The Fourier Series BET2533 ± Eng. Math. III R. M. Taufika R. Ismail FKEE, UMP
Introduction Fourier series is an expansion of a peri pe riod odic ic fu func ncti tion on f (t ) in terms of an infinite sum of cosines of cosines and sines
A
f (t ) !
a0 2
g
§ (an cos cos n[t bn sin sin n[t ) n !1
In other words, any pe peri riod odic ic fu func ncti tion on can be resolved as a summation of constant value and cosine and sine functions: f (t ) !
!
a0 2 a0 2
g
§ ( an cos n[t bn sin n[t ) n !1
( a1 cos [t b1 sin [t )
(a2 cos cos 2[t b2 sin sin 2[t ) (a3 cos cos 3[t b3 sin sin 3[t ) -
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, individually, and then recombined to obtain the solution to the original problem or an approximation to it to whatever accuracy is desired or practical.
f(t)
a0
Periodic Function
2
= t
a1 cos [t
+
b1 sin [t
+ a2 cos 2[t
+
b2 sin 2[t
+
+
«
f (t ) !
a0
!
where
2
g
§ (an cos n[t bn sin n[t )
2T T
n !1
! Fundemental frequency a0 !
2
T
f (t )d t ´ T 0
an !
2
T
f (t ) cos n[tdt ´ T 0
bn !
2
T
f (t ) sin n[tdt ´ T 0
T / 2
*we can also use the integrals limit
´
.
T / 2
Example 1 Determine the Fourier series representation of the following waveform.
Solution First, determine the period & describe the one period of the function:
T = 2
1, 0 t 1 ® f (t ) ! ¯ 0, 1 t 2 °
f (t 2) ! f (t )
Then, obtain the coefficients a0, an and bn:
a0 !
2
T
2
2
1
2
0
0
1
f (t ) d t ! ´ f (t )d t ! ´ 1d t ´ 0d t ! 1 0 ! 1 ´ T 2 0
b
Or, since ´ f (t )dt is the total area below graph a y = f (t ) over the interval [a,b], hence
a0 !
2
T
2 ¨ Area below graph ¸
f (t ) dt ! v ©© ´ T T ª 0
over [0, T ]
2
¹¹ ! v (1v 1) ! 1 º 2
an !
2
2
f (t ) cos n[t d t ´ T 0
1
« sin nT t » sin nT ! ´ 1cos nT t d t ´ 0d t ! ¬ ! ¼ nT - nT ½ 0 0 1 1
2
Notice that n is integer which leads sin nT ! 0, since sin T ! sin 2T ! sin 3T ! - ! 0
Therefore, an ! 0.
bn !
2
2
f (t ) sin n[t d t ´ T 0
1
« cos nT t » 1 cos nT ! ´ 1sin nT t d t ´ 0d t ! ¬ ! ¼ nT ½ 0 nT 0 1 1
2
Notice that cos T ! cos 3T ! cos 5T ! - ! 1 cos 2T ! cos 4T ! cos 6T ! - ! 1
or cos nT ! ( 1) n Therefore, bn !
1 ( 1)
nT
n
2 / nT ,
n odd
,
n even
!¯ °0
Finally, f (t ) !
a0 2
g
§ (an cos n[t bn sin n[t ) n !1
«1 ( 1) n » ! §¬ ¼ sin nT t 2 n !1 - nT ½ !
1
g
1
2
2
T
sin T t
2 3T
sin 3T t
2 5T
sin 5T t -
Some helpful identities sin( x) ! sin x
cos( x ) ! cos x
For n integers, sin nT ! 0
cos nT ! (1) n
sin 2nT ! 0
cos 2nT ! 1
[Supplementary]
The sum of the Fourier series terms can evolve (progress) into the original waveform From Example 1, we obtain f (t ) !
