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The CORRECT choice is (b) 8π/M

Explanation: The TRANSITION width of the main lobe in the case of Hamming window is equal to 8π/M where M is the LENGTH of the window.

Correct choice is (B) False

The best I can explain: 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. In fact, the width of each side lobes decreases with an INCREASE in M.

The CORRECT option is (a) BARTLETT window

Explanation: The Bartlett window which is ALSO called as triangular window has a time domain sequence as

h(n)=\(1-\frac{2|n-\frac{M-1}{2}|}{M-1}\), 0≤n≤M-1.

The CORRECT answer is (d) Increases

For explanation I WOULD say: The height of each side lobes increase with an increase in M such a manner that the AREA under each side lobe remains invariant to CHANGES in M.

Correct option is (a) True

Easy EXPLANATION: 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. In FACT, the width of each side lobes DECREASES with an INCREASE in M.

Right choice is (c) 4π/M

Easiest explanation: The width of the main LOBE width is MEASURED to the FIRST zero of W(ω)) is 4π/M.

The correct answer is (a) LOW PASS

Explanation: When h(n)=-h(M-1-n) and M is EVEN, we know that H(0)=0. Thus it is not USED in the design of a low pass linear phase FIR filter.

Correct answer is (b) (M-1)/2 when M is odd and M/2 when M is EVEN

The EXPLANATION is: We know that, for a anti-symmetric h(n) h(M-1/2)=0 and thus the NUMBER of filter coefficients that specify the FREQUENCY response is (M-1)/2 when M is odd and M/2 when M is even.

Right choice is (C) H(n) anti-symmetric and M odd

Easiest explanation: If h(n)=-h(M-1-n) and M is odd, we GET H(0)=0 and H(π)=0. Consequently, this is not SUITABLE as EITHER a low pass filter or a high pass filter.

Right option is (d) (M+1)/2 when M is odd and M/2 when M is EVEN

To elaborate: We know that, for a symmetric h(n), the number of FILTER coefficients that specify the FREQUENCY response is (M+1)/2 when M is odd and M/2 when M is even.

Correct choice is (b) False

Easy explanation: We KNOW that the roots of the POLYNOMIAL H(z) are IDENTICAL to the roots of the polynomial H(z^-1). This implies that if the unit sample response h(n) of the FILTER is REAL, complex valued roots must occur in complex conjugate pairs.

The correct ANSWER is (C) Reciprocal pairs

To elaborate: We know that the roots of the polynomial H(Z) are IDENTICAL to the roots of the polynomial H(z^-1). Consequently, the roots of H(z) must occur in reciprocal pairs.

Correct option is (B) ±h(M-1-n) n=0,1,2…M-1

The best explanation: An FIR filter has an LINEAR phase if its unit sample RESPONSE SATISFIES the condition

h(n)= ±h(M-1-n) n=0,1,2…M-1.

Correct answer is (a) True

To explain: We know that Z^-(M-1).H(z^-1)=±H(z). This RESULT implies that the ROOTS of the POLYNOMIAL H(z) are identical to the roots of the polynomial H(z^-1).

The correct choice is (d) z^-(M-1).H(z^-1)=±H(z)

Easy explanation: We know that H(z)=\(\sum_{K=0}^{M-1} h(k)z^{-k}\) and h(N)=±h(M-1-n) n=0,1,2…M-1

When we incorporate the symmetric and anti-symmetric conditions of the second equation into the first equation and by substituting z^-1 for z, and multiply both SIDES of the resulting equation by z^-(M-1) we get z^-(M-1).H(z^-1)=±H(z)

Right answer is (a) True

Easy explanation: We can EXPRESS the output sequence as the convolution of the unit sample RESPONSE h(n) of the system with the input signal. The lower and UPPER limits on the convolution sum REFLECT the causality and FINITE duration characteristics of the filter.

The CORRECT answer is (b) -ωP+ ωS

The best explanation: IfωP and ωS represents the PASS band edge ripple and STOP band edge ripple, then the transition WIDTH -ωP+ ωS GIVES the bandwidth of the filter.

The CORRECT ANSWER is (c) Pass band edge ripple

The EXPLANATION: Pass band edge ripple is the frequency at which the pass band starts TRANSITING to the stop band.

Correct CHOICE is (d) πδ(ω)+0.5-j0.5cot(ω/2)

The best I can EXPLAIN: Since the UNIT step function is not ABSOLUTELY summable, it has a Fourier TRANSFORM which is given by the equation

U(ω)= πδ(ω)+0.5-j0.5cot(ω/2).

The correct ANSWER is (a) True

The best I can explain: Since h(N) is completely specified by he(n), it follows that H(ω) is completely DETERMINED if we know HR(ω). Alternatively, H(ω) is completely determined from HI(ω) and h(0). In short, HR(ω) and HI(ω) are interdependent and cannot be specified independently when the SYSTEM is CAUSAL.

The correct choice is (b) H(n)=2ho(n)u(n)+h(0)δ(n), n ≥ 1

For explanation: Given h(n) is causal and h(n)= he(n)+ho(n)

=>he(n)=1/2[h(n)+h(-n)]

Now, if h(n) is causal, it is possible to RECOVER h(n) from its even part he(n) for 0≤n≤∞ or from its odd COMPONENT ho(n) for 1≤n≤∞.

=>h(n)= 2ho(n)u(n)+h(0)δ(n), n ≥ 1

since ho(n)=0 for n=0, we cannot recover h(0) from ho(n) and hence we must also know h(0).

Right OPTION is (a) True

Best explanation: ONE IMPORTANT conclusion that we made from the Paley-Wiener theorem is that the magnitude FUNCTION |H(ω)| can be zero at some frequencies, but it cannot be zero over any finite band of frequencies, since the integral then becomes infinite. Consequently, any ideal FILTER is non-causal.

