Talk:Quadrature amplitude modulation

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[edit] Two signals on one line?

How can you put two signals on a single signal line? I don't get it. How can you double the available data rate?

Thanks, --Abdull 22:03, 17 Jan 2005 (UTC)

This is possible because two independent properties of the carrier wave are being manipulated simultaneously by the transmitter: the amplitude, and the phase. (You could think of plain old amplitude modulation as being a subtype of QAM, where only the carrier's amplitude is varied while its phase is held constant.)
The receiver can also measure these two properties independently, yielding two continuously-variable quantities and, thus, two signals. In its simplest form (2QAM), this can be used either to multiplex two data streams into one transmission, or to send one data stream at twice the effective rate (i.e. two bits per baud).
(At least, this is how I understand it.)
Simon 18:30, 23 Jan 2005 (UTC)
It doesn't really double the maximum available data rate. It merely uses more of it. The maximum possible rate is determined by the Shannon-Hartley theorem, and modulation schemes try to use up as much of it as possible. (Apparently no scheme gets 100%, but some get close.) So QAM enables a higher data rate than a simple amplitude modulation of binary data, but isn't making something out of nothing. It's just using up more of the available theoretical bandwidth. (Right?) - Omegatron 19:42, Jan 23, 2005 (UTC)
Yep, that's right—although from the point of view of a terminal sending data over the line, the available data rate has increased. As you point out, there is a theoretical limit on how high this rate can get (before changes to the channel parameters are needed).
Simon 19:27, 24 Jan 2005 (UTC)
Yeah, that's probably what's confusing. It's the distinction between the maximum bandwidth available from the physical channel and the maximum bandwidth available from the transmission protocol. The transmission protocol can use up all of the channel, and the data being sent can use up all of the protocol. - Omegatron 21:03, Jan 24, 2005 (UTC)

[edit] Dialectic inconsistency

Hi. Yet another page I want to edit for English usage (my main activity on Wikipedia). I noticed the page uses the word analogue (Commonwealth English spelling) and two words (or forms of words) ending in -ize (American English spelling). The Manual of Style recommends (sensibly) that spelling and usage within an article should be uniform. But I'm not sure which we it should go. It seems that perhaps Commonwealth usage is now dominant, but looking at the page history, I see there were inconsistencies from the beginning. (The very first version has both analog and quantised; perhaps analog is creeping into Commonwealth?) I think at least some comments are necessary before unification is performed. Ddawson 01:10, 26 August 2005 (UTC)

I am the editor who rewrote the article into its current form. I speak British English, being English (and British!). However, I've adopted some Americanis(z)ations owing to the journals I write papers in insisting on American English. So I mix them up. Most of the inconsistencies will be my fault, and I do not mind to which version it is harmonized, but harmonized it certainly should be! Thanks for the vigilance. -Splash 01:25, 26 August 2005 (UTC)

Then I will go with Commonwealth, since it's dominant. Ddawson 03:03, 26 August 2005 (UTC)

-ize can also be used in British English. It's common in British academic publications. SpNeo 02:28, 29 October 2005 (UTC)
Yes, though that is largely (in my personal academic writing experience) because many of those publications are in American journals/conferences and the like. For example, the vast majority of electrical engineering journals are published by IEEE which prefers American English spelling. -Splashtalk 06:32, 29 October 2005 (UTC)

'ize is never wrong' in British English although 'ise' is sometimes wrong - so most sensible writers will always use the 'ize' form. Differential spelling is often useful where an English writer will use American spelling (for example) to distinguish technical or computing items from the 'standard' meanings - eg we can programme the writing of programs or a disk (the device) drive may be made up of a stack of discs (the shape) on a single spindle - SDB51 2006.02.02 @ 21.05 zulu

[edit] 64 QAM Constellation diagram

How can i construct a 64 QAM Constellation diagram?


    For the rectangular case, you would just use an 8x8 grid.  —Preceding unsigned comment added by 75.32.247.201 (talk) 05:24, 6 October 2007 (UTC)

[edit] Analogue QAM

Although analogue QAM is possible, this article focuses on digital QAM.

This means there's nowhere obvious on Wikipedia to add information about analogue QAM or QAM in general. Perhaps the bulk of this article can be moved to QAM (digital) and the remainder made more mixed? --Dtcdthingy 01:55, 1 June 2006 (UTC)

If you want to write about analogue QAM, just remove the 'disclaimer', add a new section heading somewhere and dive in. Unless you feel strongly that they should be in different articles. -Splashtalk 01:57, 1 June 2006 (UTC)
The problem is that most of the article has been written as if the page was called Digital QAM. You'd have to put all that under a great big Digital QAM subheading and/or rewrite lots of it to say it only applies to digital. It would be better if we just renamed it, even if that means the article here ends up relatively short. --Dtcdthingy 02:17, 1 June 2006 (UTC)

I've gone halfway and split the article into two big sections. The previous situation of completely dismissing analogue QAM was an embarrassment. --Dtcdthingy 12:42, 8 November 2006 (UTC)

[edit] Error in equation for BER of Gray coded M-ary QAM

There is a mathematical error on this page in the description of the BER of a M-ary QAM signal in AWGN, where M is even. The BER for a gray coded signal can be directly infered from the PAM signal in each quadrature phase, which are independent from each other.

Thus 64-QAM, which can be regarded as two independent 8-PAM streams, has a BER approximated by 1/3 of symbol error rate of a Gray coded 8-PAM symbol stream.


