i am reading computer networks and i am trying to solve an excercise. the excercise is

What is the theoretically the maximum rate at which data can be transmitted to a 4000Hz bandwidth communication channel having a noise of 30dB if the number of discrete signal states is 6?

I thougth that i must use the sannon hartley because the channel having noise. The the formula is 𝑅=π‘Šlog2(S/N+1). Then from the type 𝑆𝑁𝑅=10log10(𝑆/𝑁) i calculate the 𝑆/𝑁. But i cant understand

if the number of discrete signal states is 6

i must modify the formula or something else? What should i do?

  • 1
    \$\begingroup\$ I assume "having a noise of 30 dB" means that you have 1000× the signal power as noise variance? Is that correct? Because an SNR of -30 dB is defnitely a bad channel... \$\endgroup\$ Commented Dec 2, 2018 at 12:03
  • \$\begingroup\$ "number of discrete signal states": at the transmitter, or for the receiver (including noise in the signal state)? \$\endgroup\$ Commented Dec 2, 2018 at 12:05
  • \$\begingroup\$ this is the excercise i haven't any more informations about this ,and it is a little bit general. \$\endgroup\$
    – dimath
    Commented Dec 2, 2018 at 12:12
  • \$\begingroup\$ Normally, the SNR determines the number of signalling states that you can use, which is why the formula is based on that. But if the number of states is fixed, that makes the data bandwidth a direct function of the analog bandwidth. Unless you want to talk about systems that allow a certain amount of ISI, which lets you push the baud rate higher than you normally would. \$\endgroup\$
    – Dave Tweed
    Commented Dec 2, 2018 at 13:35
  • 1
    \$\begingroup\$ @DaveTweed exactly. Many commercial GPS receivers work perfectly fine with 4- or even 2-bit ADCs, simply because your channel state doesn't contain that much info about the original signal to begin with... \$\endgroup\$ Commented Dec 2, 2018 at 14:50

2 Answers 2


I am going to assume that the actual question is asking about the symbol rate, not the bit rate. Otherwise, the fact that the signal has 6 discrete states is irrelevant.

If the symbol being transmitted has 6 possible states, then each symbol conveys log2(6) = 2.58 bits of information. This is the key part you are missing.

I think the 30dB is supposed to be the channel signal to noise ratio (SNR). That means that signal power is 1000x noise power (10 ^(30/10)=1000).

Based on Shannon Hartley theorem, C = B*log2(1 + S/N) Where C is the channel capacity (maximum error-free information transmission rate) in bits/sec, and B is the channel bandwidth in Hz. S is the signal power, and N is the noise power.

So, C = 4000 * log2(1 + 1000) = 39869 bit/sec.

Since each symbol has 2.58 bits of information, the maximum symbol rate will be:

(39869 bits/sec) / (2.58 bits/symbol) = 15423 symbols / second.

So the maximum rate that data can be transmitted at is 15423 symbols/sec.

Obviously I have made a lot of assumptions. But the reason I made assumptions is because I am trying to fit the question into something that makes sense.

  • \$\begingroup\$ xmm i think bit rate ,not the symbol rate! How i can calculate the max bit rate of that channel? \$\endgroup\$
    – dimath
    Commented Dec 3, 2018 at 19:47
  • \$\begingroup\$ Shannon theorem. It is right there in my answer. 39869 bit/sec. But in that case, then the number of discrete states of the signal does not matter. Also, please note that Shannon Theorem assumes you use a modulation scheme that makes good use of the available bandwidth. \$\endgroup\$
    – user57037
    Commented Dec 3, 2018 at 23:32
  • \$\begingroup\$ thank you very much i believe thiw is the correctio solution for this excersice \$\endgroup\$
    – dimath
    Commented Dec 8, 2018 at 12:26

With binary states, you can make decisions with threshold about the center of a Noise Distribution.

With 6 states, assuming equal probability of each of the states, where do the 4? 5? thresholds go?

For simplicity, assume 6_phase modulation (like 4-phase-mod but with 6, thus 360/6 = 60 degrees apart. With thresholds at the 30 degree intervals.

Now you can assume a datapoint in vector space, and compute the RMS noise power needed to move the head of the vector by +- 30 degrees left-ish or right-is and thus move past the thresholds into the phase region allocated to the 2 adjacent phase symbols.

By the way, is the SNR not +30dB?

  • \$\begingroup\$ yes snr is 30db! \$\endgroup\$
    – dimath
    Commented Dec 3, 2018 at 19:36

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