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Unformatted text preview: Solutions Manual - Introduction to Digital Design - November 2, 2000
Exercise 7.4 103 The state diagram for this exercise is shown in Figure 7.2 on page 103.
a/0 a/0 c/0 b/1 A c/1 b/1 c/1 E b/1 a/0 c/0 D B b/1 c/1 b/1 C a/0 a/0 Figure 7.2: State Transition Diagram of Exercise 7.4 118
Exercise 7.16 Solutions Manual - Introduction to Digital Design - November 2, 2000 Based on the outputs for each state we get the rst partition P1 as: P1 = (A; D; E )(B; F; G)(C; H )
Let's call (A,D,E) as group 1, (B,F,G) as group 2 and (C,H) as group 3. We can construct a table representing the next group for each state transition: 0 2 2 1 3 3 A D E B F G C H
2 3 3 3 3 3 1 1 1 1 3 1 group 1 group 2 group 3 From the table we can see that the columns for each group of states are the same, and so, the states in each group are also 2-equivalent. P2 = P1 . Renaming the states in group 1 as , the states in group 2 as and in group 3 as , we can represent the reduced sequential system as: PS Input x=0 x=1 ;0 ;0 ;1 ;1 ;0 ;1 ...
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This note was uploaded on 04/17/2008 for the course CS 151A taught by Professor Miloseragovich during the Fall '07 term at UCLA.
- Fall '07