Question

Problem #1 (25 points) Address Space, Memory Consider a hypothetical 18-bit processor called HYP18 with all registers, including PC and SP, being 18 bits long. The smallest addressable unit in memory is an 8-bit byte. A. (4 points) What is the size of HYP18's address space in bytes and KB? How many address lines does HYP18 require? Address space: Bytes Address space: KB (KiloBytes). Address bus lines: B. (6 points) Assume that first quarter of the address space is dedicated for HYP18's RAM memory and the upper half of the address space is reserved for HYP18's Flash memory. Give address ranges for the RAM and Flash memories. Fill in the table below. What is the size of the RAM memory and the Flash memory? Start byte address End byte address RAM memory Flash memory RAM memory size (Bytes/KB]:_ Flash memory size (bytes/KB): The MSP430F20x is a microcontroller with 64 KB of address space divided between code memory (flash), RAM memory, and input/output peripherals. It has 1,024 Bytes of RAM memory starting at the address Ox0200, and 256 Bytes of address space reserved for special purpose registers and 8-bit input/output peripherals (starting at the address Ox0000) followed by 256 Bytes reserved for 16 bit input/output peripherals. The flash memory of 16 KB resides at the top of address space (highest addresses in the address space). C. (8 points) Determine the address map by filling in the following table. Address Address Sections in (hexadecimal) address space Flash Memory RAM Memory Last Flash address First Flash address Last RAM address First RAM address Last I/O address (16-bit per.) First I/O address (16-bit per.) Last 1/0 address (8-bit per.) First 1/0 address (8-bit per.) 1/0 address space 1/0 address space D. (7 points) What is the program stack (what is it, where is it located, and how we deal with it)? What is the maximum stack size in the MSP430Fx described above? What should be the initial value of SP?

Answer 1

1. As we have byte-addressable memory and we have 18 bits register. So we can address 2¹⁸ bytes with the help of this register.
So Address space: 2¹⁸ bytes.

2. Address space = 2⁸ * 2¹⁰ = 256K Bytes.(As 2¹⁰ is equal to the kilo)

3. Address bus line: As we can have 18 bits so address bus line should support 18 address lines.