power point presentation on Fourier series.

2.  There are too many functions which
can be represented in form of infinite
series but in modern science and
engineering many problem connected
with oscillation, which require to
express
in terms of sine and
cosines known as a Fourier series .

3.  It has a very wide application
infield of digital signal processing
,wave forms of electrical fields,
vibrations, and conduction of heat
and etc.
 Mathematician and physicist
has used it very
first time in theory of heat
conduction.

4. the Fourier series for the f(x)in
the interval of is given by the
trigonometric series :
}sincos{
2 0
0
nxbnxa
a
nn
n
 

 Here are called Fourier
coefficients.
 2cxc
nn baa ,,0

5. 



0
0 )sin()(
n
nn nxcAxf 
2
0
0
a
A  22
nnn bac 






 
n
n
n
b
a1
tan

6. }sincos{
2 0
0
nxbnxa
a
nn
n
 


We find Fourier series in form of:
oAs we see in the equation we have three
constants we find that constant coefficient
nn baa ,,0
oWe find the value of that
coefficients using Fourier series
nn baa ,,0

7.  







2
0
0
20
)(
1
00
2
c
c
c
c
xfa
a
x
a
}sincos{
2
)(
22
1
2
0
2
nxbnxa
a
xxf
c
c
n
c
c
n
n
c
c
c
c





Apply integration both sides in the
equation:
To find :0a
Orthogonal property of
sine and cosine

8. To find :na
xnxxfa
a
xnxa
c
c
n
n
c
c
n









cos)(
1
0cos0
2
2
2
}sincoscoscos{cos
2
cos)(
22
1
2
0
2
xnxnxbnxnxaxnx
a
xnxxf
c
c
n
c
c
n
n
c
c
c
c
 



Orthogonal property of
sine and cosine

9. To find :nb
}sinsincossin{sin
2
sin)(
22
1
2
0
2
xnxnxbnxnxaxnx
a
xnxxf
c
c
n
c
c
n
n
c
c
c
c
 



xnxxfa
b
xnxb
c
n
n
c
c
n









sin)(
1
0sin0
2
0
2
2
 20 x
c  x
Orthogonal property of
sine and cosine

11. 1. Even functions:
let f(x) be a function defined over
the symmetrical interval –L ≤ x ≤ L.
Then the function f(x) is said to be
even if
f(-x) = f(x).

12. When is a even function then is a
even and is an odd function.
Thus
)(xf nxxf cos)(
nxxf sin)(
xxfxxfa 



 

 0
0 )(
2
)(
1
0sin)(
1
cos)(
2
)(
1
0














xxfb
xnxxfxosncxfa
n
n
nxa
a
xf n
n
cos
2
)(
0
0





13. If function ƒ(x) is an even periodic function with the
period 2L (–L ≤ x ≤ L), then ƒ(x)cos(nпx/L) is even
while ƒ(x)sin(nпx/L) is odd.
Thus the Fourier series expansion of an even periodic
function ƒ(x) with period 2L (–L ≤ x ≤ L) is given by,
L
nx
a
a
xf
n
n

cos
2
)(
1
0




dxxf
L
a
L

0
0 )(
2
Where,
,2,1cos)(
2
0
  ndx
L
xn
xf
L
a
L
n

0nb

14. 2.Odd functions:
let f(x) be a function defined over
the symmetrical interval
–L ≤ x ≤ L. Then the function f(x)
is said to be odd if
f(-x) = -f(x).

15. xxfxxfb
xosncxfa
xxfa
n
n




















sin)(
2
sin)(
1
0)(
1
0)(
1
0
0
nxbxf n
n
sin)(
1




When is a odd function then
is a even and is an odd
function.
)(xf
nxxf sin)(
nxxf cos)(
ODD FUNCTIONS

16. If function ƒ(x) is an even periodic function with the
period 2L (–L ≤ x ≤ L), then ƒ(x)cos(nпx/L) is even
while ƒ(x)sin(nпx/L) is odd.
Thus the Fourier series expansion of an odd periodic
function ƒ(x) with period 2L (–L ≤ x ≤ L) is given by,
)sin()(
1 L
xn
bxf
n
n





Where,
,2,1sin)(
2
0
  ndx
L
xn
xf
L
b
L
n

ODD FUNCTIONS

17. General rules:
▪ even* even = even function
▪ odd*odd=odd function
▪ Even*odd=odd function
▪ Even+even=even
▪ Odd+odd=odd
▪ Even odd= neither even nor odd
xx cos2
xx sin5
xx sin2
xx cos
xx sin

20. Question1: Find the Fourier series of
Solution.: The Fourier series of ƒ(x) is given by,
Using above,
dxxfa 



)(
1
0
dxxx



)(
1 2


 







23
1 23
xx
 xxxxf  2
)(

21. 






