Data Representation and Digital Logic

This assignment focuses on the fundamentals of data representation in computer science. It covers topics such as binary numbers, boolean algebra, logic gates, and truth table analysis. Students are required to solve problems involving conversion between number systems, simplification of Boolean expressions, and design of simple digital circuits using logic gates. The assignment also includes a section on analyzing the behavior of digital circuits represented by truth tables.

Running head: DATA REPRESENTATION
Data Representation
Name of Student-
Name of University-
Author’s Note-

1DATA REPRESENTATION
Question 1.
(a)
I. The smallest negative number of the one’s compliment is -127 and largest positive
number of one’s compliment is 127.
II. The smallest negative number of two’s compliment is -128 and largest positive number
of two’s compliment is 127.
III. The smallest negative number of signed magnitude is -127 and largest positive number of
signed magnitude is 127.
IV. The smallest negative number of unsigned magnitude is 0 and largest positive number of
unsigned magnitude is 255.
(b)
I. (5AB)16 = (2653)8
Calculation: 5AB = (010110101011)2 = (2653)8
II. (101101.101)2 = (45.625)10
(1*25) + (0*24) + (1*23) + (1*22) + (0*21) + (1*20). (1*2-1) + (0*2-2) + (1*2-3)
= 32 + 8 + 4 + 1. 625
= (45.625)10
III. (12348)10 = (1100000111100)2

2DATA REPRESENTATION
IV. (679810)10 = (133223220)5
V. (976.6310)10 = (1111010000.10100)2
976 =

3DATA REPRESENTATION
0.6310= 0.6312 * 2 = 1.2620 num= 1
0. 2620= 0. 2620 * 2 = 0.5240 num= 0
0. 5240= 0. 5240 * 2 = 1.0480 num= 1
0. 0480 = 0. 0480 * 2 = 0.0960 num= 0
0. 0960 = 0. 0960 * 2 = 0.1920 num= 0
0.6310= 10100
VI. (1001001011)2 = (24B)16
VII. (10011110) = -(98)10
So, 16 * 6 + 2 = 98

4DATA REPRESENTATION
Since, the number given starts with 1, so the number is negative
= -(98)10
Question 2:
(a) X’(X+Y)+(XX+Y)(Y’+X)=Y+X
L.H.S. – X’(X+Y) + XXY’+ XXX+ YY’+ XY [Distributive law]
= X’X + X’Y + XXY’+ XXX + YY’+ XY [Distributive law]
= 0 + X’Y + XY’ + X + 0 + XY [Since, X’X = 0 AND XX = X]
= X’Y + XY’ + X + XY
= X’Y + XY’ + X (1 + Y)
= X’Y + XY’ + X [Since, 1 + Y = 1]
= X’Y + X (Y’ + 1)
= (X + X’) (X + Y)
= (X + Y) [Since, X + X’ = 1] = R.H.S.
(b) Let the Assignment considered as A, Blog as B, Discussion Forum as C and Quiz as D. P
denotes the number of students who fulfil the criteria for passing.
So, the truth table will be:
A B C D P
0 0 0 0 0
0 0 0 1 0
0 0 1 0 0
0 0 1 1 0
0 1 0 0 0

5DATA REPRESENTATION
0 1 0 1 0
0 1 1 0 0
0 1 1 1 0
1 0 0 0 0
1 0 0 1 0
1 0 1 0 0
1 0 1 1 1
1 1 0 0 0
1 1 0 1 1
1 1 1 0 1
1 1 1 1 1
The expression for the above table is
AB’CD + ABC’D + ABCD’ + ABCD
After minimizing the equation,
A [D (B’C + BC) + BCD’]
The circuit diagram for the equation is