Digital Filters Design MCQ [Free PDF] – Objective Question Answer for Digital Filters Design Quiz

41. If the discrimination factor ‘d’ and the selectivity factor ‘k’ of a Chebyshev I filter are 0.077 and 0.769 respectively, then what is the order of the filter?

A. 2
B. 5
C. 4
D. 3

Answer: B

We know that the order of a Chebyshev-I filter is given by the equation,

N=cosh-1(1/D./cosh-1(1/k)=4.3

Rounding off to the next large integer, we get N=5.

 

41. The equation for Heq(s) is \(\frac{\sum_{K=0}^M b_K s^K}{\sum_{K=0}^N a_K s^K}\)

A. True
B. False

Answer: A

The analog filter in the time domain is governed by the following difference equation,

\(\sum_{K=0}^N a_K y^K (t)=\sum_{K=0}^M b_K x^K (t)\)

Taking Laplace transform on both the sides of the above differential equation with all initial conditions set to zero, we get

\(\sum_{K=0}^N a_K s^K Y(s)=\sum_{K=0}^M b_K s^K X(s)\)

=> H<sub>eq</sub>(s)=Y(s)/X(s)=\(\frac{\sum_{K=0}^M b_K s^K }{\sum_{K=0}^N a_K s^K}\).

 

42. What is the first backward difference of y(n)?

A. [y(n)+y(n-1)]/T
B. [y(n)+y(n+1)]/T
C. [y(n)-y(n+1)]/T
D. [y(n)-y(n-1)]/T

Answer: D

A simple approximation to the first-order derivative is given by the first backward difference. The first backward difference is defined by
[y(n)-y(n-1)]/T.

 

43. Which of the following is the correct relation between ‘s’ and ‘z’?

A. z=1/(1+sT)
B. s=1/(1+zT)
C. z=1/(1-sT)
D. none of the mentioned

Answer: C

The correct relation between ‘s’ and ‘z’ is

s=(1-z-1)/T=> z=1/(1-sT).

 

44. What is the center of the circle represented by the image of jΩ axis of the s-domain?

A. z=0
B. z=0.5
C. z=1
D. none of the mentioned

Answer: B

Letting s=σ+jΩ in the equation z=1/(1-sT) and by letting σ=0, we get
|z-0.5|=0.5

Thus the image of the jΩ axis of the s-domain is a circle with a center at z=0.5 in the z-domain.

 

45. What is the radius of the circle represented by the image of jΩ axis of the s-domain?

A. 0.75
B. 0.25
C. 1
D. 0.5

Answer: D

Letting s=σ+jΩ in the equation z=1/(1-sT) and by letting σ=0, we get
|z-0.5|=0.5
Thus the image of the jΩ axis of the s-domain is a circle of radius 0.5 centered at z=0.5 in the z-domain.

 

46. The frequency response H(ω) will be considerably distorted with respect to H(jΩ).

A. True
B. False

Answer: A

Since the jΩ axis is not mapped to the circle |z|=1, we can expect that the frequency response H(ω) will be considerably distorted with respect to H(jΩ).

 

47. The left half of the s-plane is mapped to which of the following in the z-domain?

A. Outside the circle |z-0.5|=0.5
B. Outside the circle |z+0.5|=0.5
C. Inside the circle |z-0.5|=0.5
D. Inside the circle |z+0.5|=0.5

Answer: C

The left half of the s-plane is mapped inside the circle of |z-0.5|=0.5 in the z-plane, which completely lies in the right half z-plane.

 

48. An analog high pass filter can be mapped to a digital high pass filter.

A. True
B. False

Answer: B

An analog high pass filter cannot be mapped to a digital high pass filter because the poles of the digital filter cannot lie in the correct region, which is the left half of the z-plane(z < 0) in this case.

 

49. Which of the following is the correct relation between ‘s’ and ‘z’?

A. s=(1-z-1)/T
B. s=1/(1+zT)
C. s=(1+z-1)/T
D. none of the mentioned

Answer: A

We know that z=1/(1-sT)=> s=(1-z-1)/T.

 

50. What is the z-transform of the first backward difference equation of y(n)?

A. \(\frac{1+z^{-1}}{T}\) Y(z)
B. \(\frac{1-z^{-1}}{T}\) Y(z)
C. \(\frac{1+z^1}{T}\) Y(z)
D. None of the mentioned

Answer: B

The first backward difference of y(n) is given by the equation
[y(n)-y(n-1)]/T
Thus the z-transform of the first backward difference of y(n) is given as
\(\frac{1-z^{-1}}{T}\) Y(z).

 

51. Bilinear Transformation is used for transforming an _____ filter into a ______ filter.

  1. Analog, Digital
  2. Digital, Analog
  3. Analog, Analog
  4. Digital, Digital
Answer: A

Bilinear Transformation is used for transforming an analog filter into a digital filter.

The bilinear transformation can be regarded as a correction of the backward difference method. The bilinear transformation is used for transforming an analog filter into a digital filter.

 

52. Which of the following rule is used in the bilinear transformation?

A. Simpson’s rule
B. Backward difference
C. Forward difference
D. Trapezoidal rule

Answer: D

The bilinear transformation uses the trapezoidal rule for integrating a continuous-time function.

The bilinear transformation results from the trapezoidal rule approximation of an integral. Suppose that x(t) is the input and y(t) is the output of an integrator with a transfer function.

