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# Solution manual 8051 microcontroller by mazidi

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### Solution manual 8051 microcontroller by mazidi

1. 1. Microcontroller Solutions Chapter 2 Section 2.1:1. 8 bit 2. 8 bit 3. 8 bit 4. PSW (Program Status Word) is of 16 bit. 5. Necessary (for literal value). 6. 28H and it is kept in accumulator. 7. (a),(d),(g) are illegal and for f only 0 is required before F5H 8. (c),(d) are illegal. 9. 44H and kept in Accumulator (A). 10.1EH and kept in Accumulator (A). Section 2.4 21.0000H 22.Program counter will look for the location 0000H and if the program is not starting for that address, it will consider that there is no program written so program has to start from the location 0000H. 26.Lowest Memory is 0000H and the Highest memory is FFFFH.
2. 2. Microcontroller Solutions From Ali Akbar Siddiqui. Sir Syed University of Eng& Tech Section 2.5:29.Solved below, (Data) E A R T H (Locations) 200 201 202 203 204 9 8 7 6 5 205 206 207 208 209 20A Section 2.6:31.8 bit 32.And 33. D7 D6 CY AC D5 F0 D4 RS1 D3 RS0 D2 OV D1 -- D0 P 34.We know that in 8051 registers are of 8 bits CY Flag is raised when the carry is generated beyond past the max value that a register can store like FFH + 1. 35.AC is raised when a carry is generated from D3 to D4. Like Mov a,#0FH Add a,#1
3. 3. Microcontroller Solutions 36.CLR C ;CY=0 CPL C; CY 37.G a. CY=1 b. CY=0 c. CY=0 38.ORG 0000H MOV A,#55H ADD A,#55H ADD A,#55H ADD A,#55H ADD A,#55H END 39.RS0 and RS1. 40.On Startup Stack Location is 07H. 42.24 Bytes. 43.Register Bank 0 44.Register Bank 0 from 00H to 07H. Register Bank 1 from 08H to 0FH. Register Bank 2 from 10H to 17H. Register Bank 3 from 18H to 1FH. 45.(a) 04H (b) 00H (c) 07H (d) 05H
4. 4. Microcontroller Solutions 48. INSTRUCTIONS PUSH 0 PUSH3 PUSH 7 POP 3 POP 7 POP 0 STACK Pointer before execution. 07H 08H 09H 0AH 09H 08H Stack pointer after execution. 08H 09H 0AH 09H 08H 07H STACK 66H 7FH 5DH 49.NO. POP 7 POP 3 POP 0 REST OF THE PROGRAM WILL REMAIN THE SAME. 50.After execution of ( Mov SP,#70H ), Stack Pointer location has now become 70H instead of 07H. INSTRUCTIONS Push 5 Push 2 Push 7 Pop 7 Pop 2 Pop 5 STACK Pointer before execution. 70H 71H 72H 73H 72H 71H Stack pointer after execution. STACK 71H 72H 73H 72H 71H 70H 66H 7FH 5DH
5. 5. Microcontroller Solutions Chapter 3:Section 3.1:12.MOV R1,#100 HERE1:MOV R2,#10 HERE:DJNZ R2,HERE DJNZ R1,HERE1 13.MOV R1,#100 HERE2MOV R2,#100 HERE1:MOV R3,#10 HERE:DJNZ R3,HERE DJNZ R2,HERE1 DJNZ R1,HERE2 14.Multiplication is taking place so, 200*100=20,000 (times). 15. 16. -128 Bytes 127 Bytes Section 3.2:17. 18. 19. 20. 21. 22. 3 byte 2 byte 2 bytes 64Kb 2 bytes 1 byte
