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- Most Intel CPUs use the __________, in which each memory address is represented by two integers.3. If we have an 8 bit microcontroller that has 4kB of instruction memory starting at address 0x0000, and 2kB of data memory immediately above that, what is the next available byte in our address map?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.
- A computer employs RAM chips of 512 x 8 and ROM chips of 256 x 8. The computer system needs1K bytes of RAM, 2K bytes of ROM, and eight interface units, each with 2 registers. A memory-mapped1/0 configuration is used. The two highest-order bits of the 16 bit address bus are assigned 10 for RAM,11 for ROM, and 00 for interface registers.a. How many RAM and ROM chips are needed?b. Draw a memory-address map for the system.c. Give the address range in hexadecimal for RAM, ROM, and interface.A computer employs RAM chips of 512 x 4 and ROM chips of 256 x 8. The computer system needs 1KB of RAM, and 512 x 8 ROM and an interface unit with 256 registers each. A memory-mapped I/O configuration is used. The two higher -order bits of the address bus are assigned 00 for RAM, O1 for ROM, and 10 for interface. a) How many lines must be decoded for chip select? Specify the size of the decoder b) Draw a memory-address map for the system and Give the address range in hexadecimal for RAM, ROM c) Develop a chip layout for the above said specifications.Answer the question below based on given portion of the memory unit with a word size of 8 bits, and the four 8-bit registers: AR, BR, CR, and DR. (all values are in binary) Address Data b0100001 00001110 p0100010 01011100 00100011 00011001 The four 8-bit registers AR, BR, CR, and DR initially have the following values: AR = 00100011 BR = 00000101 CR = 00101101 DR = 01001011 Fill the missing fields in the memory unit below and determine the 8-bit values in each register after the execution of the following sequence of microoperations. DR <- MJARI, BR <- BR+2 DR <- DR+CR, AR <- AR-2 M[AR] <- DR Address Data p0100001 00100001 b0100010 00100010 00100011 00100011 AR = 00100011 BR = 00000101 CR = 00101101 DR = 01001011
- Suppose we have a byte-addressable memory of 20 bytes, built using 4 modules. Draw diagrams showing the distribution of addresses within each module, if we are using (a) highorder interleaving, and (b) low-order interleaving.Suppose a DRAM memory has 4 K rows in its array of bit cells, its refreshing period is 64 ms and 4 clock cycles are needed to access each row. What is the time needed to refresh the memory if clock rate is 133 MHz? What fraction of the memory's time is spent performing refreshes?Good explanation please
- 3. If a ROM chip in some 8-bit microcomputer system starts with an address of 97C0H while its end address is FFFFH. Compute the storage capacity of this ROM region. Draw a figure to show Fow the chip selection signal of the chip can be generated. 子信息Answer ii)Consider memory storage of a 32-bit word stored at memory word 34 in a byte addressable memory. (a) What is the byte address of memory word 34? (b) What are the byte addresses that memory word 34 spans? (c) Draw the number 0x3F526372 stored at word 342 in both big-endian and little-endian machines. Clearly label the byte address corresponding to each data byte value.