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The CORRECT option is (c) Butterworth filter

Explanation: The FREQUENCY characteristic given in the FIGURE is the magnitude RESPONSE of a 37-order Butterworth filter.

Right answer is (a) True

Best explanation: As in the analog DOMAIN, frequency transformations can be performed on a digital LOW pass filter to CONVERT it to either a band pass, band STOP or HIGH pass filter. The transformation involves the replacing of the variable z^-1 by a rational function g(z^-1).

Correct choice is (b) s→ Ωu / s

Explanation: The LOW pass-to-high pass transformation is simply achieved by replacing s by 1/s. If the desired high pass filter has the pass BAND EDGE frequency Ωu, then the transformation is

s→ Ωu / s

The correct option is (B) ΩS=Min{|A|,|B|}

The explanation: If Ωu and Ωl are the upper and lower cutoffpass BAND frequencies of the desired band stop filter and Ω1 and Ω2 are the lower and upper cutoffstop band frequencies of the desired band stop filter, then the BACKWARD design equation is

ΩS= Min{|A|,|B|}

where, =\(\FRAC{Ω_1 (Ω_u-Ω_l)}{-Ω_1^2+Ω_u Ω_l}\) and B=\(\frac{Ω_2 (Ω_u-Ω_l)}{-Ω_2^2+Ω_u Ω_l}\).

Right answer is (d) ΩS=Min{|A|,|B|}

Explanation: If Ωu and Ωl are the upper and LOWER cutoffpass band frequencies of the desired band pass FILTER and Ω1 and Ω2 are the lower and upper cutoffstop band frequencies of the desired band pass filter, then the BACKWARD design EQUATION is

ΩS=Min{|A|,|B|}

where, A=\(\frac{-Ω_1^2+Ω_u Ω_l}{Ω_1 (Ω_u-Ω_l)}\) and B=\(\frac{Ω_2^2-Ω_u Ω_l}{Ω_2 (Ω_u-Ω_l)}\).

Correct answer is (a) CHEBYSHEV filter

Best explanation: The PHASE response given in the FIGURE belongs to the FREQUENCY characteristic of a 13-order type-1 Chebyshev filter.

Correct option is (a) s → s / Ωu

The explanation is: If Ωu is the DESIRED pass BAND EDGE frequency of NEW low pass filter, then the transfer function of this new low pass filter is obtained by using the TRANSFORMATION s → s / Ωu.

Right answer is (d) ΩS=\(\frac{Ω’_S}{Ω_u}\)

For EXPLANATION: If Ωu is the DESIRED pass band EDGE frequency of new low pass filter, then the transfer function of this new low pass filter is obtained by using the transformation s → s / Ωu. If ΩS and Ω’S are the stop band frequencies of PROTOTYPE and transformed filters respectively, then the backward DESIGN equation is given by

ΩS=\(\frac{Ω’_S}{Ω_u}\) .

The correct choice is (c) 1 rad/sec

To elaborate: LET H(s) denote the TRANSFER function of a low pass analog filter with a pass BAND edge frequency ΩP EQUAL to 1 rad/sec. This filter is known as analog low pass normalized PROTOTYPE.

Correct option is (b) ΩS=\(\frac{Ω_u}{Ω’_S}\)

To elaborate: If Ωu is the DESIRED pass band edge FREQUENCY of new high pass filter, then the TRANSFER function of this new high pass filter is obtained by using the TRANSFORMATION s→Ωu/s. If ΩS and Ω’S are the STOP band frequencies of prototype and transformed filters respectively, then the backward design equation is given by

ΩS=\(\frac{Ω_u}{Ω’_S}\) .

Right CHOICE is (c) Equi-ripple in pass band and stop band

The explanation: An elliptical filter is a filter which exhibit equi-ripple BEHAVIOR in both pass band and stop band of the magnitude SQUARE RESPONSE.

Correct choice is (a) True

Explanation: An IMPORTANT characteristic of the Bessel filter is the LINEAR phase RESPONSE over the pass BAND of the filter. As a CONSEQUENCE, Bessel filters has a larger transition bandwidth, but its phase is linear within the pass band.

Right choice is (d) 3

To ELABORATE: SINCE from the magnitude square response of the type-2 chebyshev filter, it has odd number of maxima and minima in the STOP band, the ORDER of the filter is odd i.e., 3.

The correct choice is (a) Imaginary AXIS

To EXPLAIN I WOULD SAY: The ZEROS of this class of filters lie on the imaginary axis in the s-plane.

The CORRECT answer is (d) MONOTONIC BEHAVIOR in the STOP band

The best explanation: Type-2 chebyshev filters exhibit equi-ripple behavior in stop band and a monotonic characteristic in the pass band.

The CORRECT option is (B) Both poles and ZEROS

To explain: Type-1 chebyshev filters are all-pole filters where as the family of type-2 chebyshev filters CONTAINS both poles and zeros.

The CORRECT option is (d) 7

To explain: Given Ωc=1000π and Ωs=2000π

For an attenuation of 40DB, δ2=0.01. We know that

N=\(\frac{log⁡[(\frac{1}{δ_2^2})-1]}{2log⁡[\frac{Ω_s}{Ω_s}]}\)

THUS by substituting the corresponding values in the above equation, we get N=6.64

To MEET the desired SPECIFICATIONS, we select N=7.

