Approximation of Derivatives of IIR Filter design Method MCQ [Free PDF] – Objective Question Answer for Approximation of Derivatives of IIR Filter design Method Quiz

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.

The digital filter with frequency response H(ω) has the frequency response characteristics of the corresponding analog filter if the sampling interval T is selected sufficiently small to completely avoid or at least minimize the effects of aliasing.

It is clear that the impulse invariance method is inappropriate for designing high pass filters due to the spectrum aliasing that results from the sampling process.

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
ω=ΩT.

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 0<r<1.

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.

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) with period F_s.