chapter 3 the fourier series ee 207 adil s. balghonaim
TRANSCRIPT
Chapter 3 The Fourier Series
EE 207
Adil S. Balghonaim
i
j
(3,4)V =3 + 4V = i j
x
y
3 + 6 + 4V = i j k
1 2 3+ + + + N 1 2 3 NV = i i i i
1 2 N 1 2 Ni= V = V = V i i
Vectors in 2D
3
4
Vectors in 3D
N-Dimension
3
4
i = V
= V j
Jean Baptiste Joseph Fourier (21 March 1768 – 16 May 1830) was a French mathematician and physicist best known for initiating the investigation of Fourier series and their applications to problems of heat transfer and vibrations
Fourier observed that the addition of Sinusoidal functions with different frequencies and amplitude resulted in other periodical functions
= 1n = 3n
= 21n
0 0 001sin( ) sin 1si1sin5 3
775
3 n tx t t tt
Example
01( ) sin x t t03 0( ) sin 1sin3
3 x t t t
n
0
1
sin(1/ )
n oddn
n tn
0 0
10
1
( ) cos sinnn
n n
x t n ta a b n t
Fourier proposed the following:
For any periodical function of period 0T
0
0
1Tf
were
0
0
1fT
0 0 2 f
Then
Periodicals functions can be decomposed to sin and cosine functions
Similar to decompose a vector in terms of i,j,k,…
0 0
10
1
( ) cos sinnn
n n
x t n ta a b n t
The coordinate with respect , to
3
4
i j
i= V
=
V
j 0 ,We will find the in a similar man , a rn nba a
3 + 4V = i j
i
j
(3,4)V =
x
y
3
4
Similar to the 2D vector
i
j
k
, spacei j i j
( , ) spacek i j
0 0
10
1
( ) cos sinnn
n n
x t n ta a b n t
1
0
2
0 0
0
30
0
1
3
2
0
( ) cos cos cos sin
2sin
32 sin 3b b b
x t t t tt t t
a a a a
0cos n t space
0 0 0
0 0 0
0 01 1
cos cos2 cos3 ...
sin sin2 sin3 ...
{cos } {sin }n nn n
t t t
t t t
n t n t
Orthognalty ( ) in the functional space will be defined next
3
1k
kki j
i =j (1)(0) (0)(1) (0)(0) 0
In the 3D vector
Dot Product
In the functional space
0 00 0cos ,cos2 (cos )(cos 2 )To
t t t t dt 0
0 00 0
0 00 0
0 00 0
cos ,cos (cos )(cos ) 0
sin ,sin (sin )(sin ) 0
cos ,sin (cos )(sin ) 0
To
To
To
k t m t k t m t dt
k t m t k t m t dt
k t m t k t m t dt
0 0 0 0 = i j i k j k j= k = =
, spacei j i j
( , ) spacek i j
0cos n t space
1
0
2
0 0
0
30
0
1
3
2
0
( ) cos cos cos sin
2sin
32 sin 3b b b
x t t t tt t t
a a a a
We seek to find the coefficients or coordinates
Coefficients with respect to the Coefficients with respect to the
1 2 30 1 3, , 2, , , b b ba a a acos's sin's
0cos n t space
( )x t
0 0
10
1
( ) cos sinnn
n n
x t n ta a b n t
Integrating both side over one period
0
( )
T
x t dt0
0 0 01 1
cos sinn n
n nT
dtbn t n ta a
0 0 0
