1. ## Fourier of e-|t|

Hi,
I want to compute the trigonometric fourier series for x[t]= e^(-|t|) from [-1,1].
The way I see it:
I need to break down the function:
for [-1,0] x[t]=e^(t)

and
for [0,1] x[t]=e^(-t)

Compute the trigo fourier for each part and add them.

I also was thinking that the function to even, I compute only the coefficients with sin with x[t]=e^(-t)??.
what is the right way of doing it?
thanks

Hi,
I want to compute the trigonometric fourier series for x[t]= e^(-|t|) from [-1,1].
The way I see it:
I need to break down the function:
for [-1,0] x[t]=e^(t)

and
for [0,1] x[t]=e^(-t)

Compute the trigo fourier for each part and add them.

I also was thinking that the function to even, I compute only the coefficients with sin with x[t]=e^(-t)??.
what is the right way of doing it?
thanks
you are correct, it is even. for that you compute the coefficients with cosine (and of course, you do the first term as well)

we do the sine coefficients for odd functions and we leave off the first term

3. [QUOTE=Jhevon;
I forgot to ask you what is the expression of x[t]?
can I take onlu e^-t WITHOUT the absolute value??

Originally Posted by Jhevon;
I forgot to ask you what is the expression of x[t
?
can I take onlu e^-t WITHOUT the absolute value??
$x(t) = \frac {a_0}2 + \sum_{n = 1}^{\infty} a_n \cos n \pi t$

where $a_0 = \int_{-1}^{1}x(t)~dt$

and

$a_n = \int_{-1}^{1} x(t) \cos n \pi t ~dt$

now, you must split each integral into $\int_{-1}^0 ... + \int_0^1 ...$

for the [-1,0] use $x(t) = e^t$

for the [0,1] use $x(t) = e^{-t}$

or better yet, use $x(t) = e^{-t}$ and change all the integrals to $2 \int_0^1 ...$ since we are symmetric about the y-axis

5. Got it Jhevon!!!