written 7.8 years ago by | modified 2.8 years ago by |
Mumbai University > Electronics and Telecommunication Engineering > Sem 3 > Digital Electronics
Marks: 5M
Year: May 2016
written 7.8 years ago by | modified 2.8 years ago by |
Mumbai University > Electronics and Telecommunication Engineering > Sem 3 > Digital Electronics
Marks: 5M
Year: May 2016
written 7.8 years ago by |
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 ${(AB)_{16}} = {(253)_8}$