Data Representation Homework: Number Systems, Conversions, and Logic

This document presents a comprehensive solution to a data representation homework assignment. The solution addresses various aspects of data representation, including number systems such as one's complement, two's complement, signed magnitude, and unsigned magnitude, and their respective ranges. It includes conversions between binary, octal, decimal, and hexadecimal number systems. The assignment also covers logic simplification using Boolean algebra and truth tables, providing a step-by-step solution to simplify a given Boolean expression. Finally, the document provides a truth table and a minimized Boolean expression along with a circuit diagram based on the problem. This assignment solution is designed to help students understand and master the fundamental concepts of data representation and digital logic.

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