Logic and Computer Design Fundamentals M Morris Mano
- Slides: 68
逻辑与计算机 设计基础 Logic and Computer Design Fundamentals M. Morris Mano Charles R. Kime 2021/9/11 1 凌纯清: 基地 312, 13487570835, jt_lingcq@hnu. edu. cn, 155059066
3. 1 开始分层设计(2/3) 例3. 2 设计一 个 9输入偶校验 位产生器。 X 0 X 1 X 2 X 3 X 4 X 5 X 6 X 7 X 8 Top Level: 9 inputs, one output 9 -Input odd function Z (a) Symbol for circuit X 0 A 0 X 1 A 1 X 2 A 2 X 3 A 0 X 5 3 -Input A 1 odd B O function A 2 X 6 A 0 X 7 A 1 X 8 A 2 X 4 2 nd Level: Four 3 -bit odd parity trees in two levels 3 rd Level: Two 2 -bit exclusive -OR functions 3 -Input odd B O function A 0 A 1 A 2 3 -Input odd B O function (b) Circuit as interconnected 3 -input odd function blocks A 0 A 1 BO Primitives: Four 2 input NAND gates A 2 (c) 3 -input odd function circuit as interconnected exclusive-OR blocks 2021/9/11 ZO Design requires 4 X 2 X 4 = 32 2 -input NAND gates (d) Exclusive-OR block as interconnected NANDs 6
3. 2 艺映射(2/3) 2021/9/11 9
3. 2 艺映射(3/3) A B F C D E 1 C 2 X F 3 D (a) (b) A E B C F D E (c) 2021/9/11 10
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3. 5 译码(6/10) l. VHDL结构描述 2021/9/11 23
3. 5 译码(7/10) l. VHDL数据流描述 2021/9/11 24
3. 5 译码(10/10) l模型机指令译码器 l输入 8位,输出若干位用来控制 操作数的流动。 指令的�� 符号 MOV R 1,R 2 MOV M,R 2 MOV R 1,M ADD R 1,R 2 SUB R 1,R 2 AND R 1,R 2 NOT R 1 SHR R 1 SHL R 1 JMP add JZ add JC add IN R 1 OUT R 1 NOP HALT 2021/9/11 指令的二�制�� 1111 R 2 1111 R 1 11 1001 R 2 0110 R 1 R 2 1110 R 1 R 2 0101 R 1 XX 1010 R 1 00 1010 R 1 11 0001 00 00,address 0001 00 01,address 0001 00 10,address 0010 R 1 XX 0100 R 1 XX 0111 00 00 1000 00 00 27
3. 7 多路复用 0 1 1 0 A B C D 01 D 11 D 21 D 31 D 41 Out D 51 D 61 D 71 S 2 8 -to-1 S 0 MUX -应用方法 1(2/2) 0 1 1 0 0 1 Y A B C D 00 D 10 D 20 D 30 D 40 Out D 50 D 60 D 70 S 2 8 -to-1 S 0 MUX Z l数据输入为固定值的多路开关实际上就是ROM。 2021/9/11 39
3. 7 多路复用 -应用方法 2 (2/3) l重新安排表格,使得数据输入按计数顺序排列,然后用基 本函数表示各个输出。 2021/9/11 Gray ABC Binary xyz 000 001 111 010 011 100 001 110 110 010 111 101 Rudimentary Functions of C for y Rudimentary Functions of C for z y=C z=C 41
3. 10 二进制减法(5/6) 例3. 8 计算 010101002 – 010000112 01010100 – 01000011 2’s comp 01010100 + 10111101 1 0001 l有进位表示结果正确。 例3. 9 010000112 – 010101002 01000011 – 010101000011 2’s comp + 10101100 0 11101111 0001 l没有进位,应该将和变补,结果为 – 0001。 2021/9/11 2’s comp 54
3. 11 带符号数的加减法(2/8) l原码:n – 1位数字表示负数的绝对值。 l 4位原码 1111 1110 1101 1100 1011 1010 1001 1000 0001 0010 0011 0100 0101 0110 0111 -7 -6 -5 -4 -3 -2 -1 -0 +0 +1 +2 +3 +4 +5 +6 +7 l反码:n -1位数字表示负数的反码(又称基-1(退化)补码)。 l 4位反码 1000 1001 1010 1011 1100 1101 1110 1111 0000 0001 0010 0011 0100 0101 0110 0111 -7 -6 -5 -4 -3 -2 -1 -0 +0 +1 +2 +3 +4 +5 +6 +7 l补码:n -1位数字表示负数的补码(又称基数补码) 。 l 4位补码 1000 1001 1010 1011 1100 1101 1110 1111 0000 0001 0010 0011 0100 0101 0110 0111 -8 -7 -6 -5 -4 -3 -2 -1 +0 +1 +2 +3 +4 +5 +6 +7 2021/9/11 57
3. 11 带符号数的加减法(6/8) 例子 l Example 1: 0010 + 0101 l Example 2: 1011 + 1101 l Example 3: 0010 - 0100 l Example 4: 1100 - 1010 2021/9/11 60
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