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Right option is (B) [4(0.5)^n-3(0.2)^n]u(n)

The best explanation: The system function is easily determined by taking the Z-transforms of x(n) and y(n). Thus we have

H(z)=\(\FRAC{Y(z)}{X(z)} = \frac{1+0.7z^{-1}}{1-0.7z^{-1}+0.1z^{-2}} = \frac{1+0.7z^{-1}}{(1-0.2z^{-1})(1-0.5z^{-1})}\)

Upon applying PARTIAL fractions and applying the INVERSE z-transform, we get

[4(0.5)^n-3(0.2)^n]u(n).

The correct option is (d) Some, but not all of the ZEROS are outside the unit CIRCLE

To explain I would say: For an IIR filter whose SYSTEM function is defined as H(Z)=\(\frac{B(z)}{A(z)}\) to be said a mixed phase and if the system is stable and causal, then the poles are INSIDE the unit circle and some, but not all of the zeros are outside the unit circle.

Correct answer is (C) Mixed phase

Explanation: Given H(Z)= 1-5/2z^-1-3/2z^-2

By factoring the SYSTEM function we FIND the zeros for the system.

The zeros of the given system are at z=-1/2, 3

So, the system is mixed phase.

Right answer is (a) All poles and zeros are inside the UNIT circle

For EXPLANATION I WOULD say: For an IIR FILTER whose system function is defined as H(z)=\(\frac{B(z)}{A(z)}\) to be said a minimum phase,

then both the poles and zeros of the system should fall inside the unit circle.

Correct option is (b) Maximum PHASE

The BEST EXPLANATION: Given H(z)=1-z^-1-6z^-2

By factoring the system FUNCTION we find the zeros for the system.

The zeros of the given system are at z=-2,3

So, the system is maximum phase.

The CORRECT option is (d) a^n

To elaborate: Given h(n)= δ(n)-aδ(n-1)

Since h(0)=1, h(1)=-a and h(n)=0 for n≥a, we have

hI(0)=1/h(0)=1.

and

hI(n)=-AHI(n-1) for n≥1

Consequently, hI(1)=a, hI(2)=a^2,….hI(n)=a^n

Which corresponds to a CAUSAL IIR system as expected.

Correct answer is (c) [h(n)*h1(n)]*X(n)=x(n)

For EXPLANATION I would say: If h(n) is the impulse RESPONSE of an LTI system T and h1(n) is the impulse response of the inverse system T^-1, then we know that h(n)*h1(n)=δ(n)=>[h(n)*h1(n)]*x(n)=x(n).

Right option is (b) δ(n)-1/2 δ(n-1)

Explanation: Given impulse RESPONSE is h(n)=(1/2)^nu(n)

The system function corresponding to h(n) is

H(z)=\(\frac{1}{1-\frac{1}{2} z^{-1}}\) ROC:|z|>1/2

This system is both stable and CAUSAL. Since H(z) is all POLE system, its inverse is FIR and is given by the system function

HI(z)=\(1-\frac{1}{2} z^{-1}\)

Hence its impulse response is δ(n)-1/2 δ(n-1).

Right option is (a) One-to-one CORRESPONDENCE between its input and output signals

The BEST I can EXPLAIN: If we KNOW the output of a system y(n) of a system and if we can determine the input x(n) of the system uniquely, then the system is said to be INVERTIBLE. That is there should be one-to-one correspondence between the input and output signals.

The correct OPTION is (a) True

Easy explanation: The given impulse response is h(n)=ASIN(n+1)ω0u(n).

According to the above equation, the second order system with COMPLEX conjugate poles on the unit CIRCLE is a sinusoid and the system is called a digital sinusoidal oscillator or a Digital FREQUENCY synthesizer.

The correct choice is (c) All PASS filter

The explanation is: The system with the system FUNCTION given as H(Z)=z^-k is a pure DELAY system. It has a constant gain for all frequencies and hence called as All pass filter.

Correct CHOICE is (d) Notch filter

The explanation is: The given FIGURE represents the frequency RESPONSE characteristic of a notch filter with nulls at FREQUENCIES at ω0 and ω1.

