If the bit rate of the digital communication system is 'R' Kbps & the modulation used is 32-QAM, what is the minimum bandwidth required for ISI free transmission?

My understanding & analysis so far:

  1. QAM (a digital-to-analog conversion technique) is a combination of ASK & PSK. The idea behind QAM is to alter two characteristics (amplitude & phase) instead of just one to improve the bit rate.

  2. From literature (Data Communications & Networking by Forouzan), the minimum bandwidth required for QAM is same as that of ASK/PSK.

  3. BW for Binary ASK is given by B = (1+d)S, where S is the signal rate.

  4. For 32-QAM bandwidth is

       B = S = (bit rate)/(bits per signal element) = R/5 KHz

    (for 32-QAM no. of symbols per signal element is log2(32) = 5)

  • \$\begingroup\$ This sounds like homework. Show us your work so far. \$\endgroup\$ – Dave Tweed Oct 31 '14 at 19:02
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    \$\begingroup\$ Sure this looks like a homework. Actually, I am solving this as an exercise & its an assignment to myself. I am glad you asked me about my work so far. I am appending my understanding/findings so far. Thanks for the comment. \$\endgroup\$ – avngr Oct 31 '14 at 19:35

QAM really has nothing to do with PSK at all. It's just two channels of ASK that are combined together using two subcarriers that have a fixed phase shift between them.

To prevent ISI in an ASK signal, you need to be able to transmit at a minimum the main "lobe" of the power spectrum with linear phase shift. The lobe is defined by the first nulls in the PSD, which occur at ± the symbol rate/2, for a total bandwidth that's equal to the symbol rate.

So, yes, you're on the right track: The bandwidth in Hz is equal to the symbol rate, which is equal to the bitrate divided by the number of bits per symbol, or 5.

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  • \$\begingroup\$ QAM can certainly be viewed as a variation of PSK. A QAM signal can be seen as a combination of two AM signals that are 90° out of phase, or as a single carrier that is both phase modulated and amplitude modulated. If you do the math both result in the same QAM signal. \$\endgroup\$ – JanKanis Jun 15 '18 at 10:37
  • \$\begingroup\$ @JanKanis: You should write that up as an answer that explains how it helps analyze the required bandwidth. Otherwise, what's your point? \$\endgroup\$ – Dave Tweed Jun 15 '18 at 10:45
  • \$\begingroup\$ It doesn't change the analysis. My point is that I disagree with your statement that QAM has nothing to do with PSK. BPSK and QPSK are effectively 2-QAM and 4-QAM. \$\endgroup\$ – JanKanis Jun 15 '18 at 10:50

Using raised-cosine filter for ISI free transmission, the bandwidth is given as

BW = (signal-rate)*(1 + roll-off-factor)/2

The bandwidth is minimum if roll-off factor is minimum, i.e, zero.

So, the minimum bandwidth required is (R/5)*(1+0)/2 = R/10.

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