Discrete Time System Implementation MCQ Quiz – Objective Question with Answer for Discrete Time System Implementation

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).

22. Resolve the sequence into a sum of weighted impulse sequences.

A. 2δ(n)+4δ(n-1)+3δ(n-3)
B. 2δ(n+1)+4δ(n)+3δ(n-2)
C. 2δ(n)+4δ(n-1)+3δ(n-2)
D. None of the mentioned

Show Explanation

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).

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 a Convolution sum.

The four operations that are needed to be done on h(k) in order to convolute x(k) and h(k) is

Step-1:Folding
Step-2:Multiplication with x(k)
Step-3:Shifting
Step-4:Summation

First, the signal h(k) is folded to get h(-k). Then it is shifted by n to get h(n-k). Then it is multiplied by x(k) and then summed over -∞ to ∞.

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}.

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+a2.1=1+a+a2

Similarly, y(n)=1+a+a2+….an=\(\frac{1-a^{n+1}}{1-a}\).

According to the properties of convolution, the Convolution of three signals obeys the Associative property.

Let h2(n) be shifted and folded.

so, h(k)=h1(n)*h2(n)=\(\sum_{k=-\infty}^{\infty} h_1 (k)h_2 (n-k)\)

For k<0, h1(n)= h2(n)=0 since the unit step function is defined only on the right hand side.

Therefore, h(k)=\((\frac{1}{2})^k (\frac{1}{4})^{n-k}\)

=>h(n)=\(\sum_{k=0}^n (\frac{1}{2})^k (\frac{1}{4})^{n-k}\)

=\((\frac{1}{4})^n \sum_{k=0}^n(2)^k\)

=\((\frac{1}{4})^n.(2^{n+1}-1)\)

=\((\frac{1}{2})^n[2-(\frac{1}{2})^n], n>0\)

According to the properties of the convolution, convolution exhibits the distributive property.

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.

i.e h(n)=0 for n<0 .