Page 18 - DCAP108_DIGITAL_CIRCUITS_AND_LOGIC_DESIGNS

Unit 1: Number Systems

Notes

562 = ? 2
8
Solution:
Step 1: Convert each octal digit to 3

Binary digits
5 = 101 2
8
6 = 110
8 2
2 -010 2
8
Step 2: Combine the binary groups.
101 110 010
562 = 5 6 2
8
Hence, 562 = 101110010 2
8
1.5.6 Short-cut Method for Binary to Hexadecimal Conversion
The following steps are used in this method:

Step 1: Divide the binary digits into groups of four (starting from the right).
Step 2: Convert each group of four binary digits to one hexadecimal digit. Remember that
hexadecimal digits 0 to 9 are equal to decimal digits 0 to 9 and hexadecimal digits A to F are
equal to decimal values 10 to 15. Hence, for this step, we use binary to decimal conversion
procedure and represent decimal values 10 to 15 as hexadecimal A to F.

11010011 = ?
2 16
Solution:
Step 1: Divide the binary digit into groups of 4, starting from the right (LSD)
1101 0011
Step 2: Convert each group of 4 binary digits to 1 hexadecimal digit.
1101 = 1*2 + 1*2 + 0*2 + 1*2 0 0011 = 0*2 + 0*2 + 1*2 + 1*2 0
3
2
3
2
1
1
2
2
= 8 + 4 + 0 + 1 = 0 + 0 + 2 + 1
= 13 = 3 16
10
= D 16
Hence, 11010011 = D3 16
2

Convert the number 1001011010011 = ?
2 16
1.5.7 Short-cut Method for Hexadecimal to Binary Conversion

The following steps are used in this method:
Step 1: Convert decimal equivalent of each hexadecimal digit to 4 binary digits.

Step 2: Combine all resulting binary groups (4 digits each) into a single binary number.

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