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<math>= \frac{1}{k\pi} (Sin(\frac{2k\pi}{T} T_{1})</math> | <math>= \frac{1}{k\pi} (Sin(\frac{2k\pi}{T} T_{1})</math> | ||
| + | |||
| + | Now For K = 0 Condition | ||
| + | |||
| + | <math>a_{k} = 1/T \int_{T} x(t) e^{-jkw_{o}t} dt</math> | ||
| + | |||
| + | PUT THE VALUE K=0 IN ABOVE EQUATION | ||
| + | |||
| + | <math>= 1/T \int_{-T_{1}} ^ {T_{1}} 1 dt</math> | ||
| + | |||
| + | <math>= \frac{T_{1} + T_{1}}{T}</math> | ||
| + | |||
| + | <math>= \frac{2T_{1}}{T}</math> | ||
Revision as of 17:08, 30 June 2008
Determine the Fourier Series co-efficient for the following continuous time periodic signals.Show the details of your calculations and simplify your answers.
$ 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} $
