Exploring Number Conversions, Logic Gates, and Computer Architecture

This assignment solution delves into fundamental concepts of computer architecture, focusing on number system conversions and logic gate manipulation. It begins by determining the base 'x' for a given equation involving hexadecimal and an unknown base. Subsequently, it demonstrates conversions between different number systems, including hexadecimal to base-3, decimal to binary, decimal to octal, and base-8 to decimal. The assignment also explores the representation of numbers in one's complement, two's complement, and signed magnitude formats within a 3-bit computer. Furthermore, it proves the equivalence of two combinational circuits using Boolean algebra and De Morgan's Law. Finally, the solution includes a minimized circuit diagram and simplification of a Boolean expression, showcasing practical applications of logic gate principles. This document is available on Desklib, a platform offering a wealth of study resources for students.

Running head: COMPUTER ARCHITECTURE
Computer Organization
Name of Student –
Name of University –
Author’s Note –

1COMPUTER ARCHITECTURE
Answer to Question 1
a) (152)x= (6A)16
 X2 + (5 * X1) + (2 * X0) = (6 * 161) + (10 * 160)
 X2 + 5X + 2 = 106
 X2 + 5X - 104 = 0
 X2 + 13X - 8X – 104 = 0
 X(X + 13) – 8(X + 13) = 0
 (X - 8) (X + 13) = 0
 X = 8 and X = -13
So, the value of X is 8
b) Conversions
1) BED16 to base 3
BED = (B * 162) + (E * 161) + (D * 160)
= 2816 + 224 + 13
=305310
(3053)10 =

2COMPUTER ARCHITECTURE
BED16 = (11012002)3
2) 3217 changing into 2-base
3217 = (3 * 72) + (2 * 71) + (1 * 70)
= (162)10
(162)10 =
(162)10 = (10100010)2
3) 1235 converting into octal representation

3COMPUTER ARCHITECTURE
(1235) = (2323)8
4) 21.218 converting into decimal representation
21.218 = (2 * 81) + (1 * 80). (2 * 8-1) + (1 * 7=8-2)
= 17 + 0.25 + 0.015625
= 17.265625
c) i) In a computer of 3 word size, one’s Complement lowest number = 100
In a computer of 3 word size, one’s Complement lowest number of word size 3 bit = 011
ii) In a computer of 3 word size, two’s Complement lowest number of word size 3 bit =
101
In a computer of 3 word size, two’s Complement lowest number of word size 3 bit = 011
iii) In a computer of 3 word size, signed Magnitude lowest number of word size 3 bit =
111
In a computer of 3 word size, signed Magnitude lowest number of word size 3 bit = 011

4COMPUTER ARCHITECTURE
Question 2
a) L.H.S
a b c d
0 0 0 1
0 1 0 1
1 0 0 1
1 1 1 0
R.H.S.
a b c d e
0 0 1 1 1
0 1 1 0 1
1 0 0 1 1
1 1 0 0 0

5COMPUTER ARCHITECTURE
From the above discussion, it can be stated that L.H.S. is equal to R.H.S.
b) Minimized Circuit Diagram
c) X’ + Y’ + XYZ’
= X’ + Y’ + (X’ + Y’ + Z)’ [From, De Morgan Law]
= (XY (X’ + Y’ + Z))’ [From, De Morgan Law]
= (XX’Y + XYY’ + XYZ)
= (0 + 0 + XYZ)
= (XYZ)’
= X’ + Y’ + Z’ [From, De Morgan Law]
[Hence, PROVED]

6COMPUTER ARCHITECTURE
Bibliography
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26-31).