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)\, $

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett