Revision as of 21:38, 23 September 2008 by Cdleon (Talk)

CT Signal & its Fourier coefficients

Lets define the signal

$ \ x(t) = (1+2j)cos(t)+5sin(4t) $

Knowing that its Fourier series is

$ \ x(t) = (1+2j)(\frac{e^{jt} + e^{-jt}}{2}) + 5(\frac{e^{j4t} - e^{-j4t}}{2j}) $

We simplify

$ \ x(t) = \frac{1+2j}{2} e^{jt} + \frac{1+2j}{2} e^{-jt} + 5 \frac{e^{j4t}}{2j} - 5\frac{e^{-j4t}}{2j} $

Alumni Liaison

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

Dr. Paul Garrett