An introduction to Fourier Series_mathematics

Introduction to Fourier Series

1. Fourier Series
Present by
SUBHRANGSU
SEKHAR DEY
AMITY UNIVERSITY RAJASHAN
M.SC CHEMISTRY
DEPARTMENT OF
ASET

2. Joseph
Fourier(1768 -
1830)
Joseph Fourier(1768-1830), son of a
French taylor and friend of
nepolean,invented many examples of
expressions in trigonometric series in
connection with the problems of
conduction heat.His book entitled
“Theoric Analytique de le
Chaleur”(Analytical theory of heat)
published in 1822 is classical is the
theory of heat conduction for boundary
value problem.the fourier series comes
after his name for periodic function to be
expanded in trigonometric series.

3. Content
Periodic Functions
Fourier Series
Analysis of Periodic Waveforms
●Half Range Series

4. Fourier Series
Periodic Functions

5. The Mathematic Formulation
Any function that satisfies( ) ( )f t f tT
where Tis a constant and is called the period
of the function.

6. Example:4
cos
3
cos)(
tt
tf 
Find its period.)()( Ttftf  )(
4
1
cos)(
3
1
cos
4
cos
3
cos TtTt
tt

Fact:)2cos(cos  m m
T
2
3 n
T
2
4 mT6 nT8 24T
smallest T

7. Example:tttf
21coscos)( 
Find its period.)()( Ttftf  )(cos)(coscoscos
2121 TtTttt   mT2
1  nT2
2 n
m



2
1 2
1


must be a
rational number

8. Example:tttf )10cos(10cos)( 
Is this function a periodic one?



10
10
2
1
not a rational
number

9. Fourier Series
Fourier Series

10. Introduction
Decompose a periodic input signal into
primitive periodic components.
A periodic sequence
T 2T 3T
t
f(t)

11. SynthesisT
nt
b
T
nt
a
a
tf
n
n
n
n



 




2
sin
2
cos
2
)(
11
0
DC PartEven Part Odd Part
Tis a period of all the above signals)sin()cos(
2
)(
0
1
0
1
0
tnbtna
a
tf
n
n
n
n
 




Let 
0=2/T.

12. Decompositiondttf
T
a
Tt
t


0
0
)(
2
0 ,2,1 cos)(
2
0
0
0


ntdtntf
T
a
Tt
t
n ,2,1 sin)(
2
0
0
0


ntdtntf
T
b
Tt
t
n )sin()cos(
2
)(
0
1
0
1
0
tnbtna
a
tf
n
n
n
n
 





13. Proof
Use the following facts:0 ,0)cos(
2/
2/
0 

mdttm
T
T 0 ,0)sin(
2/
2/
0 

mdttm
T
T 





 nmT
nm
dttntm
T
T 2/
0
)cos()cos(
2/
2/
00 





 nmT
nm
dttntm
T
T 2/
0
)sin()sin(
2/
2/
00 nmdttntm
T
T
and allfor ,0)cos()sin(
2/
2/
00 


14. Example (Square Wave)11
2
2
0
0 



dta ,2,1 0sin
1
cos
2
2
0
0








nnt
n
ntdta
n
,6,4,20
,5,3,1/2
)1cos(
1
cos
1
sin
2
2
0
0








n
nn
n
n
nt
n
ntdtb
n





2345--2-3-4-5-6
f(t)
1

15. 11
2
2
0
0 



dta ,2,1 0sin
1
cos
2
2
0
0








nnt
n
ntdta
n
,6,4,20
,5,3,1/2
)1cos(
1
cos
1
sin
2
1
0
0

















n
nn
n
n
nt
n
ntdtb
n 2345--2-3-4-5-6
f(t)
1
Example (Square Wave)







 ttttf 5sin
5
1
3sin
3
1
sin
2
2
1
)(

16. 11
2
2
0
0 



dta ,2,1 0sin
1
cos
2
2
0
0








nnt
n
ntdta
n
,6,4,20
,5,3,1/2
)1cos(
1
cos
1
sin
2
1
0
0

















n
nn
n
n
nt
n
ntdtb
n 2345--2-3-4-5-6
f(t)
1
Example (Square Wave)-0.5
0
0.5
1
1.5 







 ttttf 5sin
5
1
3sin
3
1
sin
2
2
1
)(

17. Find the Fourier series for
In both cases note that we are integrating an odd function (xis
odd and cosine is even so the product is odd) over the
intervaland so we know that both of these integrals will be
zero.
Example

