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| − | + | == Answer == | |
| − | + | <math> a_n = \frac{1}{T}\int_{0}^{T}x(t)e^{-j\frac{2\pi}{T}n t} dt = \int_{0}^{1}x(t)e^{-j2\pi n t}</math> | |
| − | Substitute | + | Substitute <math>a=\pi, \;\; b=-j2\pi n</math> |
| − | + | <math> | |
| + | \begin{align} | ||
| + | a_n &= \int_{0}^{1}\text{sin}(at)e^{bt}dt = \left[\frac{1}{b}\text{sin}(at)e^{bt}\right]^{1}_{0} - \int_{0}^{1}frac{a\text{cos}(at)}{b}e^{bt}dt \\ | ||
| + | &= \left(\frac{e^b}{b}\text{sin}(a)\right) - \frac{a}{b}\left(\left[\frac{\text{cos}(at)}{b}e^{bt}\right]^{1}_{0} - \int_{0}^{1}\frac{-a\text{sin}(at)}{b}e^{bt}dt\right) \\ | ||
| + | &= \left(\frac{e^b}{b}\text{sin}(a)\right) - \frac{a}{b} \left( \left( \frac{\text{cos}(a)}{b}e^{b}-\frac{1}{b}\right) + \frac{a}{b}\int_{0}^{1}\text{sin}(at)e^{bt}dt \right) \\ | ||
| + | &= \frac{e^b}{a}\text{sin}(a)-\frac{a\text{cos}(a)}{b^2}e^b + \frac{a}{b^2} - \frac{a^2}{b^2}\int_{0}^{1}\text{sin}(at)e^{bt}dt \\ | ||
| + | \end{align} | ||
| + | </math> | ||
| − | As you can see | + | As you can see <math>\int_{0}^{1}\text{sin}(at)e^{bt}dt</math> term repeats, therefore it needs to be subtracted to both sides. |
| − | Then, | + | Then, <span class="texhtml">''a''<sub>''n''</sub></span> can be calculated. |
| − | + | <math> | |
| + | \frac{e^{-j2\pi n}}{-j2\pi n}\text{sin}\pi + \frac{\pi \text{cos}\pi}{4 \pi^2 n^2}e^{-j2\pi n} - \frac{\pi}{4 \pi^2 n^2} = \left(1-\frac{\pi^2}{4\pi^2 n^2}\right) \int_{0}^{1}\text{sin}(\pi t) e^{-j2\pi n t}dt = \left(1-\frac{\pi^2}{4\pi^2 n^2}\right) a_n | ||
| + | </math> | ||
| − | From this equation, | + | From this equation, <math>\frac{e^{-j2\pi n t}}{-j2\pi n}\text{sin}\pi=0</math> because of <span class="texhtml">sinπ</span>, and <math>\frac{\pi \text{cos}\pi}{4 \pi^2 n^2}e^{-j2\pi n} = \frac{-\pi}{4\pi^2 n^2}</math> |
| − | There are two terms of | + | There are two terms of <math>-\frac{\pi}{4\pi^2 n^2}</math>. |
| − | <br> | + | <br><math>a_n=\frac{-\frac{2\pi}{4\pi^2 n^2}}{1-\frac{\pi^2}{4\pi^2 n^2}} = -\frac{2\pi}{4\pi^2 n^2 - \pi^2} = \frac{2}{\pi(1-4n^2)}</math> |
Revision as of 08:36, 28 September 2010
Answer
$ a_n = \frac{1}{T}\int_{0}^{T}x(t)e^{-j\frac{2\pi}{T}n t} dt = \int_{0}^{1}x(t)e^{-j2\pi n t} $
Substitute $ a=\pi, \;\; b=-j2\pi n $
$ \begin{align} a_n &= \int_{0}^{1}\text{sin}(at)e^{bt}dt = \left[\frac{1}{b}\text{sin}(at)e^{bt}\right]^{1}_{0} - \int_{0}^{1}frac{a\text{cos}(at)}{b}e^{bt}dt \\ &= \left(\frac{e^b}{b}\text{sin}(a)\right) - \frac{a}{b}\left(\left[\frac{\text{cos}(at)}{b}e^{bt}\right]^{1}_{0} - \int_{0}^{1}\frac{-a\text{sin}(at)}{b}e^{bt}dt\right) \\ &= \left(\frac{e^b}{b}\text{sin}(a)\right) - \frac{a}{b} \left( \left( \frac{\text{cos}(a)}{b}e^{b}-\frac{1}{b}\right) + \frac{a}{b}\int_{0}^{1}\text{sin}(at)e^{bt}dt \right) \\ &= \frac{e^b}{a}\text{sin}(a)-\frac{a\text{cos}(a)}{b^2}e^b + \frac{a}{b^2} - \frac{a^2}{b^2}\int_{0}^{1}\text{sin}(at)e^{bt}dt \\ \end{align} $
As you can see $ \int_{0}^{1}\text{sin}(at)e^{bt}dt $ term repeats, therefore it needs to be subtracted to both sides.
Then, an can be calculated.
$ \frac{e^{-j2\pi n}}{-j2\pi n}\text{sin}\pi + \frac{\pi \text{cos}\pi}{4 \pi^2 n^2}e^{-j2\pi n} - \frac{\pi}{4 \pi^2 n^2} = \left(1-\frac{\pi^2}{4\pi^2 n^2}\right) \int_{0}^{1}\text{sin}(\pi t) e^{-j2\pi n t}dt = \left(1-\frac{\pi^2}{4\pi^2 n^2}\right) a_n $
From this equation, $ \frac{e^{-j2\pi n t}}{-j2\pi n}\text{sin}\pi=0 $ because of sinπ, and $ \frac{\pi \text{cos}\pi}{4 \pi^2 n^2}e^{-j2\pi n} = \frac{-\pi}{4\pi^2 n^2} $
There are two terms of $ -\frac{\pi}{4\pi^2 n^2} $.
$ a_n=\frac{-\frac{2\pi}{4\pi^2 n^2}}{1-\frac{\pi^2}{4\pi^2 n^2}} = -\frac{2\pi}{4\pi^2 n^2 - \pi^2} = \frac{2}{\pi(1-4n^2)} $
