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The correct OPTION is (a) \(\frac{|sin(ωM/2)|}{|sin(ω/2)|}\)

The explanation: We know that for a rectangular window

W(ω)=\(\sum_{n=0}^{M-1} w(n) e^{-jωn}=e^{-jω(M-1)/2} \frac{sin⁡(\frac{ωM}{2})}{sin⁡(\frac{ω}{2})}\)

THUS the window FUNCTION has a magnitude response

|W(ω)|=\(\frac{|sin(ωM/2)|}{|sin(ω/2)|}\)

The correct option is (a) True

To explain I WOULD say: For |x|≤1, |TN(x)|≤1, and it oscillates between -1 and +1 a number of times proportional to N.

The above is evident from the equation,

TN(x) = COS(Ncos^-1x), |x|≤1.

Correct choice is (a) True

The best I can explain: The amplitude of the sampling RATE converted signal should be multiplied by a FACTOR C, whose VALUE when equal to I is called as desired NORMALIZATION factor.

Right option is (d) I-fold periodic REPETITION

Easiest explanation: We observe that the SAMPLING rate INCREASE, obtained by the addition of I-1 zero samples between successive values of x(n), results in a SIGNAL whose spectrum is an I-fold periodic repetition of the INPUT signal spectrum.

Correct answer is (a) True

To ELABORATE: The spectrum of V(m) is OBTAINED by evaluating V(z) = X(z^I) on the unit circle. Thus V(ωy)= X(ωyI), where ωy denotes the FREQUENCY variable RELATIVE to the new sampling rate.

Correct ANSWER is (d) X(zI)

EXPLANATION: By TAKING the z-transform of the signal v(m), we get

V(z)=\(\sum_{m=-∞}^∞ v(m)z^{-m}\)

=\(\sum_{m=-∞}^∞ x(m)z^{-mI}\)

= X(z^-I)

Correct answer is (B) s → Ωu/s

Easiest 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.

Correct choice is (c) s→\(\frac{s(Ω_u-Ω_l)}{s^2-Ω_u Ω_l}\)

Best EXPLANATION: If Ωu and Ωl are the upper and lower cutoffpass band frequencies of the desired band STOP FILTER, then the TRANSFORMATION to be performed on the NORMALIZED low pass filter is

s→\(\frac{s(Ω_u-Ω_l)}{s^2-Ω_u Ω_l}\)

The correct CHOICE is (c) s→\(\frac{s^2+Ω_u Ω_l}{s(Ω_u-Ω_l)}\)

To ELABORATE: If Ωu and Ωl are the upper and lower cutoff pass BAND frequencies of the desired band pass filter, then the TRANSFORMATION to be performed on the normalized low pass filter is

s→\(\frac{s^2+Ω_u Ω_l}{s(Ω_u-Ω_l)}\)

Right answer is (a) \(\frac{1}{s^2+\sqrt{2} s+1}\)

Explanation: We know that the BUTTERWORTH polynomial of a 2^nd ORDER low pass filter is

B2(s)=s^2+√2 s+1

Thus the transfer FUNCTION is given as \(\frac{1}{s^2+\sqrt{2} s+1}\).

Right answer is (B) s+1

Explanation: Given that the order of the Butterworth low PASS filter is 1.

Therefore, for N=1 Butterworth POLYNOMIAL is given as B3(s)=(s-s0).

We know that, sk=\(e^{jπ(\frac{2k+1}{2N})} e^{jπ/2}\)

=> s0=-1

=> B1(s)=s-(-1)=s+1.

Right option is (d) \(\frac{π}{N} k+\frac{π}{2N}\) k=0,1,2…2N-1

The explanation: The transfer function of normalized low pass BUTTERWORTH filter is given as

HN(s).HN(-s)=\(\frac{1}{1+(\frac{s}{j})^{2N}}\)

The POLES of the above equation is obtained by equating the denominator to zero.

