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The Fourier Series
BET2533 – Eng. Math. IIIR. M. Taufika R. Ismail
FKEE, UMP
A Fourier series is an expansion of a periodic function f (t) in terms of an infinite sum of cosines and sines
Introduction
1
0 )sincos(2
)(n
nn tnbtnaa
tf
In other words, any periodic function can be resolved as a summation of constant value and cosine and sine functions:
1
0 )sincos(2
)(n
nn tnbtnaa
tf
)sincos( 11 tbta 2
0a
)2sin2cos( 22 tbta
)3sin3cos( 33 tbta
The computation and study of Fourier series is known as harmonic analysis and is extremely useful as a way to break up an arbitrary periodic function into a set of simple terms that can be plugged in, solved individually, and then recombined to obtain the solution to the original problem or an approximation to it to whatever accuracy is desired or practical.
=
+ +
+ + + …
Periodic Function2
0a
ta cos1
ta 2cos2
tb sin1
tb 2sin2
f(t)
t
1
0 )sincos(2
)(n
nn tnbtnaa
tf
where
T
dttfT
a0
0 )(2
frequency lFundementa2
T
T
n tdtntfT
a0
cos)(2
T
n tdtntfT
b0
sin)(2
*we can also use the integrals limit .
T
dttfT
a0
0 )(2
2/
2/
T
T
Example 1
Determine the Fourier series representation of the following waveform.
SolutionFirst, determine the period & describe the one period of the function:
T = 2
21,0
10,1)(
t
ttf )()2( tftf
Then, obtain the coefficients a0, an and bn:
10101)(2
2)(
2 2
1
1
0
2
00
0 dtdtdttfdttfT
aT
Or, sincey = f(t) over the interval [a,b], hence
b
a
dttf )( is the total area below graph
1)11(2
2
],0[over
graph below Area2)(
2
0
0
TT
dttfT
aT
n
n
n
tndttdtn
tdtntfT
an
sinsin0cos1
cos)(2
1
0
2
1
1
0
2
0
Notice that n is integer which leads ,since
0sin n03sin2sinsin
Therefore, .0na
n
n
n
tndttdtn
tdtntfT
bn
cos1cos0sin1
sin)(2
1
0
2
1
1
0
2
0
15cos3coscos 16cos4cos2cos
Notice that
Therefore,
even ,0
odd ,/2)1(1
n
nn
nb
n
n
or nn )1(cos
ttt
tnn
tnbtnaa
tf
n
n
nnn
5sin5
23sin
3
2sin
2
2
1
sin)1(1
2
1
)sincos(2
)(
1
1
0
Finally,
♣
Some helpful identities
For n integers,nn )1(cos 0sin n
02sin n 12cos n
xx sin)sin( xx cos)cos(
[Supplementary]
The sum of the Fourier series terms can evolve (progress) into the original waveform
From Example 1, we obtain
ttttf
5sin5
23sin
3
2sin
2
2
1)(
It can be demonstrated that the sum will lead to the square wave:
tttt
7sin7
25sin
5
23sin
3
2sin
2ttt
5sin
5
23sin
3
2sin
2
tt
3sin3
2sin
2t
sin
2
(a) (b)
(c) (d)
ttttt
9sin9
27sin
7
25sin
5
23sin
3
2sin
2
ttt
23sin23
23sin
3
2sin
2
2
1
(e)
(f)
Example 2
Given ,)( ttf 11 t
)()2( tftf
Sketch the graph of f (t) such that .33 t
Then compute the Fourier series expansion of f (t).
SolutionThe function is described by the following graph:
T = 2
T
2We find that
Then we compute the coefficients:
02
11
22
2
)(2
1
1
21
1
1
1
0
ttdt
dttfT
a
0coscos
)cos(cos0
cos)]sin([sin
sinsin
coscos)(2
22
22
1
122
1
1
1
1
1
1
1
1
n
nnn
nn
n
tn
n
nn
dtn
tn
n
tnt
tdtnttdtntfT
an
since xx cos)cos(
nnn
n
n
nn
n
n
n
tn
n
nn
dtn
tn
n
tnt
tdtnttdtntfT
b
nn
n
1
22
1
122
1
1
1
1
1
1
1
1
)1(2)1(2cos2
)sin(sincos2
sin)]cos([cos
coscos
sinsin)(2
ttt
tnn
tnbtnaa
tf
n
n
nnn
3sin3
22sin
2
2sin
2
sin)1(2
)sincos(2
)(
1
1
1
0
Finally,
♣
Example 3
Given
42,0
20,2)(
t
tttv
)()4( tvtv
Sketch the graph of v (t) such that .120 t
Then compute the Fourier series expansion of v (t).
