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The correct choice is (a) True

To explain: 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).

Right answer is (C) 0

Best explanation: SINCE we KNOW that the unit sample response of an ideal differentiator is anti-symmetric,

=>hd(0)=0.

Right answer is (a) \(\frac{cos⁡πn}{n}\)

For explanation I would say: We KNOW that, for an IDEAL differentiator, the FREQUENCY response is given as

Hd(ω)= jω ; -π ≤ ω ≤ π

Thus, we get the unit sample response CORRESPONDING to the ideal differentiator is given as

h(n)=\(\frac{cos⁡πn}{n}\).

Correct choice is (a) True

Explanation: |E(ω)|≥δ for some FREQUENCIES on the dense set, then a new set of frequencies corresponding to the L+2 largest peaks of |E(ω)| are selected and COMPUTATION is REPEATED. Since the new set of L+2 extremal frequencies are selected to increase in each iteration until it converges to the upper BOUND, this implies that when |E(ω)|≤δ for all frequencies on the dense set, the optimal solution has been found in terms of the POLYNOMIAL H(ω).

Right ANSWER is (b) Linearly proportional

For EXPLANATION: An IDEAL differentiator has a frequency response that is linearly proportional to the frequency.

Right choice is (a) WTX

To ELABORATE: FX DENOTES an ARRAY of maximum SIZE 10 that specifies the weight function in each BAND.

The correct OPTION is (C) Pass & STOP band

Explanation: The chebyshev approximation problem is viewed as an optimum design criterion on the sense that the weighted approximation error between the desired frequency RESPONSE and the actual frequency response is spread evenly ACROSS the pass band and evenly across the stop band of the filter minimizing the maximum error. The resulting filter designs have ripples in both pass band and stop band.

Correct answer is (a) True

Explanation: INITIALLY, we NEITHER know the set of external FREQUENCIES nor the parameters. To solve for the parameters, we use an iterative algorithm called the Remez exchange algorithm, in which we begin by guessing at the set of EXTREMAL frequencies.

The CORRECT choice is (b) Maximal ripple filters

To explain: In GENERAL, the filter designs that contain MAXIMUM NUMBER of alternations or ripples are called as maximal ripple filters.

The correct OPTION is (d) L+2

Best explanation: According to Alternation theorem, a necessary and sufficient condition for P(ω) to be UNIQUE, best WEIGHTED chebyshev APPROXIMATION, is that the error function E(ω) must exhibit at least L+2 extremal frequencies in S.

Right choice is (d) sinω

Explanation: If the filter has a anti-symmetric unit sample response, then we KNOW that

h(N)= -h(M-1-n)

and for M ODD in this case, Q(ω)=sin(ω).

Correct answer is (c) 1

For explanation: If the filter has a symmetric unit SAMPLE response, then we KNOW that

H(n)=h(M-1-n)

and for M odd in this case, Q(ω)=1.

Correct option is (a) 1-δ1 ≤ Hr(ω) ≤ 1+δ1; |ω|≤ωP

Best explanation: Let us consider the design of a low pass filter with the pass BAND edge frequency ωP and the ripple in the pass band is δ1, then from the general SPECIFICATIONS of the CHEBYSHEV filter, in the pass band the filter frequency response should satisfy the condition

1- δ1 ≤ Hr(ω) ≤ 1+δ1; |ω|≤ωP

The correct ANSWER is (b) FX

Easy EXPLANATION: FX denotes an ARRAY of maximum size 10 that specifies the desired frequency response in each BAND.

Right CHOICE is (d) LGRID

Easiest explanation: In Parks-McClellan PROGRAM, LGRID represents the grid density for interpolating the ERROR function. The default value is 16 if LEFT unspecified.

Correct OPTION is (b) False

Easiest explanation: If |E(ω)|≥δ for some frequencies on the dense SET, then a new set of frequencies CORRESPONDING to the L+2 LARGEST peaks of |E(ω)| are selected and COMPUTATION is repeated.

The CORRECT choice is (c) 3

Explanation: The VALUE of JTYPE=3 in the Parks-McClellan PROGRAM to select a filter that PERFORMS HILBERT transformer.

Right CHOICE is (d) 16M

Explanation: Having the solution for P(ω), we can now compute the ERROR function E(ω) from

E(ω)=W(ω)[Hdr(ω)-Hr(ω)]

on a dense set of frequency points. Usually, a NUMBER of points equal to 16M, where M is the length of the filter.

Correct choice is (c) L+3

To EXPLAIN I would say: We know that we can have at most L-1 local maxima and minima in the open interval 0<ω<π. In addition, ω=0 and π are also usually extrema. It is also maximum at ω for pass BAND and stop band frequencies. Thus the error function of a low pass filter has at most L+3 EXTREMAL frequencies.

The correct option is (a) True

The best I can explain: The weighting function on the approximation ERROR allows to choose the RELATIVE size of the errors in the different FREQUENCY bands. In particular, it is convenient to normalize W(ω) to UNITY in the stop band and set W(ω)=δ2/δ1 in the PASS band.

