《状态机考卷练习.docx》由会员分享，可在线阅读，更多相关《状态机考卷练习.docx（10页珍藏版）》请在三一办公上搜索。
1、状态机考卷练习七、综合题： 已知状态机状态图如图a所示；完成下列各题： in_a = “00”in_a = “01”st0out_a = “0101”;in_a /= “00”st1out_a = “1000”;in_a = “11”in_a /= “11”in_a /= “01”st3out_a = “1101”;in_a /= “11”st2out_a = “1100”;in_a = “11”图a 状态图 clkresetc_stateREGn_stateCOMout_ain_a图b 状态机结构图 1. 试判断该状态机类型，并说明理由。 该状态机为moore型状态机，输出数据outa和输入
2、ina没有直接逻辑关系，outa是时钟clk的同步时序逻辑。 2. 根据状态图，写出对应于结构图b，分别由主控组合进程和主控时序进程组成的VHDL有限状态机描述。 LIBRARY IEEE; USE IEEE.STD_LOGIC_1164.ALL; ENTITY MOOREB IS PORT (CLK, RESET : IN STD_LOGIC; INA : IN STD_LOGIC_VECTOR (1 DOWNTO 0); OUTA : OUT STD_LOGIC_VECTOR (3 DOWNTO 0) ); END MOOREB; ARCHITECTURE ONE OF MOOREB IS
3、 TYPE MS_STATE IS (ST0, ST1, ST2, ST3); SIGNAL C_ST, N_ST : MS_STATE; BEGIN PROCESS (CLK, RESET) BEGIN IF RESET = 1 THEN C_ST = ST0; ELSIF CLKEVENT AND CLK = 1 THEN C_ST IF INA = “00” THEN N_ST = ST0; ELSE N_ST = ST1; END IF; OUTA IF INA = “00” THEN N_ST = ST1; ELSE N_ST = ST2; END IF; OUTA IF INA =
4、 “11” THEN N_ST = ST0; ELSE N_ST = ST3; END IF; OUTA IF INA = “11” THEN N_ST = ST3; ELSE N_ST = ST0; END IF; OUTA N_ST = ST0; END CASE; END PROCESS; END ONE; 3. 若已知输入信号如下图所示，分析状态机的工作时序，画出该状态机的状态转换值和输出控制信号(out_a)； 4. 若状态机仿真过程中出现毛刺现象，应如何消除；试指出两种方法，并简单说明其原理。 方法1，添加辅助进程对输出数据进行锁存 方法2，将双进程状态机改写为单进程状态机，其输出
5、也是锁存过了，故能消除毛刺 方法3，使用状态位直接输出型状态机编码方式，其输出直接由当前状态输出，也没有毛刺 七、综合题： 根据如下所示状态图及其状态机结构图，回答问题 ina=000 ina=110 ina=100 ina /= 100 and S0 S1 ina=011 ina=101 / outa=0010 ina=111 / outa=1100 outa=1001 outa=1111 ina=101 / outa=1101 ina=011 / outa=1110 S2 ina /= “011” S3 (a) CLK RESET CLK RESET ina FSM outa ina RE
6、G SIGNAL1 SIGNAL2 COM outa (b) (c) 1.试判断该状态机类型，并说明理由。 改状态机可以为mealy型状态机，当输入ina变化时可影响输出outa立即变化 2.请问如何消除状态机输出信号毛刺？试列出至少两种方法，并说明理由。 方法1，添加辅助进程对输出数据进行锁存 方法2，将双进程状态机改写为单进程状态机，其输出也是锁存过了，故能消除毛刺 方法3，使用状态位直接输出型状态机编码方式，其输出直接由当前状态输出，也没有毛刺 3.试由b、c两图中任选一图写出其完整的VHDL程序。 LIBRARY IEEE; USE IEEE.STD_LOGIC_1164.ALL; E
7、NTITY EX7 IS PORT ( CLK, RESET : IN STD_LOGIC; INA : IN STD_LOGIC_VECTOR(2 DOWNTO 0); OUTA : OUT STD_LOGIC_VECTOR(3 DOWNTO 0) ); END EX7; - MOORE型状态机 ARCHITECTURE ONE OF EX7 IS TYPE STATE IS (S0, S1, S2, S3); SIGNAL C_ST : STATE; BEGIN PROCESS (CLK, RESET, INA) BEGIN IF RESET = 1 THEN C_ST = S0; OUT
