Computer Organization and Architecture Homework Solution

This document presents a comprehensive solution to a computer organization and architecture assignment. The solution covers several key topics including number system conversions (binary, octal, decimal, hexadecimal, and base 5), and different representations of signed integers (one's complement, two's complement, and signed magnitude). The assignment also delves into boolean algebra, simplifying expressions and constructing truth tables and circuit diagrams. The solution demonstrates the application of these concepts to solve problems related to digital logic and computer arithmetic, with references provided for further study. The assignment addresses topics such as number system conversions, boolean algebra simplification, and circuit diagram design.

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Computer Organisation and Architecture

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Contents
Computer Organisation and Architecture.............................................................................................1
QUESTION. 1..........................................................................................................................................3
A).......................................................................................................................................................3
B.)......................................................................................................................................................3
Question.2.............................................................................................................................................6
A.)......................................................................................................................................................6
B.)......................................................................................................................................................7
Truth table-....................................................................................................................................7
Boolean Expression........................................................................................................................8
Circuit Diagram:.............................................................................................................................8
REFERENCES..........................................................................................................................................9

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QUESTION. 1
A).
If a tiny computer that has a word size of 8 bits is given, then the largest positive number
will be (01111111) i.e. +127 and the smallest negative number will be (10000000) i.e. -127.
1. One’s complement
Largest positive number- (01111111) i.e. +127
Smallest negative number- (10000000) i.e. -128
2. Two’s complement
Largest positive number-(01111111) i.e. +127
Smallest negative number-(10000000) i.e. -128
3. Signed magnitude
Largest positive number-(01111111) i.e. +127
Smallest negative number-(10000000) i.e. -128
4. Unsigned magnitude
Largest positive number-(01111111) i.e. +127
Smallest negative number-(00000000) i.e. 0
B.)
i). 0X5AB into octal
(0)16 = (0000)2
(5)16 = (0101)2
(A) 16 = (1010)2
(B) 16 = (1011)2
Hence, (05AB) 16 = (010110101011)2
In octal,
010=2, 110=6, 101=5, 011=3
Hence, octal equivalent of 05AB is 2653.

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ii). 101101.1012 into Decimal
(101101.101)2 = 1*25 + 0*24 + 1*23 + 1*22 +0*21 + 1*20 +1*2-1 + 0*2-2 +1*2-3
= 32+0+8+4+0+1+0.5+0+0.125
= 45.62510
iii). 12348 into Binary
=Dividing (1234)8 by 2 till the quotient is 0,
1234/2=617(remainder is 0)
617/2=308(remainder is 1)
308/2=154(remainder is 0)
154/2=77(remainder is 0)
77/2=38(remainder is 1)
38/2=19(remainder is 0)
19/2=9(remainder is 1)
9/2=4(remainder is 1)
4/2=2(remainder is 0)
2/2=1(remainder is 0)
1/2=0(remainder is 1)
Now, reading the remainders from bottom to top.
Hence, (1234)8 is equivalent to (1010011100)2.
iv). 679810 into Base 5 number
Dividing 679810 by 5 continuously we get
(6798)10 is equal to (204143)5

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v). 976.6310 into Binary
Dividing the integral part with 2,
976/2=488(remainder is 0)
488/2=244(remainder is 0)
244/2=122(remainder is 0)
122/2=61(remainder is 0)
61/2=30(remainder is 1)
30/2=15(remainder is 0)
15/2=7(remainder is 1)
7/2=3(remainder is 1)
3/2=1(remainder is 1)
½=1(remainder is 1)
Binary of integral part of decimal is 1111010000.
Multiplying the fractional part by 2,
0.63*2=1.26(integral part is 1)
Fractional part of decimal is .1
Hence, (976.63)10 is equivalent to (1111010000.1)2.
vi) . 1001001011 into Hexadecimal
= grouping these digits into sets of four
= 0010 0100 1011
Also,
(0010)2 = (2)16
(0100)2 = (4)16

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(1011)2= (B)16
Hence, 1001001011 is equivalent to (24B) in hexadecimal.
vii). 10011110 (8-bit 2’s complement representation) to Decimal
As the first bit indicates the sign and here its 1 hence this is a negative number.
Converting the number into 1’s complement,
10011110
- 1
=10011101
Replacing 1’s with 0’s and vice-versa,
(10011101)’= (01100010)
= 0*27+1*26+1*25+0*24+0*23+0*22+1*21+0*20
=64+32+2
= (98)10
Question.2
A.)
X’(X+Y) + (XX+Y)(Y’+X)=Y+X
Taking LHS,
=X’X+X’Y+X2 Y’+X3+YY’+YX (On Multiplication)
=0+X’Y+X2Y’+X3+0+YX (By identity X.X’=0 & Y.Y’=0)
=Y (X+X’) +X2(Y’+X) (On taking common)
=Y+X (Y’+X) (By additive identity, X+X’=1 &
By multiplicative identity, X.X=X)
=Y+XY’+XX (On multiplication)

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=X+Y+XY’ (By multiplicative identity, X.X=X)
=Y+Y’X+X
=Y (1+X) +Y’X+X (By additive identity, 1+X=1)
=Y+YX+Y’X+X (On multiplication)
=Y+X (Y+Y’) +X (On taking common)
=Y+X+X (By additive identity, Y+Y’=1)
=Y+(X+X) (By additive identity, X+X=X)
=Y+X
Hence, LHS=RHS
B.)
A subject has 4 assessments, such as Assignment, Blog, Discussion Forum and Quiz. In order
to pass in the subject, a student must get as below:
• Assignment component must pass (50 or above)
• Any 2 more assessments must pass
• Fail otherwise
Truth table-
Assignment Blog Discussion
Forum
Quiz Result
Fail Fail Fail Fail Fail
Fail Fail Fail Pass Fail
Fail Fail Pass Fail Fail
Fail Fail Pass Pass Fail
Fail Pass Fail Fail Fail
Fail Pass Fail Pass Fail
Fail Pass Pass Fail Fail

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Fail Pass Pass Pass Fail
Pass Fail Fail Fail Fail
Pass Fail Fail Pass Fail
Pass Fail Pass Fail Fail
Pass Fail Pass Pass Pass
Pass Pass Fail Fail Fail
Pass Pass Fail Pass Pass
Pass Pass Pass Fail Pass
Pass Pass Pass Pass Pass
Boolean Expression
= A*(BD+DQ+QB)
= (A ∧ ((B ∧ D) ∨ (D ∧ Q) ∨ (Q ∧ B)))
Circuit Diagram:
AND
AND
AND
AND
B
D
D
Q
Q
B
A Result

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REFERENCES
Dueck, R. and Reid, K. (2012). Digital electronics. Clifton Park, N.Y.: Delmar/Cengage
Learning.
Karim, M. and Chen, X. (2008). Digital design. Boca Raton, FL: CRC Press/Taylor & Francis.