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The correct answer is (B) RHP in s-plane maps into the OUTSIDE of the unit circle in the z-plane

Explanation: In the above equation, if we substitute the VALUES of r, ℴ then we get mapping in the required way

Correct choice is (a) LHP in s-plane MAPS into the inside of the UNIT circle in the z-plane

For explanation I would say: In the above equation, if we substitute the values of R, ℴ then we get mapping in the required way

Correct choice is (c) All POINTS in the LHP & RHP of s are mapped INSIDE & outside the unit circle in the z-plane

Explanation: The BILINEAR TRANSFORMATION is a conformal mapping that transforms the jΩ-axis into the unit circle in the z-plane and all the points are linked as MENTIONED above.

The CORRECT choice is (d) \(\frac{(\frac{BT}{2})(1-z^{-1})}{1+\frac{aT}{2}-(1+\frac{aT}{2}) z^{-1}}\) & \(\frac{b}{\frac{2}{T}(\frac{1-z^{-1}}{1+z^{-1}}+a)}\)

Explanation: As we considered analog linear filter with system FUNCTION H(s) = b/s+a

Hence, we got an equivalent system function

where, s = \(\frac{2}{T}(\frac{1-z^{-1}}{1+z^{-1}})\).

Correct option is (a) \((1+\frac{aT}{2})Y(Z)-(1-\frac{aT}{2})y(n-1)=\frac{bT}{2} [X(n)+x(n-1)]\)

For EXPLANATION: When we substitute the given EQUATION in the derivative of other we get the resultant REQUIRED equation.

The correct answer is (b) y(NT) = \(\frac{T}{2} [y^{‘} (nT)+y^{‘} (nT-T)]+y(nT-T)\)

The explanation: By integrating the EQUATION,

y(t) = \(\int_{t_0}^t y^{‘} (τ)dt+y(t_0)\) at t=nT and t0=nT-T we GET equation,

y(nT) = \(\frac{T}{2} [y^{‘} (nT)+y^{‘} (nT-T)]+y(nT-T)\).

The CORRECT option is (a) True

The best explanation: Because in other techniques, only lowpass FILTERS and limited CLASS of bandpass filters are been supported. But this technique OVERCOMES the limitations of other techniques and supports more.

Correct OPTION is (B) 6.02b + 16.81 – \(20log_{10}⁡ \FRAC{R}{σ_x}\)

To ELABORATE: SQNR = \(10 log_{10}⁡\frac{P_x}{P_n}=20 log_{10} \frac{⁡σ_x}{σ_e}\)

= 6.02b + 16.81 – ⁡\(20 log_{10}\frac{R}{σ_x}\)dB.

Correct option is (c) SQNR = 6.02b+1.25 dB

The explanation: For example, if we assume that X(n) is Gaussian distributed and the RANGE o f the quantizer extends from -3σx to 3σx (i.e., R = 6σx), then less than 3 out o f every 1000 input SIGNAL amplitudes would result in an overload on the average. For R = 6σx, then the equation becomes

SQNR = 6.02b+1.25 dB.

The CORRECT option is (a) True

The best I can explain: The BILINEAR TRANSFORMATION is a conformal mapping that transforms the jΩ-axis into the unit circle in the z-plane only once, thus AVOIDING the aliasing.

The correct answer is (a) \(P_x=\sigma^2=E[x^2 (n)] \,and\, P_n=\sigma_e^2=E[e_q^2 (n)]\)

Explanation: In the EQUATION SQNR = \(10 log_{10}⁡ \frac{P_x}{P_n}\), then the terms \(P_x=\sigma^2=E[x^2 (n)]\) and \(P_n=\sigma_e^2=E[e_q^2 (n)]\).

The correct ANSWER is (b) Signal power and power of the quantization noise

For explanation I would say: In the EQUATION SQNR = \(10 log_{10}⁡\frac{P_x}{P_n}\) then the terms PX is the signal power and PN is the power of the quantization noise

Right option is (b) Signal-to-Quantization NOISE Ratio

To ELABORATE: The effect of the additive noise eq (N) on the DESIRED signal can be QUANTIFIED by evaluating the signal-to-quantization noise (power) ratio (SQNR).

