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

91. What is the expression for SQNR which can be expressed in a logarithmic scale?

A. 10 \(log_{10}⁡\frac{P_x}{P_n}\)

B. 10 \(log_{10}⁡\frac{P_n}{P_x}\)

C. 10 \(log_2⁡\frac{P_x}{P_n}\)

D. 2 \(log_2⁡\frac{P_x}{P_n}\)

Answer: A

The signal-to-quantization noise (power) ratio (SQNR), which can be expressed on a logarithmic scale (in decibels or dB. :

SQNR = 10 \(log_{10}⁡\frac{P_x}{P_n}\)

92. In the equation SQNR = 10 \(log_{10}⁡\frac{P_x}{P_n}\). what are the terms Px and Pn are called ___ respectively?

A. Power of the Quantization noise and Signal power
B. Signal power and power of the quantization noise
C. None of the mentioned
D. All of the mentioned

Answer: B

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.

93. In the equation SQNR = 10 ⁡(log_{10}frac{P_x}{P_n}), what are the expressions of Px and Pn?

A. \(P_x=\sigma^2=E[x^2 (n)] \,and\, P_n=\sigma_e^2=E[e_q^2 (n)]\)

B. \(P_x=\sigma^2=E[x^2 (n)] \,and\, P_n=\sigma_e^2=E[e_q^3 (n)]\)

C. \(P_x=\sigma^2=E[x^3 (n)] \,and\, P_n=\sigma_e^2=E[e_q^2 (n)]\)

D. None of the mentioned

Answer: A

In the equation SQNR = \(10 log_{10}⁡ \frac{P_x}{P_n}\),

then expressions of Px and Pn are \(P_x=\sigma^2=E[x^2 (n)]\) and \(P_n=\sigma_e^2=E[e_q^2 (n)]\).

94. If the quantization error is uniformly distributed in the range (-Δ/2, Δ/2), the mean value of the error is zero then the variance Pn is?

A. \(P_n=\sigma_e^2=\Delta^2/12\)
B. \(P_n=\sigma_e^2=\Delta^2/6\)
C. \(P_n=\sigma_e^2=\Delta^2/4\)
D. \(P_n=\sigma_e^2=\Delta^2/2\)

Answer: A

If the quantization error is uniformly distributed in the range (-Δ/2, Δ/2), the mean value of the error is zero then the variance Pn is

\(P_n=\sigma_e^2=\int_{-\Delta/2}^{\Delta/2} e^2 p(e)de=1/\Delta \int_{\frac{-\Delta}{2}}^{\frac{\Delta}{2}} e^2 de = \frac{\Delta^2}{12}\).

95. By combining \(\Delta=\frac{R}{2^{b+1}}\) with \(P_n=\sigma_e^2=\Delta^2/12\) and substituting the result into SQNR = 10 \(log_{10}⁡ \frac{P_x}{P_n}\), what is the final expression for SQNR = ?

A. 6.02b + 16.81 + \(20log_{10}\frac{R}{σ_x}\)

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

C. 6.02b − 16.81\(20log_{10}⁡ \frac{R}{σ_x}\)

D. 6.02b − 16.81 \(20log_{10}⁡ \frac{R}{σ_x}\)

Answer: B

SQNR = 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.

96. In the equation SQNR = 6.02b + 16.81 – (20log_{10} ⁡frac{R}{σ_x}), for R = 6σx the equation becomes?

A. SQNR = 6.02b-1.25 dB
B. SQNR = 6.87b-1.55 dB
C. SQNR = 6.02b+1.25 dB
D. SQNR = 6.87b+1.25 dB

Answer: C

For example, if we assume that x(n) is Gaussian distributed and the range of the quantizer extends from -3σx to 3σx (i.e., R = 6σx), then less than 3 out of 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.

97. In IIR Filter design by the Bilinear Transformation, the Bilinear Transformation is a mapping from

A. Z-plane to S-plane
B. S-plane to Z-plane
C. S-plane to J-plane
D. J-plane to Z-plane

Answer: B

In IIR Filter design by the Bilinear Transformation, the Bilinear Transformation is a mapping from S-plane to Z-plane

From the equation,

S=\(\frac{2}{T}(\frac{1-z^{-1}}{1+z^{-1}})\) it is clear that transformation occurs from s-plane to z-plane

98. In Bilinear Transformation, aliasing of frequency components is been avoided.

A. True
B. False

Answer: A

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.

99. Is IIR Filter design by Bilinear Transformation is the advanced technique when compared to other design techniques?

A. True
B. False

Answer: A

IIR Filter Design by Bilinear Transformation is the advanced technique because, in other techniques, only lowpass filters and a limited class of bandpass filters are been supported. But this technique overcomes the limitations of other techniques and supports more.

100. The approximation of the integral in y(t) = \(\int_{t_0}^t y'(τ)dt+y(t_0)\) by the Trapezoidal formula at t = nT and t0=nT-T yields equation?

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

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

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

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

Answer: B

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

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