This figure is similar to the previous figure, however since there are two senders there are two sets of data in these segments. Segment one, for example, includes an upper sender and a lower sender.

In the first segment of the sender's process (moving left to right), we see a row of data bits and a row of code for both the upper and lower senders. The upper sender's data bits appear as "d (1/1) = -1" and the code is: 1, 1, 1, -1, 1, -1, -1, -1. In the second segment, the data bits appear as "d (1/0) =1" and the code is: 1, 1, 1, -1, 1, -1, -1, -1. In the third segment, blue arrows point from both rows to a circle, which is labeled "Z (1/i,m) = d (1,i) o C (1/m)." The lower sender's data bits appear as "d (2/1) = 1" and the code is: 1, -1, 1, 1, 1, -1, 1, 1." In the second segment, the data bits appear as "d (2/0) = 1" and the code is: 1, -1, 1, 1, 1, -1, 1, 1." Blue arrows point from both rows to a circle, which is labeled "Z (2/i,m) = d (2/i) o C (2/m)." Blue arrows point from this equation and the equation above from the upper sender to a middle circle with a plus sign. An arrow points from this circle to the next segment, in which we see a line of code that reads: -2, 2, 2, 2. The last segment, labeled "Channel, Z* i,m," continues the code: 2, 2, 2, -2. An arrow points from here down the receiver's process, which contains two rows of code. The arrow points to the top row of code. Code continues through segment one, labeled "Time slot 1 received input" and segment two, labeled "Time slot 0 received input." The two rows of code converge in the third segment at a circle, which is labeled "d (1/i) = m(Sigma)m=1 Z(*/i,m) o C (1/m) over M. The fourth segment includes the data bits "d (1/1) = -1" and the last segment includes the data bits "d (1/0) = 1."