rst output bits would be 00. Comparing this to V where the first two output bits are 00, we say "O bit errors if parent is node 0 (time 0)." We add this 0 to the 0 number that we gave node 0 (time 0) in Figure 7.9(a), for a grand total of 0. 2. If we started at node 1 (time 0) and moved to node 0 (time 1), the first output bits would be 11. Comparing this to V where the first two output bits are 00, we say "2 bit errors if parent is node 1 (time 0)." We add this 2 to the 0 number that we gave node 1 (time 0) in Figure 7.8(a), for a grand total of 2. time L time L+1 Now, since starting at node 0 00 node A Can decide on best (time 0) and moving to node 0 parent node (A or B) (time 1) creates the fewest total bit 01 node B errors (0), we proclaim that the parent node for node 0 (time 1) is node 0 (time 0), and that it carries with it the number 0 (for zero total errors with this selection). We Figure 7.8 Underlying idea of VA Chapter Seven time O time 1 Read this first V = ( 00 O difference: Read this last Total 0+ O = O time O time 1 node O Best parent: node O node O 2 difference: lowest total node 1 Total 0+ 2 = 2 Figure 7.9 (a) Setting initial values to 0 (b) picking best parent node (with lowest total) for node 0 Channel Coding and Decoding: Part 2-Convolutional Coding and Decoding time O time 1 difference 1: node 1 Best parent: Total O + 1 = 1 node 2 lowest total node 2 difference 1: Total O + 1 = 1 node 3 time O time 1 node 2 Best parent: lowest total O node 1 lowest total 1 node 3 Best parent: node 2 Figure 7.9 (c) picking best parent node for node 1 (d) best parent node for nodes 2 4 3 Chapter Seven repeat this for node 1 (time 1). You can see this ongoing in Figure 7.9(c). There, node 2 (time 0) and node 3 (time 0) are possible parent nodes. We decided between these, as follows: 1. For node 2 (time 0) as starting node moving to node 1 (time 1), the output is 10. Comparing this to the first two bits of V, which are 00, we say "1 bit error if parent node is node 2 (time 0)." We add this 1 to the 0 number that