Unit_15 - Unit 15 Problem Solutions 15.1 (a) Implication...

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113 b c-e c d e-h e-a c-h e-a e h-f f i-e h-g i-e f-g g e-h e-b c-h e-b a-b h c-i d-h c-i f-d c-e g-d i e-f e-b c-f b-e h-f a-b h-f a b c d e f g h Unit 15 Problem Solutions 15.1 (a) Reduced state table: State Next State X = 0 X = 1 Output X = 0 X = 1 A A C 1 0 B C F 0 0 C B A 0 0 F B F 1 0 15.2 a b c e h f d g i See FLD p. 660 for reduced state table. 15.1 (b) B C D E F G H A B C D E F G F-H C-E C-D A-F C-D F-H D-E F-H B-D A-H B-E F-H B-D A-H E-G B-H F-G A-B E-F A-F B C D E F G H A B C D E F G F-H C-E C-D A-F C-D F-H D-E F-H B-D A-H B-E F-H B-D A-H E-G B-H F-G A-B E-F A-F Implication chart after one pass: Complete implication chart A H C E G B D B C because F H , (and also because C D ) F H because B H, (and also because F G ), and B H because the output differs for X = 0. So use the sequence X = 100. Input: X: 1 0 0 Starting in B : Z: 0 1 0 State: ( B ) F B Starting in G : Z: 0 1 1 State: ( G ) H H So λ 1 ( B , 100) = 010 ≠ 011 = λ 2 ( G , 100), and B G . (Alternative: λ 1 ( B , 110) = 001 ≠ 000 = λ 2 ( G , 110). Also, λ 1 ( B , 00101) ≡ λ 2 ( G , 00101), but this requires an X of length 5.
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114 15.3 S 0 S 5 a S 1 b S 5 a S 1 c S 1 S 5 a S 6 b S 5 a S 6 c S 2 S 2 a S 6 b S 2 a S 6 c S 3 S 0 a S 1 b S 4 S 4 a S 3 b S 4 a S 3 c S 5 S 0 a S 1 b S 0 a S 1 c S 6 S 5 a S 1 b a b c S 0 a S 1 b
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This note was uploaded on 09/03/2008 for the course EE 316 taught by Professor Brown during the Fall '08 term at University of Texas.

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Unit_15 - Unit 15 Problem Solutions 15.1 (a) Implication...

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