This graph appears in the textbook "Channel Codes" by Ryan & Lin (fig 1.7, section 1.5). It shows the capacity of the additive memoryless gaussian channel, with arbitrary input (Shannon formula), and with binary input, both with hard and soft decoding.

All looks ok to me except for the vertical scaling-units. If we are to believe the vertical axis labelling, the Shannon capacity for $SNR=0bB$ [*], $C=0.5$ bits per channel use, agrees with Shannon formula $C=\frac{1}{2}\log_2(1+\frac{S}{N})$. It's also ok that the capacity of the binary-input channels tend to one as $SNR$ grows. But what doesn't look right is that the capacities reach (cross?) cero at finite $SNR$.

Are we using some logarithmic vertical scale here? In that case, the labels are all wrong, aren't they?

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[*] Leaving aside the slightly objetionable confusion between $S/N$ and $E_b/N_0$ (the latter should be used for a continous time channel, which is not the case here).


The plot is correct, apart from the sloppy/confusing label stating the capacity in terms of $\mathsf{SNR}$, whereas it is plotted versus $E_b/N_0$, which is a related by different quantity.

The curve labeled as $\frac{1}{2} \log_2(1+\mathsf{SNR})$ is actually the capacity $C$ (in bits per channel use), obtained by the implicit equation

$$ C = \frac{1}{2} \log_2\left(1+\frac{E_b}{N_0}2C\right). $$

You can easily see that the above equation corresponds to a positive $C$ only for $$E_b/N_0 > \log(2) \text{ } (-1.59 \text{dB}).$$

Check out the Wikipedia article on the connection between the $\mathsf{SNR}$ and $E_b/N_0$ quantities. This article (Sec. II) is also a good read on the topic.

  • $\begingroup$ Yeah, I didn't take seriously the Eb/No label - I guessed they meant S/N - because the figure is in the context of the discrete-time channel. I think this is indeed sloppy. $\endgroup$ – leonbloy Jun 16 '17 at 13:56
  • $\begingroup$ @leonbloy. Thanks for the accept. A continuous-time channel can always be converted to an equivalent, discrete-time one by projecting the continuous-time signals onto an appropriate basis (and this is indeed what modern digital transceivers actually do). The notions of $\mathsf{SNR}$ and $E_b/N_0$ are valid for both cases, I don't see why you consider the latter as valid only in the continuous-time case. The authors indeed mean to plot capacities (rates) versus $E_b/N_0$ (judging also by the corresponding figure discussion in the book). $\endgroup$ – Stelios Jun 16 '17 at 15:09
  • $\begingroup$ Yes, I know that a continuous-time (band limited) channel can be converted to a discrete-time. What I object is that the reverse mapping seems pointless here. If we are speaking of a strictly discrete time channel (even further, we are not speaking of time at all, only computing capacities per channel use, for a plain scalar $X,Y,Z$ values), it makes little sense of speak of $N_0$ (a spectral density). $\endgroup$ – leonbloy Jun 16 '17 at 15:23
  • $\begingroup$ @leonbloy Note that when you project a (continuous) waveform of white Gaussian noise of power spectral density $N_0/2$ onto an orthonormal basis, say, $\{\phi_1(t), \phi_2(t),\ldots,\phi_m(t)\}$ you get $m$ i.i.d. "noise" scalar random variables each distributed as Gaussian with variance ("power") $\sigma^2 = N_0/2$. Now, for the discrete-time communication model, we define the $\mathsf{SNR}$ as $\mathbb{E}(x_s^2)/\sigma^2 = 2\mathbb{E}(x_s^2)/N_0$, where $x_s$ is the useful signal sample ("symbol"). This is how the term $N_0$ appears in the discrete-time case. $\endgroup$ – Stelios Jun 16 '17 at 15:43

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