CSM51Asolution_chapter8

# S4 s5 s6 s7 x0 x1 s2 s2 s2 s7 s2 s3 s3 s7 s4 s0 s4 s1

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Unformatted text preview: e de ne the following code: 138 Solutions Manual - Introduction to Digital Design - February 22, 1999 QAQB QC state 000 S0 001 S1 010 S2 011 S3 100 S4 S5 101 110 S6 111 S7 The resulting state table is PS Input Output S0 S1 S2 S3 S4 S5 S6 S7 x=0 x=1 S2 S2 S2 S7 S2 S3 S3 S7 S4 S0 S4 S1 S4 S5 S5 S5 NS z 1 0 1 0 1 0 1 0 output sequence pairs the input 0011 22237 11101 The state diagram is shown in Figure 8.20. A timing diagram can be obtained from the following input is arbitrary: xt 0 1 0 1 1 0 0 1 0 1 1 1 0 1 0 st 0 2 3 3 7 5 4 4 0 2 3 7 5 4 0 z 110000111100011 Exercise 8.18 The expressions for the ip- op inputs are TA = QA + Q0B TB = QA + QB The ip- ops change state in every clock pulse depending only on the previous state. The transition table is QAtQB t TAt TB t QA t + 1QB t + 1 00 01 10 11 1 0 1 1 0 1 1 1 10 00 01 00 PS FF inputs NS Let us de ne the following encoding: Solutions Manual - Introduction to Digital Design - February 22, 1999 139 S0/1 0,1 S7/0 1 S1/0 0 0 0,1 S6/1 0 1 1 S5/0 0 1 1 S2/1 1 S3/0 0 S0/1 0 Figure 8.20: State diagram for Exercise 8.17 QAQB 00 01 10 11 The resulting state table is S0 S2 S1 S3 S1 S2 S0 S0 PS NS The state diagram is presented in Figure 8.21. It is an autonomous modulo-3 counter. The expressions for the ip- op inputs and for the output are JA = 1 KA = 1 0 JB = QC KB = 1 JC = QB KC = 1 0 Q0 Q0 z = QA B C The sequential network does not have any input; therefore, the state register changes in each clock pulse depending only on the previous state. The transition table is Exercise 8.19 S0 S1 S2 S3 140 Solutions Manual - Introduction to Digital Design - February 22, 1999 S0 S1 S2 S3 Figure 8.21: State diagram for Exercise 8.18 PS QA tQB tQC t 000 11 11 001 11 01 11 11 010 011 11 01 11 11 100 101 11 01 110 11 11 111 11 01 Let us de ne the following encoding: JAtKA t JB tKB t JC tKC t QA t + 1QB t + 1QC t + 1 01 01 11 11 01 01 11 11 110 100 101 100 010 000 001 000 FF inputs NS Output z t 1 0 0 0 0 0 0 0 QAQB QC state 000 001 010 011 100 101 110 111 A B C D E F G H The resulting transition table is PS A B C D E F G H Let us try to reduce the number of states: P=1 1  NS G E F E C A B A z 1 0 0 0 0 0 0 0 2 A 2  B; C; D; E; F; G; H  2 222121 Solutions Manual - Introduction to Digital Design - February 22, 1999 141 3 11 32333 1 2 3 P3 = A F; H  C  B; D;4 E; G 4 11 2 4434 1 34 5 P4 = A F;2H  C  E B; D; G 5 11 2 3 445 1 345...
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## This note was uploaded on 03/26/2010 for the course CS 187154200 taught by Professor Ercegovac,m.d. during the Winter '09 term at UCLA.

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