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If the transmitted information signal takes values from a modulation constellation, then is there a way to know what the channel capacity is?

  1. Is there a connection between the modulation kind and the channel capacity?

  2. Why and when do we do $\log_2(M)$ where $M$ is the number of modulation symbols? For example, if the modulation is 16 QAM, then what does $\log_2(16)$ tell us?

  3. In equalization, when pilot symbols are sent, we get better estimation quality. This is known as non-blind equalization. But in many research articles often it is mentioned that non-blind technique wastes a lot of bandwidth due to periodically sending the training symbols. What is the meaning of bandwidth wastage and how does it occur? Does capacity decreases with using training/ pilot symbols?

These specific information is extremely hard to extract from text books and other online tutorials. Hence I have posted these problems here, hoping to get an intuitive answer. Shall be really obliged.

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1) Is there a connection between the modulation kind and the channel capacity?

The capacity of a channel indicates the upper limit of how many bits can be transmitted per second over the channel with no errors (okay, technically it is "arbitrarily low number of errors", but it's basically the same thing). We do various things to try to get as close to that upper limit as we can, such as modulation types, error correcting codes, etc., but none of them affect the channel capacity itself.

The equation for channel capacity is:

$$ C=B*\log_2(1+\frac{S}{N}) $$ where $B$ is the channel bandwidth in Hz, and $\frac{S}{N}$ is the signal to noise ratio (which is unitless). Given this, the channel capacity only depends on its bandwidth and the received signal-to-noise ratio.

Going to higher order modulation types (e.g. from QPSK to 16-QAM, from 16-QAM to 256-QAM, etc.) is one way to take advantage of the $\frac{S}{N}$ portion of the equation. If you have a very high signal-to-noise ratio it is wasteful to transmit QPSK, because the constellation points are very far apart from each other. By going to 16-QAM the points are closer together, but they are still far enough apart to avoid most errors while transmitting twice as many bits. If you still have "excessive" signal-to-noise you could go to a higher order QAM signal yet, allowing more bits per symbol to be transmitted.

Why and when do we do $\log_2(M)$ where $M$ is the number of modulation symbols? For example, if the modulation is 16 QAM, then what does $\log_2(16)$ tell us?

$M$ is not the number of modulation symbols, it is the number of constellation points. What $\log_2(M)$ tells us is the number of bits that each symbol represents. There are 16 constellation points in 16-QAM, so we could number them from 0 to 15. How many bits does it take to represent 0 to 15? 4 bits, or $\log_2(16)$.

3) In equalization, when pilot symbols are send, we get better estimation quality. This is known as non-blind equalization. But in many research articles often it is mentioned that non-blind technique wastes a lot of bandwidth due to periodically sending the training symbols. What is the meaning of bandwidth wastage and how does it occur?

What they mean is that some of the bits that could have been used to transmit information are used to transmit pilot sequences instead. In LTE, for instance, in some OFDM symbols every third subcarrier is a pilot instead of data. If each subcarrier is using QPSK modulation, then each pilot subcarrier means that there are two fewer data bits.

Does capacity decreases with using training/ pilot symbols?

No, not necessarily. Yes, the number of information bits is decreased by the pilot symbols as described above, but remember that the channel capacity refers to how many error-free bits can be transmitted over the channel. The pilots are essential for error-free communication.

A similar example is error correction codes (e.g. Turbo codes). Error correction codes use parity bits or symbols to detect and correct errors. We could replace those parity bits with information bits, but that would actually reduce the number of error-free bits that we sent through, so using the error correction code is a net gain.

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  • $\begingroup$ 1) Not necessarily. There are two ways to have a high signal-to-noise ratio: have low noise, or transmit a lot of power. Your paranthetical is correct, though- high SNR implies that the noise is lower than the transmitted signal (actually, it implies that it is higher than the received signal, but close enough). $\endgroup$ – Jim Clay Dec 12 '17 at 2:12
  • $\begingroup$ 2) All else equal, yes, more channel noise means less channel capacity. 3) No, Shannon's entropy does not tell us what the capacity is. The information theory concept that is related to channel capacity is mutual information. If $X$ is the transmitted signal and $Y$ is the received signal, the mutual information between the two is due to the channel. $\endgroup$ – Jim Clay Dec 12 '17 at 2:15
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1) Is there a connection between the modulation kind and the channel capacity?

Channel capacity is usually defined as the number of information (usually measured by the number of bits) can be sent per channel use to get arbitrarily low number of errors, but we don't know exactly how (random coding).

A channel use can be thought as a modulation symbol that symbol rate depends on modulation bandwidth. Thus AWGN channel capacity $C = \log_2(1+\mathrm{SNR})$ bits/channel-use or $C = B\log_2(1+\mathrm{SNR})$ bits/second.

A given modulation scheme (OFDM, DSSS, single carrier, etc.), under a given channel model, has its own $\mathrm{SNR}$ because of e.g. noise, interference, etc. However if we consider perfect modulation/demodulation, typical modulation schemes should give the same capacity.

2) Why and when do we do $\log_2(M)$ where $M$ is the number of modulation symbols?

Jim Clay has given an excellent answer.

3) In equalization, when pilot symbols are sent, we get better estimation quality. This is known as non-blind equalization. But in many research articles often it is mentioned that non-blind technique wastes a lot of bandwidth due to periodically sending the training symbols. What is the meaning of bandwidth wastage and how does it occur? Does capacity decreases with using training/ pilot symbols?

Pilots or training symbols can be thought as a particular kind of channel coding. Thus the bit rate one can send with pilots is smaller than the maximum bit rate (that we, again, don't know how to achieve yet) which is defined as channel capacity.

But pilots and channel estimation are essential for practical system in term of implementation; otherwise, either the complexity could be exponentially greater or the implementation is impossible because the channel code is unknown. Thus the fact whether pilots are wasteful depends on the context.

The implementation configuration of LTE, i.e. OFDM+pilots+turbo code, is a way to achieve the theoretical channel capacity. And no, channel capacity does not depends on the usage of pilots, but the practical number of bits does.

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  • $\begingroup$ Thank you for giving more explanation. When we say about channel capacity and its formula, then are we referring to digital channel or is the channel analog? $\endgroup$ – Ria George Dec 13 '17 at 0:42
  • $\begingroup$ I think you want to refer to discrete time and continuous time channel, right ? The answer would be both. Look at the unit of channel capacity, it is bits/channel-use and it is naturally related to discrete time channel. A continuous time channel can be converted to equivalent discrete time channel. $\endgroup$ – AlexTP Dec 13 '17 at 10:42
  • $\begingroup$ Thank you for your reply. Just to confirm if I understand your last comment clearly --- (1) whenever the data transmitted is discrete-in time and takes values as symbols say binary, then the channel is discrete? (2) Also, a continuous time channel can be converted to discrete time channel by sampling? I have posted a new question dsp.stackexchange.com/questions/45783/…, It would be very helpful if you can answer it since I really like all of your simple way of explanations with examples. Thank you once again. $\endgroup$ – Ria George Dec 13 '17 at 17:52

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