# Continuous Time Signal MCQ Quiz – Objective Question with Answer for Continuous Time Signal

1. When the frequency band is selected we can specify the sampling rate and the characteristics of the pre-filter, which is also called as __________ filter.

A. Analog filter
B. Anti-aliasing filter
C. Analog & Anti-aliasing filter
D. None of the mentioned

Answer: B

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.

2. What are the main characteristics of an Anti-aliasing filter?
A. Ensures that the bandwidth of the signal to be sampled is limited to the frequency range
B. To limit the additive noise spectrum and other interference, which corrupts the signal
C. All of the mentioned
D. None of the mentioned

Answer: C

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.

3. 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.

A. True
B. False

Answer: A

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.

4. 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.

A. True
B. False

Answer: A

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.

5. What is the configuration of system for digital processing of an analog signal?

A. Analog signal|| Pre-filter -> D/A Converter -> Digital Processor -> A/D Converter -> Post-filter
B. Analog signal|| Pre-filter -> A/D Converter -> Digital Processor -> D/A Converter -> Post-filter
C. Analog signal|| Post-filter -> D/A Converter -> Digital Processor -> A/D Converter -> Pre-filter
D. None of the mentioned

Answer: B

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

6. In DM, further the two integrators at encoding are replaced by one integrator placed before the comparator, and then such system is called?

A. System-delta modulation
B. sigma-delta modulation
C. Source-delta modulation
D. None of the mentioned

Answer: B

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).

7. What is the system function of the integrator that is modeled by the discrete time system?

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

B. H(z)=$$\frac{z^{-1}}{1+z^{-1}}$$

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

D. H(z)=$$\frac{z^{z^1}}{1+z^1}$$

Answer: A

The integrator is modeled by the discrete time system with system function
H(z)=$$\frac{z^{-1}}{1-z^{-1}}$$

8. What is the z-transform of sequence {dq(n)} i.e., Dq(z)= ?

A. $$H_s (z)X(z)- H_n (z)E(z)$$

B. $$H_s (z)X(z)+ H_n (z)E(z)$$

C. $$H_s (n)X(z)+ H_n (n)E(z)$$

D. $$H_n (z)X(z)- H_s (z)E(z)$$

Answer: B

the z-transform of sequence {dq(n)} i.e., Dq(z) IS

$$D_q (z)=\frac{H(z)}{1+H(z)} X(z)+\frac{1}{1+H(z)} E(z)$$

= $$H_s (z)X(z)+H_n (z)E(z)$$

9. The performance of the SDM system is determined by the noise system function Hn(z), which has a magnitude of?

A. $$|H_n (z)|=2 |sin⁡ \frac{πF}{F_s}|$$

B. $$|H_n (z)|=4 |sin⁡ \frac{πF}{F_s}|$$

C. $$|H_n (z)|=3 |sin⁡ \frac{πF}{F_s}|$$

D. $$|H_n (z)|= |sin⁡ \frac{πF}{F_s}|$$

Answer: A

The performance of the SDM system is therefore determined by the noise system function H_(n)(z), which has a magnitude frequency response: $$|H_n (z)|=2 |sin⁡ \frac{πF}{F_s}|$$.

10. The in-band quantization noise variance is given as?

A. $$\sigma_n^2=\int_{-B}^B |H_n (F)|^3 S_e (F)dF$$

B. $$\sigma_n^2=\int_{-B}^B |H_n (F)|^2 S_e (F)dF$$

C. $$\sigma_n^2=\int_{-B}^B |H_n (F)|^1 S_e (F)dF$$

D. None

Answer: B

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.

11. 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.

12. 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.

13. 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.

14. 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).

15. 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.

16. 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 $$log_{10}⁡\frac{P_x}{P_n}$$

17. 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.

18. 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 $$P_n=\sigma_e^2=E[e_q^2 (n)]$$.

19. 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

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

20. 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.

21. 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.

22. 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

23. 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.

23. 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.

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

25. 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.

26. 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.

27. 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 = $$\frac{2}{T}(\frac{1-z^{-1}}{1+z^{-1}})$$.

28. 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.

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