Convolution of Sequences

Convolution is an operation performed on two sequences to produce a third sequence that expresses how the shape of one is modified by the other. In signal processing, it allows for the analysis and modification of signals in the time domain.

The formula for convolution is as follows:

y(n) = x1(n)*x2(n) = ∑k x1(k)x2(n-k)

where y(n) is the resulting sequence, x1(n) and x2(n) are the input sequences, and k is the index of summation.

Solving the Problem

In this problem, x1(n) = x2(n) = {1,1,1}. Therefore, we can rewrite the formula as:

y(n) = x1(n)*x2(n) = ∑k x1(k)x2(n-k) = ∑k x1(k)x2(n-k) = x1(0)x2(n) + x1(1)x2(n-1) + x1(2)x2(n-2)

Substituting the values of x1(n) and x2(n), we get:

y(n) = 1*1 + 1*1 + 1*1 = 3

Therefore, the resulting sequence is {3,3,3,3,3}.

However, this is not one of the options given in the question. Therefore, we need to perform circular convolution to obtain the correct answer.

Circular Convolution

Circular convolution is a type of convolution that is performed on two finite sequences by treating them as periodic signals. It is often used in signal processing to analyze signals that have a periodic nature.

To perform circular convolution, we first need to pad the sequences with zeros to make them the same length. In this case, we can pad the sequences with one zero each to obtain:

x1(n) = {1,1,1,0}
x2(n) = {1,1,1,0}

We can then compute the circular convolution using the formula:

y(n) = ∑k x1(k)x2((n-k)modN)

where N is the length of the sequences.

Substituting the values of x1(n) and x2(n), we get:

y(0) = x1(0)x2(0) + x1(1)x2(3) + x1(2)x2(2) + x1(3)x2(1) = 1*1 + 1*0 + 1*1 + 0*1 = 2
y(1) = x1(0)x2(1) + x1(1)x2(0) + x1(2)x2(3) + x1(3)x2(2) = 1*1 + 1*1 + 1*0 + 0*1 = 2
y(2) = x1(0)x2(2) + x1(1)x2(1) + x1(2)x2(0) + x1(3)x2(3) = 1*1 + 1*1 + 1*1 + 0*0 = 3
y(3) = x1(0)x2(3) + x1(1)x2(2) + x1(2)x