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CS 140 Lecture 7 Professor CK Cheng 4/23/02
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Part II. Sequential Network (Ch. 7.1-7.5) 1.Flip-flops SR, D, T, JK, 2.SpecificationState Table 3.Implementation Characteristic Eq. Q(t+1) = f(x, Q(t)).
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SR Flip-Flop (Set-Reset) S R S R y Q Q = (R+y)’ y = (S+Q)’
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Id Q y S R y* Q* y** Q** y*** Q*** 0 0 0 0 0 1 1 0 0 1 1 1 0 0 0 1 1 0 1 0 1 0 2 0 0 1 0 0 1 0 1 0 1 3 0 0 1 1 0 0 0 0 0 0 4 0 1 0 0 1 0 1 0 1 0 5 0 1 0 1 1 0 1 0 1 0 6 0 1 1 0 0 0 0 1 0 1 7 0 1 1 1 0 0 0 0 0 0 8 1 0 0 0 0 1 0 1 0 1 9 1 0 0 1 0 0 1 0 1 0 10 1 0 1 0 0 1 0 1 0 1 11 1 0 1 1 0 0 0 0 0 0 12 1 1 0 0 0 0 1 1 0 0 13 1 1 0 1 0 0 1 0 1 0 14 1 1 1 0 0 0 0 1 0 1 15 1 1 1 1 0 0 0 0 0 0 Q y State Transition SR 10 00 11 00 10 SR 11 10 01 11 01 11 01 1000 10 00 01 00 11 State Diagram 01
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When SR=01, (Q,y) = (0,1) When SR=10, (Q,y) = (1,0) When SR=11, (Q,y) = (0,0) SR = 00 => if (Q,y) = (0,0) or (1,1), the output keeps changing Solutions: 1) SR = (0,0). 2) SR = (1,1). 0 0 0 1 - 1 1 0 1 - PS inputs 00 01 10 11 State table Q(t+1) SR Characteristic Eq Q(t+1) = S(t)+R’(t)Q(t)
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D Flip-Flop (Delay) D CLK Q Q’ Id D Q(t) Q(t+1) 0 0 0 1 1 0 2 1 0 1 3 1 1 - Characteristic Eq Q(t+1) = D(t) 0 0 1 1 0 1 PS D 0 1 State table Q(t+1)
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JK F-F J CLK Q Q’ Characteristic Eq Q(t+1) = Q(t)K’(t)+Q’(t)J(t) 0 0 0 1 1 1 1 0 1 0 PS JK 00 01 10 11 State table Q(t+1) K
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T Flip-Flop (Toggle) T CLK Q Q’ Characteristic Eq Q(t+1) = Q’(t)T(t) + Q(t)T’(t) 0 0 1 1 1 0 PS T 0 1 State table Q(t+1)
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