in Continuous-time Signals recategorized by
45 views
0 votes
0 votes

Let $y[n]$ denote the convolution of $h[n]$ and $g[n]$, where $h[n]=(\frac{1}{2})^nu[n]$ and $g[n]$ is a casual sequence. If $y[0]=1$ and $y[1]=\frac{1}{2}$, then $g[1]$ equals

  1. $0$
  2. $\frac{1}{2}$
  3. $1$
  4. $\frac{3}{2}$
in Continuous-time Signals recategorized by
15.8k points
45 views

Please log in or register to answer this question.

Welcome to GO Electronics, where you can ask questions and receive answers from other members of the community.