Can u plz draw a circuit for each operation Opcode000001010011100101110111OperationOpcode (2:0):This picks the operation to be done.O=BALU inputs:A(7:0):8 bit data to be operated on. This comes from the register file, read data a(R_data_a).ALU Outputs:0 (7:0):The result of the ALU operation.O=A^B (bitwise xor)O=A AND B (bitwise and)O=A OR B (bitwise or)O = A + B (addition)O=AB (subtract)O(7) = 0, 0(6:0) = B (7:1) (logical shift right)O=AB (multiplication)B (7:0):8 bit data to be operated on. This can come from the register file, read data b (R_data_b) or thedata input to the processor.

The answer for the above mentioned question is given in the following steps for your reference.

Opcode 000 001 010 011 100 101 110 111 Operation O=B ALU inputs: A(7:0): 8 bit data to be operated on. This comes from the register file, read data a(R_data_a). Opcode (2:0): This picks the operation to be done. O=A^B (bitwise xor) O=A AND B (bitwise and) O=A OR B (bitwise or) O= A + B (addition) O=A - B (subtract) O(7) = 0, 0(6:0) = B (7:1) (logical shift right) O=A* B (multiplication) B (7:0): 8 bit data to be operated on. This can come from the register file, read data b (R_data_b) or the data input to the processor. ALU Outputs: 0 (7:0): The result of the ALU operation.

Opcode 000 001 010 011 100 101 110 111 Operation O=B ALU inputs: A(7:0): 8 bit data to be operated on. This comes from the register file, read data a(R_data_a). Opcode (2:0): This picks the operation to be done. O=A^B (bitwise xor) O=A AND B (bitwise and) O=A OR B (bitwise or) O= A + B (addition) O=A - B (subtract) O(7) = 0, 0(6:0) = B (7:1) (logical shift right) O=A* B (multiplication) B (7:0): 8 bit data to be operated on. This can come from the register file, read data b (R_data_b) or the data input to the processor. ALU Outputs: 0 (7:0): The result of the ALU operation.

Systems Architecture

7th Edition

ISBN:9781305080195

Author:Stephen D. Burd

Publisher:Stephen D. Burd

Chapter11: Operating Systems

Section: Chapter Questions

Problem 30VE

See similar textbooks

Similar questions

The control signal Ep or iEnableoutput allows the program counter to decrement its value. True False The control signal Ep or iEnableoutput sends the content of the program counter to the W bus. True False A tri-state buffer is needed for all data going to the W bus. True False In using 'assign' statement in Verilog, the left hand side must always be a register. True False In declaring a register in Verilog, the following format must be followed: reg [LSB:MSB] identifier; True False
0001 = Load AC from memory 0010 = Store AC to memory 0101 = Add to AC from memory 0011 = Load AC (the accumulator register) from an I/O device 0111 = Store AC to an I/O device With these instructions, a particular I/O device is identified by replacing the 12-bit address portion with a 12-bit device number. Remember that a number ending with a small ‘h’ means the number is a hexadecimal number. What is the hexadecimal string that expresses the following instructions? Load AC from memory location 62h. Add the contents of memory location 451h to AC. Store AC to memory location 8h. Store AC to I/O device number 8h.
25: . Find the time delay in the following program if the crystal frequency is 1 MHz. Do not ignore the time delay due to the first and last instruction. DELAY: LDI R16, 30 AGAIN: LDI R17, 35 HERE: NOP NOP DEC R17 BRNE HERE DEC R16 BRNE AGAIN RET 26: Write a program to display 2 on 7 segment. A 7 segment is connected to PortD.
Q1- Write a program in assembly language for the 8085 microprocessor to send one byte of data located at the memory address (3000H ) using SOD at a baud rate of 1200. Information: The 8085 processor operates at a frequency of 3.072 MHz . When sending the required byte, you must adhere to the following: The two high bits of the start bits(1 1) must be sent, after that the data bits are sent, after that the low bit of the stop bit (0) is sent. The following flowchart will help you. The solution must be integrated and include the calculation of the baudrate delay time
Convert given code to LEGv8 code:int f, g, y //global 64-bit variablesint sum (int a, int b) { // at memory address X0+1000.return (a +b)} int main (void) // at memory address X0 + 800{f=2;g=3;y= sum (f, g);return y;}Convert this code, making valid assumptions about registers and register use. Notethat brackets and global variable declarations are not affecting the addresses of the instructionsin memory.
Connect a 5*7 dot-matrix display to the 8051-microcontroller (Draw the schematic on paper and upload it as a photo). Then, display the content of variable NUMBER on the display, digit by digit every one second. for example if NUMBER=249, then "2" will be displayed for one second, after that digit "4" and finally digit "9" will be displayed for a whole second. Assume that the necessary lookup table exists in Memory at address 0xF000
Assembly language programming in MIPS. Use QTSpim to run code. Write a simple Assembly Language program that has a data section declared as follows:.data.byte 12.byte 97.byte 133.byte 82.byte 236 Add the values up, compute the average, and store the result in a memory location.
Consider the following hexadecimal readout: 000000 8A00 8E00 CFA1 48BF 7900 3202 9015 AD34 000010 0218 6D30 028D 3402 AD35 0288 3102 8D35 000020 0E30 0290 DAEE 3102 4C00 0200 0040 004B Refer to the first byte of memory shown above, address 000000. Assume that this byte is used to store an 8-bit unsigned integer. What is the decimal value stored in this byte? Group of answer choices 138 -27 22,842 66
In the following code, the first printf() reached produces the output “14,” but the second printf() can cause a bus error or a segmentation fault. Why? main(){ int *p;funct(p);printf("%d\n",*p);}funct(int *p2){p2=(int *)malloc(4);*p2=14;printf("%d\n",*p2);}
The address of a byte in memory might be one of two forms. Instead of storing all of the data in memory, why not just use it as a temporary holding area to register part of it with the ALU and move the rest of it to the IR?
Conditional Execution using Bit Test (btfsc/ btfsc). 1.) Convert the provided code to the C language. Also explain the codes function. INCLUDE "P18F242.INC" CBLOCK 0x000 i, j ENDC org 0 goto main org 0x200 //start at address 0x200 main … movf i,f btfsc STATUS,Z goto end_if movf i,w addwf j,f end_if … rest of code … Dont copy from other post. Will get downvote if answer is copied.
Q2- 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 time