6. 6. Microcontroller Solutions 23.That’s because Stack works on the concept of LIFO so if the push is implied 2 times for instance, then pop must be used 2 times exactly. Section 3.3:27. T = 1.2usec F = 1/T = 833.333KHz System frequency = 833.333KHz*12 = 10MHz 28. F = 18MHz F =18MHZ/12 =1.5MHz T = 1/F =1/1.5MHz T = 0.666usec 29. F = 12MHz F =12MHZ/12 =1MHz T = 1/F =1/1.5MHz T = 1usec 30. F = 25MHz F =25MHZ/12 = 2.08MHz T = 1/F =1/2.08MHz T = 0.48usec 32. F = 11.0592MHz F =11.0592MHZ/12 = 921.6KHz T = 1/F = 1/921.6KHZ T = 1.085usec
7. 7. Microcontroller Solutions DELAY:MOV R3,#150 1 machine cycle HERE:NOP 1 mc NOP 1 NOP 1 DJNZ R3,HERE 2 mc RET 2 The time delay of the HERE loop is [150(2+1+1+1)]*1.085usec=0.813msec Now for the instruction outside the loop (mov and ret), (2+1)*1.085Usec = 3.25usec Now 0.813ms + 3.25usec = 0.8162msec 33. F = 16MHz F =16MHZ/12 = 1.33333MHz T = 1/F = 1/1.33333MHZ T = 0.75usec DELAY:MOV R3,#200 1 HERE:NOP 1 NOP 1 NOP 1 DJNZ R3,HERE 2 RET 2 The time delay of the HERE loop is [200(2+1+1+1)]*0.75usec=0.75msec Now for the instruction outside the loop (mov and ret), (2+1)*0.75usec = 2.25usec Now 0.75ms + 2.25usec = 0.75225msec 34. F = 11.0592MHz F =11.0592MHZ/12 = 921.6KHz T = 1/F = 1/921.6KHZ T = 1.085usec
8. 8. Microcontroller Solutions DELAY:MOV R5,#100 1 BACK: MOV R2,#200 1 AGAIN:MOV R3,#250 1 HERE:NOP 1 NOP 1 DJNZ R3,HERE 2 DJNZ R3,AGAIN 2 DJNZ R3,BACK 2 RET 2 The time delay of the HERE loop is [250(2+1)]*1.085usec=1.085msec The time delay of the AGAIN loop it repeats 200 times so , 1.085msec*200 = 0.217 + (3*200*1.085usec) = 0.2176s The time delay of the BACK loop, it is repeated 100 times so 0.2176*100=21.76sec Time Delay = 21.76sec 35.Try it yourself it’s just like 34 with only 2 loops instead of 3 loope. 36. To 39. For the problems from 36 to 39 everything remains the same except the time delay that is changed due to the change of Microcontroller now you must take Clock of DS89C420/30 Microcontroller i.e equal to 1 rather than 12 that was for 8051. 40.Yes it is 12 times faster because it only have 1 clock and on the other hand 8051 have 12 clocks so if we decrease the clocks our microcontroller becomes faster.
9. 9. Microcontroller Solutions Chapter 4:Section 4.1:1. 40 TH TH 2. VCC  40 PIN And GND9 3. 32 Pins 4. 8 Pins and from 32 to 39. 5. 8 Pins and from 1 to 8. 6. 8 Pins and from 21 to 28. 7. 8 Pins and from 10 to 17. 8. Input 9. P0 (Port 0) 10.P1 (Port 1) PIN 11.ORG 000H MOV P1,#0FFH; MAKE IT AN INPUT PORT MOV A,P1 MOV P2,A MOV P0,A MOV P3,A END 12.ORG 000H MOV P2,#0FFH; MAKE IT AN INPUT PORT MOV A,P2 MOV P1,A MOV P0,A END
10. 10. Microcontroller Solutions 13.P3.0 AND P3.1 14.0000H is the address upon reset. 15. (A) ORG 0000H BACK:MOV A,#0AAH MOV P1,A MOV P2,A CALL DELAY MOV A,#55H MOV P1,A MOV P2,A SJMP BACK (B) ORG 0000H MOV A,#0AAH BACK:MOV P1,A MOV P2,A CALL DELAY CPL A MOV P1,A MOV P2,A SJMP BACK Section 4.2:16.All ports are bit addressable. 17.The advantages for Bit addressable mode is that u con manipulate aa single bit without disturbing and other bits of the port by using Setb and Clr. 18.Setb P1.X OrClr P1.X where X can vary from 0 to 7.
