Convolution of Two Signals
To understand the given question, we need to first understand the concept of convolution of two signals. Convolution is a mathematical operation that combines two signals to produce a third signal. It is denoted by the symbol "*", and it is defined as follows:

y[n] = x[n] * h[n] = ∑(x[k] * h[n-k])

where y[n] is the output signal, x[n] is the first input signal, h[n] is the second input signal, and the summation is taken over all values of k.

Given Signals
In the given question, we have two signals:

x[n] = u[n - 3]
h[n] = u[n + 3]

where u[n] is the unit step function, defined as:

u[n] = 1, for n ≥ 0
u[n] = 0, for n < />

We need to find the convolution y[n] = x[n] * h[n].

Calculating the Convolution
To calculate the convolution, we substitute the given signals into the convolution formula:

y[n] = ∑(x[k] * h[n-k])

Substituting x[k] and h[n-k]:

y[n] = ∑(u[k - 3] * u[(n - k) + 3])

Simplifying the expression:

y[n] = ∑(u[k - 3] * u[n - k + 3])

We can split the summation into two parts:

y[n] = ∑(u[k - 3] * u[n - k + 3]) = ∑(u[k - 3]) * ∑(u[n - k + 3])

Now, let's analyze the two summations separately:

∑(u[k - 3]):
Since u[k - 3] is equal to 1 for k ≥ 3 and 0 for k < 3,="" the="" summation="" will="" />

∑(u[k - 3]) = 1 + 1 + 1 + ... = k - 2, for k ≥ 3

∑(u[n - k + 3]):
Similarly, u[n - k + 3] is equal to 1 for n - k + 3 ≥ 0, i.e., n ≥ k - 3, and 0 for n - k + 3 < 0,="" i.e.,="" n="" />< k="" -="" 3.="" therefore,="" the="" summation="" will="" />

∑(u[n - k + 3]) = 1 + 1 + 1 + ... = n - k + 4, for n ≥ k - 3

Final Convolution Expression
Substituting the summations back into the convolution expression:

y[n] = ∑(u[k - 3]) * ∑(u[n - k + 3]) = (k - 2) * (n - k + 4)

Simplifying further:

y[n] = (kn - 2k - nk + 4k - 2n + 8) = -n + (5k - 2)

Comparing with the given options:
a) (n - 1)u[n]
b) nu[n]
c) (n - 1)u[n]
d) u[n]

We can