Right ANSWER is (a) True

Easiest explanation: A comb FILTER can be viewed as a notch filter in which the NULLS OCCUR periodically across the frequency band, hence the analogy to an ORDINARY comb that has periodically spaced teeth.

The correct answer is (a) True

Explanation: The magnitude response of a band PASS filter with two COMPLEX poles located near the unit circle is as shown below.

The filter gas a LARGE magnitude response at the poles and hence it is called as digital RESONATOR.

Right answer is (b) 0.1/(1+0.9e^-jω)

Explanation: The difference EQUATION for the high PASS filter is

y(n)=-0.9y(n-1)+0.1x(n)

and its frequency response is GIVEN as

H(ω)=0.1/(1+0.9e^-jω).

Correct option is (a) \(0.15\frac{1-z^{-2}}{1+0.7z^{-2}}\)

To explain: Clearly, the filter MUST have poles at P1,2=re^±jπ/2 and ZEROS at z=1 and z=-1. Consequently the system function is

H(z)=\(G\frac{(z-1)(z+1)}{(z-jr)(z+jr)} = G \frac{(z^2-1)}{(z^2+R^2)}\)

The gain factor is determined by EVALUATING the frequency response H(ω) of the filer at ω=π/2. Thus we have,

H(π/2) = \(G \frac{2}{1-r^2} = 1=>G = \frac{1-r^2}{2}\)

The value of r is determined by evaluating the H(ω) at ω=4π/9. Thus we have

|H(4π/9)|^2=\(\frac{(1-r^2)^2}{4}\frac{2-2cos⁡(8π/9)}{1+r^4+2r^2 cos⁡(8π/9)}\)=1/2

On SOLVING the above equation, we get r^2=0.7.Therefore the system function for the desired filter is

H(z)=\(0.15\frac{1-z^{-2}}{1+0.7z^{-2}}\)

The correct option is (C) 0.32

Explanation: Given

H(Z)=\(\frac{b_0}{(1-pz^{-1})^2}\) and we know that z=re^jω. Here in this CASE r=1. So z=e^jω.

Given at ω=0, H(0)=1=>B0=(1-p)^2

Given at ω=π/4, |H(π/4)|^2=1/2

=>\(\frac{(1-p)^2}{(1-pe^{-jπ/4})^2}\) = 1/2

=>\(\frac{(1-p)^4}{((1-p/\sqrt 2)^2+p^2/2)^2}\) = 1/2

=> √2(1-p)^2=1+p^2-√2p

Upon solving the above quadratic EQUATION, we get the value of p as 0.32.

Correct OPTION is (a) True

Best explanation: For an IDEAL filter, in the magnitude response PLOT at the stop BAND it should have a sudden FALL which means an ideal filter should have a zero gain at stop band.

Correct answer is (b) \(\frac{1}{(1-\frac{1}{2} e^{-jω})(1-\frac{1}{4} e^{-jω})}\)

The best I can explain: The FREQUENCY response FUNCTION of the system is

H(ω) = \(\sum_{n=0}^∞ (\frac{1}{2})^n e^{-jωn}\)

=\(\frac{1}{1-\frac{1}{2} e^{-jω}}\)

SIMILARLY, the input sequence x(n) has a Fourier transform

X(ω)=\(\sum_{n=0}^∞ (\frac{1}{4})^n e^{-jωn}\)

=\(\frac{1}{1-\frac{1}{4} e^{-jω}}\)

Hence the spectrum of the signal at the output of the system is

Y(ω)=X(ω)H(ω)

=\(\frac{1}{(1-\frac{1}{2} e^{-jω})(1-\frac{1}{4} e^{-jω})}\).

Correct option is (a) True

Best explanation: If x(N) is the input of an LTI system, then we know that the output of the system y(n) is y(n)= H(ω)x(n) which means the frequency components are just amplified but no NEW frequency components are ADDED.

