Download Tutorial 6

Computer Organization and
Architecture
Tutorial 6
Kenneth Lee
For example:
2 bits 2s complement:
10 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
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
Examples: Positive overflow (2-1 + 2-2 +…+ 2-n = 1-2-n )
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
a. 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)
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
f. 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
1 bit
11 bits
52 bits
sign
Biased exponent Fraction (Significant)
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
9.27 Show how the following additions are performed.
a. 5.566 x 102 + 7.777 x 102
b. 3.344 x 101 + 8.877 x 10-2
9.27 Show how the following additions are performed.
a. 7.744 x 10-3 - 6.666 x 10-3
b. 8.844 x 10-3 – 2.233 x 10-1
B.1 Construct a truth table for following expressions:
a=ABC+ABC
b=ABC+ABC  ABC

c=A BC  BC

d=(A+B)(A+C)(A+B)


c=A BC  BC  ABC  ABC
(c is 1 when A=1,B=1,C=0 or A=1,B=0,C=1)
d=(A+B)(A+C)(A+B)
=AAA  AAB+AAC+ABC+AAB+ABB+ABC+BBC
=ABC+AB+ABC
=ABC+AB
(d is 1 when A=0,B=1,C=1, or A=1,B=0)
B.2 simplify the following expressions according to the commutative law
Commutative law: A+B = B+A; AB = BA
AB  BA+CDE+CDE+ECD
=AB  AB  CDE+CDE+CDE
=AB  CDE+CDE
AB+AC+BA
=AB+AC+AB
=AB+AC
(LMN)(AB)(CDE)(MNL)
=(LMN)(AB)(CDE)(LMN)
=(LMN)(AB)(CDE)
F(K+R)+SV+WX  VS+XW+(R+K)F
=F(K+R)+SV+WX  SV+WX+F(K+R)
=F(K+R)+SV+WX
B.4 Apply DeMorgen’s theorem
F=V+A+L=VAL
F=A+B+C+D=ABCD
B.4 Simplify the following expressions
A=ST+VW+RST
=ST+VW
A=TUV+XY+Y
=TUV+Y
A=F(E+F+G)=F
A=(PQ+R+ST)TS=ST
A=DDE
 D+D+E
=D+E
A=Y(W+X+Y+Z)Z
=YZ(W+X+YZ)
=YZ
A=(BE+C+F)C=C
B.5 Construct the operation XOR from Boolean AND, OR, and NOT
A
B
XOR
0
0
0
0
1
1
1
0
1
1
1
0
A XOR B = AB + AB
B.6 Given a NOR gate and NOT gates, draw a three input AND function
AND(A,B,C)
=ABC
=ABC
=A+B+C
=NOR(A,B,C)
B.7 Write the Boolean expression for a four-input NAND gate
NAND(A,B,C,D)
=NOT(AND(A,B,C,D))
=ABCD