Revision as of 17:40, 23 September 2008 by Amelnyk (Talk)

Define a Periodic CT signal and compute its Fourier series coefficients

Consider the following CT signal:


x(t) such that

$ ak = \frac{1}{T} \int_{0}^{T} x(t) * e^{-j*k} * \frac{2*\pi}{T} *dt $


$ x(t) = cos(2* \pi * t) * cos(4* \pi * t) $


$ = \frac{e^{j*2*\pi*t} + e^{-j*2*\pi*t}}{2} $


$ = \frac{1*e^{j*6*\pi*t}}{4} + \frac{e^{-j*2*\pi*t}}{4} + \frac{e^{j*2*\pi*t}}{4} + \frac{e^{-j*6*\pi*t}}{4} $


K = 3

K= -1

K= 1

K= -3


$ a^{3} = \frac{1}{4} $

$ a^{-1} = \frac{1}{4} $

$ a^{1} = \frac{1}{4} $

$ a^{-3} = \frac{1}{4} $

Alumni Liaison

Prof. Math. Ohio State and Associate Dean
Outstanding Alumnus Purdue Math 2008

Jeff McNeal