# Digital Signal Processing Questions and Answers – Representation of Numbers – 1

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This set of Digital Signal Processing Multiple Choice Questions & Answers (MCQs) focuses on “Representation of Numbers-1”.

1. If (101.01)2=(x)10, then what is the value of x?
a) 505.05
b) 10.101
c) 101.01
d) 5.25

Explanation: (101.01)2=1*22+0*21+1*20+0*2-1+1*2-2=(5.25)10
=>x=5.25.

2. If X is a real number with ‘r’ as the radix, A is the number of integer digits and B is the number of fraction digits, then X=$$\sum_{i=-A}^B b_i r^{-i}$$.
a) True
b) False

Explanation: A real number X can be represented as X=$$\sum_{i=-A}^B b_i r^{-i}$$ where bi represents the digit, ‘r’ is the radix or base, A is the number of integer digits, and B is the number of fractional digits.

3. The binary point between the digits b0 and b1 exist physically in the computer.
a) True
b) False

Explanation: The binary point between the digits b0 and b1 does not exist physically in the computer. Simply, the logic circuits of the computer are designed such that the computations result in numbers that correspond to the assumed location of this point.
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4. What is the resolution to cover a range of numbers xmax-xmin with ‘b’ number of bits?
a) (xmax+xmin)/(2b-1)
b) (xmax+xmin)/(2b+1)
c) (xmax-xmin)/(2b-1)
d) (xmax-xmin)/(2b+1)

Explanation: A fixed point representation of numbers allows us to cover a range of numbers, say, xmax-xmin with a resolution
Δ=(xmax-xmin)/(m-1)
where m=2b is the number of levels and ‘b’ is the number of bits.

5. What are the mantissa and exponent required respectively to represent ‘5’ in binary floating point representation?
a) 011,0.110000
b) 0.110000,011
c) 011,0.101000
d) 0.101000,011

Explanation: We can represent 5 as
5=0.625*8=0.625*23
The above number can be represented in binary float point representation as 0.101000*2011
Thus Mantissa=0.101000, Exponent=011.

6. If the two numbers are to be multiplied, the mantissa are multiplied and the exponents are added.
a) True
b) False

Explanation: 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 are multiplied and the exponents are added.

7. 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

Explanation: 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.

8. If 0<E<255, then which of the following statement is true about X?
a) Fractional number
b) Infinity
c) Mixed number
d) Zero

Explanation: According to the IEEE 754 standard, for a 32-bit machine, 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.

9. For a twos complement representation, the truncation error is ____________
a) Always positive
b) Always negative
c) Zero
d) None of the mentioned

Explanation: For a two’s complement representation, the truncation error is always negative and falls in the range
-(2-b-2-bm) ≤ Et ≤ 0.

10. Due to non-uniform resolution, the corresponding error in a floating point representation is proportional to the number being quantized.
a) True
b) False

Explanation: 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.

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