Expert: Paul Klarreich - 2/17/2009

Question
hello paul,
need some help with the following question.
Determine the fourier series of the signal f(t) = 1+t/pi for -pi<t<0.
1-t/pi for 0<t<pi.
any help would be greatly appreciated
thanks keith

Answer
Questioner:   keith
Category:  Calculus
Private:  No
 
Subject:  fourier series
Question:  hello paul,
need some help with the following question.
Determine the fourier series of the signal

f(t) = 1+t/pi for -pi<t<0.
      1-t/pi for 0<t<pi.
any help would be greatly appreciated
thanks keith

----------------------------------
OK, this appears to be an even function. That helps a bit, because you won't have any sines -- only cosines.  So you want this integral:

Two times:  1/pi times:

{pi
|(1 - t/pi) cos (nt) dt
}0

Now the '1' term:

{pi
| cos (nt) dt = sin nt / n,  0 to pi = 0.
}0

So we just do the -t/pi term:

    {pi
-1/pi| t cos (nt) dt
    }0

use IBP:

u = t,  dv = cos nt dt
du = dt,  v = 1/n sin nt

uv part is:

t/n sin nt, from 0 to pi, gives 0

-v du part is: + 1/(n pi)  times

{
| sin nt dt = - (1/n) cos nt from 0 to pi
}

= -1/n(cos n pi - cos 0)

= -1/n(-1 - 1) = + 2/n, when  n is odd

= -1/n(1 - 1) = 0,   when  n is even

Whew!  I think we have  2/(pi n^2) as the coefficient.

So it's
inf
SUM (2/pi(2k+1)^2) cos (2k+1)t
k=1

or something like that.

I hope I didn't blow too many signs here.  Wish me luck.