In general, BER=\frac{4}{k}\times \left( 1-\frac{1}{\sqrt{M}} \right) \times Q\left(\sqrt{\frac{E_b}{N_0}\times \frac{3k}{(M-1)}}\right)

Where k = log2M

Although it is true that both quadrature PAM symbols must be correct for the final QAM symbol to be entirely correct, there is no such dependancy between the bit errors occuring in each quadrature PAM channel of a QAM system!

--Crazzell 23:38, 13 September 2006 (UTC)

[edit] Appreciation

Just a word of appreciation for such a lucid presentation.

[edit] Use of first person

To be honest, I don't particularly care either way. But I'm not a fan of changing things for the sake of it (another example would be people who go around Americanising British spellings).

From the MOS:

"It is also acceptable to use “we” in mathematical derivations"

For a few counter-examples to the claim that technical articles don't use "we":

I could easily go on all night finding counter-examples... Oli Filth 23:23, 15 January 2007 (UTC)

[edit] Is only the amplitude modified in QAM?

Throughout the article, there are some unclear places whether in QAM only the amplitude of the both carriers is modulated, or both the amplitude and phase is modulated in both carriers. I believe that only the amplitude is modulated, not the phase of each carrier. In this case, I do not quite understand this sentence:

The complicating factor is that the points are no longer all the same amplitude and so the demodulator must now correctly detect both phase and amplitude, rather than just phase.

As said earlier, no change to the phase of either carrier is done. Why should a demodulator need to detect the phase? Paluchpeter 13:14, 20 July 2007 (UTC)

By altering the amplitudes of the I and Q components, the phase of the overall signal is altered. Oli Filth 13:25, 20 July 2007 (UTC)
Of course, but the phase of the combined signal is not relevant as I understand it. After receiving the combined signal and multiplying it with either sine or cosine signal and filtering the result with a low-pass filter, we will extract the corresponding I or Q component, without paying attention to the phase of the combined signal. Perhaps I am missing something... Paluchpeter 14:59, 20 July 2007 (UTC)
The phase is relevant. QAM demodulation requires a phase reference, i.e. it can only be demodulated coherently. Also, it is susceptible to phase noise. Neither of these apply to simple AM. Oli Filth 16:34, 20 July 2007 (UTC)
The phase is relevant but not because is required as phase reference for coherent demodulation, but because it actually carries information. And that is why it must be demodulated coherently, because only detecting the envelope would not result in anything (the two modulating signals are merged in the envelope and in the phase).
And regarding to simple AM, I don't understand what you say about they don't apply. Because it is possible to detect the "simple AM" (I guess AM-DSB/SSB-LC) with coherent detection rather than envelope detection, and the Rx should also be in sync with the carrier's frequency and is susceptible to phase and frequency errors as well.--Dhcpy 04:51, 1 December 2007 (UTC)


The previous comment is dead on. Regular AM can be demodulated with simple noncoherent envelope detection, but QAM requires coherent demodulation, i.e. a phase reference is required. Mathematically speaking, sine and cosine (the two orthogonal carriers that define the quadrature) are just phase-delayed (or equivalently, time-delayed) versions of each other. The receiver must somehow synchronize (typically via a phase reference) its own local oscillator so that it matches that of the transmitter. This is necessary because the signal undergoes an unknown time delay (e.g. phase delay) in the channel, and arrives at the receiver with an (at least initially) unknown phase offset (with respect to it's own local oscillator). Any non-zero phase offset (between the transmitter and receiver local oscillators) will rotate the constellation correspondingly. For instance, a rotation of 90 degrees corresponds to a misidentification (a swapping) of the sine and cosine components, which, in general, would yield a catastrophically high bit error probability, near 1/2. This requirement for phase synchronization is exactly the same as PSK, and for the same reason. In fact, the simplest way to view QAM is as a generalization of PSK, e.g. PSK without a constant envelope requirement. —Preceding unsigned comment added by 75.32.247.201 (talk) 06:28, 6 October 2007 (UTC)
Both of the carriers are only amplitude modulated, and no phase modulation is normally performed for the carriers. However, phase is still relevant at reception (although not the phase of the total signal). The receiver must synchronize to the carrier phase, beacuse the carriers can be only demodulated using coherent demodulation. Alinja 10:56, 8 October 2007 (UTC)


This is how I understand it:
If you take a look at the block diagram of the analogue QAM you'll see that modulating signal I(t) multiplies cosine and modulating signal Q(t) multiplies the sine. Those I and Q are two different signals from, lets say, two different voice channels.
At this point, the 2 resulting signals I(t)cos(wc.t) & Q(t)sin(wc.t) are, as you say, Amplitude Modulated signals, the phase of the carriers isn't changed.
But, when they are added, at that point they merge into a complex signal that varies amplitude AND phase. That can be easily shown by writing this complex signal in the "module and argument form" .
By writing the QAM signal that way, you'll see that the envelope (amplitude) varies and the phase varies as well.--Dhcpy 04:51, 1 December 2007 (UTC)


Amplitude modulating two carrier waves in quadrature is identical to amplitude and phase modulating a single carrier wave; the two are a trigonometric identity. -Dawn McGatney —Preceding unsigned comment added by McGatney (talk • contribs) 06:31, 17 January 2008 (UTC)

[edit] Time domain signal

A signal in the time domain demonstrating the transmitted signal and how its formed from its data carrying components would be really useful. ~RayLast «Talk!» 18:15, 4 April 2008 (UTC)