1
2
2
T
sin T t
2 3T
sin 3T t
2 5T
sin 5T t -
It can be demonstrated that the sum will lead to the square wave:
(a)
(b) 2
T
2
sin T t
T
(c)
2
T
sin T t
2 3T
sin 3T t
(d)
sin T t
2 3T
sin 3T t
2 5T
sin 5T t
2
T
sin T t
2 3T
sin 3T t
2 5T
sin 5T t
2 7T
sin 7T t
(e)
2
T
sin T t
2 3T
sin 3T t
2 5T
sin 5T t
2 7T
sin 7T t
(f)
1
2
2
T
sin T t
2 3T
sin 3T t -
2 23T
sin 23T t
2 9T
sin 9T t
Example 2 Given f (t ) ! t ,
1 e t e 1
f (t 2) ! f (t )
Sketch the graph of f (t ) such that 3 e t e 3. Then compute the Fourier series expansion of f (t ).
Solution The function is described by the following graph:
T = 2
We find that
[!
2T T
! T
Then we compute the coefficients: a0 !
2
1
f (t )dt ´ T 1
1
« t » 11 ! ´ tdt ! ¬ ¼ ! !0 2 1 2 - 2 ½ 1 2
1
2
an !
2
1
1
1
1
f (t ) cos n[tdt ! ´ t cos nT tdt ´ T
« t sin nT t »
!¬ !
1
sin nT t
dt ´ ¼ nT ½ 1 1 nT
sin nT [ sin( nT )]
nT
! 0 !
1
1
« cos nT t » ¬ 2 2 ¼ - n T ½ 1
cos nT cos( nT )
n T cos nT cos nT 2
2 T n 2
2
!0
since
cos( x) ! cos x
bn !
2
1
1
1
1
f (t ) sin n[t d t ! ´ t sin nT t d t ´ T 1
cos nT t « t cos nT t » d t ! ¬ ¼ nT ½ 1 1 nT 1
´
1
cos nT [ cos( nT )] « sin nT t » ! ¬ 2 2 ¼ nT - n T ½ 1 2 cos nT sin nT sin( nT ) ! 2 2 nT n T 2 cos nT 2(1) n 2(1) n 1 ! ! ! nT nT nT
Finally, f (t ) !
g
a0
§ (an cos n[t bn sin n[t )
2
n !1
g
!§ n !1
!
2
T
2(1)
n 1
nT
sin T t
sin nT t 2 2T
sin 2T t
2 3T
sin 3T t -
Example 3 Given
2 t ,
v(t ) ! ¯
°0
,
0
t 2
2
t 4
v (t 4) ! v(t )
Sketch the graph of v (t ) such that 0 e t e 12. Then compute the Fourier series expansion of v (t ).
Solution The function is described by the following graph: v (t ) 2
2
0
4
6
T = 4
We find that
[!
2T
T
!
T 2
8
10
12
t
Then we compute the coefficients: a0 !
2
4
v(t ) dt ´ T 0
2 4 ® ¾ 2 ! ¯´ (2 t )dt ´ 0dt ¿ 4 °0 2 À
!
1
2
1«
t » 2
(2 t )dt ! ¬2t ¼ ´ 2 22½ 0
2
!1 0
an !
2
4
1
4
0
2
v(t ) cos n[t d t ! ´ ( 2 t ) cos n[t d t ´ 0 ´ 2 T 0
!
2
1 « (2 t ) sin n[t »
2 ¬-
n[
2
1 sin n[t 2
¼½ 2 ´ 0 0
1 « cos n[t »
n[
d t
2
! 0 ¬ 2 2 ¼ 2 - n [ ½0 n 1 cos 2n[ 2(1 cos nT ) 2[1 (1) ] ! ! ! 2 2 2 2 2n [ n T n 2T 2
bn !
2
4
1
!
4
0
2
v (t ) sin n[t d t ! ´ ( 2 t ) sin n[t d t ´ 0 ´ 2 T 0
!
2
1 « ( 2 t ) cos n[t »
2 ¬1
1
2
¼½ 2 ´ 0 0
n[ 1 « sin n[t »
2
cos n[t
2
¬ 2 2 ¼ n[ 2 - n [ ½ 0 2 1 1 sin 2n[ ! ! ! 2 2 2n [ n[ n[ nT since sin 2n[ ! sin nT ! 0
n[
d t
Finally, v(t ) !
