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1、1 Output Sequential circuit Basic unit: FF ( gates + feedback lines ) Memory inputs former state Chapter 6 Sequential Logic Circuits 6.1 Introduction of Sequential Logic Combinational Logic Memory w Q XZ X: outside inputs Q: state of FF W: control inputs - J, K, D, T Z：outside output 2 Relationships
2、 of X，Z， W and Q: (,)ZF X Q= (,)WH X Q= 1 (,) nn QG W Q + =Characteristic equation Driving equation Output equation Sequential circuit 按照电路中输出变量是否和输入变量直接相关按照电路中输出变量是否和输入变量直接相关 Mealy-type (米里型米里型) Moore-type (莫尔型莫尔型) Output Z Qn X Output Z Qn Sequential circuits structure: Combinational Logic + Memor
3、y 3 6.2 Synchronic Sequential Circuit Analysis Example 1: Analysis the synchronic sequential circuit Analysis: Given a circuit, describe its operation and function. 1) Inputs Outputs Control inputs States X Z J0 , K0 , J1 , K1 Q1 (MSB), Q0 4 10 () nn ZXQQ= 01 n JXQ= 0 1K = 10 n JXQ= 1 1K = Character
4、istic equation Driving equation Output equation 1 0000010 () nnnnn QJ QK QXQQ + =+= 1 1111101 () nnnnn QJ QK QXQQ + =+= 2) Equations 3) State table and State diagram Given: input X , Qn Try: output Z , Qn+1 5 X=0 0 1 0 0 X=1 1 0 0 0 1 0 0 0 0 0 1 0 1 0 0 1 0 1 1 1 0 1 1 1 0 0 0 0 0 1 0 1 0 0 1 1 0 n
5、 Q 1 n Q 1 1 n Q +1 0 n Q + X Z 0 0 1 0 1 0 0 0 State table X=0 1 101 nnn QQQ + = X=1 1 010 nnn QQQ + = 10 nn ZQQ= 1 101 nnn QQQ + = 1 010 nnn QQQ + = 10 nn ZQQ= State diagram X/Z Q1 Q0 01 11 10 0/0 0/0 1/0 1/11/0 1/0 0/1 00 0/0 对应一个对应一个CLK 6 X=0, M-3 Up counter：Z1，carry output； X=1, M-3 Down counte
6、r：Z1，borrow output。 4) Circuit function 01 11 10 0/0 0/0 1/0 1/11/0 1/0 0/1 00 0/0 State diagram main circle： Modulus 3 up-down two-direction counter 7 Example 2. Analyze the following sequential circuit Neither outside input no outside output J3 = Q2n K3 = Q2nQ1n J2 = Q1n J1 = Q2n + Q3n K2 = Q3nK1
7、= Q2n +Q3n = Q2nQ3n 1= CLK 3 Q 3 Q 3 J 3 K 1 Q 1 J 1 K 1 Q 2 Q 2 Q 2 K 2 J 1 1 8 1 3 333323213 () nnnnnnnn QJ QK QQ QQQQ + =+=+ 1 2 22221232 nnnnnnn QJ QK QQ QQ Q + =+=+ 1 1 1111231231 () nnnnnnnnn QJ QK QQQQQ Q Q + =+=+ Q3n+1 Q3n = 0, Q3n = 1, Q2n+1 Q2n = 0, Q2n = 1, Q1n+1 Q1n = 0, Q1n = 1, 1 0 0 1
8、 0 1 1 1 0 1 1 1 0 0 0 0 0 1 0 1 0 0 1 1 1 n Q 2 n Q 1 3 n Q +1 2 n Q + 3 n Q 1 1 n Q + 0 0 1 1 0 1 1 0 0 1 0 1 0 0 1 1 0 1 1 1 0 0 0 1 Q2n Q2nQ1n Q1n Q3n Q2n+Q3n Q2nQ3n 9 001 011 111 110 101 010 100 000 000 is isolated state 孤立状态孤立状态 self start 自启动自启动 1 0 0 1 0 1 1 1 0 1 1 1 0 0 0 0 0 1 0 1 0 0 1 1
9、 1 n Q 2 n Q 1 3 n Q +1 2 n Q + 3 n Q 1 1 n Q + 0 0 1 1 0 1 1 0 0 1 0 1 0 0 1 1 0 1 1 1 0 0 0 1 10 6.3 Synchronous Sequential Circuit Design 3) State distribution (encoding) Known Function or State diagram Request Circuit 1) Determine state (FF) and state diagram 2) State simplification 5) Qn+1 equa
10、tions and control inputs J, K, D, T Design steps: 6) Circuit 4) Select FF 7) Self start 11 Example 1. 同步同步5进制加法计数器 进制加法计数器 (例例6.4) Design a modulus-5 synchronous addition (up) counter. 1) Determine state and state diagram 在计数脉冲在计数脉冲CLK作用 下， 作用 下， 5 states change in period, The carry out Y = 1 at the
11、 state of S4 . M-5 counter, 5 states: S0 , S1 , S2 , S3 , S4 S0S1S2 S3S4 Y=1 2) State simplificationM-5, 5 states. 不须再化简不须再化简 Method 1 12 3) State distribution (encoding) 状态分配、编码状态分配、编码 n bits binary code2n-1 states 2n S0 000 S1 001 S2 010 S3 011 S4 100 n Q2 n Q1 n Q0 1 2 +n Q 1 1 +n Q 1 0 +n Q 0 0
12、0 0 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 Y State table 4) Select FF Using JK-FFs 000001010 011100 Y=1 State diagram 5) Qn+1 equations and control inputs J, K, D, T 3 bits 13 得到得到 2 -FF control input J2 driving K-map n Q 1n Q + J K 0 0 0 1 1 0 1 1 0 1 1 0 1 22 nn QQ + JK-FF d
13、riving table (驱动表驱动表) State diagram J2 0 00 0 00 0 00 0 11 1 0X X X X X X X X X X n Q2 n Q1 n Q0 1 2 +n Q 1 1 +n Q 1 0 +n Q 0 0 0 0 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 Y 14 得到得到 1 -FF control input K1 driving K-map n Q 1n Q + J K 0 0 0 1 1 0 1 1 0 1 1 0 1 11 nn QQ + JK-FF d
14、riving table K1 0 0X 0 1X 1 10 1 01 0 0 X X X X X X X X X X State diagram n Q2 n Q1 n Q0 1 2 +n Q 1 1 +n Q 1 0 +n Q 0 0 0 0 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 Y 15 得到 各个触发器控制输入 驱动卡诺图得到 各个触发器控制输入 驱动卡诺图Control inputs 16 11 Output K-map 6) Circuit n Q2 n Q1 n Q0 1 2 +n Q 1 1
15、+n Q 1 0 +n Q 0 0 0 0 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 Y State table 17 000 001 010 011100 /0/0 /0 /0 /1 5 valid states 5 233 个个FF 直接填直接填 K-map Q3nQ2nQ1n Q3n+1Q2n+1Q1n+1 /Z Z 00 01 11 10 0 1 Q3nQ2n Q1n 001 Q3n+1 00 01 11 10 0 1 Q3nQ2n Q1n Q2n+1 00 01 11 10 0 1 Q3nQ2n Q1n Q1n+1 00 01 11 10 0 1 Q3nQ2n Q1n 001 0 10 0 11 1 00 0 00 0 0 Method 2: 先不确定用哪种先不确定用哪种 FF 18 Z 00 01 11 10 0 1 Q3nQ2n Q1n
