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CS 140 Lecture 10 Professor CK Cheng 5/02/02. Given the state table, implement with 2 JK flip flops id 0 1 2 3 4 5 6 7 Q 1 (t) 0 1 Q 0 (t) 0 1 0 1 X(t)

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Presentation on theme: "CS 140 Lecture 10 Professor CK Cheng 5/02/02. Given the state table, implement with 2 JK flip flops id 0 1 2 3 4 5 6 7 Q 1 (t) 0 1 Q 0 (t) 0 1 0 1 X(t)"— Presentation transcript:

1 CS 140 Lecture 10 Professor CK Cheng 5/02/02

2 Given the state table, implement with 2 JK flip flops id 0 1 2 3 4 5 6 7 Q 1 (t) 0 1 Q 0 (t) 0 1 0 1 X(t) 0 1 0 1 0 1 0 1 Q 1 (t+1) 0 1 0 1 - Q 0 (t+1) 1 0 1 0 - Sequential Network Synthesis (Continue)

3 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-.

4 id 0 1 2 3 4 5 6 7 Q 1 (t) 0 1 Q 0 (t) 0 1 0 1 X(t) 0 1 0 1 0 1 0 1 Q 1 (t+1) 0 1 0 1 - Q 0 (t+1) 1 0 1 0 - Back to our problem, using the excitation table to solve for the two JK F-Fs. J 1 K 1 0 – 1 – 0 – – 1 – 0 – – J 0 K 0 1 – 0 – – 1 – 0 0 – – –

5 0 2 6 4 1 3 7 5 x Q1Q1 0 1 - - 0 0 - - Q0Q0 J1:J1: 0 2 6 4 1 3 7 5 x Q1Q1 - - - 1 - - - 0 Q0Q0 K1:K1: Using K-maps to solve both JK flip flops: J 1 = Q 0 X’ K 1 = X’

6 0 2 6 4 1 3 7 5 x Q1Q1 1 - - 0 0 - - 0 Q0Q0 J0:J0: 0 2 6 4 1 3 7 5 x Q1Q1 - 1 - - - 0 - - Q0Q0 K0:K0: J 0 = Q 1 ’X’ K 0 = X’

7 Final circuit JKJK JKJK Q Q’ Q Q’ J1J1 x K1K1 K0K0 J0J0 Q’ 1 Q0Q0

8 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:

9 Implement a JK F-F with a T F-F 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 - T(t) 1 0 1 0 - T Q Q’ C1 J K Q(t+1) = f(Q(t), J(t), K(t))

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

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