!
a0 2 1
g
§ (an cos n[t bn sin n[t ) n !1
g
2[1 (1) n ]
§¯ 2 n !1 ° n 2T 2
¨ nT t ¸ 2 ¨ nT t ¸¾ cos© sin © ¹ ¹¿ ª 2 º nT ª 2 ºÀ
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. f (t ) ! f (t )
Even symmetry (cont.)
The examples of even functions are: f (t ) !| t |
f (t ) ! t 2
t
t f (t ) ! cos t
t
Even symmetry (cont.)
The integral of an even function from í A to + A is twice the integral from 0 to + A f even (t )
A
t í A
+ A
A
´ f
even
A
´
(t )d t ! 2 f even (t )d t 0
Odd symmetry
Any
function f (t ) is odd if its plot is antisymmetrical about the vertical axis, i.e. f (t ) ! f (t )
Odd symmetry (cont.)
The examples of odd functions are: f (t ) ! t 3
f (t ) ! t
t
t f (t ) ! sin t
t
Odd symmetry (cont.)
The integral of an odd function from í A to + A is zero f odd (t )
A í A
+ A
t
´ f
odd
A
(t ) dt ! 0
Even and odd functions The product properties of even and odd functions are:
(even) × (even)
=
(even)
(odd) × (odd)
=
(even)
(even) × (odd)
=
(odd)
(odd) × (even)
=
(odd)
Symmetry consideration From the properties of even and odd functions, we can show that:
for even periodic function; an !
4
T / 2
f (t ) cos n[t dt ´ T
bn ! 0
0
for odd periodic function; a0 ! a n ! 0
bn !
4
T / 2
f (t ) sin n[t dt ´ T 0
How?? [Even function] f (t )
an !
2
T / 2
4
T / 2
T
T
2
2
´ f (t ) cos n[t dt ! T ´ f (t ) cos n[t dt
T T / 2
0
t
bn !
2
T / 2
´ f (t ) sin n[tdt ! 0
T T / 2
(even) × (even)
(even) × (odd)
||
||
(even)
(odd)
How?? [Odd function] f (t )
a0 !
2
´
f (t ) d t ! 0
(odd)
an !
T
2
2
t
T / 2
T T / 2 2
T
T / 2
´ f (t ) cos n[tdt ! 0
T T / 2
bn !
2
T / 2
4
´ f (t ) sin n[t dt ! T ´ f (t ) sin n[t dt
T T / 2
0
(odd) × (odd) ||
(even)
(odd) × (even) ||
(odd)
T / 2
Example 4 1 , 2 t 1 ± Given f (t ) ! ¯ t , 1 t 1 ±1 , 1 t 2 ° f (t 4) ! f (t ) Sketch the graph of f (t ) such that 6 e t e 6. Then compute the Fourier series expansion of f (t ).
Solution The function is described by the following graph: f (t ) 1
í6
í4
í2
í1
0
T = 4
We find that
[!
2T
T
!
T 2
2
4
6
t
Then we compute the coefficients. Since f (t ) is an odd function, then a0 !
2
2
f (t ) dt ! 0 ´ T 2
and an !
2
2
f (t ) cos n[t d t ! 0 ´ T 2
bn !
2
2
4
2
f (t ) sin n[tdt ! ´ f (t ) sin n[tdt ´ T T 2
0
2 1 « » 4 ! ¬ ´ t sin n[tdt ´ 1sin n[tdt ¼ 4 -0 1 ½ 1
2
cos n[t « t cos n[t » « cos n[t » ! ¬ ´ dt ¬ ¼ n[ ½ 0 0 n[ n[ ¼½1 cos n[
1
1
« sin n[t » cos 2n[ cos n[ ! ¬ 2 2 ¼ n[ n[ - n [ ½0 cos 2n[ sin n[ 2 cos nT ! 2 2 ! n[ n[ nT since sin 2n[ ! sin nT ! 0
Finally, f (t ) !
a0 2
g
§ (an cos n[t bn sin n[t ) n !1
nT t ¨ 2 cos nT ¸ ! §© ¹ sin 2 nT º n !1 ª g (1) n 1 nT t sin ! 2§ 2 nT n !1 g
Example 5 Compute the Fourier series expansion of f (t ).
