Digital Filters MCQ Quiz in తెలుగు - Objective Question with Answer for Digital Filters - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Properties of IIR filter:

• The impulse response is of infinite duration.
• IIR system is also known as a recursive system as the present output depends on the past output because there is a feedback path from output to input.
• IIR system cannot be designed as a linear phase system.
• IIR filters are dependent on both input and output.
• IIR filters consist of zeros and poles and require less memory than FIR filters
• Stability cannot be guaranteed.
• In the IIR system, fewer multiplications and additions are reviewed, so processing speed is very fast.
• IIR filters are preferred over FIR filters when the cost is a concern.

Additional Information

Properties of FIR filter:

• Impulse Response is of finite duration
• It is a non-recursive system because there is no feedback path present in the FIR filter.
• As there is no feedback path presence in the FIR filter, it depends on input only means it is independent of output.
• FIR only consists of zeros, so they require more memory.
• They are always be designed as a linear phase. This means that the filter has no phase shift across the frequency band. Alternatively, the phase can be corrected independently on the amplitude. 
• Stability is guaranteed in the FIR system.
• In FIR, to obtain the same frequency response as IIR large number of additions & multiplication is reviewed, so the speed is very slow.
• It can be used to correct frequency-response errors in a loudspeaker to a finer degree of precision than using IIR Filter.

Bilinear Transformation:

• It is used for transforming an analog filter into a digital filter.
• It uses a trapezoidal rule for integrating a continuous-time function.
• It can be regarded as a correction of the backward difference method.
• It is a mapping from the s plane to the z plane.

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

\(s = \frac{2}{T}\frac{{\left( {1 - {z^{ - 1}}} \right)}}{{1 + {z^{ - 1}}}}\)

Properties:

• It preserves the stability and maps every point of the frequency response of the continuous-time filter to a corresponding point in the frequency response of the discrete-time filter. 
• Another valuable property of the bilinear transform is that order is preserved, i.e. an Nth-order s-plane transfer function carries over to an Nth-order z-plane transfer function.
• The bilinear transformation is a rational function that maps the left half s plane inside the unit circle and maps the jω axis in a one-to-one manner onto the unit circle.
• Bilinear Transformation only preserves the magnitude response of the analog filter.

Concept:

• FIR filters are inherently stable since the poles are found only at Z = 0 which is inside the unit circle.
• For IIR filters to be stable the necessary and sufficient condition is that all poles are inside the unit circle which is attained through system designing or filter designing.

• If the poles of the IIR filter will lie outside the unit circle of |z|=1, the filter will be unstable in nature.

Concept:

The minimum number of Taps required to realize FIR filter is given as:

\(N = \frac{{Alternation\left( {dB} \right)\; \times \;fs}}{{22\; \times \;{\rm{\Delta }}f}}\) 

Δf = f_STOP - f_pass

Given:

f_s = 200 kHz

f_stop = 15 kHz

f_pass = 10 kHz

\(N = \frac{{60\; \times \;200}}{{22\; \times \;5\; \times \;{{10}^3}}}\) 

≈ 110

Finite Impulse Response Filter (FIR):

• In signal processing, a finite impulse response filter is a filter whose impulse response is of finite duration, because it settles to zero in finite time. 
• FIR filters can be discrete-time or continuous-time, and digital or analog.

​Properties of an FIR Filter:

• FIR filters are non-recursive in nature.
• Requires no feedback.
• They are inherently stable.
• They can easily be designed to be linear phases by making the coefficient sequence symmetric.

Concept:

In the case of linear phase FIR filter, to have a one zero, z = 0 as z0 other zeros are.

\({z_0},\frac{1}{{{z_0}}},\left[ {z_0^*} \right],{\left[ {\frac{1}{{{z_0}}}} \right]^*}\)

For example:

z₀ = 3 + 4j

\(\left[ {z_0^*} \right] = 3 - 4j\)

\(\frac{1}{{{z_0}}} = \frac{1}{{3 + 4j}} = \frac{{3 - j4}}{{25}} = \frac{3}{{25}} - \frac{{j4}}{{25}}\)

\({\left[ {\frac{1}{{{z_0}}}} \right]^*} = \frac{3}{{2j}} + j\frac{4}{{25}}\)

Important Points

Given:

y(n) = 2x(n) + x(n – 1) + 2y(n – 1)

⇒ y(n) – 2y(n – 1) = 2x(n) + x(n – 1)

⇒ Y(z) – 2z^-1 Y(z) = 2 X(z) + z^-1 X(z)

⇒ Y(z) [1 – 2z^-1] = X(z) [2 + z^-1]

\(\frac{{Y\left( z \right)}}{{X\left( z \right)}} = \frac{{2 + {z^{ - 1}}}}{{1 + 2{z^{ - 1}}}}\)  

Poles: 1 – 2z^-1 = 0 ⇒ z = 2

Zeros : 2 + z^-1 = 0 ⇒ z = -0.5

As the poles lie at z = 2, which is outside the unit circle So, it is an unstable system, As the o/p consists of y(n – 1) i.e., finite pole, so given system is an IIR filter.

