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單片機原理與接口技術(shù)習題參考答案-wenkub

2023-07-12 13:49:00 本頁面
 

【正文】 ,AH9AH9: JC AH10 SUBB A,07HAH10: AJMP $ END ORG 0000H AJMP START ORG 0030HSTART: MOV R0,20H MOV R2,00H MOV R3,00H MOV R4,0FH MOV 40H,41H MOV 60H,61HLOOP1: MOV A,R0 JB ,LOOP3LOOP2: INC R2 MOV R1,40H MOV R1,A INC 40H AJMP LOOP4LOOP3: INC R3 MOV R1,60H MOV R1,A INC 60HLOOP4: INC R0 DJNZ R4,LOOP1 MOV 40H,R2 MOV 60H,R3 END BUF EQU 10H ORG 0000H AJMP START ORG 0030HSTART: MOV A,38H ADD A,2AH CLR C SUBB A,19H MOV BUF,A END DATA1 EQU 30H DATA2 EQU 50H ORG 0000H SJMP START ORG 0030HSTART: MOV R0,DATA1 MOV R1,DATA2 MOV R2,14HLOOP: MOV A,R1 XCH A,R0 XCH A,R1 INC R0 INC R1 DJNZ R2,LOOP END ORG 0000H AJMP START ORG 0030HSTART: MOV A,6 MOV DPTR,TAB DEC A RL A MOV B,A MOVC A,A+DPTR MOV R6,A MOV A,B INC A MOVC A,A+DPTR MOV R7,A ORG 1000HTAB: DW 1,4,9,16,25,36,49,64,81,100 DW 121,144,169,196,225,256,289,324,400 END ORG 0000H AJMP START ORG 0030HSTART: MOV A,04H MOV DPTR,TAB MOVC A,A+DPTR ORG 0100HTAB: DB 40H,79H,24H,30H,19H,12H,02H,78H,00H,1BH END ORG 0000H AJMP START ORG 0030HSTART: MOV R0,20H MOV R1,40H MOV R2,00HLOOP1: MOV A,R0 CJNE A,0DH,LOOP2 SJMP EDLOOP2: CJNE A,43H,LOOP3 INC R2LOOP3: INC R0 AJMP LOOP1ED: MOV A,R2 MOVX R1,A END ORG 0000H SJMP START ORG 0030HSTART: MOV A,R3 RL A MOV DPTR,JMPTAB JMP A+DPTRJMPTAB:AJMP M00 AJMP M01 AJMP M02 ...... AJMP M7F END ORG 0000H SJMP START ORG 0030HSTART: MOV R0,30H MOV R7,0FH CLR F0LOOP: MOV A,R0 MOV 10H,A INC R0 MOV 11H,R0 CLR C SUBB A,R0 JZ NEXT JC NEXT MOV R0,10H DEC R0 MOV R0,11H INC R0 SETB F0NEXT: DJNZ R7,LOOP JB F0,START END習題五一、 單項選擇題1. B2. C3. C4. C5. C6. A7. C8. A9. A10. C二、 填空題1. 22. 0CH3. 電平觸發(fā)、脈沖觸發(fā)4. 125. TL06. TF17. SCON8. 停止位9. 自動重裝入初值8位定時器/計數(shù)器10. 方式1三、簡答題 1. P0口有一個輸出鎖存器,兩個三態(tài)輸入緩沖器。 2. 讀鎖存器指令時,引腳內(nèi)的輸出鎖存器Q端狀態(tài)經(jīng)讀鎖存器輸入緩沖器進入內(nèi)部總線,由CPU根據(jù)指令要求把Q端狀態(tài)值修改后,又重新寫回鎖存器中,被稱為讀—修改—寫指令。我們把引起中斷的原因,或者能夠發(fā)出中斷請求信號的來源統(tǒng)稱為中斷源。H為使系統(tǒng)能及時響應并處理發(fā)生的所有中斷,系統(tǒng)根據(jù)引起中斷事件的重要性和緊迫程度,硬件將中斷源分為若干個級別,稱作中斷優(yōu)先級。 5. TCON、SCON、IE、IP。 8. 2個。M1M0=10,工作方式2,2個獨立8位計數(shù)器。方式2為11位移位收發(fā)器,波特率=f/32或者f/64。四、應用編程題 1. ORG 0000H LJMP MAIN ORG 0003H LJMP INT01 ORG 000BH LJMP INT ORG 0030H MAIN: MOV R0,20 MOV SP,50H MOV TMOD,01H MOV TH0,3CH MOV TL0,0B0H SETB EA SETB EX0 SETB IT0 SJMP $ INT01: SETB ET0 SETB TR0 CLR RETI INT: MOV TH0,3CH MOV TL0,0B0H DJNZ R0,DE DE0: SETB CLR ET0 CLR TR0 MOV R0,20 DE: RETI END 2. (1) 1FH,08H (2) ORG 0000H LJMP MAIN ORG 000BH LJMP INT ORG 0030H MAIN: MOV TMOD,00H MOV SP,50H MOV TH1,1FH MOV TL1,08H SETB EA SETB ET0 SETB TR0 SJMP $ INT: MOV TH1,1FH MOV TL1,08H CPL RETI END 3. ORG 0000H LJMP MAIN ORG 0030H MAIN: MOV TMOD,90H MOV TL1,00H MOV TH1,00H MOV R0,50H JB ,$ SETB TR1 JNB ,$ JB ,$ CLR TR1 MOV R0,TL1 INC R0 MOV R0,TH1 SJMP $ END 4. ORG 0000H LJMP MAIN ORG 000BH LJMP INT ORG 001BH LJMP INT11 ORG 0030H MAIN: MOV TMOD,03H MOV TH1,0C8H MOV TL1,64H MOV SP,50H SETB EA SETB ET0 SETB TR0SETB TR1 SJMP $ INT: MOV TL1,64H CPL RETI INT11: MOV TH1,0C8H CPL RETI END 5. ORG 0000H LJMP MAIN ORG 0030H MAIN: MOV R0,50H MOV R1,01H MOV R7,10H MOV TMOD,20H MOV TH1,0FEH MOV TL1,0FEH SETB TR1 CLR ES MOV SCON,40H MOV PCON,00H LOOP: MOV A,R1 MOV R0,A MOV SBUF,A WAIT: JBC TI,CONT SJMP WAIT CONT: INC R0 INC R1 DJNZ R7,LOOP END習題六一、 單項選擇題1. B2. C3. B4. D5. A二、 填空題1. 程序存儲器,數(shù)據(jù)存儲器,642.3. ,4. MOVX A ,Ri、MOVX A ,DPTR5. MOVC A ,A+PC、MOVC A ,A+ DPTR6. R0,R1,DPTR7. 接地8. 線選法、譯碼法三、 簡答題1. 對于無片內(nèi)ROM型單片機或片內(nèi)程序存儲器不夠用的情況,需要擴展片外程序存儲器。四、 設計題1.6116地址范圍:0000H——07FFH( 0000 0000 0000 0000B —— 0000 0111 1111 1111B)2.2764地址范圍: 40000H—5FFFFH 6264地址范圍: 200
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