 

53. Which of the following substitution is done in Bilinear transformations?

A. s = \(\frac{2}{T}[\frac{1+z^{-1}}{1-z^1}]\)

B. s = \(\frac{2}{T}[\frac{1+z^{-1}}{1+}]\)

C. s = \(\frac{2}{T}[\frac{1-z^{-1}}{1+z^{-1}}]\)

D. None of the mentioned

Answer: C

In bilinear transformation of an analog filter to digital filter, using the trapezoidal rule, the substitution for ‘s’ is given as

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

 

54. What is the value of \(\int_{(n-1)T}^{nT} x(t)dt\) according to trapezoidal rule?

A. \([\frac{x(nT)-x[(n-1)T]}{2}]T\)

B. \([\frac{x(nT)+x[(n-1)T]}{2}]T\)

C. \([\frac{x(nT)-x[(n+1)T]}{2}]T\)

D. \([\frac{x(nT)+x[(n+1)T]}{2}]T\)

Answer: B

The given integral is approximated by the trapezoidal rule. This rule states that if T is small, the area (integral) can be approximated by the mean height of x(t) between the two limits and then multiplying by the width. That is

\(\int_{(n-1)T}^{nT} x(t)dt=[\frac{x(nT)+x[(n-1)T]}{2}]T\)

 

55. What is the value of y(n)-y(n-1) in terms of input x(n)?

A. \([\frac{x(n)+x(n-1)}{2}]T\)

B. \([\frac{x(n)-x(n-1)}{2}]T\)

C. \([\frac{x(n)-x(n+1)}{2}]T\)

D. \([\frac{x(n)+x(n+1)}{2}]T\)

Answer: A

We know that the derivative equation is

dy(t)/dt=x(t)

On applying integrals both sides, we get

\(\int_{(n-1)T}^{nT}dy(t)=\int_{(n-1)T}^{nT} x(t)dt\)

=> y(nT)-y[(n-1)T]=\(\int_{(n-1)T}^{nT} x(t)dt\)

On applying trapezoidal rule on the right hand integral, we get

y(nT)-y[(n-1)T]=\([\frac{x(nT)+x[(n-1)T]}{2}]T\)

Since x(n) and y(n) are approximately equal to x(nT) and y(nT) respectively, the above equation can be written as

y(n)-y(n-1)=\([\frac{x(n)+x(n-1)}{2}]T\)

 

56. What is the expression for system function in z-domain?

A. \(\frac{2}{T}[\frac{1+z^{-1}}{1-z^1}]\)

B. \(\frac{2}{T}[\frac{1+z^{-1}}{1-z^1}]\)

C. \(\frac{T}{2}[\frac{1+z^{-1}}{1-z^1}]\)

D. \(\frac{T}{2}[\frac{1-z^{-1}}{1+z^{-1}}]\)

Answer: C

We know that

y(n)-y(n-1)= \([\frac{x(n)+x(n-1)}{2}]T\)

Taking z-transform of the above equation gives

=>Y(z)[1-z-1]=([1+z-1]/2).TX(z)

=>H(z)=Y(z)/X(z)=\(\frac{T}{2}[\frac{1+z^{-1}}{1-z^1}]\).

 

57. In bilinear transformation, the left-half s-plane is mapped to which of the following in the z-domain?

A. Entirely outside the unit circle |z|=1
B. Partially outside the unit circle |z|=1
C. Partially inside the unit circle |z|=1
D. Entirely inside the unit circle |z|=1

Answer: D

In bilinear transformation, the z to s transformation is given by the expression
z=[1+(T/2)s]/[1-(T/2)s].

Thus unlike the backward difference method, the left-half s-plane is now mapped entirely inside the unit circle, |z|=1, rather than to a part of it.

 

58. The equation s = \(\frac{2}{T}[\frac{1-z^{-1}}{1+z^{-1}}]\) is a true frequency-to-frequency transformation.

A. True
B. False

Answer: A

Unlike the backward difference method, the left-half s-plane is now mapped entirely inside the unit circle, |z|=1, rather than to a part of it. Also, the imaginary axis is mapped to the unit circle.

Therefore, equation s = \(\frac{2}{T}[\frac{1-z^{-1}}{1+z^{-1}}]\) is a true frequency-to-frequency transformation.

 

59. If s=σ+jΩ and z=rejω, then what is the condition on σ if r<1?

A. σ > 0
B. σ < 0
C. σ > 1
D. σ < 1

Answer: B

We know that if = σ+jΩ and z=rejω, then by substituting the values in the below expression

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

=>σ = \(\frac{2}{T}[\frac{r^2-1}{r^2+1+2rcosω}]\)

When r<1 => σ < 0.

 

60. If s=σ+jΩ and z=rejω and r=1, then which of the following inference is correct?

A. LHS of the s-plane is mapped inside the circle, |z|=1
B. RHS of the s-plane is mapped outside the circle, |z|=1
C. Imaginary axis in the s-plane is mapped to the circle, |z|=1
D. None of the mentioned

Answer: C

We know that if =σ+jΩ and z=rejω, then by substituting the values in the below expression

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

=>σ = \(\frac{2}{T}[\frac{r^2-1}{r^2+1+2rcosω}]\)

When r=1 => σ = 0.

This shows that the imaginary axis in the s-domain is mapped to the circle of unit radius centered at z=0 in the z-domain.

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