Question 9.4: Given the function (a) Sketch the function. (b) Determine it...
Given the function
f(x)=\begin{cases} 0, & for −5 ≤ x ≤ 0,\\ 3, & for 0 ≤ x ≤ 5 \end{cases} period 2l = 10
(a) Sketch the function.
(b) Determine its Fourier series.
(a)
f(x)=\begin{cases} 0, & for −5 ≤ x ≤ 0,\\ 3, & for 0 ≤ x ≤ 5 \end{cases} period 2l = 10
(b) For period 2l = 10 and l = 5, we choose the interval a to a + 2l to be −5 to 5, i.e., a =−5:
a_{n} = \frac{1}{l} \int\limits_{a}^{a+2l}{f (x) cos \frac{nπx}{l} dx} = \frac{1}{5} \int\limits_{-5}^{5}{f (x) cos \frac{nπx}{l}dx}= \frac{1}{5} \left\{\int\limits_{-5}^{0}{(0) cos \frac{nπx}{5} dx} + \int\limits_{0}^{5}{3 cos \frac{nπx}{5} dx} \right\}= \frac{3}{5} \int\limits_{0}^{5}{cos \frac{nπx}{5}dx}
= \frac{3}{5} \left\{\frac{5}{nπ} sin \frac{nπx}{5}\right\}|^{5}_{0}=0 for n ≠ 0.
For n = 0, one has a_{n} = a_{0} = (3/5) \int_{0}^{5}{cos(0πx/5)dx} = (3/5) \int_{0}^{5}{ dx} = 3.
Furthermore,
= \frac{1}{5} \left\{\int\limits_{-5}^{0}{(0) sin \frac{nπx}{5} dx} + \int\limits_{0}^{5}{3 sin \frac{nπx}{5} dx} \right\}= \frac{3}{5} \int\limits_{0}^{5}{sin \frac{nπx}{5}dx}
= \frac{3}{5} \left(− \frac{5}{nπ} cos \frac{nπx}{5}\right)|^{5}_{0} = \frac{3}{nπ} (1 −cos nπ).
Thus,
f (x) = \frac{3}{2}+ \sum\limits_{n=1}^{∞}{\frac{3}{nπ} (1 − cos nπ) sin \left(\frac{nπx}{5}\right)},i.e.,
f (x) = \frac{3}{2}+ \frac{6}{π} \left(sin \frac{πx}{5}+ \frac{1}{3}sin \frac{3πx}{5} + \frac{1}{5}sin \frac{5πx}{5}+· · ·\right).
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