SolutionThe function is described by the following graph:
T = 4
2
2 T
We find that
0 2 4 6 8 10 12 t
v (t)
2
Then we compute the coefficients:
12
22
1)2(
2
1
0)2(4
2
)(2
2
0
22
0
4
2
2
0
4
0
0
ttdtt
dtdtt
dttvT
a
222222
2
022
2
0
2
0
4
2
2
0
4
0
])1(1[2)cos1(2
2
2cos1
cos
2
10
sin
2
1sin)2(
2
1
0cos)2(2
1cos)(
2
nn
n
n
n
n
tn
dtn
tn
n
tnt
tdtnttdtntvT
a
n
n
nnn
n
n
n
tn
n
dtn
tn
n
tnt
tdtnttdtntvT
bn
21
2
2sin1
sin
2
11
cos
2
1cos)2(
2
1
0sin)2(2
1sin)(
2
22
2
022
2
0
2
0
4
2
2
0
4
0
since 0sin2sin nn
122
1
0
2sin
2
2cos
])1(1[2
2
1
)sincos(2
)(
n
n
nnn
tn
n
tn
n
tnbtnaa
tv
Finally,
♣
Symmetry Considerations
Symmetry functions:
(i) even symmetry
(ii) odd symmetry
Even symmetry
Any function f (t) is even if its plot is symmetrical about the vertical axis, i.e.
)()( tftf
Even symmetry (cont.) The examples of even functions are:
2)( ttf
t t
t
||)( ttf
ttf cos)(
Even symmetry (cont.) The integral of an even function from −A to
+A is twice the integral from 0 to +A
t
AA
A
dttfdttf0
eveneven )(2)(−A +A
)(even tf
Odd symmetry
Any function f (t) is odd if its plot is antisymmetrical about the vertical axis, i.e.
)()( tftf
Odd symmetry (cont.) The examples of odd functions are:
3)( ttf
t t
t
ttf )(
ttf sin)(
Odd symmetry (cont.) The integral of an odd function from −A to
+A is zero
t 0)(odd
A
A
dttf−A +A
)(odd tf
Even and odd functions
(even) × (even) = (even) (odd) × (odd) = (even) (even) × (odd) = (odd) (odd) × (even) = (odd)
The product properties of even and odd functions are:
Symmetry consideration
From the properties of even and odd functions, we can show that:
for even periodic function;
2/
0
cos)(4 T
n tdtntfT
a 0nb
for odd periodic function;
2/
0
sin)(4 T
n tdtntfT
b 00 naa
How?? [Even function]
2
T
2
T
2/
0
2/
2/
cos)(4
cos)(2 TT
T
n tdtntfT
tdtntfT
a
(even) × (even)
| |
(even)
0sin)(2 2/
2/
T
T
n tdtntfT
b
(even) × (odd)
| |
(odd)
)(tf
t
How?? [Odd function]
2
T
2
T
2/
0
2/
2/
sin)(4
sin)(2 TT
T
n tdtntfT
tdtntfT
b
(odd) × (odd)
| |
(even)
0cos)(2 2/
2/
T
T
n tdtntfT
a
(odd) × (even)
| |
(odd)
)(tf
t
0)(2 2/
2/
0
T
T
dttfT
a
(odd)
Example 4
Given
21,1
11,
12,1
)(
t
tt
t
tf
)()4( tftf
Sketch the graph of f (t) such that .66 t
Then compute the Fourier series expansion of f (t).
SolutionThe function is described by the following graph:
T = 4
2
2 T
We find that
0−4−6 2 4 6 t
f (t)
−2
1
−1
Then we compute the coefficients. Since f (t) is an odd function, then
0)(2 2
2
0
dttfT
a
0cos)(2 2
2
tdtntfT
an
and
n
n
n
n
n
n
n
nn
n
tn
n
n
n
tndt
n
tn
n
tnt
tdtntdtnt
tdtntfT
tdtntfT
bn
cos2sin2cos
cos2cossincos
coscoscos
sin1sin4
4
sin)(4
sin)(2
22
1
022
2
1
1
0
1
0
2
1
1
0
2
0
2
2
since 0sin2sin nn
1
1
1
1
0
2sin
)1(2
2sin
cos2
)sincos(2
)(
n
n
n
nnn
tn
n
tn
n
n
tnbtnaa
tf
Finally,
♣
Example 5
Compute the Fourier series expansion of f (t).
