Question

Obtain the Fourier series for e^ -ax from x equals to -ΠtoΠ

Answer 1

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.

Answer 2

Step-by-step explanation:

Given signal function in time domain is:

f(x) = e^{-ax}\ \ \ -\pi \le x \le \pif(x)=e

−ax

−π≤x≤π

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

\begin{gathered}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\end{gathered}

f(x)=

2

1

a

0

+ Σ

n=1

∞

a

n

Sin(nx) + Σ

n=1

∞

b

n

Sin(nx)

a

0

=

π

1

−π

∫

π

f(x)dx

a

n

=

π

1

−π

∫

π

f(x) Cos(nx)dx

b

n

=

π

1

−π

∫

π

f(x) Sin(nx)dx

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

\begin{gathered}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}\\\end{gathered}

a

0

=

π

1

−π

∫

π

e

−ax

dx=

πa

1

[−e

−ax

]

−π

π

=

πa

e

πa

−e

−πa

\begin{gathered}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} \\\end{gathered}

a

n

=

π

1

−π

∫

π

e

−ax

Cos(nx)dx

=

nπ

1

[e

−ax

Sin(nx)]

−π

π

+

nπ

a

−π

∫

π

e

−ax

Sin(nx)dx

=0−[

n

2

π

a

e

−ax

Cos(nx)]

−π

π

−

n

2

π

a

2

−π

∫

π

e

−ax

Cos(nx)dx

=−

n

2

π

a

[e

−aπ

−e

aπ

]Cos(nπ)−

n

2

a

2

a

n

a

n

=

(n

2

+a

2

)π

(−1)

n

a [e

aπ

−e

−aπ

]

\begin{gathered}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}\end{gathered}

b

n

=

π

1

−π

∫

π

e

−ax

Sin(nx)dx

=

nπ

1

[−e

−ax

Cos(nx)]

−π

π

−

nπ

a

−π

∫

π

e

−ax

Cos(nx)dx

=

nπ

(−1)

n

(e

πa

−e

−πa

)

−

n

2

π

a

[e

−ax

Sin(nx)]

−π

π

−

n

2

π

a

2

−π

∫

π

e

−ax

Sin(nx)dx

b

n

=

nπ

(−1)

n

(e

πa

−e

−πa

)

−0−

n

2

a

2

b

n

b

n

=

(n

2

+a

2

)π

(−1)

n

(e

πa

−e

−πa

) n

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) ]e

−ax

=

π

e

πa

−e

−πa

[

2

a

+Σ

n=1

∞

a

2

+n

2

(−1)

n

a

Cos(nx)+Σ

n=1

∞

a

2

+n

2

(−1)

n

n

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 \} ]=

π

e

πa

−e

−πa

[

2

a

+Σ

n=1

∞

a

2

+n

2

(−1)

n

{a Cos nx+n Sin nx}]

..done.