Computer Architecture and Organization - Study Material and Solved Assignments

This study material covers topics such as conversions, representation of values, Boolean algebra identities and more related to Computer Architecture and Organization. It also includes solved assignments and proofs. The course code is ITC544.

Running head: COMPUTER ARCHITECHTURE AND ORGANIZATION
Computer Architecture and Organization
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Subject Code: ITC544
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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

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COMPUTER ARCHITECHTURE AND ORGANIZATION
Answer to Question 1:
A.) Value of base x if (152) x = (6A) 16
The value of base is considered as X
Provided:
(152) x = (6A) 16
Or, X^2 + 5X + 2*X = 6*16 + A
Or, X^2 + 5X + 2*X = 6*16 + 10
X^2 + 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
Therefore the value of X is 8.
b) Conversions
i) BED16 converting to base-3
= B * 16*16 + E * 16 + D
= 2816 + 224 + 13
= (3053)10

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COMPUTER ARCHITECHTURE AND ORGANIZATION
(3053)10 =
Therefore,(BED)16 = (11012002)3
ii) 3217 into 2-base (binary) representation
(321)7 = (3 * 72) + (2 * 71) + (1 * 70)
= (162)10
Again, (162)10 =

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COMPUTER ARCHITECHTURE AND ORGANIZATION
Hence, (162)10 = (10100010)2
iii) (1235)10 conversion to octal representation
Solution:
Therefore, (1235)10 = (2323)8
iv) 21.218 conversion to decimal representation
21.218 = (2 * 81) + (1 * 80). (2 * 8-1) + (1 * 7=8-2)
= 17 + 0.25 + 0.015625
= 17.265625
c) Representation of value
i) Highest and lowest value for 1’s complement
Highest Value is 011
Lowest Value is 100
ii) Highest and lowest value for 1’s complement
Highest Value is 011
Lowest Value is 101
iii) Highest and lowest value signed Magnitude

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COMPUTER ARCHITECHTURE AND ORGANIZATION
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 (a.b)’
The expression of the logic diagram
a’ + b’

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COMPUTER ARCHITECHTURE AND ORGANIZATION
Truth table for a’ + b’
Hence, LHS = RHS (Proved)
b) Using basic Boolean algebra identities for Boolean variables x, y, and z, for prove
The derive expression from the above circuit is provided below:
A’. B’ + A.B = X
The given circuit that can be minimized from above circuit is:

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COMPUTER ARCHITECHTURE AND ORGANIZATION
c) Prove:
X’ + Y’ + XYZ’
= X’ + Y’ + (X’ + Y’ + Z)’ [by using De-Morgan’s Law]
= (XY (X’ + Y’ + Z))’ [by using De-Morgan’s Law]
= (XX’Y + XYY’ + XYZ)
= (0 + 0 + XYZ)
= (XYZ)’
= X’ + Y’ + Z’ [by using De-Morgan’s Law]
Therefore, X’ + Y’ + XYZ’ = X’ + Y’ + Z’ [PROVED]