11. 11. Microcontroller Solutions 19.No , a whole port cannot be complemented at a time. 20.ORG 0000H SETB P1.2 SETB P1.5 BACK:CALL DELAY CPL P1.2 CPL P1.5 SJMP BACK END 21.ORG 0000H SETB P2.5 SETB P1.7 SETB P1.3 BACK:CALL DELAY CPL P1.3 CPL P1.7 CPL P2.5 SJMP BACK END 22.ORG 0000H SETB P1.3 BACK:JB P1.3,HERE SJMP BACK HERE:MOV A,#55H
12. 12. Microcontroller Solutions MOV P2,A END 23.ORG 0000H SETB P2.7 BACK:JNB P2.7,HERE SJMP BACK HERE:MOVA,#55H MOV P0,A CALL DELAY MOC A,#0AAH MOV P0,A SJMP HERE END 24.ORG 0000H SETB P2.0 JNB P2.0,HERE MOV A,#99H MOV PI,A SJMP BACK HERE:MOV A,#66H MOV P1,A BACK: END
13. 13. Microcontroller Solutions 25.ORG 0000H SETB P1.5 AGAIN:JB P1.5,HERE SJMP AGAIN HERE:CLR PI.3 CALL DELAY SETB P1.3 CALL DELAY CLR P1.3 END 26.ORG 0000H BACK:MOV C,P1.3 ;C IS FOR CARRY FLAG. MOV P1,4,C SJMP BACK END 27. TH 5 Bit 28.ORG 0000H BACK:MOV C,P1.7 MOV P1.0,C MOV C,P1.6 MOV P1.7,C SJMP BACK
14. 14. Microcontroller Solutions Chapter 5:Section 5.1 and 5,2:3. See on page 123 Figure 5-1, and on page 124 figure 5-2. 4. Register bank 1 , 2 , 3 share the space with stack because by default stack starts from 07H and after increment data is stored in 08H on the other hand Register bank 1 address starts from 08H as well see page 123 fig 5-1. 6. It copies the contents of the location 0F0H into the accumulator rather than the value 0F0H. 7. Same as question nos 6. 8. ORG 0000H MOV R0,#50H MOV R1,#40H MOV R3,#30H PUSH 00H PUSH 01H PUSH 03H POP 1DH POP 1EH POP 1FH END 9. Registers R0 and R1. 10.ORG 0000H MOV A,#0FFH MOV R7,#32 MOV R0,#50H NAME:MOV @R0,A
15. 15. Microcontroller Solutions INC R0 CALL DELAY DJNZ R7,NAME 11.ORG 0000H MOV DPTR,#400H MOV R7,#10 MOV R0,#30H HERE:CLR A MOVC A,@A+DPTR CALL DELAY INC DPTR MOV @R0,A INC R0 DJNZ R7,HERE BACK:SJMP BACK END 12.ORG 0000 MOV P1,#0FFH MOV A,P1 MOV R0,A ; R0=x MOV B,R0 ; B=x MUL AB ; MULTIPLY A WITH B ANSWER STORE IN A=x*x DA A MOV R1,A ;R1=x*x Or x^2 MOV A,R0 ;A=x MOV B,#2 ; B=2 MUL AB ; A=2*x DA A MOV R7,#5
16. 16. Microcontroller Solutions ADD A,R1 ; A=x^2 + 2*x DA A ADD A,R7 ;A=x^2 + 2*x + 5 END 13.ORG 0000H LJMP MAIN ORG 20H MYDATA: DB 06,09,02,05,07 ORG 300H Main:MOV R0,#30H MOV DPTR,#MYDATA MOV R7,#5 HERE:CLR A MOVC A,@A+DPTR CALL DELAY INC DPTR PUSH 0E0H; Push the Accumulator into stack DJNZ R7,HERE POP 01 POP 02 POP 03 POP 04 POP 0E0H ADD A,R1 ADD A R2 ADD A,R3 ADD A,R4; A Holds the added data.
