1. Is the signal
- $ x(t) = \sum_{k = -\infty}^\infty \frac{1}{(t+2k)^{2}+1}\, $
periodic? The answer is yes because
- $ x(t+2) = \sum_{k = -\infty}^\infty \frac{1}{(t+2+2k)^2+1}\, $
- $ x(t+2) = \sum_{k = -\infty}^\infty \frac{1}{(t+2(k+1))^2+1}\, $
let $ r $ = $ k+1 $
- $ x(t+2) = \sum_{r = -\infty}^\infty \frac{1}{(t+2r)^2+1} = x(t)\, $
