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CS 140 Lecture 11 Professor CK Cheng 5/31/02. C1C2 CLK x(t) y(t) Sequential Network Implementation Mealy & Moore machine State Table  Netlist s(t) D(t)

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Presentation on theme: "CS 140 Lecture 11 Professor CK Cheng 5/31/02. C1C2 CLK x(t) y(t) Sequential Network Implementation Mealy & Moore machine State Table  Netlist s(t) D(t)"— Presentation transcript:

1 CS 140 Lecture 11 Professor CK Cheng 5/31/02

2 C1C2 CLK x(t) y(t) Sequential Network Implementation Mealy & Moore machine State Table  Netlist s(t) D(t) = g(x(t), Q(t)) y(t) = f(x(t), Q(t))

3 Input PS x Q(t) State Table NS, y NS = Q(t+1) = h(x(t),Q(t)) Output y(t) = f(x(t),Q(t))

4 (1)State Table: y(t) = f(Q(t), x(t)) Q(t+1) = h(x(t),Q(t)) (2)Excitation Table: (1)D(t) = e D (Q(t+1), Q(t)); (2)T(t) = e T (Q(t+1), Q(t)); (3)S, R, J, K (3)From 1 & 2, we derive (1)D(t) = g D (Q(t), x(t)); (2)T = g T (Q(t), x(t)); (3)S,R,J,K. (4)Use K-Map (QuineMcClusky) to derive optional combinational logic implementation. (1)D(t) = h(Q(t), x(t)) (2)Y(t) = f(Q(t), x(t))

5 State table of a JK flip flop: 00 0 1 00 0 00 1 00 1 0 0101 Q(t) Q(t+1) JK From this, we can derive the excitation table for a JK F-F: 0 0- 1 1- -0 0101 PS NS Q(t) Q(t+1) JK If Q(t) is 1, and Q(t+1) is 0, then JK needs to be 0-.

6 Excitation Tables and State Tables 0 0- 01 1 10 -0 0101 PS NS Q(t) Q(t+1) SR Excitation Tables: 0 1 0 0101 PS NS Q(t) Q(t+1) T 00 0 1 01 0 0101 PS SR Q(t) Q(t+1) SR 10 1 11 - 0 1 0 0101 PS T Q(t) Q(t+1) T State Tables:

7 0 0- 1 1- -0 0101 PS NS Q(t) Q(t+1) JK Excitation Tables: 0 1 0101 PS NS Q(t) Q(t+1) D 00 0 1 01 0 0101 PS JK Q(t) Q(t+1) JK 10 1 11 1 0 1 0101 PS D Q(t) Q(t+1) D State Tables:

8 Implement a JK F-F with a T F-F T Q Q’ C1 J K 00 0 1 01 0 0101 PS JK Q(t) Q(t+1) = h(J(t),K(t),Q(t)) = J(t)Q(t)+K(t)Q(t) JK 10 1 11 1 0 State Table

9 id 0 1 2 3 4 5 6 7 J(t) 0 1 K(t) 0 1 0 1 Q(t) 0 1 0 1 0 1 0 1 Q(t+1) 0 1 0 1 0 T(t) 0 1 0 1 0 1 0 0101 PS NS Q(t) Q(t+1) T Excitation Table T(t) = Q(t) XOR Q(t+1) T(t) = Q(t) XOR ( J(t)Q’(t) + K’(t)Q(t))

10 0 1 3 2 4 5 7 6 J K 0 0 1 0 1 0 1 1 Q(t) T: T = K(t)Q(t) + J(t)Q’(t) T Q Q’ J K


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