17. 17. Microcontroller Solutions MOV @R0,A END Section 5.3:14.INVALID 15.VALID 16.VALID 17.All ports are bit addressable. 18.See for answer on page124 figure 5-2. 19. (b),(c),(d),((f),(g),(h) are valid. 20.ORG 0000H AGAIN:SETB P1.5 CALL DELAY CALL DELAY CALL DELAY CLR P1.5 CALL DELAY SJMP AGAIN DELAY:MOV R1,#240 HERE:DHNZ R1,HERE RET END 21.ORG 0000H AGAIN:SETB P2.7 CALL DELAY CALL DELAY CALL DELAY
18. 18. Microcontroller Solutions CALL DELAY CLR P2.7 CALL DELAY SJMP AGAIN DELAY:MOV R1,#240 HERE:DJNZ R1,HERE RET END 22.ORG 0000H SETB P1.4 HERE:JNB P1.4,HERE CMD:SETB P2,7 CALL DELAY CLR P2.7 CALL DELAY SJMP CMD DELAY:MOV R1,#240 HERE:DJNZ R1,HERE RET END 23.ORG 0000H SETB P2.1 HERE:JB P1.4,HERE MOV P0,#55H SJMP \$ END
19. 19. Microcontroller Solutions 24.80H TO 87H 25.90H TO 97H 26.A0H TO A7H 27.B0H TO B7H 28.Not bit addressable register. 29.88H TO 8FH 30.E0H TO E7H 31.F0H TO F7H 32.D0H TO D7H 33.(a) P0 (b)87H (c)TCON (d)TCON (e)P1H (f)P2 (g)P2 (h)P3 (i)PSW (j)PSW (K)B 34.ORG 0000H SETB RS1; FOR SELECTING REGISTER BANK 2 SETB RS0;FOR SELECTING REGISTER BANK 2 MOV R3,A MOV R5,B END 35.CLR 0D7H 37.See example 5-14. 38.To check the carry flag there are instructions namely JC and JNC. 39. And 40 see example 5-14. 41.CY0D7H P0D0H AC0D6H OV0D2H 42.For this question see page 124 fig 5-2. 46.For this question see page 123 fir 5-1. 47.(a)20H (b)28H (c)18H (d)2DH (e)53H (f)15H (g)2CH (h)2AH (i)14H (j)37H (k)7FH
20. 20. Microcontroller Solutions 50.MOV 04,C 51.MOV 16H,0D6H ; Auxiliary carry 52.MOV 12H,0D0H 53.ORG 0000H JB ACC.0,HERE SJMP AGAIN HERE :JB ACC.1,HERE1 SJMP AGAIN HERE1:MOV B,#4 DIV AB AGAIN: END 54.ORG 0000H JB ACC.7,LCD_DISPLAY SJMP NACK LCD_DISPLAY: NACK: END 55.ORG 0000H JB 0F7H,HERE SJMP NACK LCD_DISPLAY: NACK: END
21. 21. Microcontroller Solutions 56. (A)ORG 0000H MOV R0,#24H; PROGRAM IS DONE WITH THE HELP OF FIG 51. MOV A,#0FFH MOV @R0,A MOV R0,#25H MOV A,#0FFH MOV @R0,A END (B)ORG 0000H SETB 20H SETB 21H SETB 22H SETB 23H SETB 24H SETB 25H SETB 26H SETB 27H SETB 28H SETB 29H SETB 2AH SETB 2BH SETB 2CH SETB 2DH SETB 2EH SETB 2FH END
22. 22. Microcontroller Solutions 57.ORG 0000H MOV B,#8 DIV AB CJNE B,#00,HERE SJMP FIN HERE:MOV R0,A FIN: END 58.ORG 0000H MOV R1,#8 BACK:MOV A,R2 RRC A; Rotate Right through carry means instead of 8 bit rotation it JC HERE ;include carry flag as an MSB(most significant bit). INC R0 HERE:DJNZ R1,BACK END Section 5.4:67.ORG 0000G MOV A,#55H MOV R0,#0C0H MOV R7,#16 HERE:MOV @RO,A INC R0 CALL DELAY DJNZ R7,HERE END
23. 23. Microcontroller Solutions 68.ORG 0000H MOV R7,#16 MOV R1,#60H MOV R0,#0D0H HERE:MOV A,@R1 INC R1 MOV @R0,A INC R0 CALL DELAY DJNZ R7,HERE END
24. 24. Microcontroller Solutions Chapter 6:Section 6.1:1. (a) AC=1 (b)AC=1 (c)AC=1 (e)AC=1 Cy= 0 CY=0 CY=1 CY=1 2. Already been done in the lab, see your lab files. 3. ORG 0000H MOV DPTR,#MYDATA MOV R7,#9 MOV R2,#00H MOV R3,#00H BACK:CLR A MOVC A,@A+DPTR MOV R3,A PUSH 03 INC DPTR DJNZ R7,BACK MOV R7,#9 CLR A AGAIN:POP 00 ADD A,R0 JNC HERE1 INC R2 HERE1:DJNZ R7,AGAIN MOV R3,A SJMP \$ ORG 250H