Correct choice is (c) \(5+0.888sin(\FRAC{π}{2}n-420)-1.06cos(πn+\frac{π}{4})\)

To explain I WOULD say: From the given DIFFERENCE equation, we obtain

|H(ω)|=\(\frac{|b|}{\SQRT{1-2acosω+a^2}}\)

We get |H(0)|=1, |H(π/2)|=0.074 and |H(π)|=0.053

θ(0)=0, θ(π/2)=-420 and θ(π)=0 and we know that y(n)=H(ω)x(n)

=>y(n)=\(5+0.888sin(\frac{π}{2}n-42^0)-1.06cos(πn+\frac{π}{4})\)

The correct choice is (d) \(\frac{|b|}{\sqrt{1-2acosω+a^2}}\)

To explain I would say: Given y(n)=AY(n-1)+BX(n)

=>H(ω)=\(\frac{|b|}{1-ae^{-jω}}\)

By calculating the magnitude of the above EQUATION we get

|H(ω)|=\(\frac{|b|}{\sqrt{1-2acosω+a^2}}\)

The correct answer is (a) 20-\(\frac{10}{\sqrt{5}} sin(π/2n-26.60)+ \frac{40}{3}cosπn\)

The best I can explain: The frequency RESPONSE of the system is

H(ω)=\(\sum_{n=-∞}^∞ h(n) e^{-jωn} = \frac{1}{1-\frac{1}{2} e^{-jω}}\)

For first term, ω=0=>H(0)=2

For second term, ω=π/2=>H(π/2)=\(\frac{1}{1+j\frac{1}{2}} = \frac{2}{\sqrt{5}} e^{-j26.6°}\)

For third term, ω=π=> H(π)=\(\frac{1}{1+\frac{1}{2}}\) = 2/3

Hence the response of the system to x(n) is

y(n)=20-\(\frac{10}{\sqrt{5}} sin(π/2n-26.6^0)+ \frac{40}{3}cosπn\)

The CORRECT choice is (c) \(tan^{-1}\frac{H_I (ω)}{H_R (ω)}\)

The best explanation: If h(n) is the real VALUED impulse response sequence of an LTI SYSTEM, then H(ω) can be represented as HR(ω)+j HI(ω).

=> tanθ=\(\frac{H_I (ω)}{H_R (ω)}\) => Phase of H(ω)=\(tan^{-1}\frac{H_I (ω)}{H_R (ω)}\)

Right choice is (a) –\(\sum_{k=-∞}^∞ H(k) sin⁡ωk\)

To explain I WOULD say: From the DEFINITION of H(ω), we have

H(ω)=\(\sum_{k=-∞}^∞h(k) e^{-jωk}\)

=\(\sum_{k=-∞}^∞h(k) cos⁡ωk-j\sum_{k=-∞}^∞h(k) sin⁡ωk\)

= HR(ω)+j HI(ω)

=> HI(ω)=-\(\sum_{k=-∞}^∞h(k) sin⁡ωk\)

The correct answer is (b) \(\frac{2}{\sqrt{5}} Ae^{j(\frac{nπ}{2}-26.6°)}\)

To elaborate: First we evaluate the FOURIER transform of the impulse response of the system h(n)

H(ω)=\(\sum_{n=-∞}^∞ h(n) e^{-jωn} = \frac{1}{1-1/2 e^{-jω}}\)

At ω=π/2, the above equation yields,

H(π/2)=\(\frac{1}{1+j 1/2}=\frac{2}{\sqrt{5}} e^{-j26.6°}\)

We KNOW that if the input signal is a complex exponential signal, then y(n)=x(n) . H(ω)

=>y(n)=\(\frac{2}{\sqrt{5}} Ae^{j(\frac{nπ}{2}-26.6°)}\)

The correct option is (a) True

For EXPLANATION: An Eigen FUNCTION of a system is an input SIGNAL that produces an output that DIFFERS from the input by a constant multiplicative FACTOR known as Eigen value of the system.

