Question 9.10: We will see that in the convolution sum we need to figure ou......
We will see that in the convolution sum we need to figure out how a signal x[n − k] behaves as a function of k for different values of n. Consider the signal
x[k]={\left\{\begin{array}{l l}{k}&{0\leq k\leq3,}\\ {0}&{{\mathrm{otherwise}}.}\end{array}\right.}
Obtain an expression for x[n − k] for −2 ≤ n ≤ 2 and determine in which direction it shifts as n increases from −2 to 2.
Although x[n − k], as a function of k, is reflected it is not clear if it is advanced or delayed as n increases from −2 to 2. If n = 0,
x[-k]=\left\{\begin{array}{c c c}{{-k}}&{{-3\le k\le0,}}\\ {{0}}&{{\mathrm{otherwise}.}}\end{array}\right.
For n ≠ 0, we have
x[n-k]=\left\{\begin{array}{l l}{{n-k}}&{{n-3\leq k\leq n,}}\\ {{0}}&{{\mathrm{otherwise.}}}\end{array}\right.
As n increases from −2 to 2, x[n − k] moves to the right. For n = −2 the support of x[−2 − k] is −5 ≤ k ≤ −2, while for n = 0 the support of x[−k] is −3 ≤ k ≤ 0, and for n = 2 the support of x[2 − k] is −1 ≤ k ≤ 2, each shifted to the right.