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The correct answer is (b) High PASS analog filter to high pass DIGITAL filter

The best I can explain: We know that only low pass and band pass 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.

The CORRECT CHOICE is (d) All of the mentioned

To EXPLAIN I WOULD SAY: The below diagram explains the given question

Correct OPTION is (a) \((\frac{1-z^{-1}}{T})^2\)

To explain I WOULD SAY: We know that for a second order derivative

d^2y(t)/dt^2=[y(n)-2y(n-1)+y(n-2)]/T^2

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

Right answer is (a) True

To EXPLAIN I would say: The SIGNAL processing is computationally cumbersome and appear to offer no advantages over linear phase FIR filters. CONSEQUENTLY, when an application REQUIRES a linear phase, it should be an FIR FILTER.

The correct OPTION is (a) True

To elaborate: One of the simplest methods for CONVERTING an ANALOG filter into digital filter is to approximate the differential equation by an equivalent DIFFERENCE equation.

Right choice is (d) [y(n)-y(n-1)]/T

The explanation is: 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 correct option is (a) True

For explanation: 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 correct ANSWER is (d) H(z)=\(±z^{-N} H(z^{-1})\)

The BEST I can EXPLAIN: A linear phase filter MUST have a system function that satisfies the condition

H(z)=\(±z^{-N} H(z^{-1})\)

where z^(-N) represents a delay of N units of time.

Correct choice is (a) True

Best EXPLANATION: If an IIR FILTER is STABLE and if it can be PHYSICALLY realizable, then the filter cannot have linear phase.

The correct choice is (c) INSIDE UNIT circle

Explanation: If the conversion technique is to be effective, then the LHP of s-plane should be mapped into the inside of the unit circle in the z-plane. Thus a stable analog filter will be converted to a stable digital filter.

Right answer is (a) True

The explanation is: If the CONVERSION technique is to be effective, the jΩ axis in the s-plane should map into the UNIT circle in the z-plane. Thus there will be a direct relationship between the TWO FREQUENCY VARIABLES in the two domains.

Correct option is (b) LEFT half of s-plane

Easiest explanation: An analog LINEAR time invariant SYSTEM with system function H(s) is stable if all its POLES lie on the left half of the s-plane.

The CORRECT ANSWER is (d) All of the mentioned

The explanation: There are MANY ways how we represent a SYSTEM function of which one is normal representation i.e., output/input and other ways like Laplace transform and rational system function.

Correct option is (a) Ha(s)=\( \int_{-\INFTY}^{\infty} H(t)e^{-st} dt\)

Easy explanation: We know that the impulse response h(t) and the LAPLACE TRANSFORM Ha(s) are RELATED by the equation.

Ha(s)=\( \int_{-\infty}^{\infty} h(t)e^{-st} dt\).

The correct OPTION is (d) All of the mentioned

The best I can explain: There are many techniques which are USED to convert analog filter into digital filter of which some of them are Approximation of derivatives, bilinear transformation, impulse INVARIANCE and many other methods.

Correct CHOICE is (a) True

For EXPLANATION: The estimate is used to carry out the design and if the resulting δ exceeds the specified δ2, then the length can be INCREASED until we obtain a side lobe level that meets the SPECIFICATION.

The correct option is (B) Infinite

Best EXPLANATION: DIGITAL FILTERS are the filters which can be designed from analog filters which have infinite duration unit sample response.

The CORRECT OPTION is (d) (ωs– ωp)/2π

The best I can explain: The EXPRESSION for Δf i.e., for the TRANSITION band is given as

Δf=(ωs-ωp)/2π.

Right answer is (a) True

Explanation: By spreading the approximation error over the pass band and stop band of the filter, this method results in an optimal filter DESIGN and USING this the maximum side LOBE LEVEL is MINIMIZED.

Right ANSWER is (b) Chebyshev approximation

The best explanation: The chebyshev approximation method provides total control of the filter specifications, and as a consequence, it is USUALLY PREFERABLE over the other TWO methods.

The correct answer is (a) True

To explain I would say: Frequency sampling design METHOD is PARTICULARLY attractive when the FIR is REALIZED either in the frequency domain by means of the DFT or in any of the frequency sampling realizations.

Correct choice is (b) 2π/M

To ELABORATE: In the frequency SAMPLING technique, the transition BAND is a MULTIPLE of 2π/M.

