Discrete Time Systems Implementation MCQ Quiz – Objective Question with Answer for Discrete Time Systems Implementation

61. If the two numbers are to be multiplied, the mantissa is multiplied and the exponents are added.

A. True
B. False

Answer: A

Let us consider two numbers X=M.2E and Y=N.2F
If we multiply both X and Y, we get X.Y=(M.N).2E+F
Thus if we multiply two numbers, the mantissa is multiplied and the exponents are added.

 

62. What is the smallest floating-point number that can be represented using a 32-bit word?

A. 3*10-38
B. 2*10-38
C. 0.2*10-38
D. 0.3*10-38

Answer: D

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.

Thus, the smallest floating-point number that can be represented using the 32-bit number is
(1/2)*2-127 = 0.3*10-38

Thus, the smallest floating-point number that can be represented using the 32-bit number is
(1-2-23)*2127 = 1.7*1038.

 

63. If 0<E<255, then which of the following statement is true about X?

B. Infinity
C. Mixed number
D. Zero

Answer: C

According to the IEEE 754 standard, for a 32-bit machine, a single-precision floating-point number is represented as X=(-1)s.2E-127(M).
From the above equation, we can interpret that,
If 0<E<255, then X=(-1)s.2E-127(1.M)=>X is a mixed number.

 

64. For a twos complement representation, the truncation error is _________

A. Always positive
B. Always negative
C. Zero
D. None of the mentioned

Answer: B

For a two’s complement representation, the truncation error is always negative and falls in the range

-(2-b-2-bm) ≤ Et ≤ 0.

 

65. Due to non-uniform resolution, the corresponding error in a floating-point representation is proportional to the number being quantized.

A. True
B. False

Answer: A

In floating-point representation, the mantissa is either rounded or truncated. Due to non-uniform resolution, the corresponding error in a floating-point representation is proportional to the number being quantized.

 

66. What is the binary equivalent of (-3/8)?

A. (10011)2
B. (0011)2
C. (1100)2
D. (1101)2

Answer: D

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 two’s complement of 0011=1101.

 

67. Which of the following is the correct representation of a floating-point number X?

A. 2E
B. M.2E(1/2<m<1)
C. 2M.2E(1/2<m<1)
D. None of the mentioned

Answer: B

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 1/2<M<1, multiplied by the exponential factor 2E, where the exponent E is either a negative or positive integer.

Hence a number X is represented as X= M.2E(1/2<M<1).

 

68. 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

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.

 

69. 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

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.

The largest floating-point number that can be represented using a 32-bit word is 1.7*1038

 

70. 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

According to the IEEE 754 standard, for a 32-bit machine, a single-precision floating-point number is represented as X=(-1)s.2E-127(M).
From the above equation, we can interpret that,
If E=0 and M=0, then the value of X is 0.

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