Question

5. Number conversion a. (114)dec → Doct b. (1001 1011.011)bin → ( )dec c. (1011 0101)8-bit 2's comp → ()dec d. (-39)dec → ( )8-bit 2's comp e. (1A2B.3C)hex → ( )bin f. (1100101011.11011)bin oct g. (21.2)dec → bin

Answer 1

5(a)

(114)dec--->()oct

divide 114 by 8 therefore quotient 14 and remainder is 2

again divide 14 by 8 therefore quotient is 1 and remainder is 6

dividing 1 by 8 quotient is 0 and remainder is 1 and no more division can be done .So taking remainder in reverse order--162

therefore 114dec==162oct

5(b) (10011011.011)bin-->()dec

=(1x2^7)+(0x2^6)+(0x2^5)+(1x2^4)+(1x2^3)+(0x2^2)+(1x2^1)+(1x2^0)+(0x2^-1)+(1x2^-2)+(1x2^-3)==155.375

5(c) 10110101->8bit 2's comp-->()dec

first we have to calculate 2's complement of thenumber

therefore 10110101->01001010(1's)-->01001010+1==01001011(2's complement)

therefore (0x2^7)+(1x2^6)+(0x2^5)+(0x2^4)+(1x2^3)+(0x2^2)+(1x2^1)+(1x2^0)==75(ans)

5(d)

-39dec-->()8 bits 2's complement

39-->00100111

1's complement 11011000 which is equal to(-39)

therefore 2's complement in 8 bit ==11011000+1==11011001(ans)

5(e)1A2B.3C-->bin()

1-->0001

A-->1010

2-->0010

B-->1011

3-->0011

C-->1100

therefore in binary 1101000101011.001111(ans)

5(f)(1100101011.11011)bin-->oct()

split the binary number from left to right each of group 3 bits

001 100 101 011.110 110

1 4 5 3. 6 6 (corresponding decimal value expalined in 2nd bit)

therefore 1453.66(ans)

5(g) 21.2dec-->bin

10101.0(ans)

just divide the number by 2 as explained in the first bit of the question since we have convert it to binary therefore divide it by 2.and then write the remainders in reverse order .

1. Divide the number by 2.
2. Get the integer quotient for the next iteration.
3. Get the remainder for the binary digit.
4. Repeat the steps until the quotient is equal to 0.
5. hence write the remainders in reverse order .

for 21-->10101

and after decimal .2-->0010