Correct CHOICE is (d) None of the mentioned

For explanation I would say: The values of the cutoff FREQUENCIES of a filter in general by WINDOWING technique DEPEND on the type of the filter and the length of the filter.

Correct answer is (a) Window DESIGN

The best explanation: The MAJOR disadvantage of the window design METHOD is the lack of precise CONTROL of the critical FREQUENCIES.

The correct answer is (c) WINDOWING technique

To explain I would say: The DESIGN method based on the USE of windows to TRUNCATE the impulse response h(n) and obtaining the desired spectral shaping, was the FIRST method proposed for designing linear phase FIR filters.

Right choice is (d) 1

Best explanation: The bandwidth of Hilbert transformer NEED only COVER the bandwidth of the signal to be phase shifted. Consequently, we specify the desired REAL valued frequency response of a Hilbert transformer filter is

H(ω)=1; 2π fl < ω < 2πfu

where fl and fu are the cutoff frequencies.

The correct ANSWER is (c) 0

To explain I would say: The UNIT sample response of the Hilbert transformer is given as

h(n)=\(\frac{2}{\pi} \frac{(sin(\frac{πn}{2}))^{2}}{n}\); n≠0

h(n)=0; n=0

From the above equation, it is CLEAR that h(n) becomes zero for EVEN values of ‘n’.

Right OPTION is (a) True

The explanation: We know that when h(N) is anti-symmetric, the real valued FREQUENCY response characteristic is ZERO at ω=0 for both M odd and even and at ω=π when M is odd. Clearly, then, it is IMPOSSIBLE to design an all-pass digital Hilbert transformer.

Right option is (c) Anti-symmetric

Explanation: We know that the unit sample RESPONSE of the Hilbert TRANSFORM is given as

H(n)=\(\frac{2}{\pi} \frac{(sin(\frac{πn}{2}))^{2}}{n}\); n≠0

h(n)=0; n=0

Thus from the above equation, we can TELL that h(n)=-h(-n). Thus the unit sample response of Hilbert transform is anti-symmetric in NATURE.

Right answer is (b) False

Explanation: We know that the unit sample response of the Hilbert TRANSFORM is given as

h(n)=\(\frac{2}{\pi} \frac{(SIN(\frac{πn}{2}))^{2}}{n}\); n≠0

h(n)=0; n=0

it sample response of an ideal Hilbert transform is INFINITE in duration and non-causal.

Right option is (B) False

Explanation: In view of the FACT that the ideal DIFFERENTIATOR has an anti-symmetric unit sample response, we shall confine our attention to FIR designs in which H(N)=-h(M-1-n).

The correct answer is (b) 0.45

For explanation I would say: Designs based on M odd are PARTICULARLY poor if the bandwidth EXCEEDS 0.45. The problem is BASICALLY the ZERO in the frequency response at ω=π(f=1/2). When fp < 0.45, good designs are obtained for M odd.

Correct option is (d) All of the mentioned

The explanation: The important PARAMETERS in a DIFFERENTIATOR are its LENGTH, its bandwidth and the peak relative ERROR of the APPROXIMATION. The inter relationship among these three parameters can be easily displayed parametrically.

Right CHOICE is (B) False

Best explanation: We know that the weighting function is

W(ω)=1/ω

in order that the relative ripple in the pass band be a constant. Thus, the absolute error between the desired response ω and the approximation HR(ω) increases as ω varies from 0 to 2πfp.

Correct choice is (a) 1/ω

To ELABORATE: In the design of FIR DIFFERENTIATORS based on the chebyshev approximation criterion, the weighting function W(ω) is specified in the program as

W(ω)=1/ω

in order that the RELATIVE ripple in the pass band be a constant.

Right option is (c) 0 ≤ ω ≤ 2πfp

To elaborate: In most cases of practical interest, the desired frequency RESPONSE characteristic need only be linear over the LIMITED frequency range 0 ≤ ω ≤ 2πfp, where fp is the bandwidth of the DIFFERENTIATOR.

Right answer is (c) Left UNCONSTRAINED or Constrained to be ZERO

Easiest EXPLANATION: In the frequency range 2πfp ≤ ω ≤ π, the desired response may be EITHER left unconstrained or constrained to be zero.

The correct ANSWER is (b) False

The explanation: Full band differentiators cannot be ACHIEVED with an FIR FILTERS having odd number of coefficients, since Hr(π)=0 for M odd.