250+ TOP MCQs on Representation of Numbers – 2 & Answers

Digital Signal Processing Interview Questions & Answers focuses on “Representation of Numbers”.

1. What is the binary equivalent of (-3/8)?
A. (10011)₂
B. (0011)₂
C. (1100)₂
D. (1101)₂
Answer: D
Clarification: The number (-3/8) is stored in the computer as the 2’s complement of (3/8)
We know that the binary equivalent of (3/8)=0011
Thus the twos complement of 0011=1101.

2. Which of the following is the correct representation of a floating point number X?
A. 2^E
B. M.2^E(1/2C. 2M.2^E(1/2D. None of the mentioned
Answer: B
Clarification: The binary floating point representation commonly used in practice, consists of a mantissa M, which is the fractional part of the number and falls in the range 1/2E, where the exponent E is either a negative or positive integer. Hence a number X is represented as X= M.2^E(1/2

3. What is the mantissa and exponent respectively obtained when we add 5 and 3/8 in binary float point representation?
A. 0.101010,011
B. 0.101000,011
C. 0.101011,011
D. 0.101011,101
Answer: C
Clarification: We can represent the numbers in binary float point as
5=0.101000(2⁰¹¹)
3/8=0.110000(2¹⁰¹)=0.000011(2⁰¹¹)
=>5+3/8=(0.101000+0.000011)(2⁰¹¹)=(0.101011)(2⁰¹¹)
Therefore mantissa=0.101011 and exponent=011.

4. What is the largest floating point number that can be represented using a 32-bit word?
A. 3*10³⁸
B. 1.7*10³⁸
C. 0.2*10³⁸
D. 0.3*10³⁸
Answer: B
Clarification: Let the mantissa be represented by 23 bits plus a sign bit and let the exponent be represented by 7 bits plus a sign bit.

5. If E=0 and M=0, then which of the following statement is true about X?
A. Not a number
B. Infinity
C. Defined
D. Zero
Answer: D
Clarification: According to the IEEE 754 standard, for a 32-bit machine, single precision floating point number is represented as X=(-1)^s.2^E-127(M).
From the above equation we can interpret that,
If E=0 and M=0, then the value of X is 0.

6. The truncation error for the sign magnitude representation is symmetric about zero.
A. True
B. False
Answer: A
Clarification: The truncation error for the sign magnitude representation is symmetric about zero and falls in the range
-(2^-b-2^-bm) ≤ E_t ≤ (2^-b-2^-bm).

7. What is the range of round-off error for a foxed point representation?
A. [-0.5(2^-b+2^-bm), 0.5(2^-b+2^-bm)]
B. [0, (2^-b+2^-bm)]
C. [0, (2^-b-2^-bm)]
D. [-0.5(2^-b-2^-bm), 0.5(2^-b-2-bm^-bm)]
Answer: D
Clarification: The round-off error is independent of the type of fixed point representation. The maximum error that can be introduced through rounding is 0.5(2^-b+2^-bm) and this can be either positive or negative, depending on the value of x. Therefore, the round-off error is symmetric about zero and falls in the range
[-0.5(2^-b-2^-bm), 0.5(2^-b-2-bm^-bm)].

8. What is the 2’s complement of (1100)₂?
A. (0100)₂
B. (0011)₂
C. (0111)₂
D. (1100)₂