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GEORGIA INSTITUTE OF TECHNOLOGY
School of Electrical and Computer Engineering
ECE 2030: Section C
Spring 2008
Home Work
Due: March 31, 2008
P2 P1 P0 C N2 N1 N
__ _
N 2 = P 2 P 1 + P 2 P 0 + P 2 C + P 1 P 0 C N 1 = P 2 P 1 P 0 C + P 1 P 0 + P 2 P 1 C + P 0 C
__ __ __ __
N 0 = P 1 P 0 + P 2 P 0 + P 0 C + P 2 P 1 C
b) One-hot states
P3 P2 P1 P0 X N3 N2 N1 N0 Z 0 0 0 1 0 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 1 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 1 0 0 1 0 1 __ N 3 = P 2 X
N 2 = P 1 X + P 2 X
**N 1 = P 0 X + P 3 X
N 0 = P 0 X + P 1 X + P 3 X**
Z = P 3 X
c) Gray Coded States
P1 P0 X N1 N0 Z 0 0 0 0 0 0 0 0 1 0 1 0 0 1 0 0 0 0 0 1 1 1 1 0 1 1 0 1 0 0 1 1 1 1 1 0 1 0 0 0 0 0 1 0 1 0 1 1
N 1 = XP 0 + P 1 P 0
N 0 = X
__
Z = P 1 P 0 X
Tradeoffs โ One-hot encoding: Highest space requirements (4 Flip-Flops); Low propagation time. Gray code encoding: Low space requirements (2 Flip-Flops); Low propagation time. Binary encoding: Low space requirements (2 Flip-Flops), Highest propagation time.
b) 32Mwords, address into each word. Each word is 4 bytes => Need 32M rows of 32 bits each. Use 32/4 = 8 chips. Since 32M = 2^5 .2^20 = 2^25 use the first 25 pins address pins and ground the other 3. These address pins will again be shared across all chips because we want to address into same-index rows in all chips.
c) 16Mwords, address into each word. Each word is 8 bytes => Need 16M rows of 64 bits each. Use 64/4 = 16 chips. Since 16M = 2^4 .2^20 = 2^24 use the first 24 shared address pins and ground the other 4 shared address pins.
4. Student Exercise
ECE 2030 SPIM Assignment 1
.data .globl initials initials: label1: .asciiz "***********************************" label2: .asciiz "* " label3: .asciiz " Your Name " label4: .asciiz " ECE 2030 Assignment 1 " label5: .asciiz " Section X " label6: .asciiz " 04/2008 " label7: .space 7 label8: .byte 0,1,2,3,4,5,6,7, .align 2 label9: .word 0x4400, label10: .asciiz " *" label11: .asciiz "***********************************"
newline: .asciiz "\n"
- What is the byte address of the character โ3โ in 2030? Assuming little endian: 0x Assuming big endian: 0x
- How many bytes are used in the data segment? We are using contiguous memory from 0x10010000 through 0x1001013B (both inclusive). => Exactly 0x1001013B-0x10010000 + 1 = 0x13C or 316 bytes of memory are being used.
- How many labels did you need? 11
- What are the contents of the memory location defined by Label 9? Here memory location refers to memory words (or 4 bytes) 0x00004400 and 0x000000DC
- What are the words (values) contained at addresses 0x10010020 through 0x (Little endian)
