## Determining Channel Capacity

For a particular Trellis diagram we define a connection matrix **D**. Hence for *k* states the size of the matrix is *k* × *k*. **D** is called the connection matrix because its entries *d _{ij}* = number of Trellis paths from

*S*to

_{i}*S*. For example

_{j}Notice that elements in a particular

*i*

^{th}row represent the number of paths from

*S*.

_{i}The connection matrix

**D**is therefore a transition matrix from the present state to next state. Therefore after a certain time

*t*and hence

*t*steps, for a particular state

*S*can we compute the possible number of paths (and hence possible number of sequences) that can reach the same

_{i}*S*state? The possible number of sequences at

_{i}*t*steps can be computed by applying

**D**. This is illustrated below. ❶

If

*t*= 0 ⇒ no steps are taken and hence where

*t*> 0

Then, at

*t*= 1

Similarly at *t* = 2

Therefore if

*t*= 3

*k*= 2 for the two states

*S*

_{0}and

*S*

_{1}at

*t*= 3

*t*= 3 is

We know that and hence which implies there are four possible ways (same as four possible Trellis paths) from state *S*_{0} at *t* = 0 to *S*_{0} at *t* = 3. Graphically,

*S*

_{0}at

*t*= 0 to

*S*

_{1}at

*t*= 3. Therefore the possible number of Trellis paths from

*S*

_{0}at

*t*= 0 to

*S*

_{0}and

*S*

_{1}at

*t*= 3 is eight, .

Hence

gives the **possible number of distinct** messages (sequences) that can be sent by the channel in exactly *t* steps.

*t*steps) is

*X*and

*Y*for such class of channels with memory is

## Alternative to determining for computing channel capacity

In practice it is hard to compute and , in . Therefore it is hard to compute the channel capacity by the approach of determining . An easier approach is to diagonalize **D**.

Let

*k*is the number of states. Also let the orthonormal eigenvectors of

**D**form the columns of the modal matrix

In general

*C*are constants because they are derived from

_{j}**U**

^{−1},

**U**and their constants.

If *λ _{m}* is the largest positive eigenvalue of

**D**, then for very large

*t*

*t*→ 0, thus the maximum channel capacity is

*Example*

**D**−

*λ*

**I**) we get

*C*=

_{m}*H*(

*X*) implies lossless channel. ❽

*Next:*

Summary (p:3) ➽