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a_n &= \frac{1}{T} \int_{0}^T x(t) e^{-j \frac{2\pi}{T}nt}dt \\ | a_n &= \frac{1}{T} \int_{0}^T x(t) e^{-j \frac{2\pi}{T}nt}dt \\ | ||
& =\frac{1}{2} \int_{0}^2 x(t) e^{-j \frac{2\pi}{2}nt}dt \\ | & =\frac{1}{2} \int_{0}^2 x(t) e^{-j \frac{2\pi}{2}nt}dt \\ | ||
| − | & = | + | & =\frac{1}{2} \int_{0}^1 (1) e^{-j \frac{2\pi}{2}nt}dt \\ |
| + | & =\frac{1}{2} (e^{-j \pi n}-1) \\ | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 08:54, 14 September 2010
Exercise: Compute the Fourier series coefficients of the following signal:
$ x(t)=\left\{\begin{array}{ll}1&\text{ when } 0\leq t <1 \\ 0& \text{ when } 1\leq t <2\end{array} \right. $
x(t) periodic with period two.
After you have obtained the coefficients, write the Fourier series of x(t).
Answer
Can somebody finish the following computation?
$ \begin{align} a_n &= \frac{1}{T} \int_{0}^T x(t) e^{-j \frac{2\pi}{T}nt}dt \\ & =\frac{1}{2} \int_{0}^2 x(t) e^{-j \frac{2\pi}{2}nt}dt \\ & =\frac{1}{2} \int_{0}^1 (1) e^{-j \frac{2\pi}{2}nt}dt \\ & =\frac{1}{2} (e^{-j \pi n}-1) \\ \end{align} $
