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<math>a_{0} = \frac{1}{2} \!</math> | <math>a_{0} = \frac{1}{2} \!</math> | ||
| + | |||
| + | Now using the formula for <math>a_{k} \!</math> given above, we can find the remaining <math>a_{k} \!</math>s. | ||
Revision as of 15:10, 14 October 2008
Most Difficult Problem on First Test
The problem that I found most difficult was problem number 4.
4. Compute the coefficients $ a_{k} \! $ of the Fourier series signal $ x(t) \! $ periodic with period $ T = 4 \! $ defined by
Solution
We know that
$ a_{k} = \frac{1}{T} \int_{0}^{T}x(t)e^{-jk\omega _{o}t}dt $, and since T = 4,
$ a_{k}= \frac{1}{4} \int_{0}^{4}x(t)e^{-jk\frac{\pi}{2}t}dt $
By the definition of $ a_{0} \! $, we know that it is the average of the signal over the period. In this case,
$ a_{0} = \frac{1}{2} \! $
Now using the formula for $ a_{k} \! $ given above, we can find the remaining $ a_{k} \! $s.
