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Right CHOICE is (a) \(\sum_{k=1}^N \frac{∂p_i}{∂a_k} \Delta a_k\)

The BEST explanation: The perturbation error Δpi can be expressed as

Δpi=\(\sum_{k=1}^N \frac{∂p_i}{∂a_k} \Delta a_k\)

Where \(\frac{∂p_i}{∂a_k}\), the PARTIAL derivative of pi with respect to ak, REPRESENTS the incremental change in the pole pi due to a change in the coefficient ak. Thus the total error Δpi is expressed as a sum of the incremental errors due to changes in each of the coefficients ak.

Correct choice is (b) Perturbation

The best EXPLANATION: We know that &pmacr;k = pk + Δpk, k=1,2…N and Δpk that is the ERROR RESULTING from the quantization of FILTER coefficients, which is CALLED as perturbation error.

Right answer is (d) \(\prod_{k=1}^N (1-p_k Z^{-1})\)

Easiest explanation: We know that the system function of a general IIR filter is given by the equation

H(z)=\(\frac{\sum_{k=0}^M b_k z^{-k}}{1+\sum_{k=1}^N a_k z^{-k}}\)

The DENOMINATOR of H(z) may be expressed in the form

D(z)=\(1+\sum_{k=1}^N a_k z^{-k}=\prod_{k=1}^N (1-p_k z^{-1})\)

where pk are the POLES of H(z).

Correct choice is (C) āk = ak + Δak

The BEST explanation: The quantized coefficient āk can be related to the un-quantized coefficient ak by the relation

āk = ak + Δak

where Δak represents the QUANTIZATION ERROR.

Correct option is (a) σ > 0

To explain I would say: We know that if = σ+jΩ and z=rejω, then by substituting the values in the below expression

s = \(\FRAC{2}{T}[\frac{1-z^{-1}}{1+z^{-1}}]\)

=>σ = \(\frac{2}{T}[\frac{r^2-1}{r^2+1+2rcosω}]\)

When r>1 => σ > 0.

The correct ANSWER is (b) σ < 0

Explanation: We KNOW that if = σ+jΩ and z=re^jω, then by substituting the values in the below expression

s = \(\FRAC{2}{T}[\frac{1-z^{-1}}{1+z^{-1}}]\)

=>σ = \(\frac{2}{T}[\frac{r^2-1}{r^2+1+2rcosω}]\)

When r<1 => σ < 0.

Right answer is (a) \([\frac{x(n)+x(n-1)}{2}]T\)

Best explanation: We know that the derivative equation is

dy(t)/dt=x(t)

On applying integrals both sides, we GET

\(\int_{(n-1)T}^{nT}dy(t)=\int_{(n-1)T}^{nT} x(t)dt\)

=> y(nT)-y[(n-1)T]=\(\int_{(n-1)T}^{nT} x(t)dt\)

On applying trapezoidal rule on the right hand integral, we get

y(nT)-y[(n-1)T]=\([\frac{x(nT)+x[(n-1)T]}{2}]T\)

SINCE x(n) and y(n) are approximately EQUAL to x(nT) and y(nT) RESPECTIVELY, the above equation can be written as

y(n)-y(n-1)=\([\frac{x(n)+x(n-1)}{2}]T\)

The correct option is (d) Entirely inside the unit circle |z|=1

Easy EXPLANATION: In BILINEAR transformation, the z to s transformation is given by the expression

z=[1+(T/2)s]/[1-(T/2)s].

Thus unlike the backward difference method, the left-half s-plane is now MAPPED entirely inside the unit circle, |z|=1, rather than to a part of it.

Correct choice is (b) \([\frac{x(NT)+x[(n-1)T]}{2}]T\)

The BEST EXPLANATION: The given integral is approximated by the trapezoidal rule. This rule states that if T is SMALL, the area (integral) can be approximated by the MEAN height of x(t) between the two limits and then multiplying by the width. That is

\(\int_{(n-1)T}^{nT} x(t)dt=[\frac{x(nT)+x[(n-1)T]}{2}]T\)

The CORRECT answer is (c) \(\frac{T}{2}[\frac{1+z^{-1}}{1-z^1}]\)

Explanation: We know that

y(n)-y(n-1)= \([\frac{x(n)+x(n-1)}{2}]T\)

Taking z-transform of the above EQUATION gives

=>Y(z)[1-z^-1]=([1+z^-1]/2).TX(z)

=>H(z)=Y(z)/X(z)=\(\frac{T}{2}[\frac{1+z^{-1}}{1-z^1}]\).