18. Next here is the integral for
In this case we’re integrating an even function (x and sine are both
odd so the product is even) on the interval and so we can
“simplify” the integral as shown above.The reason for doing this
here is not actually to simplify the integral however.It is instead
done so that we can note that we did this integral back in the
Fourier sine series section and so don’t need to redo it in this
section.Using the previous result we get,

19. In this case the Fourier series is,

20. Fourier Series
Analysis of
Periodic Waveforms

21. Waveform Symmetry
Even Functions
Odd Functions)()( tftf  )()( tftf 

22. Decomposition
Any function f(t)can be expressed as the
sum of an even function f
e(t)and an odd
function f
o(t).)()()( tftftf
oe )]()([)(
2
1
tftftf
e  )]()([)(
2
1
tftftf
o 
Even Part
Odd Part

23. Example






00
0
)(
t
te
tf
t
Even Part
Odd Part






0
0
)(
2
1
2
1
te
te
tf
t
t
e 






0
0
)(
2
1
2
1
te
te
tf
t
t
o

24. Half-Wave Symmetry)()( Ttftf 
and 2/)( Ttftf 

25. Quarter-Wave Symmetry
Even Quarter-Wave Symmetry
TT/2T/2
Odd Quarter-Wave Symmetry
T
T/2T/2

26. Hidden Symmetry
The following is a asymmetry periodic function:
Adding a constant to get symmetry property.
A
TT
A/2
A/2
TT

27. Fourier Coefficients of
Symmetrical Waveforms
The use of symmetry properties simplifies the
calculation of Fourier coefficients.
–Even Functions
–Odd Functions
–Half-Wave
–Even Quarter-Wave
–Odd Quarter-Wave
–Hidden

28. 0,cos)(
2
cos
2
)(
0
1
0






ndx
l
xn
xf
l
a
where
l
xn
a
a
xf
l
n
n
n

 HALF RANGE SERIES
COSINE SERIES
Afunction definedin canbeexpanded
asaFourierseriesofperiodcontainingonly
cosinetermsbyextending suitablyin .
(Asanevenfunction))(xf )(xf ),0(l l2 )0,(l

29. SINE SERIES
Afunctiondefinedin canbeexpanded
asaFourierseriesofperiodcontaining
onlysinetermsbyextending suitablyin
[Asanoddfunction]1,sin)(
2
sin)(
0
1






ndx
l
xn
xf
l
b
where
l
xn
bxf
l
n
u
n

 )(xf )(xf ),0(l ).0,(l l2

30. Expand in half range
(a) sine Series (b) Cosine series.
SOLUTION
(a)
Extend the definition of given function to that of
an odd function of period 4
i.e








20;
02;
)(
xx
xx
xf 20,)(  xxxf
Example of
Half Range Series

32. Here dx
l
xn
xf
l
b
a
l
n
n



0
sin)(
2
0
 




nn
xn
n
xn
x
dx
xn
xfb
n
n
)1(4
)
2
2
sin
(1)
2
2
cos
(
2
sin)(
2
2
2
0
2
22
2
0




















33. 2
sin
)1(4
)(
1
xn
n
xf
n
n






 (b)
Extend the definition of given function to that of
an even function of period 4








20;
02;
)(
xx
xx
xf

35. dx
l
xn
xf
l
a
b
l
n
n



0
cos)(
2
0
  
0
;
1)1(4
)
2
2
cos
(1)
2
2
sin
(
2
cos)(
2
2
22
2
0
2
22
2
0


















n
nn
xn
n
xn
x
dx
xn
xfa
n
n






36. 

2
0
0 2xdxa  
2
cos
1)1(4
1)(
1
22
xn
n
xf
n
n








37. Thank You