SolutionThe function is described by
T = 3
3
22 T
and
32,1
21,2
10,1
)(
t
t
t
tf
)()3( tftf
T = 3
Then we compute the coefficients.
3
81
2
32)01(
3
421
3
4)(
4)(
2 2/3
1
1
0
2/3
0
3
0
0
dtdtdttf
Tdttf
Ta
3
8)23()12(2)01(
3
2121
3
2)(
2 3
2
2
1
1
0
3
0
0
dtdtdtdttf
Ta
Or, since f (t) is an even function, then
Or, simply
3
84
3
2
period ain
graph below area Total2)(
2 3
0
0
T
dttfT
a
3
2sin
2
3
2sinsin2
2
sin2
3sin2
3
4
sin2
3sin2sin
3
4
sin2
3
4sin
3
4
cos2cos13
4
cos)(4
cos)(2
2/3
1
1
0
2/3
1
1
0
2/3
0
3
0
n
n
nn
n
nn
n
nn
nn
n
tn
n
tn
tdtntdtn
tdtntfT
tdtntfT
an
;3
2
1
1
1
0
3
2cos
3
2sin
12
3
4
3
2cos
3
2sin
2
3
4
)sincos(2
)(
n
n
nnn
tnn
n
tnn
n
tnbtnaa
tf
Finally,
♣
and 0nb since f (t) is an even function.
Function defines over a finite interval
Fourier series only support periodic functions In real application, many functions are non-
periodic The non-periodic functions are often can be
defined over finite intervals, e.g.
y = t
y = 1 y = 1
y = 2
y = t2
Therefore, any non-periodic function must be extended to a periodic function first, before computing its Fourier series representation
Normally, we prefer symmetry (even or odd) periodic extension instead of normal periodic extension, since symmetry function will provide zero coefficient of either an or bn
This can provide a simpler Fourier series expansion
)(ty
t
)(tf
t
)(even tf
)(odd tf
t
t
lttytf 0,)()(
)()( tfltf
lT
0,)(
0,)()(
tlty
lttytf
)()2( tfltf
lT 2
0,)(
0,)()(
tlty
lttytf
)()2( tfltf
lT 2
l0
l0 l2ll2
l0 l2ll2
l3l3
l3l3
l0 l2ll2 l3l3
T
T
T
Periodic extension
Even periodic extension
Odd periodic extension
Non-periodicfunction
Half-range Fourier series expansion
The Fourier series of the even or odd periodic extension of a non-periodic function is called as the half-range Fourier series
This is due to the non-periodic function is considered as the half-range before it is extended as an even or an odd function
If the function is extended as an even function, then the coefficient bn= 0, hence
1
0 cos2
)(n
n tnaa
tf
which only contains the cosine harmonics. Therefore, this approach is called as the
half-range Fourier cosine series
If the function is extended as an odd function, then the coefficient an= 0, hence
1
sin)(n
n tnbtf
which only contains the sine harmonics. Therefore, this approach is called as the
half-range Fourier sine series
Example 6
ttf 0,1)(
Compute the half-range Fourier sine series expansionof f (t), where
SolutionSince we want to seek the half-range sine series,the function to is extended to be an odd function:
T = 2π
12
T
0 π t
f (t)
1
−1
−π 2π−2π0 π t
f (t)
1
Hence, the coefficients are
00 naa
0
2/
0
sin12
4sin)(
4ntdttdtntf
Tb
T
n
and
even ,0
odd ,/4)cos1(
2cos2
0 n
nnn
nn
nt
Therefore,
ntn
ntnn
tf
nnn
sin4
sin)cos1(2
)(
odd 11
♣
Example 7
10,12)( tttf
Determine the half-range cosine series expansionof the function
Sketch the graphs of both f (t) and the periodic function represented by the series expansion for −3 < t < 3.