Correct CHOICE is (d) Trapezoidal rule

Easy explanation: Bilinear TRANSFORMATION uses trapezoidal rule for INTEGRATING a CONTINUOUS time function.

Right option is (c) s = \(\frac{2}{T}[\frac{1-z^{-1}}{1+z^{-1}}]\)

To explain I would say: In BILINEAR transformation of an analog FILTER to digital filter, using the trapezoidal rule, the SUBSTITUTION for ‘s’ is given as

s = \(\frac{2}{T}[\frac{1-z^{-1}}{1+z^{-1}}]\).

Right option is (b) \(\frac{1-Z^{-1}}{T}\) Y(z)

Explanation: The FIRST backward difference of y(n) is GIVEN by the EQUATION

[y(n)-y(n-1)]/T

Thus the z-transform of the first backward difference of y(n) is given as

\(\frac{1-z^{-1}}{T}\) Y(z).

The correct option is (B) False

The explanation: An analog high pass filter cannot be MAPPED to a DIGITAL high pass filter because the poles of the digital filter cannot lie in the correct region, which is the left-half of the z-plane(z < 0) in this case.

The correct answer is (a) True

The EXPLANATION is: Since jΩ axis is not mapped to the circle |z|=1, we can EXPECT that the frequency RESPONSE H(ω) will be considerably DISTORTED with respect to H(jΩ).

Right option is (d) 0.5

The explanation: LETTING s=σ+jΩ in the equation z=1/(1-sT) and by letting σ=0, we get

|z-0.5|=0.5

Thus the image of the jΩ AXIS of the s-domain is a circle of RADIUS 0.5 centered at z=0.5 in z-domain.

The CORRECT OPTION is (c) Inside the circle |z-0.5|=0.5

For explanation: The left HALF of the s-plane is mapped inside the circle of |z-0.5|=0.5 in the z-plane, which completely lies in the RIGHT half z-plane.

Right CHOICE is (b) z=0.5

To explain: LETTING s=σ+jΩ in the equation z=1/(1-sT) and by letting σ=0, we get

|z-0.5|=0.5

Thus the IMAGE of the jΩ AXIS of the s-domain is a circle with CENTRE at z=0.5 in z-domain.

Correct choice is (b) 5

To explain: We know that the ORDER of a chebyshev-I filter is given by the equation,

N=cosh^-1(1/d)/cosh^-1(1/k)=4.3

Rounding off to the next large INTEGER, we GET N=5.

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

Explanation: A SIMPLE approximation to the FIRST ORDER derivative is given by the first backward difference. The first backward difference is defined by

[y(n)-y(n-1)]/T.

The correct option is (d) Ellipse

The BEST I can explain: The POLES of HN(s).HN(-s) is GIVEN by the characteristic equation 1+ϵ^2TN^2(s/j)=0.

The ROOTS of the above characteristic equation lies on ellipse, thus the poles of HN(s).HN(-s) are found to lie on ellipse.

Correct OPTION is (a) 1+ϵ^2TN^2(s/j)=0

For EXPLANATION: We know that for a chebyshev filter, we have

|H(jΩ)|=\(\frac{1}{\sqrt{1+ϵ^2 T_N^2(\frac{Ω}{Ω_P})}}\)

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

By replacing jΩ by s and consequently Ω by s/j in the above equation, we get

=>|HN(s)|^2=\(\frac{1}{1+ϵ^2 T_N^2 (s/j)}\)

The poles of the above equation is given by the equation 1+ϵ^2TN^2(s/j)=0 which is called as the characteristic equation.

Correct answer is (a) True

Explanation: In the PASS band of the frequency response of the low pass chebyshev-I FILTER, the sum of number of maxima and minima is EQUAL to the order of the filter.

Right option is (c) 3

Explanation: In the magnitude FREQUENCY response of a LOW pass chebyshev-I filter, the pass BAND has 3 maxima and 2 minima(ORDER 5=3 maxima+2 minima).

The CORRECT choice is (b) 2

To EXPLAIN: In the magnitude FREQUENCY RESPONSE of a low pass chebyshev-I filter, the pass BAND has 2 maxima and 2 minima(order 4=2 maxima+2 minima).

The correct option is (d) \(\FRAC{1}{\sqrt{1+ϵ^2 T_N^2 (\frac{Ω}{Ω_P})}}\)

The explanation is: The magnitude frequency response of a low pass CHEBYSHEV-I filter is GIVEN by

|H(jΩ)|=\(\frac{1}{\sqrt{1+ϵ^2 T_N^2(\frac{Ω}{Ω_P})}}\)

where ϵ is a parameter of the filter related to the ripple in the pass band and TN(x) is the N^th ORDER chebyshev polynomial.

Correct OPTION is (a) True

Explanation: There are TWO types of chebyshev filters. The Chebyshev-I filter is equi-ripple in the PASS band and monotonic in the stop band and the chebyshev-II filter is QUITE opposite.

The correct CHOICE is (b) NC
Explanation: The equi-ripple property of the chebyshev filter YIELDS a narrower transition band compared with that obtained when the magnitude response is monotone. As a consequence of this, the order of a chebyshev filter needed to ACHIEVE the GIVEN frequency DOMAIN specifications is usually lower than that of a Butterworth filter.

The CORRECT option is (c) 1

Explanation: We know that a chebyshev polynomial of degree N is DEFINED as

TN(x) = cos(Ncos^-1x), |x|≤1

cosh(Ncosh^-1x), |x|>1

Thus |TN(&PM;1)|=1.

The CORRECT option is (d) ±1

To explain I would say: We know that a CHEBYSHEV POLYNOMIAL of degree N is defined as

TN(X) = cos(Ncos^-1x), |x|≤1

cosh(Ncosh^-1x), |x|>1

For x=0, we have TN(0)=cos(Ncos^-10)=cos(N.π/2)=±1 for N even.

Right answer is (d) 4x^3-3x

The best I can explain: We know that a chebyshev POLYNOMIAL of degree N is defined as

TN(x) = cos(Ncos^-1x), |x|≤1; TN(x) = cosh(Ncosh^-1x), |x|>1

And the recursive formula for the chebyshev polynomial of order N is given as

TN(x)=2xTN-1(x)-TN-2(x)

Thus for a chebyshev filter of order 3, we obtain

T3(x)=2xT2(x)-T1(x)=2x(2x^2-1)-x=4x^3-3x.

The correct option is (b) x

Explanation: We know that a CHEBYSHEV POLYNOMIAL of degree N is DEFINED as

TN(x) = cos(Ncos^-1x), |x|≤1

cosh(Ncosh^-1x), |x|>1

For a degree 1 chebyshev filter, the polynomial is obtained as

T0(x)=cos(cos^-1x)=x.

Correct CHOICE is (b) 3

The EXPLANATION: Given information is

Ω1=2π*20=125.663 rad/sec

Ω2=2π*45*10^3=2.827*10^5 rad/sec

Ωu=2π*20*10^3=1.257*10^5 rad/sec

Ωl=2π*50=314.159 rad/sec

We know that

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

=> A=2.51 and B=2.25

Hence ΩS=Min{|A|,|B|}=> ΩS=2.25 rad/sec.

The order N of the NORMALIZED low pass Butterworth filter is COMPUTED as follows

N=\(\frac{log⁡[(10^{-K_P/10}-1)(10^{-K_s/10}-1)]}{2 log⁡(\frac{1}{Ω_S})}\)=2.83

Rounding off to the next large integer, we get, N=3.

Correct option is (c) 2.25 rad/sec

Explanation: Given INFORMATION is

Ω1=2π*20=125.663 rad/sec

Ω2=2π*45*103=2.827*105 rad/sec

Ωu=2π*20*103=1.257*105 rad/sec

Ωl=2π*50=314.159 rad/sec

We know that

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

=> A=2.51 and B=2.25

Hence ΩS=Min{|A|,|B|}=>ΩS=2.25 rad/sec.