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Right CHOICE is (c) y(n)=-a1y(n-1)-A2Y(n-2)+ b0x(n)+b1x(n-1)+b2x(n-2)

The best explanation: The equation of the difference equation of any system is DEFINED as

y(n)=\(-\sum_{k=1}^Na_k y(n-k)+\sum_{k=0}^{M}b_k X(n-k)\)

In the GIVEN diagram, N=M=2

So, substitute the values of the N and M in the above equation.

We get, y(n)=-a1y(n-1)-a2y(n-2)+b0x(n)+b1x(n-1)+b2x(n-2)

The correct choice is (b) PURELY RECURSIVE system

Easy EXPLANATION: SINCE the output of the system depends only on the present value of the INPUT and the past values of the output, the system is a purely recursive system.

Right OPTION is (b) rhh(L)*RXX(l)

To explain I WOULD SAY: ryy(l)=y(l)*y(-l)

=[h(l)*x(l)]*[h(-l)*x(-l)]

=[h(l)*h(-l)]*[x(l)*x(-l)]

=rhh(l)*rxx(l).

Correct answer is (C) ryx(L)=h(l)*rxx(l)

The BEST I can explain: Let x(n) be the input signal and y(n) be the output signal with impulse response h(n).

We know that y(n)=x(n)*h(n)=\(\sum_{K=-\INFTY}^{\infty} x(k)h(n-k)\)

The cross correlation between the input signal and output signal is

ryx(l)=y(l)*x(-l)=h(l)*[x(l)*x(-l)]=h(l)*rxx(l).

Right choice is (a) True

Best EXPLANATION: Let US CONSIDER a signal x(N) WHOSE auto correlation is defined as rxx (l).

We know that, for auto correlation sequence rxx (l)=rxx (-l).

So, auto correlation sequence is an even sequence.

Right choice is (d) All of the mentioned

Explanation: RXX(l)=\(\sum_{n=-\infty}^{\infty} x(n)x(n-l)\)

For l≥0, rxx(l)=\(\sum_{n=l}^{\infty} x(n)x(n-l)\)

=\(\sum_{n=l}^{\infty} a^n a^{n-l}\)

=\(a^{-l}\sum_{n=l}^{\infty} a^{2N}\)

=\(\frac{1}{1-a^2}a^l\)(l≥0)

For l<0, rxx(l)=\(\sum_{n=0}^{\infty} x(n)x(n-l)\)

=\(\sum_{n=0}^\infty a^n a^{n-l}\)

=\(a^{-l}\sum_{n=0}^{\infty} a^{2n}\)

=\(\frac{1}{1-a^2}a^{-l}\)

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

Right CHOICE is (a) \(\frac{r_{xx}(L)}{r_{xx}(0)}\)

The best explanation: If the SIGNAL involved in auto CORRELATION is scaled, the shape of auto correlation does not change, only the amplitudes of auto correlation sequence are scaled accordingly. Since scaling is unimportant, it is often desirable, in practice, to normalize the auto correlation sequence to the range from -1 to 1. In the case of auto correlation sequence, we can simply divide by rxx (0). Thus the NORMALIZED auto correlation sequence is defined as ρxx(l)=\(\frac{r_{xx}(l)}{r_{xx}(0)}\).

Right answer is (C) a^2rxx(0)+b^2ryy(0)+2abrxy(1)

For explanation I would say: The energy SIGNAL of the signal AX(n)+by(n-l) is

\(\sum_{n=-\INFTY}^{\infty}[ax(n)+by(n-l)]^2\)

= \(a^2 \sum_{n=-\infty}^{\infty}x^2(n)+b^2 \sum_{n=-\infty}^{\infty}y^2(n-l)+2ab \sum_{n=-\infty}^{\infty}x(n)y(n-l)\)

= a^2rxx(0)+b^2ryy(0)+2abrxy(l)

Correct OPTION is (d) |RXY(l)|≤\(\SQRT{r_{xx}(0).r_{yy}(0)}\)

For explanation: We know that, a^2rxx(0)+b^2ryy(0)+2abrxy(l) ≥0

=> (a/b)^2rxx(0)+ryy(0)+2(a/b)rxy(l) ≥0

Since the quadratic is nonnegative, it follows that the DISCRIMINATE of this quadratic MUST be non positive, that is 4[r^2xy(l)- rxx(0) ryy(0)] ≤0 => |rxy(l)|≤\(\sqrt{r_{xx}(0).r_{yy}(0)}\).

