Computer Organization and Architecture

1. Computer Organization and Architecture Tutorial 6 Kenneth Lee

3. For example: • 2 bits 2s complement: • 11 00 01 • -2 -1 0 +1 • 4 bits 2s complement: • 1000 1001 1010 1011 1100 1101 1110 1111 0000 0001 0010 0011 0100 0101 0110 0111 • -8 -7 -6 -5 -4 -3 -2 -1 0 +1 +2 +3 +4 +5 +6 +7 • Addition of two 2-bit 2s complements generate a 2-bit result • 11 • +01 • 100 • Means 3+1=4 or -1+1=0 • Multiplication of two 2-bit in 2s complements generate a 4-bit result • 11 • X 01 • 0011 • 0000 • 0011 • In unsigned integer, it means 3x1=3; but in 2s complement, it means (-1)x(+1)=+3. • It is due to the unsigned integer and 2s complement have different extension pattern • 11 • X 01 • 1111 • 0000 • 1111

5. Biased representation example: 4-bit biased representation: 4-bit binary unsigned integer is from 0000 (0) ~ 1111 (15) Bias = 2k-1 – 1 = 23 – 1 = 7 4-bit biased representation is from (0 – 7) ~ (15 – 7) 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 -7 -6 -5 -4 -3 -2 -1 0 +1 +2 +3 +4 +5 +6 +7 +8 The smallest is 0000 and the largest is 1111 The same with unsigned integer

6. Examples: Positive overflow (2-1 + 2-2 +…+ 2-n = 1-2-n )

7. Examples: Exponent overflow For 4-bit exponent in biased representation, the range is -7 ~ +8, so the largest exponent is +8 Examples: significant overflow 0.11 + 0.01 = 1.00

8. e is in 0~X with bias q, so the exponent is in –q~X-q • the largest positive significant is 1-b-p • (e.g. if b is 2 and significant is 3 digits, the largest positive is (0.111)2; • if b is 10 and significant is 3 digits, the largest positive is (0.999)10 ) • so the largest positive value is (1-b-p)x(bX-q) • the smallest positive significant is b-p • (e.g. if b is 2 and significant is 3 digits, the largest positive is (0.001)2; • if b is 10 and significant is 3 digits, the largest positive is (0.001)10 ) • so the smallest positive value is b-px(b-q) • b. For the normalized floating-point numbers, the difference with above is that • the first bit can not be 0, so the largest value will keep the same. • But the smallest positive value will be b-1 • (e.g. if b is 2 and significant is 3 digits, the largest positive is (0.1)2; • if b is 10 and significant is 3 digits, the largest positive is (0.1)10 ) • so the smallest positive significant is b-1x(b-q)

9. c. Minus means the sign is 1 • (-1.5)10 = (-1.1)2x20 so E is 0 and its biased representation is 01111111 (0+27-1) • (0.5)10 = (0.1)2 so the significant is 100000000000000000000000 • d. 384 = 110000000 = 1.1x28 = 1.1x21000 • So E is 8 and its biased representation is 8+27-1 = 135 = 10000111 • The significant is 0.1 and represented as 100000000000000000000000 • e. 1/16 = (0.0001)2 = 1.0x2-4 • So E is -4 and its biased representation is -4+27-1=123=01111011 • The significant is 0.0 and represented as 000000000000000000000000 • Minus means the sign is 1 • 1/32 = 0.0001 = 1.0x2-5 • So E is -5 and its biased representation is -5+27-1=122=01111010 • The significant is 0.0 and represented as 000000000000000000000000

10. 1 bit 11 bits 52 bits The exponent value is in 1~2046 (0 and 2047 are kept for special use) The bias is 1023 so the biased exponent is in -1022~+1023 So the largest positive is (2-2-52)x21023, the smallest positive is 2-1022

12. 9.27 Show how the following additions are performed. • 5.566 x 102 + 7.777 x 102 b. 3.344 x 101 + 8.877 x 10-2

13. 9.27 Show how the following additions are performed. • 7.744 x 10-3 - 6.666 x 10-3 • 8.844 x 10-3 – 2.233 x 10-1

15. B.1 Construct a truth table for following expressions:

17. B.2 simplify the following expressions according to the commutative law Commutative law: A+B = B+A; AB = BA

18. B.4 Apply DeMorgen’s theorem

19. B.4 Simplify the following expressions

20. B.5 Construct the operation XOR from Boolean AND, OR, and NOT

21. B.6 Given a NOR gate and NOT gates, draw a three input AND function

22. B.7 Write the Boolean expression for a four-input NAND gate