解:(R0)=30H,(A)=40H,(R1)=40H,(B)=10H,(30H)=0CAH,(P3)=0CAH, (10H)=20H,(30H)=20H 3-16 解:1、(R6)=55H
2、(8FH)=55H 3、(A)=90H 4、(A)=0F0H 5、(80H)=81H
6、片外(8FH)=0F0H 7、(SP)=0B1H 8、(A)=0FH 9、(A)=8FH,(R)=55H 3-18 解:(1)(A)=55H,(CY)=1,(OV)=0,(AC)=1,(P)=0 (2)(A)=56H,(Cy)=1,(OV)=0,(AC)=1,(P)=1 (3)(A)=00H,(CY)=0,(OV)=0,(AC)=0,(P)=0 (4)(R1)=79H,(CY)=1,(OV)=0,(AC)=0,(P)=0
(5)(78H)=0DCH,(CY)=1,(OV)=0,(AC)=0,(P)=0 (6)(B)=01H,(A)=E0H,(OV)=1,(CY)=0 (7)(A)=1EH,(B)=00H,(OV)=1,(CY)=0,(P)=1 (8)(78H)=58H (9)(A)=7FH,(P)=1 (10)(80H)=14H 3-21
ORG 0000H LJMP START ORG 1000H
START:MOV R0,#30H MOV R1,#32 NEXT: MOV A,@R0 MOVX DPTR,#2000H MOVX @DPTR,A
INC R0 INC DPTR
DJNZ R1, NEXT END
3-23 ORG 0000H LJMP START
ORG 1000H START: CLR C
MOV R0,#ST1
MOVX A,@R0;取ST1
MOV R1,A;ST1保存在R1里 INC R0
MOVX A,@R0 ;取ST2 MOV R2,A SUBB A, R1 INC R0
JC COMP ;ST1>ST2 MOV A,R2 SJMP ODD COMP:MOV A,R1 ODD:MOVX @R0,A END
3-27 ORG 0000H LJMP START ORG 0100H
START:MOV A,#01H LOOP1:MOV R2,#10 LOOP2:MOV P1,A ACALL DELAY MOV P1,#00H
ACALL DELAY DJNZ R2,LOOP2 RL, A
AJMP LOOP1
DELAY: MOV R7,#10 DL2:MOVR6,#200 DL1:MOV R5,#250
DJNZ R5,$
DJNZ R6,DL1
DJNZ R7,DL2 RET END
4-9 (1)MOV IE,#83H
(2) MOV IE,#92H (3)MOV IE,#09FH (4)MOV IE,#00H
4-11 ORG 0000H AJMP MAIN ORG 0003H AJMP INTT0 ORG 0100H
MAIN: MOV SP,#60H
SETB EA SETB EX0 SETB IT0 MOV A,#00H MOV P1,A HALT:SJMP HALT
ORG 0200H INTT0:PUSH,ACC MOV R0,#08H
MOV A,#01H MOV P1,A
LOOP1: ACALL DELAY1 RL A
DJNZ R0,LOOP1 RETI
DELAY1:1S延时子程序
RET END
5-5 解:
ORG 0000H AJMP MAIN ORG 000BH AJMP INQP ORG 0030H MAIN: MOV SP,#60H MOV TMOD,#00H MOV TH0,#FFH
MOV TL0,#1BH SETB TR0 SETB ET0 SETB EA HERE: AJMP HERE INQP: MOV TH0,#0FFH MOV TL0,#1BH
CPL P1.0 RETI END
5-7解:
, 中断10次实现360us的定时
ORG 0000H AJMP MAIN ORG 000BH LJMP int0 ORG 0030H MAIN: MOV SP,#60H SETBP1.0
MOV R0,#0AH;连续中断10次 MOV TMOD,#02H;T0工作模式2
SETB ET1 SETB EA
MOV TL0,#0DCH;
MOV TH0,#0DCH;36us初值
SETB TR0;启动T0 SJMP $ ORG 0100H
INT0:JNB P1.0, LOOP;低电平转 DJNZ R0, LOOP1 MOV R0,#0AH CPL P1.O
MOV TH0, #0DCH AJMP BACK
LOOP1:CJNE R0,#1,BACK
MOV TH0,#0D8H;提前将低电平定时初值装入计数器 AJMP BACK LOOP: CPL P1.0 BACK:RETI END
6-5 解:1帧数据有10位,1分钟传送9600*60=576000bit,即576000/10=57600个字节。
6-6解:1帧有10位,1分钟可传1800*10=18000位,每秒钟可传18000/60=300bit/s
6-7设发送位于20H首地址的数据,接收的数据存放到50H开始的单元 解:scon:01010000B=50H;允许接收
波特率=
ORG 0000H AJMP MAIN 0RG 0023H LJMP COM ORG 0100H
MAIN:MOV TMOD,#20H MOV TL1,#0E8H MOV TH1,#0E8H MOV SCON,#50H MOV IE,90H
SETB TR1 MOV R0,#20H MOV R7,#0AH MOV R1,#50H MOV A,@R0 MOV SBUF,A HERE:SJMP HERE COM:JBC TI,SEND CLR RI
MOV A,SBUF MOV @R1,A INC R1 AJMP BACK SEND:INC R0 MOV A,@R0 MOV SBUF,A BACK:RETI END
8-1 解:分辨率=
8-2 解:
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