25. 25. Microcontroller Solutions MYDATA: DB 53,94,56,92,74,65,43,23,83 END 4. Just use DA( Decimal Adjust instruction in question 3) 5. (a) AND (b) ORG 0000H MOV R0,#40H MOV R7,#16 MOV A,#55H HERE:MOV @R0,A INC R0 DJNZ R7,HERE MOV R7,#16 MOV R1,#60H MOV R0,#40H CLR A BACK:ADD A,@R0 JNC HERE1 INC @R1 HERE1:INC R0 DJNZ R7,BACK MOV R0,#61H MOV @R0,A SJMP \$ END 9. ORG 0000H MOV R4,#00H
26. 26. Microcontroller Solutions MOV A,#48H MOV R0,#9AH ADD A,R0 MOV R7,A MOV A,#0BCH MOV R0,#7FH ADDC A,R0 ;ADC is add with carry, if by adding A and R0 MOV R6,A ;Carry generates .What will it do , it will add both A and R0 With MOV A,#34H;the Upside to it, it will also add the carry flag if it generates. MOV R0,#89H ADDC A,R0 MOV R5,A JNC HERE INC R4 HERE:MOV R0,#40H MOV A,R4 MOV @R0,A INC R0 MOV A,R5 MOV @R0,A INC R0 MOV A,R6 MOV @R0,A INC R0 MOV A,R7 MOV @R0,A END Here R5=BE,R6=3BH,R7=E2.The and is BE3BE2H.
27. 27. Microcontroller Solutions 12.ORG 0000H mov a,#77 mov b,#34 mulab end 13.ORG 0000H mov a,#77 mov b,#3 divab end 14.No, Only on A and B. 15.ORG 0000H MOV DPTR,#MYDATA MOV R0,#30H CALL TRANSFER CALL ADDITION CALL AVERAGE LJMP FIN TRANSFER: MOV R7,#9 HERE: CLR A MOVC A,@A+DPTR MOV @R0,A INC DPTR INC R0 DJNZ R7,HERE
28. 28. RET
29. 29. Microcontroller Solutions ADDITION: MOV R7,#9 MOV R0,#30H CLR A HERE1: ADD A,@R0 INC R0 DJNZ R7,HERE1 RET AVERAGE: MOV B,#9 DIV AB MOV R7,A RET ORG 250H MYDATA: DB 3,9,6,9,7,6,4,2,8 FIN: END Section 6.3:23.(a) A=40h (b)A=F6H (c)A=86H Rest do it yourself just use Keil write instruction and see the result in project window. 24.Just as in Question no 23 write those instruction in Kiel and view the result of accumulator in Project window.
30. 30. Microcontroller Solutions 27.There is no such instruction like CJE. 28.In this question you must monitor the status of the carry flag after the execution of CJNE. Write a program below and check the status of the carry flag in the PSW resister. (a) ORG 0000H BACK:MOV A,#25H ;Here the carry flag will go high.Always remember CJNE A,#44H,over ; that carry will only go high when the value of source SJMP BACK ; of CJNE instruction is greater than its destination. OVER: END ; (b) ORG 0000H back:mov a,#0ffh ;Here carry flag will not go high as the value of the cjne a,#6fh,over ;destination is greater than that of the source. sjmp back over: end Rest of the parts of question 28 now you can do them on your own. 30.(a)MOV A,#56H SWAP A; What swap do is swap the upper and lower nibble now A becomes ; A=65H RR A RR A
31. 31. Microcontroller Solutions (C)CLR C MOV A,#4DH ; A=0100 1101B SWAP A; A=D4 OR A= 1101 0100B RRC A ;9 BIT ROTATION, A= 0 0110 1010B. You can see the zero before ;8-bit that is the carry bit that you included through RRC instructin. RRC A ; A= 0 0011 0101b RRC A ; A= 1 0001 1010b 32.ORG 0000H MOV P1,#0FFH MOV R7,#8 MOV A,P1 AGAIN:RRC A JC HERE INC R0 HERE:DJNZ R7,AGAIN END 33.ORG 0000H MOV R7,#8 MOV A,#68H AGAIN:RRC A JC HERE INC R0 HERE:DJNZ R7,AGAIN END 34.ORG 0000H MOV R7,#8 MOV A,#68H
