This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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 \} ]


1 5 1
please click on thanks azur blue button above