Solution The function is described by 1
0 t 1
,
T = 3
± f (t ) ! ¯2 , 1 t 2 ± 1 , 2 t 3 ° f (t 3) ! f (t ) T = 3
and [ !
2T T
!
2T 3
Then we compute the coefficients. » 2 8 a0 ! ´ f (t ) d t ! ¬ ´ 1d t ´ 2d t ´ 1d t ¼ ! ?(1 0) 2(2 1) (3 2)A! T 0 3 -0 3 2 1 ½ 3 2
2«
3
1
2
3
Or, since f (t ) is an even function, then » 4« ¨ 3 ¸» 8 a0 ! ´ f (t ) dt ! f (t )dt ! ¬ ´ 1dt ´ 2dt ¼ ! ¬(1 0) 2© 1¹¼ ! ´ T 0 T 0 3 -0 ª 2 º½ 3 1 ½ 32
3
4
4«
3/ 2
1
3/ 2
Or, simply a0 !
2
3
´
T 0
f (t ) dt !
2 ¨ Total area below graph ¸
v ©© T ª
in a per iod
2
8
¹¹ ! v 4 ! 3 º 3
an !
2
3
4
3/ 2
f (t ) cos n[tdt ! f (t ) cos n[tdt ´ ´ T T 0
0
1 3/ 2 « » 4 ! ¬ ´ 1 cos n[tdt ´ 2 cos n[tdt ¼ 3 -0 1 ½
!
4 « sin n[t »
1
4 « 2 sin n[t »
¬ ¬ ¼ 3 - n[ ½ 0 3 -
n[
3/ 2
¼½ 1
4 « ¨ 3n[ ¸ » sin n[ 2© sin ! sin n[ ¹ ¼ ¬ 3n[ 2 ª º ½ 4 ¨ 3n[ ¸ ! sin n[ ¹ © 2 sin 3n[ ª 2 º
;
[!
2 ¨ 2nT ¸ 2 2nT sin ! © 2 sin nT sin ¹! nT ª 3 º nT 3
2T 3
and bn ! 0
since f (t ) is an even function.
Finally, f (t ) !
a0 2
g
§ ( an cos n[t bn sin n[t ) n !1
2nT ¸ 2nT t ¨ 2 sin ! §© ¹ cos 3 n !1 ª nT 3 º 3 4 2 g ¨1 2nT ¸ 2nT t ! § © sin ¹ cos 3 T n !1 ª n 3 º 3 4
g
Function defines over a finite interval
Fourier series only support periodic functions In real application, many functions are nonperiodic The non-periodic functions are often can be defined over finite intervals, e.g. y = 2 y = 1
y=1
Therefore, any non-periodic function must be extended to a periodic f unction 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
Periodic extension
Non-periodic function
f (t ) ! y (t )
y (t )
f (t l ) ! f (t )
0 t l
,
3l 2l l 0
T ! l
0
l
f (t )
l
2l 3l
t
T
Even periodic extension f even (t )
t
y (t ) f (t ) ! ¯ y (t ) °
,
0 t l
,
l t 0
f (t 2l ) ! f (t )
3l 2l l 0
T ! 2l
y (t )
,
f odd (t )
0 t l
° y (t ) , l t 0 f (t 2l ) ! f (t ) T ! 2l
2l 3l
T
Odd periodic extension f (t ) ! ¯
l
t
3l 2l l 0 T
l
2l
3l
t
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 f (t ) !
a0 2
g
§ an cos n[t n !1
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 g
f (t ) !
§b
n
sin n[t
n !1
which only contains the sine harmonics.
Therefore, this approach is called as the half-range Fourier sine series
Example 6 Compute the half-range Fourier sine series expansion of f (t ), where f (t ) ! 1 ,
0 t T
Solution Since we want to seek the half-range sine series, the function to is extended to be an odd function: f (t )
f (t )
1
1 0
t
í2
í
0
í1 T = 2
[!
