Homework Solution: Computer Architecture, Number Systems & Logic

This homework assignment solution covers fundamental concepts in Computer Organization and Architecture. It includes problems related to number system conversions, such as converting between decimal, binary, octal, hexadecimal, and other bases. The solution demonstrates step-by-step conversions and arithmetic operations in different number systems. Additionally, the assignment explores logic circuits, including proving the equivalence of combinational circuits using Boolean algebra and De Morgan's laws. Simplification of logic expressions and circuit reduction techniques are also addressed, providing a comprehensive overview of basic digital logic principles. This resource is useful for students studying computer architecture and digital design, and Desklib offers more solved assignments and past papers.

Running head: COMPUTER ORGANIZATION AND ARCHITECTURE
Computer Organization and Architecture
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COMPUTER ORGANIZATION AND ARCHITECTURE
Answer to Question 1:
a) Given (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
Since, the value of X cannot be negative; so value of X is 8
(152)8= (6A) 16
b) i) BED = (B * 162) + (E * 161) + (D * 160)
= 2816 + 224 + 13
= (3053)10
(3053)10 =
Divider Dividend Remainder
3 3053 2
3 1017 0
3 339 0
3 113 2
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COMPUTER ORGANIZATION AND ARCHITECTURE
3 37 1
3 12 0
3 4 1
1
BED16 = (11012002)3
ii) (321)7 = (3 * 72) + (2 * 71) + (1 * 70)
= (162)10
So, (162)10 =
Divider Dividend Remainder
2 162 0
2 81 1
2 40 0
2 20 0
2 10 0
2 5 1
2 2 0
1
(162)10 = (10100010)2
iii) Conversion
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COMPUTER ORGANIZATION AND ARCHITECTURE
Divider Dividend Remainder
8 1235 3
8 154 2
8 19 3
2
So, (1235)10 = (2323)8
iv) For given 21.218 = (2 * 81) + (1 * 80). (2 * 8-1) + (1 * 7=8-2)
= 17 + 0.25 + 0.015625
= 17.265625
(21.21)8 = 17.265625
c) i) One’s compliment negative number= 100
One’s compliment positive number = 100
ii) Two’s compliment negative number = 101
Two’s compliment positive number = 011
iii) Signed magnitude negative number = 111
Signed magnitude positive number = 011
Answer to Question 2:
a) To prove both the circuits are equal or not, following steps are shown
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COMPUTER ORGANIZATION AND ARCHITECTURE
A B C D
0 0 0 1
0 1 0 1
1 0 0 1
1 1 1 0
The output of 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
So, the output of L.H.S. is equal to the output of R.H.S.
Hence, both the circuits are equal.
b)
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COMPUTER ORGANIZATION AND ARCHITECTURE
From the circuit, it can be said that
(A' . B') + (A . B)
= A'B' + AB
= ((AB)'(A'B')’)' … From De Morgan’s theorem
= ((A' + B')(A + B))' … From Distributive Law
= (A'A + B'A + A'B + B'B)'
= (1 + AB' + A'B + 1)'
= (AB' + A'B)'
= (A ^ B)' that is A XNOR B
c) X’ + Y’ + XYZ’
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COMPUTER ORGANIZATION AND ARCHITECTURE
= X’ + Y’ + (X’ + Y’ + Z)’ [From De-Morgan’s Law]
= (XY (X’ + Y’ + Z))’ [From De-Morgan’s Law]
= (XX’Y + XYY’ + XYZ)
= (0 + 0 + XYZ)
= (XYZ)’
= X’ + Y’ + Z’ [From De-Morgan’s Law]
= X’ + Y’ + XYZ’ = X’ + Y’ + Z’ [HENCE, PROVED]
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COMPUTER ORGANIZATION AND ARCHITECTURE
Bibliography
Harris, S., & Harris, D. (2015). Digital Design and Computer Architecture: ARM Edition.
Morgan Kaufmann.
Wang, S., & ZHANG, C. (2016). Computer architecture.
Impagliazzo, J., Pow-Sang, J. A., Trejos, I., Meier, R., & Nunes, D. J. (2017, March). Latin and
American perspectives on the computer engineering (CE2016) report. In World Engineering
Education Conference (EDUNINE), IEEE (pp. 9-13). IEEE.
Tanenbaum, A. S. (2016). Structured computer organization. Pearson Education India.
Comer, D. (2017). Essentials of computer architecture. CRC Press.
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