The CORRECT choice is (a) rxy(l)=x(l)*x(-l)

Best explanation: We know that, the correlation of TWO SIGNALS x(n) and y(n) is rxy(l)=\(\sum_{n=-\infty}^{\infty} x(n)y(n-1)\)

Let x(n)=y(n) => RXX(l)=\(\sum_{n=-\infty}^{\infty}x(n)x(n-l)\) = x(l)*x(-l)

Correct CHOICE is (d) All of the mentioned

Explanation: LET us consider x(n) be the input reference signal and y(n) be the reflected signal.

Now, the relation between the two signals is given as y(n)=αx(n-D)+w(n)

Where α-attenuation factor representing the signal loss in the round-trip transmission of the signal x(n)

D-time delay between the time of projection of signal and the reflected back signal

w(n)-NOISE signal generated in the electronic parts in the front end of the RECEIVER.

The CORRECT option is (d) Non-stable

Easiest explanation: An arbitrary relaxed system is said to be BIBO stable if it has a bounded output for EVERY value in the bounded input. So, the system GIVEN in the QUESTION is a Non-stable system.

The correct option is (c) Causal

To elaborate: A SYSTEM is said to be causal if the output of the system is defined as the FUNCTION shown below

y(N)=F[X(n),x(n-1),x(n-2),…]

So, according to the conditions GIVEN in the question, the system is a causal system.

The CORRECT choice is (b) False

To elaborate: For n=-1, y(-1)=X(1)

That is, the output of the SYSTEM at n=-1 is depending on the FUTURE value of the INPUT at n=1. So the system is a non-causal system.

The CORRECT choice is (b) False

The explanation: If the INPUT is DELAYED by k units then the output will be y(N,k)=x(n-k)-x(n-k-1)

If the output is delayed by k units then y(n-k)=x(n-k)-x((n-k)-1)

=>y(n,k)=y(n-k). Hence the system is time-invariant.

The correct answer is (a) Static SYSTEM

The best I can explain: Since the output of the system y(n) depends only on the present value of the input X(n) but not on the past or the FUTURE values of the input, the system is called as static or memory-less system.

The CORRECT OPTION is (a) Accumulator

For explanation I would say: From the equation given, y(n)=x(n)+x(n-1)+x(n-2)+…. .This system calculates the RUNNING sum of all the past input values TILL the present time. So, it acts as an accumulator.

Right CHOICE is (b) \(\frac{n(n+1)}{2}\)

The best explanation: GIVEN that the system is initially relaxed, that is y(-1)=0

According to the equation of the accumulator,

y(n)=\(∑_{k=-∞}^n X(n)\)

=\(∑_{k=-∞}^{-1} x(n)+∑_{k=0}^n x(n)\)

=\(y(-1)+ ∑_{k=0}^n n*u(n)\)

=\(0+∑_{k=0}^n n\)(since u(n)=1 in 0 to n)

=\(\frac{n(n+1)}{2}\)

The correct answer is (d) Impulse function

For explanation: ACCORDING to the definition of the impulse function, it is DEFINED only at n=0 and is not defined elsewhere which is as PER the signal GIVEN.

The CORRECT choice is (d) None of the mentioned

To explain I WOULD SAY: For example, the value of the output at the time n=0 is y(0)=x(1), that is the system is advanced by one unit.

Right choice is (C) (0,1,2,3)

Explanation: An identical system is a system whose output is same as the INPUT, that is it does not perform any OPERATION on the input and TRANSMITS it.

The CORRECT option is (b) 01

Easiest EXPLANATION: When the value of ‘r’ lies between 0 and 1 then the value of X(n) goes on decreasing exponentially with INCREASE in value of ‘n’. So, the signal is called as decaying exponential signal.

The CORRECT choice is (a) Down-sampling

Explanation: If the signal x(N) was originally obtained by sampling a signal xa(t), then x(n)=xa(nT). Now, y(n)=x(2n)(say)=xa(2nT). Hence the time scaling operation is equivalent to changing the sampling rate from 1/T to 1/2T, that is to DECREASE the rate by a factor of 2. So, time scaling is ALSO CALLED as down-sampling.

