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PLEASE ANSWER IT ASAP MOD = 11 EFFECTIVE ADDRESS CALCULATION MOD=00 MOD=01 R/M W=0 W=1 R/M MOD = 10 000 001 010 011 100 101
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MOD = 11 EFFECTIVE ADDRESS CALCULATION MOD=00 MOD=01 R/M W=0 W=1 R/M MOD = 10 000 001 010 011 100 101 110 111 AL CL DL BL AH CH DH BH AX DX BX SP BP SI DI 000 001 010 011 100 101 110 111 (BX) + (SI) (BX) + (DI) (BP)+(SI) (BP)+(DI) (SI) (DI) DIRECT ADDRESS (BX) (BX) + (SI) + DB (BX) + (DI) + DB (BP)+(SI) + D8 (BP)+(DI) + D8 (SI) + D8 (DI) + D8 (BP)+D8 (BX) + D8 (BX) + (SI) + D16 (BX) + (DI) +D16 (BP)+(SI) + D16 (BP)+(DI) +D16 (SI) + D16 (DI) + D16 (BP)+D16 (BX) + D16 Figure: R/M vs MOD Chart for MOV: 100010 instruction REGISTER SELECT (SEE FIGURE 3-7) BYTE 3 BYTE 4 BYTE 1 1000 10 OP CODE DW MOD ol LOW DISPLACEMENT HIGH DISPLACEMENT REG RM OR DIRECT ADDRESS LOW BYTE DIREST ADDRESS HIGH BYTE (5 BITS) ADDRESSING MODE (SEE FIGURE 3.8) BYTENORD DATA O = BYTE 1 - WORD DIRECTION TO FROM REG O = FROM 1 - TO OPERATION CODE 3 a) Let us assume that the CS and IP values of the base and offset of an ISR are 1210h and 1010h. The ISR is 8 lines of code where each line takes 2 bits to be stored. At which address, shall we find the IRET instruction? (2.5 marks) b) Describe the process that would take place if an interrupt arises at the IRO pin of 8259A while servicing an interrupt that arose at IR3 pin of 8259A. You are expected to refer to the 3 registers in particular. (2.5 marks) You have been given two instructions, viz i) MOV DX, [10101 H) and ii) MOV AX, [10000 H]. Now answer the following questions based on these two instructions 5. a) Suppose you have an 8086 which is running at 60 MHz having a duty cycle of 35%. Now how many nanoseconds in total would the 8086 take to complete the two given "MOV operations". Also find out for how long (in nanoseconds) does each clock pulse stay low? Show all workings. (2+1 = 3 marks) b) Write overall in which bus cycle the pins AO and BHE' take in what values for the given two instructions. (1+1 = 2 marks)Step by Step Solution
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