Right answer is (c) H(ω)x(N)

To explain I would say: If x(n)= AE^jωn is the input and h(n) is the response o the system, then we KNOW that

y(n)=\(\sum_{k=-∞}^∞ h(k)x(n-k)\)

=>y(n)=\(\sum_{k=-∞}^∞ h(k)Ae^{jω(n-k)}\)

= A \([\sum_{k=-∞}^∞ h(k) E^{-jωk}] e^{jωn}\)

= A. H(ω). e^jωn

= H(ω)x(n)

The CORRECT choice is (b) \(\frac{1}{1-2acosω+a^2}\)

The EXPLANATION: Given x(n)= a^nu(n), |a|<1

The auto correlation of the above SIGNAL is

rxx(l)=\(\frac{1}{1-a^2}\) a^|l|, -∞< l <∞

According to Wiener-Khintchine Theorem,

Sxx(ω)=F{rxx(l)}=\([\frac{1}{1-a^2}]\).F{a^|l|} = \(\frac{1}{1-2acosω+a^2}\)

Right answer is (B) \(\frac{1-a^2}{1-2acosω+a^2}\)

Explanation: First we observe x(N) can be expressed as

x(n)=x1(n)+x2(n)

where x1(n)= a^n, n>0

=0, elsewhere

x2(n)=a^-n, n<0

=0, elsewhere

Now applying Fourier transform for the above TWO signals, we get

X1(ω)=\(\frac{1}{1-ae^{-jω}}\) and X2(ω)=\(\frac{ae^{jω}}{1-ae^{jω}}\)

Now, X(ω)=X1(ω)+ X2(ω)=\(\frac{1}{1-ae^{-jω}}+\frac{ae^{jω}}{1-ae^{jω}}=\frac{1-a^2}{1-2acosω+a^2}\).

Right option is (c) A\(\FRAC{sin⁡(M+\frac{1}{2})ω}{sin⁡(\frac{ω}{2})}\)

For explanation: CLEARLY, x(n)=x(-n). THUS the signal x(n) is real and even signal. So, we know that

\(X(ω)=X_R(ω)=A(1+2∑_{n=1}^∞ cos⁡ωn)\)

On simplifying the above equation, we obtain

X(ω)=A\(\frac{sin⁡(M+\frac{1}{2})ω}{sin⁡(\frac{ω}{2})}\).

Correct answer is (d) \(\frac{-asinω}{1-2acosω+a^2}\)

EASIEST explanation: Given, X(ω)=\(\frac{1}{1-ae^{-jω}}\), |a|<1

By multiplying both the numerator and DENOMINATOR of the above equation by the complex conjugate of the denominator, we obtain

X(ω)=\(\frac{1-ae^{jω}}{(1-ae^{-jω})(1-ae^{jω})} = \frac{1-acosω-jasinω}{1-2acosω+a^2}\)

This expression can be SUBDIVIDED into REAL and imaginary parts, THUS we obtain

XI(ω)=\(\frac{-asinω}{1-2acosω+a^2}\).

Right CHOICE is (C) \(\FRAC{1-acosω}{1-2acosω+a^2}\)

The explanation is: Given, X(ω)=\(\frac{1}{1-ae^{-jω}}\), |a|<1

By multiplying both the numerator and DENOMINATOR of the above EQUATION by the complex conjugate of the denominator, we obtain

X(ω)=\(\frac{1-ae^{jω}}{(1-ae^{-jω})(1-ae^{jω})} = \frac{1-acosω-jasinω}{1-2acosω+a^2}\)

This expression can be subdivided into real and imaginary parts, thus we obtain

XR(ω)=\(\frac{1-acosω}{1-2acosω+a^2}\).

The CORRECT choice is (b) –\(\frac{1}{π} \int_0^π\)[XI(ω) sinωn] dω

The EXPLANATION: If x(N) is REAL and odd then, x(n)cosωn is odd and x(n) sinωn is even. Consequently

XR(ω)=0

XI(ω)=\(-2\sum_{n=1}^∞ x(n) sin⁡ωn\)

=>x(n)=-\(\frac{1}{π} \int_0^π\)[XI(ω) sinωn] dω

Right answer is (a) True

To explain I WOULD say: We know that if x(n) is a REAL signal, then XI(n)=0 and xR(n)=x(n)

We know that, xR(n)=x(n)=\(\FRAC{1}{2π}\int_0^{2π}\)[XR(ω) cosωn- XI(ω) sinωn] dω

Since both XR(ω) cosωn and XI(ω) sinωn are even, x(n) is also even

=> x(n)=\(\frac{1}{π} \int_0^π\)[XR(ω) cosωn- XI(ω) sinωn] dω