Boolean Algebra Solutions

This homework assignment presents solutions to problems related to Boolean algebra. The first problem involves interpreting single-precision IEEE 754 floating-point representations and converting 5-bit words into signed magnitude, one's complement, and two's complement representations. The second problem focuses on designing a logic circuit to represent a 24-hour clock and determine when a door is open based on specific binary inputs. The solution includes a truth table, logic expressions, and a simplified logic diagram. The assignment demonstrates a strong understanding of binary representation, logic gates, and Boolean algebra simplification techniques. The provided solutions are detailed and well-explained, making them a valuable resource for students studying digital logic and computer science.

Running head: BOOLEAN ALGEBRA
Boolean Algebra
Name of the Student:
Name of the University:
Author Note

1
BOOLEAN ALGEBRA
Answer to question 1
a)
Single precision IEEE 754 format- 0 01111110 10100000000000000000000
Sign bit is 0
Exponent is 01111110
Mantissa is 10100000000000000000000
Decimal value -.8125
Hence the value is 8.125x10^(-1)
b)
5-bit word- 00110
Signed magnitude- (-6)
One’s complement- 11001
Two’s complement- 11010
Answer to question 2
a)
The table below is used for the representation of the clock
Hours Clock Decimal Representation
A B C D E
1 0 0 0 0 1
2 0 0 0 1 0
3 0 0 0 1 1
4 0 0 1 0 0
5 0 0 1 0 1

2
BOOLEAN ALGEBRA
6 0 0 1 1 0
7 0 0 1 1 1
8 0 1 0 0 0
9 HIGH(1) 0 1 0 0 1
10 HIGH(1) 0 1 0 1 0
11 HIGH(1) 0 1 0 1 1
12 HIGH(1) 0 1 1 0 0
13 HIGH(1) 0 1 1 0 1
14 HIGH(1) 0 1 1 1 0
15 HIGH(1) 0 1 1 1 1
16 HIGH(1) 1 0 0 0 0
17 1 0 0 0 1
18 1 0 0 1 0
19 1 0 0 1 1
20 1 0 1 0 0
21 1 0 1 0 1
22 1 0 1 1 0
23 1 0 1 1 1
24 1 1 0 0 0
In the above table the 24-hour clock is represented by the 5 bit binary values.
Now for the High or 1 represents that the door is open.
Hence the values that represents that the door is open are:

3
BOOLEAN ALGEBRA
The logic of the diagram is:
A'BC'D'E+A'BC'DE'+A'BC'DE+A'BCD'E' = A’B(C’D’E+C’DE’+C’DE+CD’E’)
= A’B(D(C’E’+C’E) +D’(C’E+CE’)) = A’B(D(C’E’+C’E) +D’(C’E+CE’)) = A’B (D’C’+D (C
xor E))
A'BCD'E’+A'BCDE'+A'BCDE+AB'C'D'E' = A’BCD + A'BCD’E’+AB'C'D'E' = A’BCD + D’E’
(A’BC + AB’C’)
Hence, the expression of the problem is minimized and the diagram is provided below. This
diagram represents the clock and the output represents the intervals in the which the doors are
open.

4
BOOLEAN ALGEBRA
Fig: Logic diagram

5
BOOLEAN ALGEBRA
(Source: Created by the author)
b)
X’Y+XYZ’+Y’+XZ (Y+Y’)
= X’Y+XYZ’+Y’+XZY+XZY’
= X’Y+XY (Z+Z’) +Y’+XZY’
=X’Y+XY+Y’+XZY’
=Y+Y’+XZY’
=1+XZY’
=1 (PROVED)