ITC544 - Computer Architecture and Organization: Base Conversion Proof

This assignment focuses on computer architecture and organization, covering base conversions and Boolean algebra. It includes finding the value of a base, converting numbers between different bases (decimal, binary, octal, hexadecimal), and representing values using one's complement, two's complement, and signed magnitude. The assignment also involves proving Boolean algebra identities and minimizing a given circuit. Desklib provides a platform where students can access these solutions and explore a wide range of study materials.

Running head: COMPUTER ARCHITECHTURE AND ORGANIZATION
Advance computer Architecture
Full name:
Student ID:
Subject Code: ITC544
Author’s Note

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COMPUTER ARCHITECHTURE AND ORGANIZATION
Table of Contents
Answer to Question 1:.....................................................................................................................2
a.) Finding 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
Answer to Question 1:
a.) Finding the value of base x if (152) x = (6A) 16
Let the value of base be X,
Provided,
(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.
Therefore, 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

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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

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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 above logic diagram is: (a.b)’
Truth table for the expression is given below:
A b a.b (a.b)’
0 0 0 1

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COMPUTER ARCHITECHTURE AND ORGANIZATION
0 1 0 1
1 0 0 1
1 1 1 0
The expression for the a above logic diagram is provided below
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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COMPUTER ARCHITECHTURE AND ORGANIZATION
b) Using basic Boolean algebra identities for Boolean variables x, y, and z, for prove
The expression that is derived from the given circuit is provided below:
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]