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3. Write the difference between 1’s complement and 2’s complement method.
Ans.
One’s complement Two’s complement
1. 1’s complement of a binary negative number is 1. 2’s complement of a binary negative number is
obtained by replacing 0 with 1 and 1 with 0. obtained by adding 1 to LSB of its 1’s complement.
2. 0 has two different representation +0 as 00000000 2. 0 has only one representation 00000000.
as -0 as 11111111.
3. A n-bit register has positive largest number as 3. A n-bit register has positive largest number as
2 (n-1) - 1 and negative smallest number as -2 (n-1) - 1. 2 (n-1) - 1 and negative smallest number as -2 (n-1) .
4. Find the value of:
a. 6 >> 1
b. -6 >> 1
c. 3 << 1
Ans. a. Considering 8-bit code, 6 is represented as 00000110.
Shifting 1 bit to the right and filling the left with a sign bit we get:
0 0 0 0 0 0 1 1
The decimal value of 0000011 is 3.
b. Considering 8-bit code, 6 is represented as 00000110.
Number is negative. So one’s complement is 11111001.
Two’s complement is 11111010.
Shifting 1 bit to the right and filling the left with 1, we get:
1 1 1 1 1 1 0 1
One’s complement is 00000010.
Two’s complement is 00000011.
Decimal equivalent = -3.
c. Considering 8-bit code, 3 is represented as 00000011.
Shifting 1 bit to the left and filling the right bit with 0, we get:
0 0 0 0 0 1 1 0
2
1
Decimal value is 1 × 2 + 1 × 2 = 6
5. Write the decimal value of the floating point single precision number.
1 10000011 10110110000000000000000
Ans. Sign bit is 1, so the number is negative.
0
7
1
Exponent bit 10000011 = 1 × 2 + 1 × 2 + 1 × 2 = 131
Adjust bias by subtracting excess = 131 - 127 = 4
Fractional part = 1.1011011
Number = 1.1011011 × 2 4
= 11011.011
4
-2
1
3
0
= 1 × 2 + 1 × 2 + 1 × 2 + 1 × 2 + 1 × 2 + 1 × 2 -3
= 16 + 8 + 2 + 1 + 1/4 + 1/8
= 27.375
Adding sign bit -27.375
6. Write the difference between fixed point notation and floating point notation.
Ans. Fixed point Floating point
1. This representation has fixed number of digits 1. A formula representation of real numbers as an
after radix point. approximation to support a trade off between range
and precision.
2. It is used to represent limited range of values. 2. It is used to represent wide range of values.
67
Encoding 67

