Quantization Errors Analysis MCQ Quiz – Objective Question with Answer for Quantization Errors Analysis

1. 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?

A. Granular noise
B. Overload noise
C. Particulate noise
D. Heavy noise

Answer: A

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.

2. If the input analog signal falls outside the range of the quantizer (clipping), eq (n) becomes unbounded and results in _____________
A. Granular noise
B. Overload noise
C. Particulate noise
D. Heavy noise

Answer: B

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.

3. In the mathematical model for the quantization error eq (n), to carry out the analysis, what are the assumptions made about the statistical properties of eq (n)?

A. The error eq (n) is uniformly distributed over the range — Δ/2 < eq (n) < Δ/2.
B. The error sequence is a stationary white noise sequence. In other words, the error eq (m) and the error eq (n) for m≠n are uncorrelated.
C. The error sequence {eq (n)} is uncorrelated with the signal sequence x(n).
D. All of the above

Answer: B

In the mathematical model for the quantization error eq (n). To carry out the analysis, the following are the assumptions made about the statistical properties of eq (n).
i. The error eq (n) is uniformly distributed over the range — Δ/2 < eq (n) < Δ/2.
ii. The error sequence is a stationary white noise sequence. In other words, the error eq (m)and the error eq (n) for m≠n are uncorrelated.
iii. The error sequence {eq (n)} is uncorrelated with the signal sequence x(n).
iv. The signal sequence x(n) is zero mean and stationary.

4. What is the abbreviation of SQNR?

A. Signal-to-Quantization Net Ratio
B. Signal-to-Quantization Noise Ratio
C. Signal-to-Quantization Noise Region
D. Signal-to-Quantization Net Region

Answer: B

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).

5. What is the scale used for the measurement of SQNR?

A. DB
B. db
C. dB
D. All of the mentioned

Answer: C

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), which can be expressed on a logarithmic scale (in decibels or dB.

6. 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 \(log_{10}⁡\frac{P_x}{P_n}\)

7. 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.

8. 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 the terms \(P_x=\sigma^2=E[x^2 (n)]\) and \(P_n=\sigma_e^2=E[e_q^2 (n)]\).

9. 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

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

10. 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.

11. 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.

12. 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

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

13. 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.

14. Is 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.

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

16. We use y{‘}(nT)=-ay(nT)+bx(nT) to substitute for the derivative in y(nT) = \(\frac{T}{2} [y^{‘} (nT)+y^{‘} (nT-T)]+y(nT-T)\) and thus obtain a difference equation for the equivalent discrete-time system. With y(n) = y(nT) and x(n) = x(nT), we obtain the result as of the following?

A. \((1+\frac{aT}{2})Y(z)-(1-\frac{aT}{2})y(n-1)=\frac{bT}{2} [x(n)+x(n-1)]\)

B. \((1+\frac{aT}{n})Y(z)-(1-\frac{aT}{n})y(n-1)=\frac{bT}{n} [x(n)+x(n-1)]\)

C. \((1+\frac{aT}{2})Y(z)+(1-\frac{aT}{2})y(n-1)=\frac{bT}{2} (x(n)-x(n-1))\)

D. \((1+\frac{aT}{2})Y(z)+(1-\frac{aT}{2})y(n-1)=\frac{bT}{2} (x(n)+x(n+1))\)

Answer: A

When we substitute the given equation in the derivative of another we get the resultant required equation.

17. The z-transform of below difference equation is?

\((1+\frac{aT}{2})Y(z)-(1-\frac{aT}{2})y(n-1)=\frac{bT}{2} [x(n)+ x(n-1)]\)

A. \((1+\frac{aT}{2})Y(z)-(1-\frac{aT}{2}) z^{-1} Y(z)=\frac{bT}{2} (1+z^{-1})X(z)\)

B. \((1+\frac{aT}{n})Y(z)-(1-\frac{aT}{2}) z^{-1} Y(z)=\frac{bT}{n} (1+z^{-1})X(z)\)

C. \((1+\frac{aT}{2})Y(z)+(1-\frac{aT}{n}) z^{-1} Y(z)=\frac{bT}{2} (1+z^{-1})X(z)\)

D. \((1+\frac{aT}{2})Y(z)-(1+\frac{aT}{2}) z^{-1} Y(z)=\frac{bT}{2} (1+z^{-1})X(z)\)

Answer: A

By performing the z-transform of the given equation, we get the required output/equation.

