S3 Sequential Circuit Implementation

# S3 Sequential Circuit Implementation - Sequential Circuit...

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Sequential Circuit Implementation The example that I will develop here is also covered in the text. I will elaborate on the procedure that a designer would take in obtaining an implementation from some state table. Encoding the State Table Here is the state table that models the circuit that we want to implement. Presumably, this state table has already been reduced. PS x 0 1 A A / 0 B / 0 B A / 0 C / 1 C B / 0 D / 0 D C / 1 D / 0 NS / z With a reduced state table in hand, we must now make a state assignment. Normally, we would follow the state assignment procedure that we saw in class. Instead, we will use an arbitrary assignment: A = 00, B = 01, C = 11, D = 10. Applying the state assignment yields an encoded state table . y 1 y 2 x 0 1 00 00 / 0 01 / 0 01 00 / 0 11 / 1 11 01 / 0 10 / 0 10 11 / 1 10 / 0 y 1 + y 2 + / z All instances of each state have been replaced by the corresponding state code . At this point, we now know the next state for each total present state code. Once we choose a flip-flop, we can obtain the appropriate instances of excitation for all combinations of the total present state. D Flip-Flops Using a D flip-flop is easy, because the excitation input is the same as the next state. If the maps shown below represent the next states, then the same maps show the D flip-flop excitation. x x 0 1 0 1 y 1 y 2 00 0 0 y 1 y 2 00 0 1 01 0 1 01 0 1 11 0 1 11 1 0 10 1 1 10 1 0 y 1 + y 2 +

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We can show the groups and the minimization. x x 0 1 0 1 y 1 y 2 00 0 0 y 1 y 2 00 0 1 01 0 1 01 0 1 11 0 1 11 1 0 10 1 1 10 1 0 D 1 = xy 2 + y 1 y 2 D 2 = xy 1 + x y 1 When we apply these equations to the D inputs of the flip-flops, we have implemented the circuit whose behavior is modeled by the state table. D C S R Q Q D C S R Q Q +5V 0 1 0 1 0 1 +5V D1 D2 CLK CLR y2 y2' x y1' y1 JK Flip-Flops: A First Approach As a first attempt to complete the circuit using JK flip-flops, we will expand the state table to incorporate the JK flip-flop excitation. On the face of it, the JK flip-flop is unlike the D flip-flop in that knowledge of the next state does not provide direct knowledge of the JK flip-flop excitation. Instead, we must determine J and K for each combination of the total present state.
y 1 y 2 x x x 0 1 0 1 0 1 00 00 / 0 01 / 0 0 × 0 × 0 × 1 × 01 00 / 0 11 / 1 0 × 1 × × 1 × 0 11 01 / 0 10 / 0 × 1 × 0 × 0 × 1 10 11 / 1 10 / 0 × 0 × 0 1 × 0 × y 1 + y 2 + / z J 1 K 1 J 2 K 2 We derived J 1 and K 1 by comparing y 1 to y 1 + for each combination of x , y 1 , and y 2 . We derived J 2 and K 2 by comparing y 2 to y 2 + for each combination of x , y 1 , and y 2 . With J 1 , K 1 , J 2 , and K 2 in hand, we can now generate Karnaugh maps for the excitation inputs. x x x x 0 1 0 1 0 1 0 1 y 1 y 2 00 0 0 y 1 y 2 00 × × y 1 y 2 00 0 1 y 1 y 2 00 × × 01 0 1 01 × × 01 × × 01 1 0 11 × × 11 1 0 11 × × 11 0 1 10 × × 10 0 0 10 1 0 10 × × J 1 = xy 2 K 1 = x y 2 J 2 = xy 1 + x y 1 K 2 = x y 1 + xy 1 When we apply these equations to the J and K inputs of the flip-flops, we have implemented the circuit whose

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## S3 Sequential Circuit Implementation - Sequential Circuit...

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