COA-ITC544 Assignment 1 - Desklib

This article provides answers to COA-ITC544 Assignment 1 questions on conversions, representation of values, Boolean algebra identities, and more. The subject code is ITC544, and the course is Computer Architecture and Organization. The article is relevant to students studying in CSU-Melbourne.

COA-ITC544 Assignment 1
Computer Architecture and Organization
Student name:Wasif Numan
Student ID Number:11640426
Subject Code: ITC544
CSU-Melbourne

COA-ITC544-Assignment 1
Table of Contents
Answer to Question 1:.................................................................................................................2
A.) Value of base x if (152)x = (6A) 16......................................................................................2
b) Conversions.............................................................................................................................2
c) Representation of value..........................................................................................................4
Question 2:.......................................................................................................................................5
a) Prove........................................................................................................................................5
b)Using basic Boolean algebra identities for Boolean variables x, y, and z, for prove...............6
c) Prove:......................................................................................................................................7
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COA-ITC544-Assignment 1
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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COA-ITC544-Assignment 1
(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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COA-ITC544-Assignment 1
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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COA-ITC544-Assignment 1
Highest Value is 011
Lowest Value is 111
Question 2:
a) Prove
The expression for the above logic diagram is: (a.b)’
Truth tablefor (a.b)’
The expression of the logic diagram
a’ + b’
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COA-ITC544-Assignment 1
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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COA-ITC544-Assignment 1
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]
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