Question

Consider the discrete-time signal $x[n]=u[-n+5]-u[n+3]$, where $u[n]=\begin{cases}1,& n\ge 0\\0,&n<0\end{cases}$. The smallest $n$ for which $x[n]=0$ is ______.

Hint:

Turn each shifted unit step into an inequality, intersect the conditions, and then read off where the expression is zero/one.

Answer 1

Correct Answer: -3

Evaluate the unit steps: - $u[-n+5]=1$ when $-n+5\ge 0 \Rightarrow n\le 5$, else $0$.
- $u[n+3]=1$ when $n+3\ge 0 \Rightarrow n\ge -3$, else $0$.
Hence \[ x[n]=u[-n+5]-u[n+3]= \begin{cases} 0, & \text{if } n\le 5 \text{ and } n\ge -3\ (\!-3\le n\le 5),
1, & \text{if } n<-3,\\ -1, & \text{if } n>5. \end{cases} \] Therefore $x[n]=0$ on the interval $[-3,5]$. The {smallest} such $n$ is $-3$.