12
Digital Electronics
Solution
(a) The decimal equivalent of (12LM) is given by
M × 5
0 + L × 5
1 + 2 × 5
2 + 1 × 5
3 = 4 × 5
0 + 3 × 5
1 + 2 × 5
2 + 1 × 5
3 L = 3 M = 4
= 4 + 15 + 50 + 125 = 194
The decimal equivalent of (L1) is given by
L × 5
−1 + 1 × 5
−2 = 3 × 5
−1 + 5
−2 = 064
Combining the results, (12LM.L1) 5 = (194.64) 10 .
(b) The total number of possible four-digit combinations = 5
4
= 625.
Example 1.10
The 7’s complement of a certain octal number is 5264. Determine the binary and hexadecimal
equivalents of that octal number.
Solution
• The 7’s complement = 5264.
• Therefore, the octal number = (2513) 8 .
• The binary equivalent = (010 101 001 011) 2 = (10101001011) 2 .
• Also, (10101001011) 2 = (101 0100 1011) 2 = (0101 0100 1011) 2 = (54B) 16 .
• Therefore, the hex equivalent of (2513) 8 = (54B) 16 and the binary equivalent of (2513) 8 =
(10101001011) 2 .
1.17 Floating-Point Numbers
Floating-point notation can be used conveniently to represent both large as well as small fractional
or mixed numbers. This makes the process of arithmetic operations on these numbers relatively much
easier. Floating-point representation greatly increases the range of numbers, from the smallest to the
largest, that can be represented using a given number of digits. Floating-point numbers are in general
expressed in the form
N = m × b
e
(1.1)
where m is the fractional part, called the significand or mantissa, e is the integer part, called the
exponent, and b is the base of the number system or numeration. Fractional part m is a p-digit number
of the form (±d.dddd dd), with each digit d being an integer between 0 and b – 1 inclusive. If the
leading digit of m is nonzero, then the number is said to be normalized.
Equation (1.1) in the case of decimal, hexadecimal and binary number systems will be written as
follows:
Decimal system
N = m × 10
e
(1.2)
Digital Electronics
Solution
(a) The decimal equivalent of (12LM) is given by
M × 5
0 + L × 5
1 + 2 × 5
2 + 1 × 5
3 = 4 × 5
0 + 3 × 5
1 + 2 × 5
2 + 1 × 5
3 L = 3 M = 4
= 4 + 15 + 50 + 125 = 194
The decimal equivalent of (L1) is given by
L × 5
−1 + 1 × 5
−2 = 3 × 5
−1 + 5
−2 = 064
Combining the results, (12LM.L1) 5 = (194.64) 10 .
(b) The total number of possible four-digit combinations = 5
4
= 625.
Example 1.10
The 7’s complement of a certain octal number is 5264. Determine the binary and hexadecimal
equivalents of that octal number.
Solution
• The 7’s complement = 5264.
• Therefore, the octal number = (2513) 8 .
• The binary equivalent = (010 101 001 011) 2 = (10101001011) 2 .
• Also, (10101001011) 2 = (101 0100 1011) 2 = (0101 0100 1011) 2 = (54B) 16 .
• Therefore, the hex equivalent of (2513) 8 = (54B) 16 and the binary equivalent of (2513) 8 =
(10101001011) 2 .
1.17 Floating-Point Numbers
Floating-point notation can be used conveniently to represent both large as well as small fractional
or mixed numbers. This makes the process of arithmetic operations on these numbers relatively much
easier. Floating-point representation greatly increases the range of numbers, from the smallest to the
largest, that can be represented using a given number of digits. Floating-point numbers are in general
expressed in the form
N = m × b
e
(1.1)
where m is the fractional part, called the significand or mantissa, e is the integer part, called the
exponent, and b is the base of the number system or numeration. Fractional part m is a p-digit number
of the form (±d.dddd dd), with each digit d being an integer between 0 and b – 1 inclusive. If the
leading digit of m is nonzero, then the number is said to be normalized.
Equation (1.1) in the case of decimal, hexadecimal and binary number systems will be written as
follows:
Decimal system
N = m × 10
e
(1.2)