2T T
!1
2
t
Hence, the coefficients are a0 ! a n ! 0
and bn !
4
T / 2
4
T
f (t ) sin n[tdt ! 1sin ntdt ´ ´ T 2T 0
0
T
®4 / nT , 2 « cos nt » 2 (1 cos nT ) ! ¯ ! ¬ ! ¼ , T n ½ 0 nT °0
n odd n even
Therefore, g
f (t ) !
2
g
4
§ nT (1 cos nT ) sin nt ! § nT sin nt n !1
n !1 n odd
Example 7 Determine the half-range cosine series expansion of the function f (t ) ! 2t 1 ,
0 t 1
Sketch the graphs of both f (t ) and the periodic function represented by the series expansion for í3 < t < 3.
Solution Since we want to seek the half-range cosine series, the function to is extended to be an even function: f (t )
f (t )
1
1 1
1
2
1
t
2 3
2
1
1
1 T = 2
[!
2T
T
! T
3
t
Hence, the coefficients are T / 2 1 1 4 4 2 (2t 1) dt ! 2 t t 0 ! 0 a0 ! f (t )dt ! T 0 20
´
an !
4
? A
´
T / 2
4
1
f (t ) cos n[t d t ! ´ ( 2t 1) cos nT t d t ´ 2 T 0
0
1
sin nT t « ( 2t 1) sin nT t » ! 2¬ 2´ 2d t ¼ nT nT ½0 0
!
2 sin nT
nT
1
1
« cos nT t » 4¬ 2 2 ¼ - n T ½ 0
2 2 ® / 8 T n 4(cos nT 1) ! !¯ 2 2 n T ° 0
,
n odd
,
n even
bn ! 0
Therefore, g
f (t ) ! a0
§a
n
cos n[t
n !1
g
8 ¸ 8 ¨ ! 0 § © 2 2 ¹ cos nT t ! 2 n T º T n !1 ª n odd
g
1
§n
n !1 n odd
2
cos nT t
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
f(t)
P dc
P avg
a0 2
=
P a1
t
P b1
a1 cos [t
+
b1 sin [t
+
P a2
P b2
a2 cos 2[t
+
b2 sin 2[t
+
+
«
For sinusoidal (cosine or sine) signal, 2
2
P !
V rms R
¨V peak ¸ © ¹ 2 V 1 peak 2 º ª ! ! 2
R
R
For simplicity, we often assume R = 1, which yields P !
1
2
2 V peak
For sinusoidal (cosine or sine) signal, P avg ! P dc P a P b P a2 P b2 1
1
2
¨ a0 ¸ 1 2 1 2 1 2 1 2 ! © ¹ a1 b1 a2 b2 2 2 2 ª 2 º 2 1
@ P avg ! a02 4
1
g
§ 2 n !1
(an2 bn2 )
Exponential Fourier series
Recall that, from the Euler¶s identity, e s jx ! cos x s j sin x
yields e e jx
cos x !
2
jx
and
e e jx
sin x !
j 2
jx
Then the Fourier series representation becomes f (t ) !
a0 2
g
§ (an cos n[t bn sin n[t ) n !1
« ¨ e jn[t e jn[t ¸ ¨ e jn[t e jn[t ¸» ¹¹ bn ©© ¹¹¼ ! § ¬an ©© 2 n !1 - ª 2 j 2 º ª º½ g « ¨ e jn[t e jn[t ¸ ¨ e jn[t e jn[t ¸» a0 ¹¹ jbn ©© ¹¹¼ ! § ¬an ©© 2 n !1 - ª 2 2 º ª º½ a0
g
«¨ an jbn ¸ jn[t ¨ an jbn ¸ jn[t » ! § ¬© ¹e © ¹e ¼ 2 n !1 -ª 2 º 2 ª º ½ g a0 ¨ an jbn ¸ jn[t g ¨ an jbn ¸ jn[t ! §© ¹e § © ¹e 2 n !1 ª 2 º 2 º n !1 ª a0
g
Here, let we name
and
c0 !
a0 2
cn !
an jbn 2
,
c n !
an jbn 2
. Hence,
¨ an jbn ¸ jn[t g ¨ an jbn ¸ jn[t §© f (t ) ! ¹e § © ¹e 2 n !1 ª 2 º 2 º n !1 ª cín c0 cn g
a0
g
g
n !1
n !1
g
g
n !1
n ! 1
! c0 § cn e jn[t § c n e jn[t ! c0 § cn e jn[t § cn e jn[t !
