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In a first course in Information Theory, when the operational interpretation of channel capacity is introduced, it is said to be the highest data rate (in bits/channel-use) of reliable communication. While reading a few papers, I came across channel capacity being expressed in units of bits/s/Hz. So I was thinking about the connection between the two units and came up with the following explanation. Please let me know if this is wrong.

For a bandlimited channel (bandwidth = $W$ Hz), you can transmit at $2W$ symbols/sec by the Nyquist sampling theorem. So the rate "per bandwidth" (spectral efficiency) can be written as 2 symbols/sec/Hz. If each symbol is 1 bit, then you are transmitting 1 bit in each of the samples. So is 1 bit/channel-use equivalent to 2 bits/sec/Hz?

What is one "channel-use"?

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    $\begingroup$ You are talking of capacity of two different types of channel. In one case, the channel inputs and outputs are discrete in time,and so bits per channel use is the natural metric. If units are attached to the discrete time instants (e.g. one use per microsecond), then one can use bits per second too. In the second case, the inputs and outputs are continuous-time signals which occupy bandwidth and so the natural measure is bits per second per Hertz. $\endgroup$
    – Dilip Sarwate
    Feb 21, 2012 at 4:15
  • $\begingroup$ Thanks! So as an example, for the AWGN channel with power constraint but no bandwidth constraint, it makes sense to talk about capacity in terms of bits/channel-use since we could in principle transmit as rapidly as desired (or as you said, in bits/sec if we know the rate of transmission). But for the bandlimited case, the formula for capacity in bits/sec can be restated in units of bits/sec/Hz (normalizing by the bandwidth). $\endgroup$
    – rk2
    Feb 21, 2012 at 9:23
  • $\begingroup$ You might want to look at Prof. Pramod Viswanath's lecture notes here. $\endgroup$
    – Dilip Sarwate
    Feb 21, 2012 at 15:57
  • $\begingroup$ @Dilip: I like your comment; I'd convert it to an answer. $\endgroup$
    – Jason R
    Feb 22, 2012 at 3:31
  • $\begingroup$ @Jason R OK, done! I expanded the material slightly $\endgroup$
    – Dilip Sarwate
    Feb 22, 2012 at 21:47

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You are talking of capacity of two different types of channel.

In one case, the channel inputs and outputs are discrete in time. At the $i$-th time instant, the received signal is $X_i+N_i$ where $X_i$ is the received symbol of average energy $E$ and $N_i$ is the noise (typically modeled as a sequence of i.i.d. $\mathcal N(0,\sigma^2)$ random variables). The channel capacity of this discrete-time Gaussian channel $~$is $$C = \frac{1}{2}\log_2\left ( 1 + \frac{E}{\sigma^2}\right) ~\text{bits per channel use}$$ and so bits per channel use is the natural metric. If we are told how far apart in time the discrete time instants are, e.g. one channel use per microsecond, then a capacity $C$ bits per channel use can be stated as bits per second, e.g. $C$ Mbps for our one microsecond example.

In the second case, the inputs and outputs are continuous-time signals which occupy bandwidth and so the natural measure is bits per second per Hertz. There are more complications involved in making the transition from the continuous-time channel to the discrete model, and in connecting the bandwidth $W$, the received signal $P$ and the noise spectral density $N_0$ to $E$ and $\sigma^2$ (see here for some details), but when all this is done we get Shannon's celebrated formula $$C = W\cdot \log_2\left ( 1 + \frac{P}{N_0W}\right) ~ \text{bits per second}$$ for the capacity of the additive white Gaussian noise (AWGN) channel of bandwidth $W$. This capacity can also be expressed as $C/W$ bits per second per Hertz.

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  • $\begingroup$ Hi, the link in your answer seems to point to a 404 now -- will it be possible for you to update it? $\endgroup$
    – Avijit
    Jan 31, 2018 at 12:24
  • $\begingroup$ @Avijit ${}{}{}$ Done! $\endgroup$
    – Dilip Sarwate
    Jan 31, 2018 at 16:14

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