Are capacity and spectral efficiency identical for a dicrete-time digital signal?

Question

I saw this answer and saw on the wikipedia page that both spectral efficiency and capacity can be measured in bpcu i.e. 'bits/symbol'. I just want to confirm then that they are identical for a discrete-time digital signal.

This diagram is plotting $\log_2(1+SNR)$ but referring to it as 'capacity', which confirms my suspicion because this should typically be spectral efficiency before being multiplied with bandwidth to get capacity. I consistently see diagrams showing 'spectral efficiencies' and 'capacities' showing this range of values.

I do not know where the $1/2$ factor came from in the answer I linked, or why they've used variance of the discrete noise in the SNR, but it appears that capacity and spectral efficiency are the same for a digital signal.

My guess would be that 'channel use' means per sample. Imagine an OFDM with 64 subcarriers using 16QAM and pretend none of the carriers are used for pilots or guard bands. The total number of bits per symbol will be 4*64 i.e. 256, but because there are 64 samples in a symbol and hence 64 frequencies used, if you divide that by 64 you get 4. Therefore spectral efficiency = channel use in this scenario?

If 48 of the 64 subcarriers are used to transmit data and the rest are pilots and guard bands, then I'm guessing that the spectral efficiency changes to 192/64 and changes to 192/80 if you include the cyclic prefix i.e. 2.4 bpcu before taking into account error coding of 3/4 where it becomes 1.8bpcu. And modulation efficiency is 4 bpcu now? i.e. 192/48.

Is this because there is no concept of a capacity as distinct from spectral efficiency in a discrete-time signal?

If you calculate the spectral efficiency of the OFDM signal using the bps/Hz method, you get the same result i.e. $\log_2\left(1+\frac{S}{N}\right)$, but the capacity is instead $B\log_2\left(1+\frac{S}{N}\right)$ i.e. 20MHz x 4 bits = 80Mbps. You can also calculate this capacity if you know the capacity in bpcu and know the time interval of the channel use, then you do 4*sample freuquency = 4 x 20MHz = 80Mbps

Answer 1

The capacity is a property of a channel. A channel can transport no more than a certain amount of information per channel use. When we use a fixed modulation and coding scheme ("discrete-time signal"), this scheme will be able to support a certain amount of information per channel use (which for the channel to support it must be $\leq$ its capacity). It is thus strictly speaking incorrect to speak of the capacity of a certain modulation and coding scheme.

For a given modulation and coding scheme such as a QAM, you can of course compute its achievable data rate in bits per channel use (i.e., per unit bandwidth). You can also interpret this as a spectral efficiency, since it measures how efficient the scheme uses bandwidth (how many Mbit/s do I get for each MHz of bandwidth I invest?). And the results you get may be reminiscent of the capacity since under certain conditions, the schemes can get somewhat close to it.

It is still not the same. Capacity is a property of a channel, spectral efficiency is a property of a transmission scheme (modulation and coding scheme).

Answer 2

I do not know where the 1/2 factor came from in the answer I linked, or why they've used variance of the discrete noise in the SNR, but it appears that capacity and spectral efficiency are the same for a digital signal. No, they are not identical.

The $1/2$ comes from the fact that we do not consider complex signal. Specifically, a given channel with bandwidth $B$ can be used to transmit up to $2B$ samples in AWGN channels. Differently speaking, the degree of freedom of a continuous AWGN channel having bandwidth $B$ is $2B$.

Is this because there is no concept of a capacity as distinct from spectral efficiency in a discrete-time signal?

No. Loosely speaking,

• Capacity: we look at a channel (which is characterized by its bandwidth, noise model, noise power and other impairments) denoted $H$, capacity is the supremum of data rate that one can send data with arbitrarily small error probability over that very channel. The transmission schemes to achieve that rate supremum are denoted $S$, which are often unknown.
• Spectral efficiency: always the channel $H$, but this time we fix a transmission scheme $s$, then the efficiency is data rate divided by bandwidth. Here, we "forget" the error probability. In your cited figure, the QAM curves are not spectral efficiency by normal definition, but rather spectral efficiency multiplied by correct decoding probability. Some will say that this metric makes more senses

Therefore, if the error probability of $s$ is arbitrarily small, spectral efficiency coincides with channel capacity. Otherwise, it is safe to claim that channel capacity is the upper-bound of spectral efficiency.

Answer 3

With the help of comments and the other answer, you will certainly resolve your terminological confusion about spectral efficiency vs. capacity. In my answer, I address the "1/2 factor and variance" issues of your question.

Modern textbooks derive Shannon's formula for the maximum channel capacity of a continuous-time channel from the channel capacity of a discrete-time Gaussian channel.