18. What is the system function of the equivalent digital filter? H(z) = Y(z)/X(z) = ?

A. \(\frac{(\frac{bT}{2})(1+z^{-1})}{1+\frac{aT}{2}-(1-\frac{aT}{2}) z^{-1}}\)

B. \(\frac{(\frac{bT}{2})(1-z^{-1})}{1+\frac{aT}{2}-(1+\frac{aT}{2}) z^{-1}}\)

C. \(\frac{b}{\frac{2}{T}(\frac{1-z^{-1}}{1+z^{-1}}+A.}\)

D. \(\frac{(\frac{bT}{2})(1-z^{-1})}{1+\frac{aT}{2}-(1+\frac{aT}{2}) z^{-1}}\) & \(\frac{b}{\frac{2}{T}(\frac{1-z^{-1}}{1+z^{-1}}+A.}\)

Answer: D

As we considered analog linear filter with system function H(s) = b/s+a
Hence, we got an equivalent system function

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

18. In the Bilinear Transformation mapping, which of the following are correct?

A. All points in the LHP of s are mapped inside the unit circle in the z-plane
B. All points in the RHP of s are mapped outside the unit circle in the z-plane
C. All points in the LHP & RHP of s are mapped inside & outside the unit circle in the z-plane
D. None of the mentioned

Answer: C

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.

19. In Nth order differential equation, the characteristics of bilinear transformation, let z=rejw,s=o+jΩ Then for s = \(\frac{2}{T}(\frac{1-z^{-1}}{1+z^{-1}})\), the values of Ω, ℴ are

A. ℴ = \(\frac{2}{T}(\frac{r^2-1}{1+r^2+2rcosω})\), Ω = \(\frac{2}{T}(\frac{2rsinω}{1+r^2+2rcosω})\)

B. Ω = \(\frac{2}{T}(\frac{r^2-1}{1+r^2+2rcosω})\), ℴ = \(\frac{2}{T}(\frac{2rsinω}{1+r^2+2rcosω})\)

C. Ω=0, ℴ=0

D. None

Answer: A

s = \(\frac{2}{T}(\frac{z-1}{z+1}) \)

= \(\frac{2}{T}(\frac{re^jw-1}{re^jw+1})\)

= \(\frac{2}{T}(\frac{r^2-1}{1+r^2+2rcosω}+j \frac{2rsinω}{1+r^2+2rcosω})(s = ℴ+jΩ)\)

20. In equation ℴ = \(\frac{2}{T}(\frac{r^2-1}{1+r^2+2rcosω})\) if r < 1 then ℴ < 0 and then mapping from s-plane to z-plane occurs in which of the following order?

A. LHP in s-plane maps into the inside of the unit circle in the z-plane

B. RHP in s-plane maps into the outside of the unit circle in the z-plane

C. All of the mentioned

D. None of the mentioned

Answer: A

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

In equation ℴ = \(\frac{2}{T}(\frac{r^2-1}{1+r^2+2rcosω})\) if r < 1 then ℴ < 0 and then mapping from s-plane to z-plane occurs in which LHP in s-plane maps into the inside of the unit circle in the z-plane.

21. In equation ℴ = \(\frac{2}{T}(\frac{r^2-1}{1+r^2+2rcosω})\), if r > 1 then ℴ > 0 and then mapping from s-plane to z-plane occurs in which of the following order?

A. LHP in s-plane maps into the inside of the unit circle in the z-plane
B. RHP in s-plane maps into the outside of the unit circle in the z-plane
C. All of the mentioned
D. None of the mentioned

Answer: B

In equation ℴ = \(\frac{2}{T}(\frac{r^2-1}{1+r^2+2rcosω})\), if r > 1 then ℴ > 0 and then mapping from s-plane to z-plane occurs in which RHP in s-plane maps into the outside of the unit circle in the z-plane

In the above equation, if we substitute the values of r, ℴ then we get mapping in the required way.

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