Chapter 7

# Chapter 7 - Chapter 7 Problem Solutions 7.1 Excitation and...

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- 7.1 - Chapter 7 Problem Solutions 7.1. Excitation and output expressions: J 1 = xQ - 2 + y K 1 = yQ - 2 + x - Q 2 J 2 = x - y - K 2 = xy + Q 1 z = Q 1 Q - 2 + x - Q - 1 Excitation table: Present state (Q 1 Q 2 ) Excitation (J 1 K 1 ,J 2 K 2 ) Output (z) Inputs (xy) Inputs (xy) 00 01 10 11 00 01 10 11 00 00,10 11,00 10,00 11,01 1100 01 01,10 11,00 00,00 10,01 10 00,11 11,01 10,01 11,01 1111 11 01,11 11,01 00,01 10,01 0000 Transition table: Present state (Q 1 Q 2 ) Next state (Q + 1 Q + 2 ) Output (z) Inputs (xy) Inputs (xy) 00 01 10 11 00 01 10 11 00 01 10 10 10 1 1 0 0 01 01 11 01 10 1 1 0 0 10 11 00 10 00 1 1 1 1 11 00 00 10 10 0 0 0 0

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- 7.2 - 7.1. (continued) State table: Present state Next state Output (z) Inputs (xy) Inputs (xy) 00 01 10 11 00 01 10 11 00 A B C C C 1 1 0 0 01 B B D B C 1 1 0 0 10 C D A C A 1 1 1 1 11 D A A C C 0 0 0 0 State diagram:
- 7.3 - 7.2. Excitation and output expressions: D 1 = x - Q 2 D 2 = x - Q - 1 Q 2 + xQ - 1 Q - 2 z = Q 1 Q 2 Excitation table: Present state (Q 1 Q 2 ) Excitation (D 1 D 2 ) Output (z) Input (x) 01 00 00 01 0 01 11 00 0 10 00 00 0 11 10 00 1 Transition table (Same as excitation table): Present state (Q 1 Q 2 ) Next state (Q + 1 Q + 2 ) Output (z) Input (x) 00 00 01 0 01 11 00 0 10 00 00 0 11 10 00 1

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- 7.4 - 7.2. (continued) State table: Present state Next state Output (z) Input (x) 01 00 A A B 0 01 B D A 0 10 C A A 0 11 D C A 1 State diagram:
- 7.5 - 7.3. Excitation and output expressions: T 1 = xQ 2 + Q - 1 Q 2 T 2 = x + Q - 1 Q 2 z 1 = xQ - 1 z 2 = x - Q 2 Excitation table: Present state (Q 1 Q 2 ) Excitation (T 1 T 2 ) Output (z 1 z 2 ) Input (x) Input (x) 0101 00 00 01 00 10 01 11 11 01 10 10 00 01 00 00 11 00 11 01 00 Transition table: Present state (Q 1 Q 2 ) Next state (Q + 1 Q + 2 ) Output (z 1 z 2 ) Input (x) Input (x) 00 00 01 00 10 01 10 10 01 10 10 10 11 00 00 11 11 00 01 00

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- 7.6 - 7.3. (continued) State table: Present state Next state Output (z 1 z 2 ) Input (x) Input (x) 0101 00 A A B 00 10 01 B C C 01 10 10 C C D 00 00 11 D D A 01 00 State diagram:
- 7.7 - 7.4. Excitation and output expressions: J 1 = Q 2 K 1 = x D 2 = x(Q - 1 +Q - 2 ) z = Q 2 + x - Q 1 Excitation table: Present state (Q 1 Q 2 ) Excitation (J 1 K 1 ,D 2 ) Output (z) Input (x) Input (x) 0101 00 00,0 01,1 0 0 01 10,0 11,1 1 1 10 00,0 01,1 1 0 11 10,0 11,0 1 1 Transition table: Present state (Q 1 Q 2 ) Next state (Q + 1 Q + 2 ) Output (z) Input (x) Input (x) 00 00 01 0 0 01 10 11 1 1 10 10 01 1 0 11 10 00 1 1

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- 7.8 - 7.4. (continued) State table: Present state Next state Output (z) Input (x) Input (x) 0101 00 A A B 0 0 01 B C D 1 1 10 C C B 1 0 11 D C A 1 1 State diagram:
- 7.9 - 7.5. A: A multiple of three 1's have occurred (where none is a multiple of three). B: One greater than a multiple of three 1's have occurred. C: Two 1's greater than a multiple of three 1's have occurred. Present state Next state Output (z) Input (x) Input (x) 0101 * A A B 0 0 B B C 0 0 C C A 0 1 7.6. A: Last input was 0. B: One 1 has occurred. C: Two consecutive 1's have occurred. D: Three consecutive 1's have occurred. E: More than three consecutive 1's have occurred. Present state Next state Output (z) Input (x) Input (x) * A A B 0 0 B A C 1 0 C A D 0 0 D A E 1 0 E A E 0 0

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- 7.10 - 7.7. A: Waiting for a single 0 (or, equivalently, 1's occurring after more than one 0 has occurred). B: Exactly one 0 has occurred. C: Two or more consecutive 0's have occurred.
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Chapter 7 - Chapter 7 Problem Solutions 7.1 Excitation and...

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