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       <math>= \frac{2T_{1}}{T}</math>
 
       <math>= \frac{2T_{1}}{T}</math>
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[[Problem 7 Part b_OldKiwi]]

Latest revision as of 16:25, 3 July 2008

Determine the Fourier Series co-efficient for the following continuous time periodic signals.Show the details of your calculations and simplify your answers.

Figure. OldKiwi.jpg

$ a_{k} = 1/T \int_{T} x(t) e^{-jkw_{o}t} dt $

     $ = 1/T \int_{-T_{1}} ^ {T_{1}} 1*e^{-jk2\frac{\pi}{T} t} dt $  (x(t)=1)
     
     $ = 1/T \int_{-T_{1}} ^ {T_{1}} e^{-jk2\frac{\pi}{T} t} dt $
     $ = 1/T [\frac{e^{-jk2\frac{\pi}{T} t}}{-jk2\frac{\pi}{T}}]_{-T_{1}} ^ {T_{1}} $
     $ = \frac{-1}{jk2\pi} (e^{-jk2\frac{\pi}{T} T_{1}} - e^{jk2\frac{\pi}{T} T_{1}}) $
     $ = \frac{1}{k\pi} (Sin(\frac{2k\pi}{T} T_{1}) $

Now For K = 0 Condition

$ a_{k} = 1/T \int_{T} x(t) e^{-jkw_{o}t} dt $

     PUT THE VALUE  K=0 IN ABOVE EQUATION
     $ = 1/T \int_{-T_{1}} ^ {T_{1}} 1 dt $
    
     $ = \frac{T_{1} + T_{1}}{T} $
     $ = \frac{2T_{1}}{T} $

Problem 7 Part b_OldKiwi

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