CS302 Assignment
Question no. 1:
Perform the Following Operations with
complete steps.
1. (725157)0+ (F12ADE) H
— (101101010101011111001110101)2 — (9925)0
Answer.
(725157)0
Covert this octal numb into decimal number
(725157)0 = 7 x 85+2 x 84+5
x 83+1 x 82+5 x 81+7 x 80
(725157)0
=229376+8192+2560+64+40+7
(725157)0
= (240238) D
(F12ADE)H
Convert this hex
number into decimal number
(F12ADE)H=F
x 165+1 x 164+2 x 163+A x 162+D x
161+E x 16°
(F12ADE)H=
(15) x 1048576+1 x 65536+2 x 4096+ (10) x 256+ (13) x 164(14) x 1
(F12ADE)H
=15728640+65536+8192+2560+208+14
(F12ADE)H= (15805150) D
(101101010101011111001110101)2
Convert binary into decimal number
First we converting it into Octal Number
System.
= (552537165)0
Now converting it into Decimal Numbers
System.
(552537165)0
= 5 x 88+5 x 87+2 x 86+5 x 85+3 x 84+7
x 83+1 x 82+6 x 81+5 x 8°
(552537165)0
= 83886080+10485760+524288+163840+12288+3584+64+48+5 (552537165)0= (95075957)
D
Now add all these four values
= (240239) D-(15805150)
D-(95075957) D-(9925) D
=-(79040493) D Answer
Question no. 2:
Step l: Convert the Hexadecimal
number to binary by replacing the each hexadecimal symbol with appropriate four
bits binary number. So we get.
|
Hexadecimal
|
E
|
A
|
C
|
1
|
2
|
9
|
|
number Binary
|
1110
|
1010
|
1100
|
0001
|
0010
|
1001
|
Step 2: Now convert the each binary number to gray code by
the grayest digit is sane as the left most binary digit and then add the
adjacent pair of binary digits to find the next gray code.
|
number Binary
|
1110
|
1010
|
1100
|
0001
|
0010
|
1001
|
|
Gray Code
|
1001
|
1111
|
1010
|
0001
|
0011
|
1101
|
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