1
§
cn e jn[t c0
n ! g
g
§ n !1
cn e jn[t !
g
§
cn e jn[t
n ! g
Then, the coefficient cn can be derived from cn !
!
an jbn 2 1 2
T
j 2
T
f (t ) cos n[t d t f (t ) sin n[t d t ´ ´ 2 T 2 T 0
0
T T « » 1 ! ¬ ´ f (t ) cos n[t d t j ´ f (t ) sin n[t d t ¼ T - 0 0 ½
!
1
T
f (t )[cos n[t j sin n[t ]d t ´ T 0
!
1
T
´
T 0
f (t )e jn[t d t
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: c0 ! cn !
a0 2
!
1
T
f (t ) d t t ´ T
cn !
0
an jbn
c n !
2 an jbn 2
or
c n ! c n
1
T
´
T 0
f (t )e jn[t d t t
Example 8 Obtain the complex Fourier series of the following function f (t ) e 2T
1
4T
2T
0
2T
4T
t
Solution Since T ! 2T ,
c0 !
1
[ ! 1 . Hence
T
t f (t )d t ´ T 0
! !
1
2T
2T
´
e t d t t
0
? eA 2T 1
t 2T 0
!
e 2T 1 2T
cn !
1
T
f (t )e ´ T
jn[t
d t t
0
!
1
2T
2T
´
t jnt
d t t !
ee
0
1 «e
1
2T
2T
´
e (1 jn ) t d t t
0
2T
» ! ¬ ¼ 2T - 1 jn ½ 0 (1 jn ) t
2T (1 jn )
since e j 2 nT
e e
1
2T
1 ! ! ! 2T (1 jn) 2T (1 jn) 2T (1 jn) e
1
2T j 2 nT
e
cos 2nT j sin sin 2nT ! 1 0 ! 1 ! cos
cn
n!0
!
e 2T 1 2T (1 jn)
! n!0
e 2T 1 2T
! c0
Therefore, the complex Fourier series of f (t ) is g
f (t ) !
§
cn e jn[t !
n ! g
g
e 2T 1
§ 2T (1 jn)
e jnt
n ! g
*Notes: Even though c0 can be found by substituting cn with n = 0, sometimes it doesn¶t works (as shown in the next example). Therefore, it is always better to calculate c0 alone.
cn is a complex term, and it depends on n. Therefore, we may plot a graph of |cn| vs n.
cn !
e 2T 1 2T 1 n
2
In other words, we have transformed the function f (t ) in the time domain ( t ), to the function cn in the frequency domain ( n).
Example 9 Obtain the complex Fourier series of the function in Example 1.
Solution [ ! T c0 !
1
T
1
1
f (t ) dt ! ´ 1dt ! ´ T 2 2 0
cn !
1
1
0
T
f (t )e ´ T
jn[t
dt !
0
1«e
1
2
1
1e ´ 2
jnT t
0
1
» j jnT e ( ! ¬ ! 1) ¼ 2 - jnT ½ 0 2nT jnT t
´
dt 0 1
But
e
jnT
! cos nT j sin nT ! cos nT ! (1)
Thus, cn !
!
j 2nT j 2nT
(e
jnT
1)
[(1) 1] ! ¯ n
Therefore,
§
n odd n even
,
{ c0 .
n!0
g
f (t ) !
j / nT ,
° 0
*Here notice that cn
n
cn e
n ! g
jn[t
!
1
2
g
j
§ nT
n ! g n{0 n odd
e
jnT t
