Example of Computation of Fourier series of a CT SIGNAL

A practice problem on "Signals and Systems"


Let the signal be $ \ x(t) = \cos(2t\pi /3) \sin(2t \pi /9) $

which has a fundamental frequency of 9 Now computing its coefficients:

$ \ x(t)= (\frac{e^{j2t\pi /3} + e^{-j 2t\pi/ 3}}{2}) (\frac{e^{j 2t\pi /9} - e^{-j2t \pi /9}}{2j}) $


$ \ x(t)= \frac{e^{-j4t\pi /9} - e^{-j8t\pi /9} + e^{j8t\pi /9} - e^{j4t\pi /9}}{4j} $

Now rearranging the terms:

$ \ x(t)= \frac{-e^{j4t\pi /9} + e^{-j4t\pi /9} + e^{j8t\pi /9} - e^{j8t\pi /9}}{4j} $

$ \ x(t)= \frac{-e^{j(2)2t\pi /9} + e^{-j(2)2t\pi /9} + e^{j(4)2t\pi /9} - e^{j(4)2t\pi /9}}{4j} $

Now, the Fourier coefficients can clearly be seen:

$ \ a_{2}= \frac{-1}{4j} = \frac{j}{4} $

$ a_{-2}= \frac{1}{4j} = \frac{-j}{4} $

$ a_{4}= \frac{1}{4j} = \frac{-j}{4} $

$ a_{-4}= \frac{-1}{4j} = \frac{j}{4} $


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Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang