The Fourier's half range expansion in cosine series of a
function f(t) defined over (0 ,L) is given by:
f(t) = a0/2
+ summation {an cos (npt/L) }, where p stands for pi and n
=1,2,3....
Where a0 = 2/L* Integral {f(t)} dt , t from0to
L,
where an = 2/L* integral { f(t) cos(npt/L) }dt , t from
0 to L.
In this case we have
:
g(t) = f(t) = 2t+3 , fort in (0, L), where L =
2.
g(t) = 3-2t , for t in (-L,0), where L =
2
g(t+2L) = g(t+4) = g(t). So g(t) is an even function in
(-L L) and is equal to f(t) in (0,L) and so we can expand
itas:
g(x) = a0/2 + summation an*(cos
(npt/L)).
a0 = (2/L) Integral { 2t+3)dt, t from 0 to L,
but L=2.
= 2/2{[t^2+3t] t=2 - {t^2+2t] t= 0} = (4+6)=
10.
an = (2/L) *Integral (3t+2)cos(npt/L) dt , t=0 to
L
We use Leibnitz' rule that : Integral (UV)dt = UV1
-U'V2+U''V3-u'''V4+...., where ' stands for differentiation and suffix's denote
Integration. So
an = (2/2) *(2t+3)(cos(npt/L)1 -
(2t+3)'(cos(npt/L)2+... t from 0 to 2
=(2/(np){[
(2t+3)sin(npt/2) , t=2] - [(2t+3)sin(npt/2), t=0]]
+
-{(2/np) (sin(npt/2)2}
=0
- {[(2/np) (-cos (npt)/2)/ (np/2) ] at t =2 ]- [(2/np)(-cos(np/2)/(np/2) at t
=0}
=(2/np)^2 { (-1)^n -1}
an
= 4/p^2n^2 {(-1)^n - 1}
Therefore the required half range
Fourier cosine series is :
g(t) = 10/2 + (4/p^2)* summation
{[((-1)^n -1)/n^2] cos (npt/2)} for n =1, 2,3,.....
= 10/2
+ (4/p^2) {( -2/1^2) cos (pt/2) - 2/3^2 cos(3pt/2)
-(2/5^2)cos(5pt/2)-.................}
= 10/2 -(8/p^2){
(1/1^2)cos(pt/2)+(1/3^3)cos(3pt/2)+(1/5^2)cos(1/^2)(5pt/2)+cos(7pt/2)+.............}
No comments:
Post a Comment