Correct choice is (d) h(n)=2he(n)u(n)-he(0)δ(n), n ≥ 0

Easiest explanation: GIVEN h(n) is CAUSAL and h(n)= he(n)+ho(n)

=>he(n)=1/2[h(n)+h(-n)]

Now, if h(n) is causal, it is POSSIBLE to RECOVER h(n) from its even part he(n) for 0≤n≤∞ or from its odd component ho(n) for 1≤n≤∞.

=>h(n)= 2he(n)u(n)-he(0)δ(n), n ≥ 0.

Right option is (a) True

The explanation is: We know that the ideal low pass filter is non-causal. HENCE, a ideal low pass filter cannot be REALIZED in PRACTICE.

Correct option is (C) Causal

The explanation is: If |H(ω)| is square integrable and if the integral \(\int_{-\PI}^\pi |ln⁡|H(ω)||dω\) is FINITE, then we can associate with |H(ω)| and a PHASE response θ(ω), so that the resulting filter with the frequency response H(ω)=|H(ω)|e^jθ(ω) is causal.

The correct answer is (b) \(\int_{-π}^π|ln⁡ |H(ω)||dω \LT \infty\)

Easy EXPLANATION: If h(n) has finite energy and h(n)=0 for n<0, then according to the Paley-Wiener THEOREM, we have

\(\int_{-π}^π|ln⁡ |H(ω)||dω \lt \infty\)

Correct CHOICE is (d) Ideal low pass filter at ω=π/4

To elaborate: At N=0, the equation for ideal low pass filter is GIVEN as H(n)=ω/π.

From the given figure, h(0)=0.25=>ω=π/4.

Thus the given figure represents the unit SAMPLE response of an ideal low pass filter at ω=π/4.

Right choice is (b) (-1/8,1/8)

EASIEST explanation: We know that

|v(n-1)| ≤ \(\frac{(\frac{1}{2}).2^{-b}}{1-|a|}\)

GIVEN |a|=3/4 and b=4 => |v(n-1)| ≤ 1/8=> The DEAD band is (-1/8,1/8).

The correct choice is (a) True

To elaborate: There is an possible LIMIT cycle mode with zero input, which occurs as a result of rounding the multiplications, corresponds to an equivalent second order system with poles at z=±1. In this case the two pole filter exhibits OSCILLATIONS with an amplitude that falls in the DEAD band BOUNDED by 2^-b/(1-|a1|-a2).

The correct ANSWER is (c) |a1|+|a2|<1

To explain: It can be easily shown that a NECESSARY and sufficient condition for ensuring that no zero-input overflow LIMIT cycles occur is |a1|+|a2|<1

which is extremely restrictive and hence an unreasonable constraint to IMPOSE on any second order section.

The correct choice is (d) (-1/16,1/16)

The explanation is: We KNOW that

|v(n-1)| ≤ \(\frac{(\frac{1}{2}).2^{-b}}{1-|a|}\)

Given |a|=1/2 and b=4 => |v(n-1)| ≤ 1/16=> The DEAD BAND is (-1/16,1/16).

Correct ANSWER is (C) |v(n-1)| ≤ \(\frac{(1/2).2^{-b}}{1-|a|}\)

The best explanation: Since the QUANTIZATION product av(n-1) is obtained by rounding, it follows that the quantization error is BOUNDED as

|Qr[av(n-1)]-av(n-1)| ≤ \((\frac{1}{2}).2^{-b}\)

Where ‘b’ is the number of bits (exclusive of SIGN) used in the representation of the pole ‘a’ and v(n). Consequently, we get

|v(n-1)|-|av(n-1)| ≤ \((\frac{1}{2}).2^{-b}\)

and hence

|v(n-1)| ≤ \(\frac{(\frac{1}{2}).2^{-b}}{1-|a|}\) .

The correct answer is (a) True

For explanation: When the input sequence X(n) to the filter becomes zero, the output of the filter then, after a number of iterations, enters into the limit cycle. The output remains in the limit cycle until another input of SUFFICIENT size is APPLIED that drives the SYSTEM out of the limit cycle. Similarly, zero input limit cycles occur from non-zero initial conditions with the input x(n)=0.

Correct ANSWER is (c) Both of the mentioned

The BEST EXPLANATION: The oscillations in the output of the recursive system are CALLED as limit CYCLES and are directly attributable to round-off errors in multiplication and overflow errors in addition.

Right option is (d) Periodic oscillations in the output

The best I can explain: In the RECURSIVE SYSTEMS, the nonlinearities due to the finite-precision arithmetic operations often cause periodic oscillations to occur in the output EVEN when the input sequence is zero or some non zero constant value.

Right option is (b) False

To elaborate: In the realization of a digital filter, either in digital HARDWARE or in software on a digital computer, the QUANTIZATION inherent in the FINITE precision arithmetic operations RENDER the system linear.

Right choice is (B) ZEROS

Explanation: The SENSITIVITY analysis MADE on the poles of a system results on the zeros of the IIR FILTERS.

Correct option is (B) False

Best explanation: If the POLES are tightly clustered as they are in a narrow BAND filter, the lengths of |pi-pl| are small for the poles in the vicinity of pi. These small lengths will contribute to large errors and hence a large perturbation ERROR results.

Right option is (c) \(\frac{-p_i^{N-k}}{\prod_{l=1}^n p_i-p_l}\)

For explanation I WOULD SAY: The EXPRESSION for \(\frac{∂p_i}{∂a_k}\) is GIVEN as follows

\(\frac{∂p_i}{∂a_k}=\frac{-p_i^{N-k}}{\prod_{l=1}^n p_i-p_l}\)