Question 9.9: A triangular discrete pulse is defined as x[n] = { n 0 ≤ n ≤......
A triangular discrete pulse is defined as
x[n]={\left\{\begin{array}{l l}{n}&{0\leq n\leq10,}\\ {0}&{{\mathrm{otherwise}}.}\end{array}\right.}
Find an expression for y[n] = x[n + 3] + x[n − 3] and z[n] = x[−n] + x[n] in terms of n and carefully plot them.
Replacing n by n + 3 and by n − 3 in the definition of x[n] we get the advanced and delayed signals
x[n+3]=\left\{\begin{array}{l l}{{n+3}}&{{-3\leq n\leq7,}}\\ {{0}}&{{\mathrm{otherwise},}}\end{array}\right.
and
x[n-3]=\left\{\begin{array}{l l}{{n-3}}&{{3\leq n\leq13,}}\\ {{0}}&{{\mathrm{otherwise},}}\end{array}\right.
so that when added we get
y[n] = x[n + 3] + x[n − 3]=\left\{\begin{array}{l l}{{n+3}}&{{-3\leq n\leq2,}}\\ {{2n}}&{{3\leq n\leq7,}}\\ {{n-3}}&{{8\leq n\leq13,}}\\ {{0}}&{{\mathrm{otherwise.}}}\end{array}\right.
Likewise, we have
z[n] = x[n] + x[−n]=\left\{\begin{array}{r l}{{n}}&{{1\leq n\leq10,}}\\ {{0}}&{{n=0,}}\\ {{-n}}&{{-10\leq n\leq-1,}}\\ {{0}}&{{\mathrm{otherwise.}}}\end{array}\right.
The results are shown in Fig. 9.2.
