CT Fourier Series

If input is $ (1 + j)cos(\frac{\pi}{2}t + \pi)sin(\frac{\pi}{4}t) $


$ = (1 + j)(\frac {e^{j \frac{\pi}{2}t + \pi j} + e^{-j \frac{\pi}{2}t - \pi j}}{2})(\frac{e^{j \frac{\pi}{4}t} - e^{-j \frac{\pi}{4}t}}{2j}) $

$ = \frac{1 + j}{4j}(e^{j \frac{\pi}{2}t}e^{\pi j} + e^{-j \frac{\pi}{2}t}e^{ \pi j})(e^{j \frac{\pi}{4}t} - e^{-j \frac{\pi}{4}t}) $

$ = \frac{1 + j}{4j}(e^{j \frac{\pi}{2}t}(1) + e^{-j \frac{\pi}{2}t}(-1))(e^{j \frac{\pi}{4}t} - e^{-j \frac{\pi}{4}t}) $

$ = \frac{1 + j}{4j}(e^{j \frac{3 \pi}{4}t} - e^{j \frac{\pi}{4}t} - e^{-j \frac{\pi}{4}t} + e^{-j \frac{3 \pi}{4}t}) $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood