Q1. a) Let us assume that the CS and IP values of the base and offset of an ISR are 1230h and 2000h. The ISR is 10 lines of code where each line takes 4 bits to be stored. At which address, shall we find the IRET instruction? b) Describe the complete process of how an interrupt of type n (INTn) is serviced.
Q: 3 a) Let us assume that the CS and IP values of the base and offset of an ISR are 1210h and 1010h.…
A: The Answer to the given question is given below:
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Q1. a) Let us assume that the CS and IP values of the base and offset of an ISR are 1230h and 2000h. The ISR is 10 lines of code where each line takes 4 bits to be stored. At which address, shall we find the IRET instruction?
b) Describe the complete process of how an interrupt of type n (INTn) is serviced.
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- II. IMPLEMENTATION OF SIMPLE SYSTEMS. Implement the given RTL below using bus connection and tri-state buffers. Assume that the given control signals are mutually exclusive, and all registers are 4-bit wide. You may include additional flip-flop pins if necessary. V: c€ E, DEE W: DEB, E €1 (high) X: DEC, BEC, A€c Y: cEA, E E A Z: AD, B E0 (low)a. Find the address accessed by each of the following instructions. If DS = 0100H, BX= 0120H, DATA = 0140H, and SI = 0050H and real mode operation:1. MOV DATA[SI], ECX2. MOV BL, [ BX+SI]b. Descriptor contains a base address of 00260000H, a limit of 00110H, and G = 1,determine starting and ending locations are addressed by the descriptor for aCore2.3 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? b) Describe the process that would take place if an interrupt arises at the IR0 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.
- Question 24 Consider instruction in which the instruction does nothing. It just occupies space. Choose the correct sequence of circuits which will correctly implement this instruction. TMP below is a temporary register. NOP IP->MAR MEM(MAR)-> MDR MDR->IR TMP->AX AX->TMP IP-> ADDR(IR) IP->MAR MEM(MAR)-> MDR MDR->IR AX->BX BX->AX IP -> ADDR(IR) IP->MAR MEM(MAR)-> MDR MDR->IR IP -> ADDR(IR) IP->MAR MEM(MAR)-> MDR MDR->IR IP -> IP+212. In a BSL instruction block B3:10 holds the word 8ECOH, and the bit starting address is B3:0/15. What word is held in the B3:10 after one low-to-high transition signals are sent to this BSL instruction block's input?Please complete the binary encoding for the following BL instruction at address Ox2000 using the given information. Please write the answer in a group of four bits as follows: 0010 0010 0010 1100. Address Instructions 0x2000 bl sub 0x2004 add r3, r2, r1 0x2008 sub r6, r5, r3 sub: stmfd r13!, {r0-r12, 0X200C r14] 0x2010 add r7, r2, r5 Ox2014 mov r3, r2
- 1) Jump and Jump and Link instructions have a six-bit op-code and a 26-bit jump address field. How is the full 32-bit Jump Address calculated? The jump address field is sign-extended six bits to the left. The leftmost four bits of the PC are concatenated with the jump address field, and two bits of zero are concatenated to the end of that. The leftmost six bits of the PC are concatenated with the jump address field. The jump address is ANDed with Register 0 to produce the jump address. -------------------------------------------------------------------------------- 2) How is a branch target address calculated? Sign-extend bits 15:0 of the instruction, append two bits of zeros, and add the result to the PC. Zero extend bits 15:0 of the instruction, append two bits of ones, and subtract the result from the PC. Add the immediate operand to register zero, and sign extend the result. Concatenate the leftmost 16 bits of the PC with the rightmost 16 bits of the instruction.…Q1: Determine the physical address of the source operand base on the Based Indexed Addressing Mode. The MOV instruction MOV AX, [BX].[16A0] [SI] The contents of IP, CS and DS are 0120, A342p, and 2C60, respectively. Also, the contents of BX and SI are 6752, and C344, respectively. Explain by the draw the registers and the logical of the system memory in the before execution and after execution. Note: the content of PA is 2B7C and the coding of this instruction is XXXXQ1- Write a program in assembly language for the 8085 microprocessor to receive 10 bytes of data via the SID and store it at the memory address (3000H to 3009H) using a baud rate of 1200. Information: The 8085 processor operates at a frequency of 3.072 MHz. When you receive each byte of the required bytes, you must adhere to the following: The bits of two high bits will be received at the beginning of the reception (start bits), after that the data bits will be received, after that the low bit of the stop bit will be received (stop bit). The following flowchart will help you, but you should notice that this flowchart deals with one byte, and you are required to deal with 10 bytes The solution must be integrated and include the calculation of the baudrate delay time Of+CD!HID+[00 Yes SIDATA Read SID Start Bit? Wait for Half-Bit Time Set up Bit Counter Wait Bit Time Read SID Save Bit Decrement Bit Counter All Bits Received? Add Bit to Previous Bits Go Back to Get Next Bit Return IMUNI
- 02:- (A) Find the phicycal address if (BP) = 0100H, (SI) = 0200H , (SS) = 2000H and a displacement of 10H, of the instruction MOV AL. (BP+S+10H]. Which the name of Addressing Modes? 02:- (B) Find the result of the following: 1111-iIH+110010111011B 03: Write a piece of code to find the number of negative integers in an array of size 1024 bytes contain signed numbers stored at addresses starting at 21000H, store the result in a location 51000HQ2- Write a program in assembly language for the 8085 microprocessor to receive one byte of data via the SID and store it at the memory address (3000H to 3009H) using a baud rate of 1200. Information: The 8085 processor operates at a frequency of 3.072 MHz . When receive the required bytes, you must adhere to the following: The bits of two high bits will be received at the beginning of the reception(start bits 1 1 ), after that the data bits will be received, after that the low bit of the stop bit will be received (stop bit 0 ). The following flowchart will help you. The solution must be integrated and include the calculation of the baudrate delay time3. Let say BX = 1000H, SI = 2000H, DISP = 1234H, DS =1200H. %3D Determine the effective address of the Word that will be transferred to DX when executing this instruction: MOV DX,[BX+SI+1234]