The correct answer is (B) x(n)=-x(-n)

EASY explanation: ACCORDING to the definition of anti-symmetric signal, the signal x(n) should be symmetric over origin. So, for the signal x(n) to be symmetric, it should SATISFY the condition x(n)=-x(-n).

The CORRECT ANSWER is (d) (1/2)*(x(t)-x(-t))

To elaborate: Let x(t)=xe(t)+XO(t)

=>x(-t)=xe(-t)-xo(-t)

By subtracting the above two equations, we get

xo(t)=(1/2)*(x(t)-x(-t)).

Correct answer is (c) x(K)*δ(n-k)

EASIEST EXPLANATION: The given signal is defined only when n=k by the definition of delta FUNCTION. So, x(n)*δ(n-k)= x(k)*δ(n-k).

Right choice is (c) Unit ramp signal

Best explanation: When we PLOT the GRAPH for the given function, we get a straight line PASSING through ORIGIN with a unit positive SLOPE. So, the function is called a unit ramp signal.

The correct answer is (b) nθ

Best explanation: Given X(n)=a^n=(r.e^jθ)^n = r^n.e^jnθ

=>x(n)=r^n.(cosnθ+jsinnθ)

PHASE FUNCTION is TAN^-1(cosnθ/sinnθ)=tan^-1(tan nθ)=nθ.

Right OPTION is (d) Not defined

Easy EXPLANATION: For a DISCRETE time signal, the value of x(n) exists only at integral VALUES of n. So, for a non- integer value of ‘n’ the value of x(n) does not exist.

Right choice is (b) False

To explain: Let S=\(\sum_{N=-{\INFTY}}^{\infty}|H(n)|\)

=\(\sum_{n=-{\infty}}^{\infty}2^n u(n-1)\)

=\(\sum_{n=-{\infty}}^{\infty}2^n\)

=2+4+8+…=∞

So, the system is not STABLE.

The correct ANSWER is (d) Impulse response is zero for negative values of n

To explain: Let us consider a LTI system having an output at time n=n0 GIVEN by the convolution formula

y(n)=\(\sum_{k=-{\infty}}^{\infty}h(k)X(n_0-k)\)

We split the summation into two INTERVALS.

=>y(n)=\(\sum_{k=-{\infty}}^{-1}h(k)x(n_0-k)+\sum_{k=0}^{\infty}h(k)x(n_0-k)\)

=(h(0)x(n0)+h(1)x(n0-1)+h(2)x(n0-2)+….)+(h(-1)x(n0+1)+h(-2)x(n0+2)+…)

As PER the definition of the causality, the output should depend only on the present and past values of the input. So, the coefficients of the terms x(n0+1), x(n0+2)…. should be equal to zero.

that is, h(n)=0 for n<0 .

Right CHOICE is (a) True

Easiest EXPLANATION: ACCORDING to the properties of convolution, Convolution of three signals OBEYS Associative PROPERTY.

Right choice is (a) True

Best EXPLANATION: ACCORDING to the properties of the CONVOLUTION, convolution exhibits distributive property.

Right CHOICE is (B) X(n-n0)

Easy explanation: x(n)*δ(n-n0)=\(\sum_{K=-{\infty}}^{\infty} x(k)\delta(n-k-n_0)\)

=x(k)|k=n-n0

=x(n-n0)

Correct option is (a) \(\frac{1-a^{N+1}}{1-a}\)

Easiest explanation: Now fold the signal x(n) and shift it by ONE UNIT at a time and SUM as follows

y(0)=x(0)h(0)=1

y(1)=h(0)x(1)+h(1)x(0)=1.1+a.1=1+a

y(2)=h(0)x(2)+h(1)x(1)+h(2)x(0)=1.1+a.1+a^2.1=1+a+a^2

Similarly, y(n)=1+a+a^2+….a^n=\(\frac{1-a^{n+1}}{1-a}\).

The correct OPTION is (c) {1,3,6,5,3}

The EXPLANATION: Let y(n)=x(n)*h(n)(‘*’ symbol indicates CONVOLUTION symbol)

From the formula of convolution we get,

y(0)=x(0)h(0)=1.1=1

y(1)=x(0)h(1)+x(1)h(0)=1.1+2.1=3

y(2)=x(0)h(2)+x(1)h(1)+x(2)h(0)=1.1+2.1+3.1=6

y(3)=x(1)h(2)+x(2)h(1)=2.1+3.1=5

y(4)=x(2)h(2)=3.1=3

Therefore, y(n)=x(n)*h(n)={1,3,6,5,3}.

