Here, I want to simulate an M-ary [pulse-amplitude modulation] system. Anyhow, I chose my binary source to generate information at a rate of 1Mbps. The data "virtually" has a zero intersymbol interference (a.k.a zero ISI). If the channel of bandwidth 125 kHz. What is the minimum possible value of M? I found it like this:

M = 1 MBPS/125KHZ = 8 Bits

But the simulation fails to finish the work. But when I chose M=20, it works fine. Am I missing something here?


Thanks everyone, I think I found the answer: Since we have:

\$ B_T = R_B/[2* log_2(M)]\$ .........(1)


\$ R_B\$ = Bit Rate = 1 MBPS ==> \$ R_B/2 = 500*10^3\$ symbols per second

\$ B_T\$ = The transmission bandwidth= \$125*10^3\$ symbols per second

So, substituting the values in equation (1) will lead us to know that the minimum value of M is 16.

Thank you everyone

  • \$\begingroup\$ Your going to have to explain what you are doing and particularly define your terms a lot better than this. Unless someone here happens to be in the same class you are taking, you aren't likely to find someone that knows what all that stuff means. \$\endgroup\$ – Olin Lathrop Jan 4 '13 at 23:16
  • 1
    \$\begingroup\$ Hint: In a bandpass channel of bandwidth 125 kHZ with no ISI, you are essentially limited to 125,000 symbols per second depending on the details of your system. If M-ary PAM is in use, each symbol carries 8 bits of information (using the 125 kilosymbols/second figure), and so each symbol must be capable of taking on one of \$2^8 = 256\$ amplitudes to convey 8 bits. It is not at all surprising that choosing \$M=8\$ does not work. What is surprising is your claim that using \$M=20\$ gives you answers that agree with something since you say "it works fine." \$\endgroup\$ – Dilip Sarwate Jan 5 '13 at 3:12
  • \$\begingroup\$ @Dilip Sarwate: Why is it surprising? I am using Simulink to simulate my system. I designed my block diagram so that \$ log_2(M)\$ is always an integer. In other words, if I put M=20, the diagram will show me that \$ log_2(20) = 5\$. \$\endgroup\$ – James Mitch Jan 6 '13 at 19:28
  • \$\begingroup\$ @JamesMitch Please don't assert that \$\log_2(20) = 5\$ even if Simulink tells you so. \$2^5 = 32\$, not \$20\$. \$\endgroup\$ – Dilip Sarwate Jan 6 '13 at 22:13
  • \$\begingroup\$ Yeah, I know. I just randomly chose M to be 20 instead of 8. Then, the output were as I expected. When I gave M = 8, the software gave me an error. \$\endgroup\$ – James Mitch Jan 6 '13 at 22:23

This looks like homework, around here people are encouraged to learn from the experience of getting stuck.

You seem to be mixing up terms.

M = typically used to represent # of symbols and having zero ISI is a given, otherwise you've not chosen wisely. That is the point of constellation/trellis decoding system.

Your initial calculation should read 1 MBps = 125 KHz * 8 bits/symbol.

So the question is, why does a constellation of 20 symbols work when a constellation of 8 doesn't. Why isn't 20 a power of 2? Given that 20 fits into 5 bits what has happened to the other 12 symbols?

  • there are some hints in here...

But your question is also lacking, you're not explaining how you are simulating the system, what is meant by fails to "finish the work" etc.

You will need to clearer, more concise and reread/understand the basic theory before asking a more detailed question, with information and partial analysis. Please come back and update the question.

  • \$\begingroup\$ Sorry to tell you, this is not a homework. Anyway, thanks =) \$\endgroup\$ – James Mitch Jan 6 '13 at 19:31

The edited version of the question and the calculations shown therein indicate that what is being used is a \$M\$-ary quadrature amplitude modulation (\$M\$-QAM) system, not a \$M\$-ary pulse amplitude modulation (PAM) system. As per the specification, \$8\$ bits are being transmitted per symbol at a symbol rate of \$125,000\$ symbols per second for a data rate of \$1\$ Megabit per second. \$4\$ of the \$8\$ bits are carried on the in-phase carrier and the other \$4\$ on the quadrature carrier. Each carrier is modulated using \$16 = 2^4\$-ary pulse amplitude modulation; the over-all signal is a \$256\$-QAM signal carrying \$8\$ bits per QAM symbol (which, by the way, is generally regarded as a complex-valued symbol with amplitudes represented as \$x+jy\$ where \$x, y \in \{\pm 1, \pm 3, \pm 5, \pm 7, \pm 9, \pm 11, \pm 13, \pm 15\}\$ and \$j = \sqrt{-1}\$).


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