Mumbai University > Electronics and Telecommunication Engineering > Sem 3 > Digital Electronics

Marks: 5M

Year: May 2016

i) Conversion to BCD:

The hexadecimal number is $(AB)_{16}$

First convert it into decimal number: $(AB)_{16}$ = $10 \times {16^1}$ + $11 \times {16^0}$ $ = 160 + 11 = 171 = {(171)_{10}}$

BCD code for ${(171)_{10}}$ is 0001 0111 0001

Combine all binary bits by removing spaces, 000101110001

Hence ${(AB)_{16}} = {(000101110001)_{BCD}}$

ii) Conversion to Octal code:

The conversion takes place as follows

Covert ${(AB)_{16}}$ into binary:

Combine 4 – bit binary sections by removing spaces:

${(AB)_{16}} = {(10101011)_2}$

Group these binary bits into group of 3 – bits:

Each 3 bit group is converted to an octal digit

Thus ${(AB)_{16}} = {(253)_8}$