32. 32. Microcontroller Solutions AGAIN:RLC A JC HERE INC R0 HERE:DJNZ R7,AGAIN END 40.ORG 0000H MOV R7,#9 MOV P1,#0FFH AGAIN1:MOV A,P1 ANL A,#0FH ORL A,#30H MOV R1,A MOV R4,#34H HERE:CJNE A,#30H,HERE1 SJMP BACK HERE1:CJNE A,#31H,HERE2 SJMP BACK HERE2:CJNE A,#32H,HERE3 SJMP BACK HERE3:CJNE A,#33H,HERE4 SJMP BACK HERE4:CJNE A,#34H,HERE5 SJMP BACK HERE5:CJNE A,#35H,HERE6 SJMP BACK HERE6:CJNE A,#36H,HERE7 SJMP BACK HERE7:CJNE A,#37H,HERE8 SJMP BACK
33. 33. Microcontroller Solutions HERE8:CJNE A,#38H,HERE9 SJMP BACK HERE9:CJNE A,#39H,HERE10 SJMP BACK HERE10:ANL A,#0FH ADD A,#37H CJNE A,#41H,HERE11 SJMP BACK HERE11:CJNE A,#42H,HERE12 SJMP BACK HERE12:CJNE A,#43H,HERE13 SJMP BACK HERE13:CJNE A,#44H,HERE14 SJMP BACK HERE14:CJNE A,#45H,HERE15 SJMP BACK HERE15:CJNE A,#46H,AGAIN1 BACK: MOV P2,A SJMP \$ END
34. 34. Microcontroller Solutions 43.This program is same as the check-sum program, right next to 6-36 example. The only difference is here you have to find the checksum byte of a whole sentence and in the program you had to find the check sum of HEX values. I point out the difference that has to make for this program the rest of the program will remain the same, -----------------------------------------------------DATA_ADDR EQU 400H COUNT EQU 31 ; Nos of characters in the whole sentence. RAM_ADDR EQU 20H ORG 0000H CALL COPY_DATA CALL CAL_CHKSUM CALL TEST_CHKSUM COPY_DATA: --- ;THERE SUBROUTINES ARE PRESENTIS THE BOOK Pg 172. --------------------------------------RET CAL_CHKSUM:------------------------------------------RET TEST_CHKSUM:--------------------------------------------RET ORG 400H MYBYTE: DB ‘Hello, my fellow World citizens’ END
35. 35. Microcontroller Solutions 46.ORG 0000H MOV R7,#9 MOV P1,#0FFH AGAIN1:MOV A,P1 ANL A,#0FH ORL A,#30H MOV R1,A MOV R4,#34H HERE:CJNE A,#30H,HERE1 SJMP BACK HERE1:CJNE A,#31H,HERE2 SJMP BACK HERE2:CJNE A,#32H,HERE3 SJMP BACK HERE3:CJNE A,#33H,HERE4 SJMP BACK HERE4:CJNE A,#34H,HERE5 SJMP BACK HERE5:CJNE A,#35H,HERE6 SJMP BACK HERE6:CJNE A,#36H,HERE7 SJMP BACK HERE7:CJNE A,#37H,HERE8 SJMP BACK HERE8:CJNE A,#38H,HERE9 SJMP BACK HERE9:CJNE A,#39H,AGAIN1 BACK: MOV P2,A
36. 36. Microcontroller Solutions SJMP \$ END
37. 37. Microcontroller Solutions Chapter 9:1. 2. 3. 4. 5. 6. 7. 2 timers 16 bit, Timer 0 and Timer 1. TH0 AND TL0. TH1 AND TL1. NO,These register are not bit addressable. 8 bit TMOD is used to initialize the Timer0 or Timer1 and also Mode of timer to which we have to use.it also let us to select that weather we have to use Timer or Counter. 8. No 9. Use the Figure 9-3 of TMOD register. Gate C/T M1 M0 Gate C/T M1 M0 0 1 1 0 0 1 1 0 10.Just Divide the XTAL values with 12 for frequencies and for the time period take the inverse of frequence. 11.(a)13 bit (b)16 bit (c)8 bit 12.(a)Mode 08192 in decimal you can find out by 2^(13)=8192, that is because our Mode is of 13 bit and 2000H (b)Mode 165536 in decimal 2^(16)=65536, it is 16 bit and HEX Value is FFFFH. (c)Mode 2256 in decimal 2^(8)-256, it is 8 bit and HEX value i