8、A 0); ELSIF RISING_EDGE(CLK) THEN CASE C_ST IS WHEN S0 = IF INA = 101 THEN OUTA = 0010; ELSIF INA = 111 THEN OUTA = 1100; C_ST = S1; ELSIF INA = 110 THEN C_ST = S2; OUTA IF INA = 000 THEN C_ST IF INA = 100 THEN C_ST = S2; ELSIF INA = 011 THEN C_ST = S1; OUTA = 1111; ELSE C_ST IF INA = 101 THEN OUTA
9、= 1101; ELSIF INA = 011 THEN OUTA = 1110; C_ST C_ST = S0; END IF; OUTA 0); END CASE; END IF; END PROCESS; END ONE; - MEALY型状态机 ARCHITECTURE TWO OF EX7 IS TYPE STATE IS (S0, S1, S2, S3); SIGNAL C_ST, N_ST : STATE; BEGIN REG : PROCESS (CLK, RESET) BEGIN IF RESET = 1 THEN C_ST = S0; ELSIF CLKEVENT AND
10、CLK = 1 THEN C_ST N_ST = S1; IF INA = 101 THEN OUTA = 0010; ELSE OUTA = ELSIF INA = 111 THEN OUTA OUTA = 1001; IF INA = 000 THEN N_ST = S1; ELSIF INA = 110 THEN N_ST = S2; ELSE N_ST OUTA = 1111; IF INA = 100 THEN N_ST = S2; ELSIF INA = 011 THEN N_ST = S1; ELSE N_ST IF INA = 101 THEN OUTA = 1101; ELS
11、IF INA = 011 THEN OUTA = 1110; ELSE OUTA = 0000;END IF; N_ST N_ST = S0; END CASE; END PROCESS; OUTA 0); END TWO; 已知一个简单的波形发生器的数字部分系统框图如下图所示 图中lcnt、lrom都是在MAX+PlusII中使用MegaWizard调用的LPM模块，其VHDL描述中Entity部分分别如下： ENTITY lcnt IS PORT ( clock q ); : IN STD_LOGIC ; : OUT STD_LOGIC_VECTOR (9 DOWNTO 0) END lc
12、nt; ENTITY lrom IS PORT ( address : IN STD_LOGIC_VECTOR (9 DOWNTO 0); q : OUT STD_LOGIC_VECTOR (9 DOWNTO 0) ); END lrom; 试用VHDL描述该系统的顶层设计。 LIBRARY IEEE; USE IEEE.STD_LOGIC_1164.ALL; ENTITY MYSG IS PORT (CLK : IN STD_LOGIC; TO_DA : OUT STD_LOGIC_VECTOR (9 DOWNTO 0) ); END MYSQ; ARCHITECTURE ONE OF MY
13、SQ IS SIGNAL ADDR : STD_LOGIC_VECTOR (9 DOWNTO 0); COMPONENT LCNT PORT (CLOCK : IN STD_LOGIC; Q : OUT STD_LOGIC_VECTOR (9 DOWNTO 0) ); END COMPONENT; COMPONENT LROM PORT (ADDRESS : IN STD_LOGIC_VECTOR (9 DOWNTO 0); Q : OUT STD_LOGIC_VECTOR (9 DOWNTO 0) ); END COMPONENT; BEGIN U1 : LCNT PORT MAP (CLOCK = CLK, Q = ADDR); U2 : LROM PORT MAP (ADDRESS = ADDR, Q = TO_DA); END ONE;
