# About channel capacity and sub-channels

The Shannon-Hartley theorem describes the maximum rate at which information can be sent over a bandwidth-limited AWGN channel. This rate is called the channel capacity. If $$B$$ is the bandwidth of the channel in Hz, $$S$$ is the signal power (in units of W), and $$N_0$$ is the noise power spectral density (in units of W/Hz), then the channel capacity $$C$$ in units of bits per second is$C = B \log_2\left(1 + \frac{S}{N_0B}\right).$

Let us now consider that we make $$n$$ “sub-channels”, by selecting $$n$$ disjoint bandwidth intervals contained in the total bandwidth of the channel. We denote the bandwidth of these sub-channels by $$B_j$$, $$j = 1,\ldots,n$$. Clearly, we have the constraint $$\sum_{j=1}^n B_j \leq B$$. Likewise, we divide our total transmit power $$S$$ into the $$n$$ sub-channels, allocating power $$S_j$$ to the signal in the sub-channel $$j$$. We have $$\sum_{j=1}^n S_j = S$$. Under these conditions, each sub-channel will have capacity $$C_j$$, given by the formula above with $$B_j$$ and $$S_j$$ in place of $$B$$ and $$S$$.

The natural question regards using the $$n$$ sub-channels in parallel to transmit data: what is the maximum of the sum $$\sum_{j=1}^n C_j$$ under these conditions and how can it be achieved? It is probably clear from the definition of channel capacity that this sum is always smaller or equal than $$C$$. After all, by dividing the channel into sub-channels we cannot do any better than by considering it as a whole.

People used to communications theory might find intuitive that we can achieve $$\sum_{j=1}^n C_j = C$$, and that this happens if and only if we use all the bandwidth ($$\sum_{j=1}^n B_j = B$$) and the SNRs of the sub-channels, defined by $$S_j/(N_0B_j)$$, are all equal, so that $$S_j = SB_j/B$$. After all, this is pretty much how OFDM and other channelized communication methods work. In this post I give an easy proof of this result.

First of all, since the expression $$B_j \log_2(1+S_j/(N_0B_j))$$ is increasing in $$B_j$$, if $$\sum_{j=1}^nB_j < B$$, then we can increase some of the values of the $$B_j$$ until we have the equality, increasing as a consequence the sum $$\sum_{j=1}^n C_j$$. Therefore, we can restrict ourselves to the case when $$\sum_{j=1}^nB_j = B$$.

Now we consider that $$B_j$$ are fixed and apply Langrange multipliers to find the extrema of the function$f(S_1,\ldots,S_n) = \sum_{j=1}^n B_j \log\left(1 + \frac{S_j}{N_0B_j}\right)$subject to the condition $$\sum_{j=1}^n S_j = S$$. This gives the system of equations$B_j\left(1+\frac{S_j}{N_0B_j}\right)^{-1}\frac{1}{N_0B_j} + \lambda = 0,$where $$\lambda$$ is the Langrange multiplier. These equations can be rearranged as$\frac{S_j}{B_j} = -N_0 – \frac{1}{\lambda}.$ It follows that the quotients $$S_j/B_j$$ are all equal.

Now we write $$S_j/B_j = K$$. We have$S = \sum_{j=1}^n S_j = \sum_{j=1}^n K B_j = KB.$Therefore, $$S_j/B_j = S/B$$. This implies$\begin{split}\sum_{j=1}^n C_j &= \sum_{j=1} B_j \log_2\left(1 + \frac{S_j}{N_0B_j}\right) = \sum_{j=1} B_j \log_2\left(1 + \frac{S}{N_0B}\right) \\&= B \log_2\left(1 + \frac{S}{N_0B}\right) = C.\end{split}$This is what we wanted to show.

The attentive reader will perhaps have noticed that we also have the restrictions $$S_j \geq 0$$, so we need to be a bit careful when applying Lagrange multipliers. However, this is not a problem. If the maximum was found at a point in which some of the $$S_j$$ are zero, we can remove these variables (their corresponding capacities $$C_j$$ are zero) and consider a problem with a smaller $$n$$ in which all the $$S_j$$ are strictly greater than zero.

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