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

For a linear phase filter, we know that

H(z)=\(±z^{-N} H(z^{-1})\)

where z(-N) represents a delay of N units of time. But if this were the case, the filter would have a mirror image pole outside the unit circle for every pole inside the unit circle. Hence the filter would be unstable.

The order of operations to be performed in order to realize linear phase IIR filter

(i) Passing x(-n) through a digital filter H(z)
(ii) Time reversing the output of H(z)
(iii) Time reversal of the input signal x(n)
(iv) Passing the result through H(z)

If the restriction on physical reliability is removed, it is possible to obtain a linear phase IIR filter, at least in principle. This approach involves performing a time-reversal of the input signal x(n), passing x(-n) through a digital filter H(z), time reversing the output of H(z), and finally, passing the result through H(z) again.

The signal processing is computationally cumbersome and appears to offer no advantages over linear phase FIR filters. Consequently, when an application requires a linear phase, it should be an FIR filter.

One of the simplest methods for converting an analog filter into a digital filter is to approximate the differential equation by an equivalent difference equation.

For the derivative dy(t)/dt at time t=nT, we substitute the backward difference [y(nT)-y(nT-T)]/T. Thus
dy(t)/dt =[y(nT)-y(nT-T)]/T
=[y(n)-y(n-1)]/T

where T represents the sampling interval and y(n)=y(nT).

The analog differentiator with output dy(t)/dt has the system function H(s)=s, while the digital system that produces the output [y(n)-y(n-1)]/T has the system function H(z) =(1-z-1)/T. Thus the relation between s-domain and z-domain is given as
s=(1-z-1)/T.

We know that dy(t)/dt =[ y(n)-y(n-1)]/T

Second order derivative of y(t) is d(dy(t)/dt)/dt=[y(n)-2y(n-1)+y(n-2)]/T2.

We know that for a second order derivative

d2y(t)/dt²=[y(n)-2y(n-1)+y(n-2)]/T2

=>s2 = \(\frac{1-2z^{-1}+z^{-2}}{T^2} = (\frac{1-z^{-1}}{T})^2\)

We know that
s=(1-z-1)/T
=> z=1/(1-sT)
Given s= jΩ => z = 1/(1- jΩT)
Thus from the above equation if Ω varies from -∞ to ∞, then the corresponding locus of points in the z-plane is a circle of radius 1/2 with a center at z=1/2.

The mapping is true between the s-plane and z-domain when

A. Points in LHP of the s-plane into points inside the circle in the z-domain
B. Points in RHP of the s-plane into points outside the circle in the z-domain
C. Points on an imaginary axis of the s-plane into points onto the circle in z-domain

The possible location of poles of the digital filter are confined to relatively small frequencies and as a consequence, the mapping is restricted to the design of low pass filters and bandpass filters having relatively small resonant frequencies.

22. Which of the following filter transformation is not possible?

A. High pass analog filter to low pass digital filter
B. High pass analog filter to high pass digital filter
C. Low pass analog filter to low pass digital filter
D. None of the mentioned

Show Explanation

We know that only low pass and bandpass filters with low resonant frequencies in the digital can be designed. So, it is not possible to transform a high pass analog filter into a corresponding high pass digital filter.

By proper choice of the coefficients of {αk}, it is possible to map the jΩ-axis into the unit circle.