The correct answer is (a) CONVOLUTION sum

Easiest EXPLANATION: The input X(n) is CONVOLUTED with the impulse response h(n) to yield the OUTPUT y(n). As we are summing the different values, we call it as Convolution sum.

The correct ANSWER is (b) 2δ(n+1)+4δ(n)+3δ(n-2)

To ELABORATE: We know that, x(n)δ(n-K)=x(k)δ(n-k)

x(-1)=2=2δ(n+1)

x(0)=4=4δ(n)

x(2)=3=3δ(n-2)

Therefore, x(n)= 2δ(n+1)+4δ(n)+3δ(n-2).

Correct option is (d) [-\(\frac{1}{5}\) (-1)^N+\(\frac{6}{5}\) (4)^n]U(n)

Easy explanation: The homogenous solution of the GIVEN equation is yh(n)=C1(-1)^n+C2(4)^n—-(1)

To find the impulse response, x(n)=δ(n)

now, for n=0 and n=1 we get

y(0)=1 and

y(1)=3+2=5

From equation (1) we get

y(0)=C1+C2 and

y(1)=-C1+4C2

On solving the above two set of equations we get

C1=-\(\frac{1}{5}\) and C2=\(\frac{6}{5}\)

=>h(n)= [-\(\frac{1}{5}\) (-1)^n + \(\frac{6}{5}\) (4)^n]u(n).

The correct answer is (b) yp(N)+YH(n)

For explanation I would say: The linearity property of the linear constant coefficient difference equation allows us to add the HOMOGENEOUS and particular SOLUTION in order to obtain the total solution.

Correct choice is (c) \(\frac{8}{5}\) 2^n

To explain I would say: The assumed solution of the difference equation to the forcing equation x(n), called the particular solution of the difference equation is

yp(n)=Kx(n)=K2^nu(n) (where K is a scale factor)

Upon substituting yp(n) into the difference equation, we obtain

K2^nu(n)=\(\frac{5}{6}\)K2^n-1u(n-1)-\(\frac{1}{6}\) K2^n-2u(n-2)+2^nu(n)

To DETERMINE K we must evaluate the above equation for any n>=2, so that no term vanishes.

=> 4K=\(\frac{5}{6}\)(2K)-\(\frac{1}{6}\) (K)+4

=> K=\(\frac{8}{5}\)

=> yp(n)=(8/5) 2^n.

Correct choice is (b) (-1)^n+1 + (4)^n+2

To ELABORATE: Given difference equation is y(n)-3Y(n-1)-4y(n-2)=0—-(1)

Let y(n)=λ^n

Substituting y(n) in the given equation

=> λ^n – 3λ^n-1 – 4λ^n-2 = 0

=> λ^n-2(λ^2 – 3λ – 4) = 0

the roots of the above equation are λ=-1,4

Therefore, general form of the solution of the HOMOGENOUS equation is

yh(n)=C1 λ1^n+C2 λ2^n

=C1(-1)^n+C2(4)^n—-(2)

The zero-input response of the system can be calculated from the homogenous solution by evaluating the constants in the above equation, given the INITIAL conditions y(-1) and y(-2).

From the given equation (1)

y(0)=3y(-1)+4y(-2)

y(1)=3y(0)+4y(-1)

=3[3y(-1)+4y(-2)]+4y(-1)

=13y(-1)+12Y(-2)

From the equation (2)

y(0)=C1+C2 and

y(1)=C1(-1)+C2(4)=-C1+4C2

By equating these two set of relations, we have

C1+C2=3y(-1)+4y(-2)=15

-C1+4C2=13y(-1)+12y(-2)=65

On solving the above two equations we get C1=-1 and C2=16

Therefore the zero-input response is Yzi(n) = (-1)^n+1 + (4)^n+2.

The correct choice is (c) c(-a)^N

To explain I would SAY: The assumed solution obtained by assigning X(n)=0 is

YH(n)=λ^n

=>y(n)+ay(n-1)=0

=>λ^n+a λ^n-1=0

=>λ^n-1(λ+a)=0

=>λ=-a

=>yh(n)=cλ^n=c(-a)^n