0 01 1
0
cos sinn n
n nT T T
dt n tdt na tdtba
0 0 0
0 0
1 1
0
sin cos n
n
n
n
T T T
dt n tdt da b n t ta
0 0
0
0
T
dta 00 a T
0
0
0
( ) 1 =
T
x t dtT
a
0Finding a
0cos n t space
( )x t
i
j
(3,4)V = 3 + 4V = i j
x
y
3
43
4
i = V
= V j
1
0
2
0 0
0
30
0
1
3
2
0
( ) cos cos cos sin
2sin
32 sin 3b b b
x t t t tt t t
a a a a
By comparison to the 2D-vector ,
0dot product
1 ( ) cosa x t t
02 ( ) c 2ost ta x 0( ) cosn x ta t nSimilarly
0( ) sinn x tb t n
0Projection of x(t) on the direction of cos t
0Projection of x(t) on the direction of sinn t
0 0
10
1
( ) cos sinnn
n n
x t n ta a b n t
Multiplying both side by cos5w0t and Integrating over one period
0
0
( )c 5os
T
x t tdt0
00 001 1
cos si 5n cosn n
nn
T
dtn t n t ta ba
0 0
0
0 0 01
0 01
0
cos cos cos
co
5 5
5s sin
n
n
n
n
T T
T
t
b
a adt t n tdt
t n tdt
5Finding a
0 0
10
1
( ) cos sinnn
n n
x t n ta a b n t
0
0
( )c 5os
T
x t tdt0 0
0
0 0 01
0 01
0
cos cos cos
co
5 5
5s sin
n
n
n
n
T T
T
t
b
a adt t n tdt
t n tdt
0
Integrating sinusoidal over one period
take it inside summation
take it inside summation
0 0 0
00 00 01 1
( )cos cos cos sin cos5 5 5n
n
n
n
T T T
x t tdt n t tdt n t tb tda
0 0
10
1
( ) cos sinnn
n n
x t n ta a b n t
0 0 0
00 00 01 1
( )cos cos cos sin cos5 5 5n
n
n
n
T T T
x t tdt n t tdt n t tb tda
0 0 0 0
0 0 0 0
1 12 21 12 2
cos cos cos( ) cos( )
si
5 5 5
5 5n cos sin( ) sin( )5
n t t n t n t
n t t n t n t
From Trigonometric Identity
0 0
10
1
( ) cos sinnn
n n
x t n ta a b n t
0 0
0
00 01
001
( )cos cos cos
sin c s5o
5 5n
n
n
n
T T
T
x t tdt n t tdt
b t
a
n t dt
0
0
0 01
0 01
1 1cos( ) cos( )2 2
1 1si
5 5
5n( ) sin( )2 2
5
n
nn
n
T
T
a
b
n t n t dt
n t n t dt
0 0
10
1
( ) cos sinnn
n n
x t n ta a b n t
0 0
0
0 001
0 01
1 1( )cos cos( ) cos( )2 2
1 1 sin( ) sin( )2
5 5 5
5 52
n
n
n
n
T T
T
a
b
x t tdt n t n t dt
n t n t dt
0 01 1
since n nTT
0 0
10
1
( ) cos sinnn
n n
x t n ta a b n t
0 0 0
0 0
0 001
0 01
1( )cos cos( ) cos( )2
1 sin
5
( ) sin( )2
5 5
5 5
n
nn
n
T
b
ax t tdt n tdt n t dt
T T
n tdt n t dt
T T
0 for all n except n = 5
cos(0)
at n = 5 cos(0) = 1
odt T
To
0 for all n 0 for all n
0 for all n
0
0
55( )cos 1( )2
T
ox t tdt a T 0
0
5 ( )cos52oT
x t tdtaT
05 is the projection of ( ) in the direction of cos 5x ta t
0 0
10
1
( ) cos sinnn
n n
x t n ta a b n t
0
0
5 ( )cos52oT
x t tdtaT
05 is the projection of ( ) in the direction of cos 5x ta t
We can repete the same procedure for any n
0
0
( )cos2oT
