微机原理试卷(B)答案.doc

微机原理及接口技术（B）答案一、 填空1. E0H,-542. 1000, 3E8H, 1111101000B, 313030303. 50H4. 37820H5. 由高地址向低地址伸展，字6. NEG AX,；AND AX, 7FFFH7. (1) AX=568AH, AX=5690H (2) AX=0011H, AX=0107H (3) AX=005AH, AX=0090H, AX=0908H8. 高速缓冲存储器、内存储器、辅助存储器9. 独立的I/O寻址方式、存储器映像的I/O寻址方式， 独立的I/O寻址方式。S1二、 存储形式：1234S2531323334NB3522NW288FFFBPFFFEFF三、编写程序段1. MOV CL, 4 2. MOV AL, BXSHL AX, CL XCHG AL, BX+3 SHL BL, CL MOV BX, AL MOV AL, BL MOV AL, 1BX SHR DH, CL XCHG AL, 2BX OR AL, DH MOV 1BX, AL3. MOV AH, AL AND AL, 0FH CMP AL, 0AH JB G1 ADD AL, 7 G1: ADD AL, 30H MOV CL, 4 AND AH, 0F0H SHR AH, CL CMP AH, 0AH JB G2 ADD AH, 7 G2: ADD AH, 30H 四、编写程序。 stack segment stack stack dw 32 dup(0) stack ends data segment IBUF DB PLEAST INPUT A HEX NUMBER$,0AH, 0DH OBUF DB 5,0,5 DUP(0) OBUF1 DB 5 DUP (0), $ data ends code segment begin proc far assume ss:stack,cs:code,ds:data push ds sub ax, ax push ax mov ax, data mov ds, ax MOV DX, OFFSET IBUF MOV AH, 9 INT 21H MOV DX, OFFSET OBUF MOV AH, 10 INT 21H MOV BL, OBUF+1 MOV BH, 0 MOV AX, 0 MOV CL, 4 MOV SI, OFFSET OBUF+2 MOV DI, OFFSET OBUF1+4LOP: SHL AX, CL MOV DH, SI CMP DH, A JB LOP1 SUB DH, 7LOP1: SUB DH, 30H OR AL, DH INC SI DEC BX JNZ LOP MOV CX, 10LOP2: MOV DX, 0 DIV CX OR DL, 30H MOV SI, DL DEC DI AND AX, AX JNZ LOP2 MOV DX, OFFSET OBUF2 MOV AH, 9 INT 21H retbegin endpcode ends end begin五、16K8/2k8=8,即共需8片存储器芯片。16K=16384=,所以组成16K存储器共需要14根地址线。2K=2048=,即11根作为字选择线，选择存储器芯片片内的单元。14-11=3，3根作为片选择线，选择8片存储器芯片。芯片的11根地址线为,余下的高位地址线为，所以译码电路对，进行译码。译码电路及译码输出线的选址范围如图所示。六、(1) 控制电路(2)控制程序：stack segment stack stack dw 32 dup(0)stack endsdata segment data endscode segmentbegin proc far assume ss:stack,cs:code,ds:data push ds sub ax, ax push ax mov ax, data mov ds, axAGAIN:MOV DX, 280H IN AL, DX OUT DX, AL MOV CL, 4 MOV AH, AL AND AH, 0F0H SHR AH, CL CMP AH, A JC ADD30 ADD AH, 7ADD30:ADD AH, 30H MOV DL, AH MOV AH, 2 INT 21H AND AL, 0FH CMP AL, A JC ADD130 ADD AL, 7ADD130:ADD AL, 30H MOV DL, AL MOV AH, 2 INT 21H MO