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} $
Comments:
For $ {\color{red}n\neq 0} $:
$ \begin{align} a_n &= \frac{1}{T} \int_{0}^T x(t) e^{-j \frac{2\pi}{T}nt}dt {\color{OliveGreen}\surd}\\ & =\frac{1}{2} \int_{0}^2 x(t) e^{-j \frac{2\pi}{2}nt}dt {\color{OliveGreen}\surd} \\ & =\frac{1}{2} \int_{0}^1 (1) e^{-j \frac{2\pi}{2}nt}dt {\color{OliveGreen}\surd}\\ &=\frac{1}{2} \left. \frac{e^{-j \frac{2\pi}{2}nt}}{{\color{red} -j \pi n }} \right|_0^1\text { (note that one should not divide by n if }n=0)\\ & =\frac{1}{2} \frac{(e^{-j \pi n}-1)}{ {\color{red} -j \pi n }} \\ \end{align} $
Now deal with the case $ {\color{red}n= 0} $: separately:
$ {\color{red} \begin{align} a_0 &= \frac{1}{T} \int_{0}^T x(t) e^{-j \frac{2\pi}{T} 0 t}dt \\ & =\frac{1}{T} \int_{0}^T x(t) dt \\ &= \text{ average of } x(t) \text{ over one period} \\ &= \frac{1}{2} \end{align} } $
Comments:
Then, the Fourier series expression of the x(t) will be
$ x(t) = \sum_{n=-\infty}^{\infty} a_n e^{j \frac{2\pi}{T} n t } = \frac{1}{2} + \sum_{n=-\infty,n\neq0}^{\infty} \frac{(e^{-j \pi n}-1)}{2(-j \pi n)} e^{j \pi n t } $
--Jaemin 15:21, 22 September 2010 (UTC)
