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 $0$ $\frac{1}{2}$ $1$ $\frac{3}{2}$ Continuous-time Signals gate2012-ec continuous-time-signals signals-and-systems convolution + – Milicevic3306 asked Mar 25, 2018 • recategorized Nov 15, 2020 by gatecse Milicevic3306 16.0k points 91 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.