CSCE 211: Digital Logic Design

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Presentation transcript:

CSCE 211: Digital Logic Design Chin-Tser Huang huangct@cse.sc.edu University of South Carolina

Chapter 7: The Design of Sequential Systems

Review: Design Process for Combinational Systems Step 1: Represent each of the inputs and output in binary. Step 1.5: If necessary, break the problem into smaller subproblems. Step 2: Formalize the design specification either in the form of a truth table or of an algebraic expression. Step 3: Simplify the description. Step 4: Implement the system with the available components, subject to the design objectives and constraints. 11/19/2009

Design Process for Sequential Systems Step 1: From a word description, determine what needs to be stored in memory, that is, what are the possible states. Step 2: If necessary, code the inputs and outputs in binary. Step 3: Derive a state table or state diagram to describe the behavior of the system. Step 4: Choose a state assignment, that is, code the states in binary. Step 5: Choose a flip flop type and derive the flip flop input maps or tables. Step 6: Produce the logic equation and draw a block diagram (as in the case of combinational systems). 11/19/2009

Revisit Continuing Example 6 CE6. A system with one input x and one output z such that z = 1 iff x has been 1 for at least three consecutive clock times. 11/19/2009

State Assignment of CE 6 11/19/2009

K-map for Next State q1* = x q2 + x q1 q2* = x q2´ + x q1 11/19/2009

K-map for Output z = q1 q2 11/19/2009

Design with D Flip Flops Therefore, D1 = x q2 + x q1 D2 = x q´2 + x q1 11/19/2009

Implementation using D Flip Flops 11/19/2009

Design with JK Flip Flops 11/19/2009

Design with JK Flip Flops J1 = xq2 K1 = x´ z = q1q2 J2 = x K2 = x´ + q´1 11/19/2009

Design with T Flip Flops T1 = x´q1 + xq´1q2 z = q1q2 T2 = x´q2 + xq´2 + xq´1q2 11/19/2009

Another Example: Up/Down Counter A counter that can count up or down according to a control input Counts up when x=0 Counts down when x=1 11/19/2009

Design with JK Flip Flops JA = KA = 1 JB = KB = x´A + xA´ JC = KC = x´BA + xB´A´ 11/19/2009