The correct choice is (d) -δ2 ≤ Hr(ω) ≤ δ2;|ω|≥ωs

The explanation: Let US CONSIDER the design of a low pass FILTER with the stop band edge FREQUENCY ωs and the ripple in the stop band is δ2, then from the general SPECIFICATIONS of the chebyshev filter, in the stop band the filter frequency response should satisfy the condition

-δ2 ≤ Hr(ω) ≤ δ2;|ω|≥ωs

Correct ANSWER is (b) Chebyshev APPROXIMATION

The best EXPLANATION: The filter design method described in the design of optimum equi RIPPLE linear phase FIR filters is formulated as a chebyshev approximation problem.

The correct OPTION is (b) Reduce side lobe

For EXPLANATION: To reduce side LOBES, it is desirable to optimize the frequency specification in the transition band of the FILTER.

Right choice is (d) \(\frac{2π}{M}\)(k+α)

The BEST explanation: In the frequency sampling method for FIR filter design, we SPECIFY the DESIRED frequency response Hd(ω) at a SET of EQUALLY spaced frequencies, namely

ωk=\(\frac{2π}{M}(k+α)\)

where k=0,1,2…M-1/2 and α=0 0r 1/2.

The correct choice is (c) Poles and ZEROS at equally SPACED POINTS on the unit CIRCLE

To elaborate: There is a POTENTIAL problem for frequency sampling realization of the FIR linear phase filter. The frequency sampling realization of the FIR filter introduces poles and zeros at equally spaced points on the unit circle.

Right ANSWER is (a) True

Easy EXPLANATION: Although the frequency sampling method provides us with another means for designing LINEAR phase FIR filters, its major ADVANTAGE LIES in the efficient frequency sampling structure, which is obtained when most of the frequency samples are zero.

The correct option is (b) False

To EXPLAIN I would say: The symmetry CONDITION, along with the symmetry conditions for {h(n)}, can be used to reduce the frequency specifications from M points to (M+1)/2 points for M odd and M/2 for M EVEN. Thus the linear equations for determining {h(n)} from {H(k+α)} are CONSIDERABLY SIMPLIFIED.

Correct CHOICE is (a) \(\FRAC{1}{M} \sum_{k=0}^{M-1}H(k+α)E^{j2π(k+α)n/M}\); n=0,1,2…M-1

The BEST explanation: We know that

H(k+α)=\(\sum_{n=0}^{M-1} h(n)e^{-j2π(k+α)n/M}\); k=0,1,2…M-1

If we multiply the above equation on both sides by the exponential exp(j2πkm/M), m=0,1,2….M-1 and sum over k=0,1,….M-1, we GET the equation

h(n)=\(\frac{1}{M} \sum_{k=0}^{M-1}H(k+α)e^{j2π(k+α)n/M}\); n=0,1,2…M-1

Right answer is (d) H*(M-k-α)

Explanation: SINCE {h(N)} is real, we can EASILY show that the frequency SAMPLES {H(k+α)} satisfy the symmetry condition

H(k+α)= H*(M-k-α).

The correct CHOICE is (d) \(\sum_{n=0}^{M-1} h(n)e^{-jωn}\)

The explanation is: The desired output of an FIR filter with an INPUT h(n) and using a window of length M is given as

H(ω)=\(\sum_{n=0}^{M-1} h(n)e^{-jωn}\)

The correct option is (c) Transition BAND

Best explanation: To reduce the side lobes, it is desirable to OPTIMIZE the frequency SPECIFICATION in the transition band of the filter. This optimization can be accomplished numerically on a digital computer by means of linear PROGRAMMING TECHNIQUES.

Right CHOICE is (b) False

To elaborate: SINCE the WIDTH of the MAIN LOBE is inversely proportional to the value of M, if the value of M increases then the main lobe becomes narrower.

The CORRECT OPTION is (a) True

To elaborate: In the frequency sampling method, we specify the frequency response Hd(ω) at a set of equally SPACED FREQUENCIES, namely ωk=\(\FRAC{2π}{M}(k+\alpha)\)

The CORRECT choice is (d) Hanning window

Easy explanation: The Hanning window has a time DOMAIN sequence as

h(N)=\(\FRAC{1}{2}(1-cos⁡\frac{2πn}{M-1})\), 0≤n≤M-1

Correct ANSWER is (b) Increase with increase in M

To explain I would say: The FREQUENCY of oscillations in the pass BAND of a low pass filter INCREASES with an increase in the value of M, but they do not diminish in amplitude.

The CORRECT answer is (c) 12π/M

Best EXPLANATION: The TRANSITION width of the main lobe in the CASE of Blackman window is EQUAL to 12π/M where M is the length of the window.

The correct OPTION is (c) KAISER window

The explanation is: The frequency RESPONSE SHOWN in the figure is the frequency response of a low pass filter designed using a Kaiser window of length M=61 and α=4.

Right answer is (a) True

For explanation I would say: The multiplication of hd(n) with a rectangular window is identical to TRUNCATING the FOURIER series representation of the desired filter characteristic Hd(ω). The truncation of Fourier series is known to introduce ripples in the frequency RESPONSE characteristic H(ω) DUE to the non-uniform convergence of the Fourier series at a discontinuity. The OSCILLATORY behavior near the band edge of the low pass filter is known as Gibbs phenomenon.

Correct option is (c) -32

Easiest EXPLANATION: The peak SIDE lobe in the CASE of Hanning WINDOW has a value of -32dB.

Correct OPTION is (a) -13

Explanation: The peak side lobe in the CASE of RECTANGULAR window has a value of -13dB.