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Why is the allocated bandwith smaller with QAM-16 or 64 compared to e.g. QAM-4? I am not a professional so I'm looking for an intuitive explanation.

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NxN-QAM means Quadrature Amplitude Modulation and it is a modulation scheme where the transmitted signal is the "mix" of two quadrature carriers whose amplitude is digitally modulated independently so as to give N different possible amplitude levels per each carrier. Therefore 64-QAM is 8x8-QAM, for example.

The total bandwidth of such a signal is proportional to the baud rate \$\dfrac 1 T\$ where T is the symbol time, i.e. the time needed to transmit a symbol. Note that each simbol in NxN-QAM carries \$\log_2(NxN)\$ bits of information, therefore 64-QAM carries 6 bit of information per symbol, whereas 4-QAM carries only 2 bit per symbol and 16-QAM carries 4.

If you consider a constant information transmission rate, i.e. a constant bit rate, you can see that increasing the number of bits per symbol makes the symbol time increase, hence the required bandwidth decreases.

To be more explicit, imagine you have to transmit a message with a bit rate of 64kbit/s. If you use 4-QAM you can transmit 2 bit per symbol, so you need to transmit at 32kSymbols/s (32kBaud). If you use 64-QAM you can transmit 6 bit per symbol, hence your baud rate drops to ~10.6kSymbols/s (10.6kBaud). Since we said that bandwidth is proportional to baud rate you see how the required bandwidth dropped using 64-QAM for a constant bit rate.

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  • \$\begingroup\$ @ Lorenzo Donati What is the name of the formula which sasy that bandwidth is proportional to baud rate? Another question: in your example, the T equals to 1/32kSymbols, right? \$\endgroup\$ – Quirik Jul 5 '15 at 8:47
  • \$\begingroup\$ @Navi: refer to my "answer" below for this. Less work for you Lorenzo Donati! \$\endgroup\$ – Mister Mystère Jul 5 '15 at 12:06
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This is a complement to Lorenzo Donati rather than an answer

The bandwidth is the range of frequency beyond which a sinusoid is considered to be attenuated (conventionally, by -3dB or a factor of $1/\sqrt{2}$).

However, digital signals are not sinusoids. The Fourier transform shows that any digital signal (any signal for that matter) is made of a defined superposition of sinusoids, with defined amplitude=f(frequency) characteristics. At least instantaneously, because Fourier transforms strictly speaking apply only to periodic signals (i.e. clocks, not comms signals).

The very sharp edges of digital signals require high frequency sinusoids to exist; as they are low-pass-filtered, the corners will degrade into ringing, reducing the plateau representative of the binary value.

Meaning: for a digital signal of X Hz, you need the medium to have a bandwidth much greater than X Hz, "much greater" meaning at the point where the distortions at the corners of the signal are negligible for your application.

In your case, distortion means a lower quality signal, therefore either an increased error bit rate for the same datarate, or a decreased datarate for the same error bit rate. This phenomena is more important for QAM64 than it is for QAM16 for example because the levels to resolve are much smaller.

That's certainly what Lorenzo Donati meant by "bandwidth proportional to baudrate". Again, this depends on the requirements, but a gross guesstimate could be bandwidth >= 100*baudrate.

Edit: Here is an illustration of how the digital signals are constructed from sine waves. Imagine applying a low pass filter to these sine waves, and your will understand why the bandwidth cannot be simply equal to the baudrate.

enter image description here

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