2323
1 22 33

 0
3
3
2
a

nxdxxfan cos)(
1




Now,
xxxf  2
)(
2
22
22
32
2
)1(4
)1(
)12(
)1(
)12(
1
cos
)12(
cos
)12(
1
sin
)2(
cos
)12(
sin
)(
1
n
nn
n
n
n
n
n
nx
n
nx
x
n
nx
xx
n
nn







 





















 





 



















22.  
n
n
nn
n
nx
n
nx
x
n
nx
xx
n
n
nn
)1(2
)1(
)1(
)(
)1(
)(1
cos
)2(
sin
)12(
cos
)(
1
22
22
32
2

















































nxdxxx sin)(
1 2




nxdxxfbn sin)(
1




Now,
Hence fourier series of, f(x) = x2+x,








 



1
2
2
2
sin
)1(2
cos
)1(4
3 n
nn
nx
n
nx
n
xx


23. Expand in a Fourier series.
Solution








xx
x
xf
0,
0,0
)(
22
1
)(0
1
)(
1
0
2
0
0
0




















   
x
x
dxxdxdxxfa

24. 














22
0
0
0
0
0
)1(11cos
cos1
sin
1sin
)(
1
cos)(0
1
cos)(
1
nn
n
n
nx
n
dxnx
nn
nx
x
dxnxxdx
dxnxxfa
n
n

















 






←cos n = (-1)n

25. we have
Therefore
n
nxdxxbn
1
sin)(
1
0
 





 








1
2
sin
1
cos
)1(1
4
)(
n
n
nx
n
nx
n
xf



26. Expand f(x) = x, −2 < x < 2 in a Fourier series.
Solution
it is an odd function on (−2, 2) and p = 2.
Thus


n
dxx
n
xb
n
n
1
2
0
)1(4
2
sin


 






1
1
2
sin
)1(4
)(
n
n
x
n
n
xf



27. Ch12_27
 If a function f is defined only on 0 < x < L, we
can make arbitrary definition of the function on
−L < x < 0.
 If y = f(x) is defined on 0 < x < L,
(i) reflect the graph about the y-axis onto −L < x <
0; the function is now even. See Fig 1.
(ii) reflect the graph through the origin onto −L < x
< 0; the function is now odd. See Fig 2
(iii) define f on −L < x < 0 by f(x) = f(x + L). See
Fig 3

28. Fig 1

29. Fig 2

30. Fig 3

31. Expand f(x) = x2, 0 < x < L, (a) in a cosine
series, (b) in a sine series (c) in a Fourier series.
Solution
The graph is shown in Fig

32. Ch12_32
Then
22
2
0
2
2
0
2
0
)1(4
cos
2
,
3
22


n
L
dxx
L
n
x
L
a
Ldxx
L
a
n
L
n
L










1
22
22
cos
)1(4
3
)(
n
n
x
L
n
n
LL
xf


na :

33. (b)
]1)1[(
4)1(2
sin
2
33
212
0
2





n
n
L
n
n
L
n
L
dxx
L
n
x
L
b















1
23
12
sin]1)1[(
2)1(2
)(
n
n
n
x
L
n
nn
L
xf



34. (c) With p = L/2, n/p = 2n/L, we have
Therefore




n
L
xdx
L
n
x
L
b
n
L
xdx
L
n
x
L
aLdxx
L
a
L
n
L
n
L
2
0
2
22
2
0
22
0
2
0
2
sin
2
2
cos
2
,
3
22







 




 
1
2
22
2
sin
12
cos
1
3
)(
n
x
L
n
n
x
L
n
n
LL
xf


Contd……

35.  http://en.wikipedia.org/wiki/Fourier_series
 Advance engineering mathematics by : Atul
books publication(GTU)