Nt1110 Unit 3 Assignment 1 Question Paper

ASSESSMENT ITEM 3 - ASSIGNMENT 1
SUDHEER KUMAR NANDIGAM
11622446

1. -29
Binary Format of 29: 11101
16 bit binary representation of 29 is 00000000 00011101
16 bit signed magnitude representation of -29 is 10000000 00011101
2’s complement form of -29 is NOT (00000000 00011101) +1
NOT (00000000 00011101) is 11111111 11100010
11111111 11100010 +1
11111111 11100011

2’s complement form of -29 is 11111111 11100011

2. 0 01111110 10100000000000000000000

Sign bit: 0 i.e. positive number
Exponent: 01111110 i.e. 8 bit
Significand: 10100000000000000000000 i.e. 23 bit

X= (-1)s*(1+significand)*2(exponent-127)

S=0
Exponent value in decimal= (01111110)2 =126
Significand = 101

X= (-1)0*(1.101) *2(126-127) =1*(1.101)
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X’Y + XYZ’ + Y’ + XZ (Y+Y’) = 1

Left Hand Side:
X’Y + XYZ’ + Y’ + XZ (Y+Y’)
X’Y + XYZ’ + Y’ + XZY+XZY’ //Distributive law//
X’Y + XYZ’ + XZY + Y’ +XZY’ //Commutative Law//
X’Y+XY (Z’+Z) +Y’+XZY’ //Distributive law//
X’Y+XY (1) +Y’+XZY’ //Inverse Law//
X’Y+XY+Y’ (1) +XZY’ //Identity Law//
X’Y+XY+Y’ (Z+Z’) +XZY’ //Inverse Law//
X’Y+XY+Y’Z+Y’Z’+XZY’ //Distributive law//
X’Y+XY+Y’Z+XZY’+Y’Z’ //Commutative Law//
Y (X’+X) +Y’Z (1+X) +Y’Z’ //Distributive law//
Y (1) +Y’Z (1+X) +Y’Z’ //Inverse Law//
Y+Y’Z (1) +Y’Z’ //Null Law//
Y+Y’Z+Y’Z’
Y+Y’ (Z+Z’) //Distributive law//
Y+Y’ (1) //Inverse Law//
(Y+Y’) //Inverse Law//
1
Output for Left Hand Side is 1 which is equals to Right Hand side value
Therefore Left hand side is equal to Right hand side
i.e. X’Y + XYZ’ + Y’ + XZ (Y+Y’) = 1 4. Assumptions

X as Core subject1
Y as Core subject 2
Z as Elective subject
1 as High Distinction (HD)
0 as Not High Distinction

Below is the truth table for above mentioned
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* Based on point 3
* Based on point 3
* Based on point 2

Expression based on above truth table is XY’Z+ XYZ+X’YZ +XYZ’

XY’Z+ XYZ+X’YZ +XYZ’
XY’Z+YZ (X+X’) +XYZ’ //Distributive Law//
XY’Z+YZ (1) +XYZ’ //Inverse Law//
XY’Z+YZ+XYZ’
Z (XY’+Y) +XYZ’ //Distributive Law//
Z(Y+Y’X) +XYZ’ //Commutative Law//
Z ((Y+Y’)(Y+X))+XYZ’ //Distributive over AND//
Z ((1)(Y+X))+XYZ’ //Inverse Law//
Z(Y+X) +XYZ’ //Identity Law //
ZY+ZX+XYZ’ //Distributive Law//
ZY+X (Z+YZ’) //Distributive Law//
ZY+X (Z+Z’Y) //Commutative Law//
ZY+X ((Z+Z’)(Z+Y)) //Distributive over AND//
ZY+X ((1)(Z+Y)) //Inverse Law//
ZY+X(Z+Y) //Identity Law//
ZY+XZ+XY //Distributive Law//
XY+YZ+XZ //Commutative Law//

OR
XY’Z+ XYZ+X’YZ +XYZ’
XY’Z+ XYZ+XYZ+X’YZ +XYZ’ //Idempotent Law//
XY’Z+ XYZ+XYZ+XYZ+X’YZ +XYZ’ //Idempotent Law//
XY’Z+XYZ+XYZ+X’YZ+XYZ+XYZ’ //Commutative Law//
XZ (Y’+Y) +YZ (X+X’) +XY (Z+Z’) //Commutative Law//
XZ (1) +YZ (1) +XY (1) //Inverse Law//
XZ+YZ+XY //Identity

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