Question

The convolution of $y\left[n\right]=\frac{1}{4^n}u\left[n\right]\ast u[n+2]$ is

• A. $(\frac{16}{3}-\frac{1}{4^n})u[n+2]$
• B. $(\frac{1}{3}-\frac{1}{4^n})u\left[n\right]$
• C. $(\frac{1}{3}-\frac{12}{4^n})u[n+2]$
• D. $(\frac{4}{3}-\frac{1}{48}{\left(\frac{1}{4}\right)}^n)u[n+2]$

Answer 1

The correct option is D $(\frac{4}{3}-\frac{1}{48}{\left(\frac{1}{4}\right)}^n)u[n+2]$

$y\left[n\right]={\sum }_{k=-\infty }^{\infty }{\left(\frac{1}{4}\right)}^{k}u\left[k\right]u[n-k+2]$

$\because u\left[k\right]=1fork\ge 0$

and u[n-k+2] = 1,

For n - k + 2 $\ge$ 0 (or) k $\ge$ n + 2

Hence, $y\left[n\right]={\sum }_{k=0}^{n+2}{\left(\frac{1}{4}\right)}^k=\frac{1-{\left(\frac{1}{4}\right)}^{n+3}}{1-\frac{1}{4}},n+2\ge 0$

$=\frac{4}{3}-\frac{1}{48}\times \frac{1}{4^n}$ $n\ge -2$

$\therefore y\left[n\right]=(\frac{4}{3}-\frac{1}{48}{\left(\frac{1}{4}\right)}^n)u[n+2]$