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2015-05-03T01:32:24+05:30

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Given signal function in time domain is:
f(x) = e^{-ax}\ \ \ -\pi \le x \le \pi

Fourier series for a function f(x) is defined as:

f(x)=\frac{1}{2}a_0\ +\ \Sigma_{n=1}^{\infty}\ a_n\ Sin(nx)\ +\ \Sigma_{n=1}^{\infty}\ b_n\ Sin(nx)\\\\a_0=\frac{1}{\pi} \int\limits^{\pi}_{-\pi} {f(x)} \, dx\\\\a_n=\frac{1}{\pi} \int\limits^{\pi}_{-\pi} {f(x)\ Cos(nx)} \, dx\\\\b_n=\frac{1}{\pi} \int\limits^{\pi}_{-\pi} {f(x)\ Sin(nx)} \, dx

Hence, let us compute now the coefficients of various frequency components:

a_0=\frac{1}{\pi} \int\limits^{\pi}_{-\pi} {e^{-ax}} \, dx=\frac{1}{\pi a}[ -e^{-ax} ]_{-\pi}^{\pi}=\frac{e^{\pi a}-e^{-\pi a}}{\pi a}\\

a_n=\frac{1}{\pi} \int\limits^{\pi}_{-\pi} {e^{-ax}\ Cos(nx)} \, dx\\\\= \frac{1}{n\pi} [ e^{-ax} Sin(nx)]_{-\pi}^{\pi}  + \frac{a}{n\pi} \int\limits^{\pi}_{-\pi} {e^{-ax}\ Sin(nx)} \, dx\\\\=0- [ \frac{a}{n^2 \pi}\ e^{-a x}\ Cos(n x) ]_{-\pi}^{\pi} - \frac{a^2}{n^2 \pi} \int\limits^{\pi}_{-\pi} {e^{-ax}\ Cos(nx)} \, dx \\\\=-\frac{a}{n^2 \pi}\ [ e^{-a \pi}- e^{a\pi}] Cos(n\pi)-\frac{a^2}{n^2}\ a_n\\\\a_n=\frac{(-1)^n a\ [ e^{a \pi}- e^{-a\pi}]}{(n^2+a^2)\pi} \\

b_n=\frac{1}{\pi} \int\limits^{\pi}_{-\pi} {e^{-ax}\ Sin(nx)} \, dx\\\\= \frac{1}{n\pi} [- e^{-ax} Cos(nx)]_{-\pi}^{\pi}  - \frac{a}{n\pi} \int\limits^{\pi}_{-\pi} {e^{-ax}\ Cos(nx)} \, dx\\\\=\frac{(-1)^n(e^{\pi a}-e^{-\pi a})}{n\pi}-\frac{a}{n^2\pi} [ e^{-ax} Sin(nx) ]_{-\pi}^{\pi}\ -\frac{a^2}{n^2\pi} \int\limits^{\pi}_{-\pi} {e^{-ax}\ Sin(nx)} \, dx\\\\b_n=\frac{(-1)^n(e^{\pi a}-e^{-\pi a})}{n\pi}-0-\frac{a^2}{n^2}\ b_n\\\\b_n=\frac{(-1)^n(e^{\pi a}-e^{-\pi a})\ n}{(n^2+a^2)\pi}

e^-ax  Over  [-π,  π] :  the Fourier  series is :

e^{-ax}= \frac{e^{\pi a}-e^{-\pi a}}{\pi}\ [ \frac{a}{2} + \Sigma_{n=1}^{\infty} \frac{(-1)^n a}{a^2+n^2}\ Cos(nx) + \Sigma_{n=1}^{\infty} \frac{(-1)^n n}{a^2+n^2} \ Sin(nx) ]

 =\frac{e^{\pi a}-e^{-\pi a}}{\pi}\ [ \frac{a}{2} + \Sigma_{n=1}^{\infty} \frac{(-1)^n}{a^2+n^2}\ \{a\ Cos\ nx + n\ Sin\ nx \} ]

..done.

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