ITC544 Computer Architecture: Number Conversion & Boolean Algebra
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Homework Assignment
AI Summary
This assignment focuses on computer architecture and organization, covering number system conversions and Boolean algebra proofs. It includes converting between different bases such as decimal, binary, octal, and hexadecimal, along with proving Boolean algebra identities using basic identities and De-Morgan's Law. The assignment provides step-by-step solutions for number conversions and truth tables to demonstrate the equivalence of logical expressions. Desklib offers this solved assignment and many other resources to aid students in their studies.

Running head: COMPUTER ARCHITECHTURE AND ORGANIZATION
Computer Architecture and Organization
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Subject Code: ITC544
Author’s Note
Computer Architecture and Organization
Full name:
Student ID:
Subject Code: ITC544
Author’s Note
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COMPUTER ARCHITECHTURE AND ORGANIZATION
Table of Contents
Question 1:.......................................................................................................................................2
a.) Determining the value of base x if (152) x = (6A) 16............................................................2
b) Conversions.............................................................................................................................2
c) Representation of value..........................................................................................................5
Question 2:.......................................................................................................................................5
a) Prove........................................................................................................................................5
b) Using basic Boolean algebra identities for Boolean variables x, y, and z, for prove..............7
c) Prove:......................................................................................................................................7
COMPUTER ARCHITECHTURE AND ORGANIZATION
Table of Contents
Question 1:.......................................................................................................................................2
a.) Determining the value of base x if (152) x = (6A) 16............................................................2
b) Conversions.............................................................................................................................2
c) Representation of value..........................................................................................................5
Question 2:.......................................................................................................................................5
a) Prove........................................................................................................................................5
b) Using basic Boolean algebra identities for Boolean variables x, y, and z, for prove..............7
c) Prove:......................................................................................................................................7

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COMPUTER ARCHITECHTURE AND ORGANIZATION
Question 1:
a.) Determining the value of base x if (152) x = (6A) 16
Let the value of base be X,
Given,
(152) x = (6A) 16
Or, X^2 + 5X + 2*X = 6*16 + A
Or, X^2 + 5X + 2*X = 6*16 + 10
X2 + 5X + 2 = 106
X^2 + 5X- 104 = 0
X^2 + 13X- 8X – 104 = 0
X (X + 13) – 8(X + 13) = 0
(X - 8) (X + 13) = 0
X = 8 and X = -13
Hence, X is 8.
The value of the base is 8.
b) Conversions
i) BED16 converting to base-3
Solution:
COMPUTER ARCHITECHTURE AND ORGANIZATION
Question 1:
a.) Determining the value of base x if (152) x = (6A) 16
Let the value of base be X,
Given,
(152) x = (6A) 16
Or, X^2 + 5X + 2*X = 6*16 + A
Or, X^2 + 5X + 2*X = 6*16 + 10
X2 + 5X + 2 = 106
X^2 + 5X- 104 = 0
X^2 + 13X- 8X – 104 = 0
X (X + 13) – 8(X + 13) = 0
(X - 8) (X + 13) = 0
X = 8 and X = -13
Hence, X is 8.
The value of the base is 8.
b) Conversions
i) BED16 converting to base-3
Solution:
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COMPUTER ARCHITECHTURE AND ORGANIZATION
= B * 16*16 + E * 16 + D
= 2816 + 224 + 13
= (3053)10
(3053)10 =
3 3053
3 1017 2
3 339 0
3 113 0
3 37 2
3 12 1
3 4 0
3 1 1
So, (BED)16 = (11012002)3
ii) 3217 into 2-base (binary) representation
Solution:
(321)7 = (3 * 72) + (2 * 71) + (1 * 70)
= (162)10
Again, (162)10 =
2 162
COMPUTER ARCHITECHTURE AND ORGANIZATION
= B * 16*16 + E * 16 + D
= 2816 + 224 + 13
= (3053)10
(3053)10 =
3 3053
3 1017 2
3 339 0
3 113 0
3 37 2
3 12 1
3 4 0
3 1 1
So, (BED)16 = (11012002)3
ii) 3217 into 2-base (binary) representation
Solution:
(321)7 = (3 * 72) + (2 * 71) + (1 * 70)
= (162)10
Again, (162)10 =
2 162
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COMPUTER ARCHITECHTURE AND ORGANIZATION
2 81 0
2 40 1
2 20 0
2 10 0
2 5 0
2 2 1
2 1 0
Hence, (162)10 = (10100010)2
iii) (1235)10 conversion to octal representation
Solution:
8 1235
8 154 3
8 19 2
8 2 3
Hence, (1235)10 = (2323)8
iv) 21.218 conversion to decimal representation
Solution:
21.218 = (2 * 81) + (1 * 80). (2 * 8-1) + (1 * 7=8-2)
= 17 + 0.25 + 0.015625
COMPUTER ARCHITECHTURE AND ORGANIZATION
2 81 0
2 40 1
2 20 0
2 10 0
2 5 0
2 2 1
2 1 0
Hence, (162)10 = (10100010)2
iii) (1235)10 conversion to octal representation
Solution:
8 1235
8 154 3
8 19 2
8 2 3
Hence, (1235)10 = (2323)8
iv) 21.218 conversion to decimal representation
Solution:
21.218 = (2 * 81) + (1 * 80). (2 * 8-1) + (1 * 7=8-2)
= 17 + 0.25 + 0.015625

