Signed Binary Numbers - Free MCQ Practice Test with solutions, GATE EE

Given Number is 10100
The Right shift of the content in register is same as the content divided by 2 
Apply Right Shift ⇒ 11010              
Operation right shift is equivalent to divided by 2

To represent a given 5-bit number using 6- bits in a 2's complement representation, we simply copy the MSB bit as it is till we get the required 6 bits, i.e.
X = 01110 = 001110 
Y = 11001 =111001
Ignoring the carry, we get the addition of the two in 2's complement number as:∴ x + y = 000111

Concept:

• If number is positive; MSB = 0
• Then 2’s complement will be the same
• If number is negative; MSB = 1
• Then 2’s complement will be different from its obtained result

Calculation:
Given,

X = 00110
since, the MSB = 0
∴ it is a positive number.
Decimal equivalent: 0 + 1 x 2² + 1 x 2¹ + 0 x 2⁰ = + 6
Y = 10011
since, the MSB = 1 
∴ it is a negative number,
We need to take the 2's complement of Y, that is.
1's complement (Y) + 1
01100 + 1
⇒ 01101
Decimal equivalent → 0 + 1 × 2³ + 1 × 2² + 0 + 1 x 2⁰ → -13
The sum of X and Y is
+6 - 13 = - 7
The 2's complement of - 7
→ 1's complement of 7 + 1
→ 1's complement of 00111 + 1
→ 11000 + 1 → 11001

The smallest negative number is the largest binary value.

1111 is -1, 1110 is -2, 1101 is -3, etc down to 1000 which represents -8.

Signed magnitude representation uses the most significant bit (MSB) a sign bit.

1. If the sign bit is ‘0’ then the number is positive.
2. If the sign bit is ‘1’ then the number is negative.

The remaining bits represent the magnitude of the binary number.

The dynamic range in 
In fixed point, unsigned integer representation using N-bit, the range of Max to Min is 2^N to 1

In fixed-point signed integer representation using N-bit, the range of Max to Min number is 2^N-1 to 1

Calculation:
The dynamic range of a 32-bit binary number B is:


 

Exponent = e – 127 varies from –127 to 128,
e = 0 = all bits (b₃₀ ---- b₂₃) are zero.
Now, the dynamic range is determined by the size of the exponent, which is
1 × 2^-127(min) to 2¹²⁸ (max).
Dynamic range:

Difference in Dynamic Ranges = 6.03 (2⁸ – 31)
So, Option (2) is correct.

Concept:

1’s complement representation of a binary number is obtained by toggling all the bits, i.e. replacing 1 with 0, and 0 with 1.
2’s complement representation of a binary number is obtained by adding 1 to the 1’s complement representation.

Application:

(127)₁₀ = (01111111)₂
1’s complement representation will be:
1’s complement = 10000000
Number of 1’s is the 1’s complement is, n = 1
Now, the two (2’s) complement representation will be:
2’s complement = 10000000 + 1 = 10000001
Number of 1’s in 2’s complement is, m = 2
∴ The required ratio is m : n = 2 : 1

Concept: 
1's complement of Binary: 1's complement of a Binary number is defined by the value obtained by inverting all the bit, i.e, 0 as 1 and 1 as 0.

Calculation:
The 1's complement of the given binary digit will be:
(101100) → (010011)

64 in binary form is represented as:
6410 = (1000000)₂
Taking the 1's complement of the above, we get 0111111
Adding 1 to the 1's complement, we get the 2's complement representation of the number, i.e. 1000000.
Since there is a 1 in the LSB, the number is a negative number with value 64.
∴ The 2's complement of -64₁₀ contains 7 bits.