Digital Signal Processing Representation of Number MCQ Quiz – Objective Question with Answer for Digital Signal Processing Representation of Number MCQ

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

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.

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.

A fixed point representation of numbers allows us to cover a range of numbers, say, x_max-x_min with a resolution

Δ=(x_max-x_min)/(m-1)

where m=2b is the number of levels and ‘b’ is the number of bits.

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.

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.

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.

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

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

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.

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.

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.2^E(1/2<M<1).

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.

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.

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

The truncation error for the sign-magnitude representation is symmetric at about zero and falls in the range
-(2^-b-2^-bm) ≤ E_t ≤ (2^-b-2^-bm).

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 at about zero and falls in the range

[-0.5(2^-b-2^-bm), 0.5(2^-b-2-bm^-bm)]

The one’s complement of (1100)2 is (0011)2. Thus the two complements of this number are obtained as (0011)₂+(0001)₂=(0100)₂.

Since the binary digit b-A is the first bit in the representation of the real number, it is called as the most significant bit(MSB. of the number.

According to the IEEE 754 standard, for a 32-bit machine, a 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=255 and M≠0, then X is not a number.

Once the desired frequency band is selected we can specify the sampling rate and the characteristics of the pre-filter, which is also called an anti-aliasing filter. The anti-aliasing filter is an analog filter that has a twofold purpose.

The anti-aliasing filter is an analog filter that has a twofold purpose. First, it ensures that the bandwidth of the signal to be sampled is limited to the desired frequency range.

Using an antialiasing filter is to limit the additive noise spectrum and other interference, which often corrupts the desired signal. Usually, additive noise is wideband and exceeds the bandwidth of the desired signal.

Analog signal processing operations cannot be done very precisely either since electronic components in analog systems have tolerances and introduce noise during their operation. In general, a digital system designer has better control of tolerances in a digital signal processing system than an analog system designer who is designing an equivalent analog system.

Once we have specified the pre-filter requirements and have selected the desired sampling rate, we can proceed with the design of the digital signal processing operations to be performed on the discrete-time signal.

The selection of the sampling rate Fs=1/T, where T is the sampling interval, not only determines the highest frequency (Fs/2) that is preserved in the analog signal but also serves as a scale factor that influences the design specifications for digital filters and any other discrete-time systems through which the signal is processed.

The anti-aliasing filter is an analog filter that has a twofold purpose.
Analog signal|| Pre-filter -> A/D Converter -> Digital Processor -> D/A Converter -> Post-filter

In DM, Furthermore, the two integrators at the encoder can be replaced by a single integrator placed before the comparator. This system is known as sigma-delta modulation (SDM).