e 0.02 number that we gave node 0 (time 1) in Figure 8.15(b), for a grand total of 4.87. 3. If we started at node 1 (time 1) and moved to node 0 (time 2) using the top branch, the second output symbol would be (0, 1). Comparing this to r where the second sent signal is 2=(1.1, 0.1), we say "d= 2.02 distance if parent is node 1 (time 1), top branch." We add this 2.02 to the 0.5 number that we gave node 1 (time 1) in Figure 8.15(c), for a grand total of 2.52. Trellis-Coded Modulation (TCM): The Wisdom of Modulator and Coder Togetherness r2=(1.1,1.0.1) time 1 time 2 so=(1,0) S4=(-1,0) node 0 (time 2) Sc=(0,-1) Figure 8.16 Moving through the trellis using the VA (from time 1 to time 2). Only movement to top node is drawn. 4. If we started at node 1 (time 1) and moved to node 0 (time 2) using the bottom branch, the second sent symbol would be (0, -1). Comparing this to r where the second output symbol is (1.1, 0.1), we say "d4= 2.42 distance if parent is node 1 (time 1), bottom branch." We add this 2.42 to the 0.5 number that we gave node 0 (time 1) in Figure 8.15(b), for a grand total of 2.92. Since node 0 (time 1) to node 0 (time 2), along the top branch, creates the small- est total (0.04), we proclaim node 0 (time 1) and the "top" branch the best parent and branch. We continue this process for node 1 (time 2), node 2 (time 2) and node 3 (time 2). And that ends our next move from left to right through the trellis. We repeat the process at time 3, and then at time 4. Now, at the end, we have four end nodes with four numbers. For example, end node 0 (time 4) comes with value 0.08. We choose the end node with the smallest value. From here, we know the history of parent nodes and branches, SO we can "back- track" through the trellis, and determine the s' and the m'. And we're done. In our case, we choose node 0 (time 4) with a final value of 0.08, and we "back- track," leading us to the path shown in Figure 8.14. This leads us to the output m' = 00 00 00 00). Now you know how the TCM decoder works to undo the effects a