n x t tdtnT
a 5We can repete the same procedure for and we will get b
0
0
5 ( )sin52oT
x t tdtbT
Then for any n we get
0
0
( )cos2oT
n x t tdtnT
b
0 0
10
1
( ) cos sinnn
n n
x t n ta a b n t
0
0
0
1 ( )T
x t dtT
a The average of x(t)
0
00
cos2 ( ) 0n
T
a n tx t dt nT
00
0sin2 ( )n
T
b nx t dt tT
Therefor
Example 3-4
The average value of x(t) = 00 0 a
00
0cos2 ( )n
T
a nx t dt tT
20
0
0 0 20
0
0
0
2 2co c)s o( s T T
T
A dt A dtT T
n t n t
0 0
0 0
20 0
0 0 20
sin sin2 T T
T
n t n tn n
AT
0
Thus all the coefficients are ro zena
Note : ( ) oddx t 0( ) is oddcos n tx t 0na
00T
0
2
T
A
A
( )x t
t
00
0sin2 ( )n
T
b nx t dt tT
20
0
0 0 20
0
0
0
2 2si s)n i( n T T
T
A dt A dtT T
n t n t
2 1 cosnAn
2 odd
even0
An
n
n
0 0 01 14( ) sin sin3 sin53 5
Ax t t t t
oddn cos 1n
evenn cos 1n
2 1 cosnbn
nA
= 1n = 3n = 21n
00
/ 2sin ( 1) ( 1)2
nn / 2sin ( 1) ( 1)
2nn
Now for n =`1
/ 2
2
11
2
2( 1) even
( 1)
0 odd
n
n
n
a nn
n
00 0cos sin
jn te n t j n t
00 0cos sin
jn te n t j n t
0 0
0cos 2
jn t jn te en t
0 0
0sin 2
jn t jn te etj
0 0
10
1
( ) cos sinnn
n n
x t n ta a b n t
0 0 0 0
1 10( )
2 2
jn t jn t jn t jn t
n
nn
n
e e e ex t ba a
j
0 0
10
1
1 1
2 2( ) ( ) ( )jn t jn t
n n n
n n
n ba a ax t j e j eb
0 0
1 11 2
0
1 1
2 2( ) ( ) ( )nn
jn t jn n
n t
n n
x t b bj e j ea a a
term 1 and term 2 are complex conjugate of each other
Then we can write x(t) as
1 1 2 21 21 1( ) ( )2 2
j jX a X ab b
11 221 21 1( ) ( )2 2
a aj jX b X b
1( )2
n n nja bX
1( )2
n nnX a bj
0 0
X a
0 010 20 02 12 2 {( )
{ }} jj t tj jt tX e X e X X et X ex
Where
0 0
1 11 2
0
1 1
2 2( ) ( ) ( )nn
jn t jn n
n t
n n
x t b bj e j ea a a
0 0100 202 1
2 2( )
{ } { } j t jj t j t tXX e X ex et XX e
0 00
1 1
( ) jn t jn t
n
n
n
nXx e X eXt
0jn t
n
nX e
0
1
jn tn
n
X e
0
1
jn tn
n
eX
0
1
jn t
n
neX
0 0
10
1
( ) cos sinnn
n n
x t n ta a b n t
0jn t
n
nX e
0 0
10
1
( ) cos sinnn
n n
x t n ta a b n t
0jn t
n
nX e
How to find ? nX
1( )2
n n njbX a Since
0
00 00
021 2 ( ) ( )cos sin
2 TT
n n tx t dt j x tn tTT
X dt
00
00
1 ( ) scos inT
n tx t n tj dtT
0
00
1 ( ) jn t
T
x t e dtT
0 0
10
1
( ) cos sinnn
n n
x t n ta a b n t
0jn t
n
nX e
1( )2
n n njbX a 0
00
1 ( ) jn t
T
x t e dtT
Another method to find ? nX0jm t
0Multiplying both side of x(t) by e and integrating over T
0
0
( ) jm t
T
x t e dt
0
0
0 jn t
T
njm t
ne e dtX
0
( ) 0 j n m t
nT
n e dtX
0( ) jn tn
n
Xx t e
1( )2
n n njbX a 0
00
1 ( ) jn t
T
x t e dtT
0
0
( ) jm t
