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- 2. There are interrupt in 8086 Mp, user, starting from interrupt number_ of them are available for the 3. IVT is a table that contain the of ISR of each interrupt. The size of the table is 4. The low bank (even bank) in 8086Mp memory has a size of bytes, starting from address: and ending with address:_Write a service routine which resets all elements of an array that resides in memory location from A000 H to A0FF H with DS equal to 0000 H. The service routine address is CS:IP where CS is 2000 H and IP is 0100H. Assume the interrupt type that is called is 50 (x8086- nano)1. In memory-mapped I/O… a. The I/O devices and the memory share the same address space b. The I/O devices have a seperate address space c. The memory and I/O devices have an associated address space d. A part of the memory is specifically set aside for the I/O operation
- LDR uses an address called a "Register Offset" address. Why? Group of answer choices The data value loaded is at an address in the memory. The data value loaded is the results of adding a register and immediate to obtain the reference for the data value Memory is accessed at a location that is a numeric value, starting at (offset from) 0 The ALU can only use register (not memory) values All data access is relative to a memory address or 0. The LDR command uses what units in the CPU? Select all that apply. Group of answer choices Data Memory Multiplication Unit Barrel shifter ALU Addition is done in what CPU unit? Group of answer choices The multiplication unit The registers The ALU The barrel shifterAnswer the following: A section of the memory space with the address from 0x6000 to 0x9FFF needs to be fully partitioned to two peripherals with the same memory size. The first peripheral's address starts at Ox6000 and ends at Ox . The second peripheral's address starts at Ox ends at Ox9FFF. and In embedded programming, let i = 13, then j =i<< 2; we have j = 0x wise operation, let i = 0x00ff and then i &= ~0x0010, we have i = Ox . And for bit-What is the difference between isolated I/O and memory-mapped I/O? What are theadvantages and disadvantages of each?
- Question 4: There is an application that requires the hardware: an Intel 8031, a Program ROM of 8Kx8, a Data ROM of 4Kx8 for look-up tables, TWO Data RAMs of 8Kx8. The memory map of the design should be: Program ROM should start at address 0000μ. Then, the Data ROM should come above the Program ROM. Finally the Data RAMs must go to the top of the memory map. There should be no gaps between the memory addresses of the external ROMs or RAMS. A. Using logic gates, draw the pin connections of the design. Label your diagram fully. B. Calculate the address space of the ROMs and RAMs of your design. C. Show the design's address space on a memory map, starting with 0000μ at the bottom and FFFFH at the top.What's the main difference between Memory-mapped I/O and Isolated I/O?There is an application that requires the following hardware: an Intel 8031, a Program ROM of 8Kx8, a Data ROM of 4Kx8 for look-up tables and a Data RAMs of 8Kx8. The memory map of the design: Program ROM should start at address 0000H. Then, the Data ROM should come above the Program ROM. Finally the Data RAM must go to the top of the memory map. There should be no gaps between the memory addresses of the external ROMs. Calculate the address space of the ROMs and RAMs of your design.
- Can the virtual addresses 0x100000 and 0x100008 be mapped to 2 far apart locations in physical RAM?There is an application that requires the following hardware: an Intel 8031, a Program ROM of 8Kx8, a Data ROM of 4Kx8 for look-up tables and a Data RAMs of 8Kx8. The memory map of the design: Program ROM should start at address 0000H. Then, the Data ROM should come above the Program ROM. Finally the Data RAM must go to the top of the memory map. There should be no gaps between the memory addresses of the external ROMs. Show the design’s address space on a memory map, starting with 0000H at the bottom and FFFFH at the top.Compare and contrast the differences between Isolated I/O and Memory Mapped I/O