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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 $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva