0 0 votes A signal $m(t)$ with bandwidth $500 \mathrm{~Hz}$ is first multiplied by a signal $g(t)$ where \[ g(t)=\sum_{\mathrm{R}=-\infty}^{\infty}(-1)^{k} \delta\left(t-0.5 \times 10^{-4} k\right) \] The resulting signal is then passed through an ideal lowpass filter with bandwidth $1 \; \mathrm{kHz}$. The output of the lowpass filter would be $\delta(t)$ $m(t)$ $0$ $m(t) \delta(t)$ Continuous-time Signals gate2006-ec signals-and-systems analog-communications continuous-time-signals + – admin 275 views answer comment Share Follow Add Sync Questions Print 0 reply Please log in or register to add a comment.
