Digital Signal Processing MCQ [Free PDF] – Objective Question Answer for Digital Signal Processing 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

The second difference that is used to replace theSecond 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.

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.

In the impulse invariance method, our objective is to design an IIR filter having a unit sample response h(n) that is the sampled version of the impulse response of the analog filter. That is

h(n)=h(nT); n=0,1,2…

where T is the sampling interval.

When a continuous-time signal x(t) with spectrum X(F) is sampled at a rate F_s=1/T samples per second, the spectrum of the sampled signal is periodic repetition.

When a continuous-time signal x(t) with spectrum X(F) is sampled at a rate F_s=1/T samples per second, the spectrum of the sampled signal is a periodic repetition of the scaled spectrum F_s.X(F).

We know that z=e^sT

Now substitute s=σ+jΩ and z=r.e^jω, which represents ‘z’ in the polar form

On equating both sides, we get

r=e^σT

Thus when σ=0, the value of ‘r’ varies from r=1.

In the impulse invariance method, the normalized frequency f is given by

f= F/F_s.

Aliasing occurs if the sampling rate F_s is less than twice the highest frequency contained in X(F).

The above-given frequency response depicts the frequency response of a low-pass analog filter and the frequency response of the corresponding digital filter.