∴ The system is unstable IIR.

No. of stages \( = S = Integer\;\left( {\frac{{K \times {F_s}}}{{{\rm{\Delta }}f}}} \right)\) 

Also, x(n) → (0 – 2.4 MHz) with f_s = 400 MHz = 1/T_s

\(x\left( n \right) \xrightarrow{sampled} x\left( {n{T_s}} \right)\)

\( = x\left( {\frac{{n.1}}{{{f_s}}}} \right) = x\left( {\frac{n}{{400 \times {{10}^6}}}} \right)\)

For, LPF (1)

Input x(n) after sampling,

\({S_1} = \frac{{k \times {F_s}}}{{{\rm{\Delta }}f}},\;with\;k = 3,\;{f_s} = 400\;MHz\)

Tradition band = |Pass band (f) – Stop band (f)|

Δf = 4 – 1.8 = 2.2 MHz

Putting values,

\({S_1} = \frac{{3 \times 400 \times {{10}^6}}}{{2.2 \times {{10}^6}}}\)

S₁ = (545.45)_integer = 545

For (LPF) (2):

Input x(n) is decimated by 50.

Decimation is nothing but reducing the number of samples i.e. compression.

\(x (n)\xrightarrow[by \ a]{decimination} x(an)\)

with a > 1.

Decimation by 50 means,

\(x\left( \frac{n}{{{f}_{s}}} \right)=x\left( \frac{n}{400\times {{10}^{6}}} \right)\begin{matrix} \xrightarrow[by \ 50]{decimated} \end{matrix}x\left( \frac{n\times 50}{400\times {{10}^{6}}} \right)\)

\(=x\left( \frac{n}{8\times {{10}^{6}}} \right)\)

The sampling time of \(x\left( \frac{n}{8\times {{10}^{6}}} \right)\) is

therefore \({{T}_{s}}=\frac{1}{8\times {{10}^{6}}}\) sec

So, f_s = 8 × 10⁶ Hz

\(\therefore {{S}_{2}}=\frac{3\times 8\times {{10}^{6}}}{0.2\times {{10}^{6}}}=3\times 40=120\)

Concept:

A digital filter is stable if and only if all the poles of the filter transfer function lie inside the unit circle in the z plane.

\({y[n+1]-y[n]\over T}+0.1y[n]=x[n]\)

Calculation:

Taking z transform:

\({z^{-1}y[z]-y[z]}+0.1T\space y[z]=x[z]\)

\({y[z]\over x[z]}={T\over z^{-1}-1+0.1T}\)

Pole is given by:

\({ z^{-1}-1+0.1T}=0\)

\(z={1\over 1-0.1T}\)

For stability, 0 < z < 1

\(0<{1\over 1-0.1T}<1\)

0 < T < 20

Correct answer is option 1): -3 \(ω\)

Concept:

A finite impulse response (FIR) filter is a filter whose impulse response (or response to any finite length input) is of finite duration because it settles to zero in finite time. The impulse response of the FIR filter is symmetric over the period of time. Infinite impulse Response filter ( IIR) is having internal feedback so It will respond for Infinite period of time. Consider FIR filter of order N. The filter will take N+1 samples to settle down. The FIR design is Given below.

Calculation:

h[n] = \(\sum_{k=0}^{6} \)(4 − |k − 3|) δ[n − k]

there seven samples as K=0 to 6 symmetric over K=3

h(0) = h(6)

h(1) = h(5)

h(2) = h(4)

h(3) =h(3)

The transfer function of this filter can be written as

H(z) = h[0]+ h[1]z^-1+ h[2] z^-2 + h[3] z^-3 + h[4] z^-4 + h[5] z^-5 + h[6] z^-6 

 The frequency response of the filter can be written as

H(e^jω) = e^-3jω  [ h[3] + \(\sum_{n=0}^{2} \) 2h[n] cos [ω(n-3)] ]

so the phase value is -3ω