=> \(1+(\frac{s}{j})^{2N}\)=0

=> s=(-1)^1/2N.j

=> sk=\(E^{jπ(\frac{2k+1}{2N})} e^{jπ/2}\), k=0,1,2…2N-1

The poles are THEREFORE on a circle with radius unity and are placed at angles,

θk=\(\frac{π}{N} k+\frac{π}{2N}\) k=0,1,2…2N-1

Right option is (d) \(\FRAC{1}{1+Ω^{2N}}\)

Explanation: We know that the MAGNITUDE response of a low PASS Butterworth filter of order N is given as

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

For a normalized filter, ΩC =1

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

Thus the magnitude SQUARED response of the normalized low pass Butterworth filter of order N is given by the equation,

|H(jΩ)|^2=\(\frac{1}{1+Ω^{2N}}\).

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

To elaborate: 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.

Right ANSWER is (d) All of the mentioned

Easiest explanation: Butterworth filters have a very SMOOTH pass BAND, which we pay for with a RELATIVELY wide transmission region.

Correct option is (b) -20log(δS)

For explanation I would SAY: δS is known as the stop BAND ATTENUATION, and its VALUE in dB is GIVEN as -20log(δS).

Correct option is (a) -20log(1- δP)

To EXPLAIN I would say: 1-δP is KNOWN as the PASS band RIPPLE or the pass band attenuation, and its value in dB is given as -20log(1-δP).

Right choice is (d) 1-δP≤|H(jΩ)|≤1

Easiest EXPLANATION: From the magnitude frequency response of the LOW pass filter, the hatched region in the pass band INDICATE forbidden magnitude value WHOSE value is GIVEN as

1- δP≤|H(jΩ)|≤1.

Right answer is (c) 0≤|H(jΩ)|≤ δS

The explanation: From the magnitude frequency response of the LOW pass filter, the HATCHED region in the stop band indicate forbidden magnitude VALUE whose value is GIVEN as

0≤|H(jΩ)|≤ δS.

The correct ANSWER is (C) Transition band

Best explanation: From the magnitude frequency response of a low PASS filter, we can state that the region between pass band and stop band frequencies is KNOWN as ‘transition band’ where no SPECIFICATIONS are provided.

Right option is (a) STOP band

To explain: From the magnitude FREQUENCY RESPONSE of a LOW pass filter, we can STATE that the region after stop band frequency is known as ‘stop band’ where the signal is stopped or restricted.

Correct option is (b) False

The EXPLANATION is: The major difficulty with any ITERATIVE procedure that searches for the parameter values that minimize a non-linear function is that the process may converge to a local MINIMUM INSTEAD of a global minimum.

Correct choice is (a) True

For EXPLANATION I WOULD say: Due to the non-linear nature of E(P,G), its minimization over the remaining 4K parameters is performed by an iterative NUMERICAL optimization method.

Right ANSWER is (d) Non-Linear FUNCTION of 4K+1 PARAMETERS

Easy EXPLANATION: The squared error function E(p,G) is a non-linear function of 4K+1 parameters.

Correct answer is (b) 1/2

Best EXPLANATION: The emphasis on the ERRORS AFFECTING the design may be equally WEIGHTED between magnitude and delay by taking the VALUE of λ as 1/2.

Correct OPTION is (c) 4K-dimension vector

The explanation: In the ERROR FUNCTION ‘p’ DENOTES the 4K dimension vector of the filter coefficients.

The CORRECT OPTION is (a) True

The explanation: We know that the error in delay is defined as TG(ωk) – TD(ωk). HOWEVER, the choice of Td(ωk) for error in delay is complicated by the difficulty in assigning a nominal delay of the filter.

Right CHOICE is (a) Tg(ωk)-TD(ωk)

EXPLANATION: Similarly as in the previous question, the error in DELAY at ωk is defined as Tg(ωk)-Td(ωk), where Td(ωk) is the desired delay RESPONSE.

Right option is (a) True

Explanation: INSTEAD of DEALING with the phase response ϴ(ω), it is more convenient to DEAL with the ENVELOPE DELAY as a function of frequency.

The CORRECT option is (b) G.A(ωk) – AD(ωk)

To EXPLAIN: The error in magnitude at the FREQUENCY ωk is G.A(ωk) – Ad(ωk) for 0 ≤ |ω| ≤ π, where Ad(ωk) is the desired magnitude response at ωk.