Wikipedia's articles on 'AWGN', 'Channel capacity', 'Mutual information', and 'Differential entropy' explain the 1/2 factor issue in full. To keep the narrative uninterrupted, I concatenate the pieces from these articles and add some comments.

The Gaussian channel is a discrete-time channel with an AWGN noise model and constrained power. X is input, Z is noise, and Y is output: $$ Y_i = X_i + Z_i $$ Z_i is i.i.d. random variables drawn from a zero-mean normal distribution with variance N₀ W/Hz; Z_i are uncorrelated with X_i. For a codeword (x₁, ... x_k) transmitted through the channel, the power constraint inequality is $$ {1\over{k}}\sum\limits_{j=1}^k {x_i}^2 \leq P $$ The channel capacity for the power-constrained channel is, quite understandable (at least seemingly plausible), the maximum of a mutual information taken over all possible realizations of noise transformed by the channel.

For the rigorous definition of emphasized words, consult the references. You may want to delve into the depths of probability theory in order to understand the concept of mutual information in full, but for the first reading it is sufficient to grasp the skeleton of derivation and proofs.

Following the definition, the discrete-time Gaussian channel capacity is the supremum of a mutual information over all possible choices of the marginal distribution p_X(x): $$ C = \sup\limits_{p_X(x)}I(X;Y) $$ Mutual information is expressed as I(X;Y) = H(Y) - H(Y|X), H denotes the differential entropy. For Y = X + Z, we re-write H(Y|X) as H(X + Z|X) = H(Z|X), and, because X and Z are independent, H(Z|X) = H(Z).

For a given variance, differential entropy is maximized with a normal distribution of the signal sample values -- any other distribution gives lesser differential entropy. This fact is important for our derivation, but it also results in that the most effective codes are those that transform signals to coded data similar to white noise as close as possible.

With this upper bound on H(Y), we express H(Y) and H(Z) via the differential entropy for the normal distribution, which we take from Table of differential entropies: $$ {\rm H}(Z) = {\rm ln}(σ\sqrt{2πe}){\rm [nats]} = {1\over2}(1+log(2σ^2 π)){\rm [bits]} \\ {\rm H}(Y) = {\rm ln}(ν\sqrt{2πe}){\rm [nats]} = {1\over2}(1+log(2ν^2 π)){\rm [bits]} $$ where the noise variance is σ² = N₀, the output signal variance is ν² = (P + N₀·W)/W, W is the channel bandwidth. $$ {\rm I}(X;Y) = {\rm H}(Y) - {\rm H}(Z) = \\ {1\over{2}}(1+log(2ν^2 π)) - {1\over{2}}(1+log(2σ^2 π)) = \\ {1\over{2}}log({(P + N)\over{N}}) = {1\over{2}}log(1 + {P\over{N}}) $$ The 1/2 factor originates in the differential entropy expression for normal distribution and subsequently enters into the formula for the channel capacity of a discrete-time Gaussian channel. In this derivation, you can also see the role of a variance variable.

Because the sampling (Nyquist) frequency is two times the channel bandwidth W, we can represent the continuous channel as a discrete time Gaussian channel transmitting 2W times a second. The channel capacity becomes $$ C_{continuous} = {1\over{2}}log(1 + {P\over{N}})·2W = W·log(1 + {P\over{N}}) $$

This material is a prerequisite for understanding Eb/N0 and the "gross" link spectral efficiency. Do not let advanced math to discourage you: for the first steps, you can take the formulas for granted, only make sure your reference materials and textbooks are trustworthy. Very important is to compute: write the computation programs in a computer language of your choice, run the simulations, and compare the results with the data taken from measurements or from published reports and papers.

Answer 4

In the end I came to the conclusion that the maximum spectral efficiency a channel allows with its SNR is $\log_2(1+\frac S N)$. This is a property of an analogue channel with an SNR and is measured in $bps/Hz$. It can be modelled in the digital domain with the same value but units $bpcu$.

A digital signal modulation scheme has a spectral efficiency $\log_2(M)$, where $M$ is the number of discrete symbols (modulation order), and is measured in $bpcu$, but when converting this to the analogue domain, the spectral efficiency becomes $\frac F B \log_2(M)~~~(bps/Hz)$, where $F$ is the baud rate.

The capacity of the complex digital domain channel is $\log_2(1+\frac S N)~~~(bpcu)$ because there's no concept of bandwidth. The capacity of the complex analogue channel is $B\log_2(1+\frac S N)~~~(bps)$, where $B = 2W$. In both cases, the real version prepends a 1/2 factor.