T
x t e dt
0
0
0 jn t
T
njm t
ne e dtX
0
( ) 0 j n m t
nT
n e dtX
0
( ) 0
0
0 if m n
if m nj n m t
T
e dtT
0
0
0
( ) ( )jnm t
m
T
x t e dt X T 0
00
1 ( ) jm t
m
T
X x t e dtT
Since it is true for all m then it is true for all n
0
00
1 ( ) jn t
n
T
X x t e dtT
0
( )x t
0
2
T0T t
A
00
00
sin 02( )
02
TA t t
x tT
t T
Example 3-6 Find the complex Fourier series coefficients for A half-rectified sine wave
0
2
T0T
00
00
sin 02( )
02
TA t t
x tT
t T
0
00
1 ( ) jn tn
T
X x t e dtT
1n
0 0
0since sin 2
j t j te etj
0 02 /T
00
00
sin 02( )
02
TA t t
x tT
t T
(1 ) 1since j n j j ne e e
1 ( 1)n
( 1)n
2
0 odd
n 1 evenA
(1 )
n
n
Xn
n
00
00
sin 02( )
02
TA t t
x tT
t T
2
0 odd
n 1 evenA
(1 )
n
n
Xn
n
0 / 2
0 00
1
00
2
Tj t j t j tAX e e e dt
jT
0 / 2
2 0
00
1 2
Tj tA e dt
jT
4
Aj
Similarly
1 4AXj
0
00
1 ( ) jn tn
T
X x t e dtT
00
00
sin 02( )
02
TA t t
x tT
t T
First Entry in Table 3-1
Table 3-1(old Book)
0jn tn
n
X e
0
0
1 ( ) jn tnX x t eT
0 0
10
1
( ) cos sinnn
n n
x t n ta a b n t
0
00
cos2 ( ) 0n
T
a n tx t dt nT
00
0sin2 ( )n
T
b nx t dt tT
1( )2
nnnX a jb
00X a
1( )2
n nn aX bj *nX
*n n na X X
*n nnb XjX
Symmetry Properties of Fourier Series coefficients
0
0
0
1 ( )T
x t dtT
a
2Re[ ]n na X 2Im[ ]n nb X
Line Spectra
00 02 1 0 01 2
2 2
{ }
{
}
j t j tt t jj X X eX e X eX e
0( ) jn tn
n
Xx t e
where
| |n nnX X In general a complex number that can be represented as a phasor
0tn
jnX e Is a rotating phasor of frequency 0n
Therefore, x(t) consists of a summation of rotating phasors
| |k kkC C
| |kC
10
0 1 2 3
1.52.5
2.01.52.5
2.0
123 k
10
0 1 2
3
12
3 k
30o
30o
90o
90o
k
since ( ) is an odd function 0o
x t C
kC
Time domain Fourier domain or Frequency domain
one to one
Let ( ) = ( ) + y t Ax t B
Known
0
kyjk t
k
C e
unknown
what are the coefficients interms of the coefficientsky kxC CQuestion unknown known
Writing y(t) as
0
y
C
kyC
0 0y xC AC B 0
ky kxC AC k
Let ( ) be as shownx t
Let ( ) = ( ) + y t Ax t B
Let ( ) be as showny t
one to one
one to onewhat are and
oy kyC C
0( ) kyjk t
k
Cy t e
unknown
We wish to find the Fourier series for the sawtooth signal ( ) y t
First, note that the total amplitude variation of ( ) is while the total variation of ( ) is 4. x t X y to4Also note that we invert x(t) to get y(t), yielding =o
AX
4 ( ) = ( ) + ( ) + 1o
y t Ax t B x tX
one to one
one to onewhat are and
oy kyC C
0( ) kyjk t
k
Cy t e
unknown
4 ( ) = ( ) + ( ) + 1o
y t Ax t B x tX