Maximum rate Shanon Hartley

Question

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=𝑊log2(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?

Answer 1

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.

Answer 2

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?