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COMPUTER ARCHITECHTURE AND ORGANIZATION
= 17.265625
c) Representation of value
i) One's complement
Highest Value is 011
Lowest Value is 100
ii) Two's complement
Highest Value is 011
Lowest Value is 101
iii) Signed Magnitude
Highest Value is 011
Lowest Value is 111
Question 2:
a) Prove
The expression for the logic diagram is: (a.b)’
Truth table of the expression is provided below:
A b a.b (a.b)’
0 0 0 1
COMPUTER ARCHITECHTURE AND ORGANIZATION
= 17.265625
c) Representation of value
i) One's complement
Highest Value is 011
Lowest Value is 100
ii) Two's complement
Highest Value is 011
Lowest Value is 101
iii) Signed Magnitude
Highest Value is 011
Lowest Value is 111
Question 2:
a) Prove
The expression for the logic diagram is: (a.b)’
Truth table of the expression is provided below:
A b a.b (a.b)’
0 0 0 1
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0 1 0 1
1 0 0 1
1 1 1 0
The expression of the logic diagram
a’ + b’
The truth table for the expression is provided below:
A b a’ b’ a’ + b’
0 0 1 1 1
0 1 1 0 1
1 0 0 1 1
1 1 0 0 0
Hence, LHS = RHS (Proved)
COMPUTER ARCHITECHTURE AND ORGANIZATION
0 1 0 1
1 0 0 1
1 1 1 0
The expression of the logic diagram
a’ + b’
The truth table for the expression is provided below:
A b a’ b’ a’ + b’
0 0 1 1 1
0 1 1 0 1
1 0 0 1 1
1 1 0 0 0
Hence, LHS = RHS (Proved)
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b) Using basic Boolean algebra identities for Boolean variables x, y, and z, for prove
The expression derived from the given circuit
A’. B’ + A.B = X
The given circuit that can be minimized from given circuit is:
c) Prove:
X’ + Y’ + XYZ’
= X’ + Y’ + (X’ + Y’ + Z)’ [ by De-Morgan’s Law]
= (XY (X’ + Y’ + Z))’ [ by De-Morgan’s Law]
= (XX’Y + XYY’ + XYZ)
= (0 + 0 + XYZ)
COMPUTER ARCHITECHTURE AND ORGANIZATION
b) Using basic Boolean algebra identities for Boolean variables x, y, and z, for prove
The expression derived from the given circuit
A’. B’ + A.B = X
The given circuit that can be minimized from given circuit is:
c) Prove:
X’ + Y’ + XYZ’
= X’ + Y’ + (X’ + Y’ + Z)’ [ by De-Morgan’s Law]
= (XY (X’ + Y’ + Z))’ [ by De-Morgan’s Law]
= (XX’Y + XYY’ + XYZ)
= (0 + 0 + XYZ)

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COMPUTER ARCHITECHTURE AND ORGANIZATION
= (XYZ)’
= X’ + Y’ + Z’ [ by De-Morgan’s Law]
Hence, X’ + Y’ + XYZ’ = X’ + Y’ + Z’ [PROVED]
COMPUTER ARCHITECHTURE AND ORGANIZATION
= (XYZ)’
= X’ + Y’ + Z’ [ by De-Morgan’s Law]
Hence, X’ + Y’ + XYZ’ = X’ + Y’ + Z’ [PROVED]
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