Correct answer is (b) Overload noise

Easy EXPLANATION: In the statistical APPROACH, we assume that the quantization ERROR is random in nature. We model this error as noise that is added to the original (unquantized) signal. If the input analog signal falls OUTSIDE the range of the quantizer (clipping), eq (n) BECOMES unbounded and results in overload noise.

Correct choice is (a) Granular noise

The best explanation: In the statistical approach, we assume that the quantization error is random in nature. We MODEL this error as noise that is added to the original (unquantized) SIGNAL. If the INPUT analog signal is within the range of the quantizer, the quantization error eq (n) is bounded in magnitude

i.e., |eq (n)| < Δ/2 and the RESULTING error is CALLED Granular noise.

Correct answer is (B) \(\sigma_n^2=\int_{-B}^B |H_n (F)|^2 S_e (F)DF\)

The best I can explain: The in-band quantization NOISE variance is given as: \(\sigma_n^2=\int_{-B}^B |H_n (F)|^2 S_e (F)dF\) where \(S_e (F)=\frac{\sigma_e^2}{F_(s)}\) is the power spectral DENSITY of the quantization noise.

Right option is (a) \(|H_n (z)|=2 |sin⁡ \frac{πF}{F_s}|\)

For EXPLANATION I would say: The PERFORMANCE of the SDM system is therefore DETERMINED by the noise system function H_(n)(z), which has a magnitude frequency response: \(|H_n (z)|=2 |sin⁡ \frac{πF}{F_s}|\).

Right option is (a) H(z)=\(\frac{z^{-1}}{1-z^{-1}}\)

To elaborate: The INTEGRATOR is modeled by the DISCRETE TIME system with system function

H(z)=\(\frac{z^{-1}}{1-z^{-1}}\)

The correct answer is (B) \(H_s (Z)X(z)+ H_n (z)E(z)\)

To explain: \(D_q (z)=\frac{H(z)}{1+H(z)} X(z)+\frac{1}{1+H(z)} E(z)\)

= \(H_s (z)X(z)+H_n (z)E(z)\).

The CORRECT CHOICE is (b) Analog signal|| Pre-filter -> A/D Converter -> Digital Processor -> D/A Converter -> Post-filter

Best EXPLANATION: The anti-aliasing filter is an analog filter which has a TWOFOLD purpose.

Analog signal|| Pre-filter -> A/D Converter -> Digital Processor -> D/A Converter -> Post-filter

The correct choice is (a) True

To explain: 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 rateFs=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 correct answer is (a) True

Explanation: ANALOG signal processing operations cannot be done very PRECISELY EITHER since ELECTRONIC components in analog systems have tolerances and they 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.

The correct choice is (b) Anti ALIASING filter

Explanation: 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 which has a twofold purpose.

The correct answer is (a) Not a number

To elaborate: 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=255 and M≠0, then X is not a number.

The correct answer is (c) All of the mentioned

To explain I would say: The anti aliasing filter is an analog filter which 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.

Correct choice is (C) MSB

Explanation: 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.

Right option is (a) (0100)2

To elaborate: The ones complement of (1100)2 is (0011)2. THUS the two complement of this number is obtained as (0011)2+(0001)2=(0100)2.

Right OPTION is (d) Zero

The best explanation: 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.

Right option is (C) 0.101011,011

The BEST I can explain: We can represent the numbers in BINARY float POINT as

5=0.101000(2^011)

3/8=0.110000(2^101)=0.000011(2^011)

=>5+3/8=(0.101000+0.000011)(2^011)=(0.101011)(2^011)

THEREFORE mantissa=0.101011 and exponent=011.

The correct option is (b) M.2^E(1/2
The explanation is: 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

Correct choice is (a) True

Easiest 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.

Right ANSWER is (B) Always negative

Best explanation: For a two’s complement representation, the truncation error is always negative and FALLS in the range

-(2^-b-2^-bm) ≤ ET ≤ 0.

The correct option is (c) Mixed number

To explain I would say: 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 0X is a mixed number.

Correct answer is (d) 0.3*10^-38

Best 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)*2^127=1.7*10^38.

Correct option is (d) 0.101000,011

The best I can explain: We can represent 5 as

5=0.625*8=0.625*2^3

The above NUMBER can be REPRESENTED in BINARY float POINT representation as 0.101000*2^011

Thus Mantissa=0.101000, Exponent=011.

Correct answer is (b) False

For 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.

Right CHOICE is (a) True

Easiest 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.