A new PIC18 has been developed with a program memory of 32 MB. What should be the length (number of bits) of the program counter (PC) of that PIC18? How many different program memory locations can be addressed by the new PIC18? o a. PC is 25 bits and is able to address 32000000 different locations o b. PC is 25 bits and is able to address 33554432 different locations C. PC is 21 bits and is able to address 32000000 different locations o d. PC is 21 bits and is able to address 33554432 different locations
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- If a microprocessor has a cycle time of 0.5 nanoseconds, what’s the processor clock rate? If the fetch cycle is 40% of the processor cycle time, what memory access speed is required to implement load operations with zero wait states and load operations with two wait states?Question 6 Not yet answered A new PIC18 has been developed with a program memory of 32 MB. What should be the length (number of bits) of the program counter (PC) of that PIC18? How many different program memory locations can be addressed by the new PIC18? o a. PC is 25 bits and is able to address 32000000 different locations o b. PC is 25 bits and is able to address 33554432 different locations O C. PC is 21 bits and is able to address 32000000 different locations o d. PC is 21 bits and is able to address 33554432 different locations.A digital computer has a memory unit with 24 bits per word. The instruction set consists of 150 different operations. All instructions have an operation code part (opcode) and an address part (allowing for only one address). Each instruction is stored in one word ofmemory.a. How many bits are needed for the opcode?b. How many bits are left for the address part of the instruction?c. What is the maximum allowable size for memory?d. What is the largest unsigned binary number that can be accommodated in one word of memory?
- By assuming that X = 3, and 33 is a two digit number, consider memory storage of a 64-bit word stored at memory word 33 in a byte-addressable memory (a) What is the byte address of memory word 33? (b) What are the byte addresses that memory word 33 spans? (c) Draw the number 0xF1234567890ABCDE stored at word 33 in both big endian and little-endian machines. Clearly label the byte address corresponding to each data byte value.b. How many bits are required to address a 4M × 16 main memory if main memory is byte-addressable?c. How many bits are required to address a 1M × 8 main memory if main memory is byte-addressable?Suppose that the CPU is tasked to read data "X" from RAM with an address "C7FFF" and write the same data to RAM with an address "C8FFF" to replace data "Y." Describe in your own words how data X overwrites Y in RAM.
- 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 is the highest and the lowest address in the 8086’s memory address? Is memory in the 8086 microprocessor organized as byte, words or double words ? What is the value of the double word stored in memory starting at address B000316 if the contents of memory location B000316, B000416, B000516 and B000616 are 1116, 2216, 3316 and 4416 respectively ? What is the function of the stack?Suppose the RAM for a certain computer has 256M words, where each word is 16 bits long. a. What is the capacity of this memory expressed in bytes? b. If this RAM is byte addressable, how many bits must an address contain? c. If this RAM is word addressable, how many bits must an address contain?
- 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.Suppose that you have a computer with a memory unit of 24 bits per word. In this computer, the assembly program's instruction set consists of 198 different operations. All instructions have an operation code part (opcode) and an address part (allowing for only one address). Each instruction is stored in one word of memory. a. How many bits are needed for the opcode? b. How many bits are left for the address part of the instruction? c. How many additional instructions could be added to this instruction set without exceeding the assigned number of bits? Discuss and show your calculations.A digital computer has a memory unit with 24 bits per word. The instruction set consists of 150 different operations. All instructions have an operation code part (opcode) and an address part (allowing for only one address, similar to Marie). Each instruction is stored in one word of memory. a) How many bits are needed for the opcode? b) How many bits are left for the address part of the instruction? c) What is the maximum allowable size for memory?