SolutionSince we want to seek the half-range cosine series,the function to is extended to be an even function:
T = 2
T
2
t
f (t)
t
f (t)
122
33 1
1
11
11
2
1
Hence, the coefficients are
02)12(2
4)(
4 1
02
1
0
2/
0
0 ttdttdttfT
aT
1
022
1
0
1
0
1
0
2/
0
cos4
sin2
2sin
2sin)12(
2
cos)12(2
4cos)(
4
n
tn
n
n
dtn
tn
n
tnt
tdtnttdtntfT
aT
n
even ,0
odd ,/8)1(cos4 22
22 n
nn
n
n
0nb
Therefore,
odd 1
22
odd 1
22
10
cos18
cos8
0
cos)(
nn
nn
nn
tnn
tnn
tnaatf
♣
Parseval’s Theorem
Parserval’s theorem states that the average power in a periodic signal is equal to the sum of the average power in its DC component and the average powers in its harmonics
=
+ +
+ + + …
20a
ta cos1
ta 2cos2
tb sin1
tb 2sin2
f(t)
t
Pavg
Pdc
Pa1 Pb1
Pa2 Pb2
For sinusoidal (cosine or sine) signal,
R
V
R
V
R
VP
2peak
2
peak2
rms
2
12
For simplicity, we often assume R = 1Ω,
which yields2
peak2
1VP
For sinusoidal (cosine or sine) signal,
22
22
21
21
2
0
dcavg
2
1
2
1
2
1
2
1
2
2211
babaa
PPPPPP baba
1
2220avg )(
2
1
4
1
nnn baaP
Exponential Fourier series
Recall that, from the Euler’s identity,
xjxe jx sincos
yields
2cos
jxjx eex
2sin
j
eex
jxjx and
Then the Fourier series representation becomes
11
0
1
0
1
0
1
0
1
0
222
222
222
222
)sincos(2
)(
n
tjnnn
n
tjnnn
n
tjnnntjnnn
n
tjntjn
n
tjntjn
n
n
tjntjn
n
tjntjn
n
nnn
ejba
ejbaa
ejba
ejbaa
eejb
eea
a
j
eeb
eea
a
tnbtnaa
tf
Here, let we name
11
0
222)(
n
tjnnn
n
tjnnn ejba
ejbaa
tf
2nn
n
jbac
,
2nn
n
jbac
Hence,
n
tjnn
n
tjnn
n
tjnn
n
tjnn
n
tjnn
n
tjnn
n
tjnn
ececcec
ececc
ececc
10
1
110
110
and .2
00
ac
c0c−ncn
Then, the coefficient cn can be derived from
Ttjn
T
TT
TT
nnn
dtetfT
dttnjtntfT
tdtntfjtdtntfT
tdtntfT
jtdtntf
T
jbac
0
0
00
00
)(1
]sin)[cos(1
sin)(cos)(1
sin)(2
2cos)(
2
2
1
2
In fact, in many cases, the complex Fourier series is easier to obtain rather than the trigonometrical Fourier series
In summary, the relationship between the complex and trigonometrical Fourier series are:
2nn
n
jbac
2nn
n
jbac
T
dttfT
ac
0
00 )(
1
2 T
tjnn dtetf
Tc
0
)(1
nn cc or
Example 8
Obtain the complex Fourier series of the followingfunction
2 44 2 0f (
t) =
e t
2e
1
)(tf
t
Since , . Hence
Solution
2
1
2
1
2
1
)(1
22
0
2
0
0
0
ee
dte
dttfT
c
t
t
T
12T
)1(2
1
)1(2
1
)1(2
1
12
1
2
1
2
1
)(1
222)1(2
2
0
)1(
2
0
)1(2
0
0
jn
e
jn
ee
jn
e
jn
e
dtedtee
dtetfT
c
njjn
tjn
tjnjntt
Ttjn
n
since 1012sin2cos2 njne nj
jnt
nn
tjnn e
jn
eectf
)1(2
1)(
2
Therefore, the complex Fourier series of f (t) is
♣
0
2
0
2
0 2
1
)1(2
1c
e
jn
ec
nnn
*Notes: Even though c0 can be found by substitutingcn with n = 0, sometimes it doesn’t works (as shownin the next example). Therefore, it is always better tocalculate c0 alone.
2
2
12
1
n
ecn
cn is a complex term, and it depends on nω.Therefore, we may plot a graph of |cn| vs nω.
In other words, we have transformed the functionf (t) in the time domain (t), to the function cn in thefrequency domain (nω).
Example 9
Obtain the complex Fourier series of the function inExample 1.
Solution
2
11
2
1)(
1 1
00
0 dtdttfT
cT
)1(22
1
012
1)(
1
1
0
2
1
1
00
jn
tjn
tjnT
tjnn
en
j
jn
e
dtedtetfT
c
)1(2
jn
n en
jc
But njn nnjne )1(cossincos
Thus,
even ,0
odd ,/]1)1[(
2 n
nnj
n
j n
Therefore,
odd 0
2
1)(
nn
n
tjn
n
tjnn e
n
jectf
♣
*Here notice that .00cc
nn
even ,0
odd ,1
n
nncn
The plot of |cn| vs nω is shown below
2
10 c
0.5