The CORRECT option is (a) BUTTERWORTH

To explain I would say: The MAGNITUDE square response shown in the given FIGURE is for Butterworth FILTER.

The correct option is (c) 1/(1+ε^2)

To explain: 1/(1+ε^2) gives the BAND edge VALUE of the magnitude square RESPONSE |H(Ω)|^2.

Correct CHOICE is (d) \(\frac{1}{1+(\frac{s}{j})^{2N}}\)

For explanation I would say: We know that the magnitude squared frequency response of a normalized low PASS Butterworth filter is given as

|H(jΩ)|^2=\(\frac{1}{1+Ω^{2N}}\) => HN(jΩ).HN(-jΩ)=\(\frac{1}{1+Ω^{2N}}\)

REPLACING jΩ by ‘s’ and hence Ω by s/j in the above equation, we get

HN(s).HN(-s)=\(\frac{1}{1+(\frac{s}{j})^{2N}}\) which is called the transfer function.

Right option is (b) All-pole filter

Easiest explanation: LOW pass BUTTERWORTH filters are also CALLED as all-pole filters because it has only non-zero POLES.

Right ANSWER is (a) \(\FRAC{1}{\sqrt{1+(\frac{Ω}{Ω_C})^{2N}}}\)

The explanation is: A Butterworth is CHARACTERIZED by the magnitude FREQUENCY response

|H(jΩ)| = \(\frac{1}{\sqrt{1+(\frac{Ω}{Ω_C})^{2N}}}\)

where N is the order of the filter and ΩC is defined as the cutoff frequency.

Correct ANSWER is (c) Sufficiently small

For EXPLANATION: Aliasing in this MATCHED z-transformation can be avoided by selecting the sampling INTERVAL T sufficiently small.

The correct choice is (b) IMPULSE INVARIANCE

The best I can EXPLAIN: We observe that the poles OBTAINED from the matched z-transform are identical to the poles obtained with the impulse invariance method.

Right option is (a) True

Explanation: To preserve the FREQUENCY response characteristic of the ANALOG filter, the sampling interval in the matched z-transformation must be properly selected to yield the POLE and zero LOCATIONS at the equivalent position in the z-plane.

The correct CHOICE is (b) 1-e^aTz^-1

To elaborate: If T is the sampling INTERVAL, then each FACTOR of the form (s-a) in H(s) is mapped into the factor (1-e^aTz^-1) in the z-domain.

Right answer is (a) True

To ELABORATE: In this method of TRANSFORMING ANALOG filter into an equivalent digital filter is to map the poles and zeros of H(s) directly into poles and zeros in the z-plane.

The correct ANSWER is (d) Matched Z-transform

For explanation: In this METHOD of transforming analog filter into an EQUIVALENT DIGITAL filter is to map the poles and zeros of H(s) directly into poles and zeros in the z-plane.

The correct choice is (c) R>1

Best explanation: We know that z=e^sT

Now substitute s=σ+jΩ and z=r.e^jω, that is represent ‘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.

The correct ANSWER is (d) Fs

For explanation: When a continuous time signal x(t) with spectrum X(F) is SAMPLED at a rate Fs=1/T samples PER SECOND, the spectrum of the sampled signal is periodic repetition of the scaled spectrum Fs.X(F) with period Fs.

The correct choice is (a) 0
Explanation: We know that z=e^sT

Now SUBSTITUTE s=σ+jΩ and z=r.e^jω, that is represent ‘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

The CORRECT option is (C) ω=ΩT

The best I can explain: We know that z=e^sT

Now SUBSTITUTE s=σ+jΩ and z=r.e^jω, that is REPRESENT ‘z’ in the polar form

On equating both SIDES, we get

ω=ΩT.

The correct answer is (d) High pass

To EXPLAIN: It is clear that the impulse INVARIANCE method is in -appropriate for DESIGNING high pass filter due to the SPECTRUM ALIASING that results from the sampling process.

Correct choice is (B) False

Easy explanation: 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.

Correct answer is (a) True

Best explanation: 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 correct choice is (c) Digital filter with aliasing

To explain: In the given diagram, the CONTINUOUS line is the FREQUENCY response of ANALOG filter and dotted line is the frequency response of the CORRESPONDING digital filter with aliasing.

The correct choice is (b) False

The EXPLANATION is: Aliasing occurs if the sampling rate FS is less than twice the highest FREQUENCY contained in X(F).

The CORRECT option is (a) F/Fs

Explanation: In the impulse invariance method, the normalized FREQUENCY f is GIVEN by

f= F/Fs.

Correct option is (b) Fs.X(F)

For explanation: When a continuous time signal x(t) with spectrum X(F) is sampled at a rate Fs=1/T samples PER second, the spectrum of the sampled signal is PERIODIC REPETITION of the scaled spectrum Fs.X(F).

Right ANSWER is (C) Periodic repetition

Best EXPLANATION: When a continuous TIME signal x(t) with SPECTRUM X(F) is sampled at a rate Fs=1/T samples per second, the spectrum of the sampled signal is periodic repetition.

The CORRECT choice is (a) True

The best I can explain: 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.

Correct choice is (a) True

The explanation is: By proper choice of the COEFFICIENTS of {αk}, it is POSSIBLE to MAP the jΩ-axis into the unit CIRCLE.