Mathematically, a baseband modulated signal may be approximated by a sinusoidal continuous wave signal with a frequency f_m". But what is a baseband modulated signal? Up till this point, a baseband signal is equated with a modulating signal. The "frequency f_m", as opposed to f_c, seems to confirm that we're talking about a baseband signal, i.e. the modulating signal, not the transmitted signal. Huttarl (talk) 15:24, 5 April 2018 (UTC)[reply]

The article says: contrast this with FM audio broadcasting, where the ratio is around 10,000. FM broadcasting, in the US at least, is around 100MHz. 10000:1 would be 10kHz. I believe without a stereo subcarrier, broadcast FM can go to 18KHz or so, as long as it is low enough at 19KHz not to trigger the stereo demodulator. With subcarriers such as SCA, it is up to 70KHz or more, so closer to 1400:1. Gah4 (talk) 22:45, 16 July 2018 (UTC)[reply]