Transcript Chapter4-FourierSeries-updated

EE 207
Chapter 3 The Fourier Series
Adil S. Balghonaim
Vectors in 2D
y
V = (3,4)
4
V = 3i + 4j
j
i
Vectors in 3D
N-Dimension
3
x
3= V i
4=V j
V = 3i + 6 j + 4k
V = 1i 1 +  2i 2 + 3i 3 +  +  N i N
1 = V i 1  2 = V i 2   N = V i N
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
Example
x (t )  sin0t  1sin 30t  1sin50t  1 sin 70t
3
5
7
n =1
x 1(t )  sin0t
n = 21
n =3
x 3(t )  sin0t  1sin30t
3
n 

  (1/ n ) sin n0t
n 1
n odd
Fourier proposed the following:
For any periodical function of period
were
1
T0 
f0
1
 f 0
T0
T0
 0  2 f 0
Then


n 1
n 1
x (t )  a0   an cos n 0t  b n sin n0t
Periodicals functions can be decomposed to sin and cosine functions
Similar to decompose a vector in terms of i,j,k,…


n 1
n 1
x (t )  a0   an cos n 0t  b n sin n0t
Similar to the 2D vector
y
V = (3,4)
V = 3i + 4j
4
j
i
3
x
The coordinate with respect to i , j
3=V i
4 =V j
We will find the a0, an, bn in a similar manar
k
i
k  (i , j) space
j
i , j space
i j


n 1
n 1
x (t )  a0   an cos n 0t  b n sin n0t
x (t )  a0 a1 cos0t a2 cos 20t a3 cos30t 
b1sin0t b2 sin 20t b3 sin 30t 
cos0t  cos20t  cos30t...
sin0t  sin20t  sin30t...
n 
{cos n0t}nn
1 {sin n0t}n 1
cos n0t space
Orthognalty (  ) in the functional space will be defined next
In the 3D vector
k  (i , j) space
3
i j =  ik jk  (1)(0)  (0)(1)  (0)(0)  0
k 1
i , j space
i j
Dot Product
i j=0
i k=0
j k=0 j k=0
In the functional space
 cos0t,cos20t    (cos 0t)(cos 20t)dt  0
To
 cos k0t,cos m0t    (cos k0t)(cos m0t)dt  0
To
 sin k0t,sin m0t    (sin k0t)(sin m0t)dt  0
To
 cos k0t,sin m0t    (cos k0t)(sin m0t)dt  0
To
cos n0t space
x (t )  a0 a1 cos0t a2 cos 20t a3 cos30t 
b1sin0t b2 sin 20t b3 sin 30t 
x (t )
cos n0t space
We seek to find the coefficients or coordinates
a0,
a1,a2 ,a3
b1,b2 ,b3
Coefficients with respect to the cos's
Coefficients with respect to the sin's
Finding a0


n 1
n 1
x (t )  a0   an cos n 0t  b n sin n0t
Integrating both side over one period


x (t ) dt

a0dt 
T0



n 1
n 1

a   a n cos n  0t   b n sin n  0t  dt


 a n cos n  0tdt 
n 1
T0

a0dt 

a0dt
T0


n 1
T0


T0
T0




 0

T0
 b n sin n  0tdt
n 1
T0
0



an sinn0tdt 


n 1
0

bn cosn0tdt
T0
a0 T 0
a0 = 1
T0

T0
x (t ) dt
x (t )  a0 a1 cos0t a2 cos 20t a3 cos30t 
b1sin0t b2 sin 20t b3 sin 30t 
x (t )
y
V = (3,4)
4
cos n0t space
j
i
3
x
V = 3i + 4j
3= V i
4=V j
By comparison to the 2D-vector ,
a1  x (t )  cos0t
Projection of x(t) on the direction of cos0t
dot product
a2  x (t )  cos 20t
Similarly
an  x (t )  cos n0t
bn  x (t ) sin n0t
Projection of x(t) on the direction of sin n0t
Finding a5


n 1
n 1
x (t )  a0   an cos n 0t  b n sin n0t
Multiplying both side by cos50t and Integrating over one period

T0
x(t )cos 50tdt

 a cos5 tdt  cos5 t  a cos n tdt




 0




n 1
n 1


0 

a   a n cos n 0 t   bn sin n t cos 50tdt
T0

0
T0
0
0
n
n 1
0
T0


T0

cos 50t  bn sin n 0 tdt
n 1


n 1
n 1
x (t )  a0   an cos n 0t  b n sin n0t

x(t )cos 50tdt

0
 a0 cos50tdt 
T0
T0

cos 50t  a n cos n 0 tdt
T0
Integrating
sinusoidal
over one period



n 1
take it inside summation

cos 50t  bn sin n 0 tdt
n 1
T0

T0
x(t )cos 50tdt 

T0

 a n cos n 0 t cos 5 0 tdt 
n 1
take it inside summation

T0

 bn sin n 0 t cos 50tdt
n 1


n 1
n 1
x (t )  a0   an cos n 0t  b n sin n0t

T0
x(t )cos 50tdt 


 a n cos n 0 t cos 5 0 tdt 
n 1
T0


 bn sin n 0 t cos 50tdt
n 1
T0
From Trigonometric Identity
cos n0t cos 50t  1 cos(n  5)0t  1 cos(n  5)0t
2
2
1
1
sin n0t cos 50t  sin(n  5)0t  sin(n  5)0t
2
2


n 1
n 1
x (t )  a0   an cos n 0t  b n sin n0t

x(t )cos 50tdt 
T0


 a n cos n 0 t cos 5 0 tdt
n 1
T0



 bn sin n 0 t cos50tdt
n 1
T0


T0


T0


1
1
cos(n 5) 0 t  cos(n  5) 0 t dt
 an
 2

n 1
2


 bn
n 1
 1 sin(n 5) t  1 sin(n  5) t dt
0
0
 2

2


n 1
n 1
x (t )  a0   an cos n 0t  b n sin n0t

x(t )cos 50tdt 
T0

 1 cos(n 5) t
a
0
 n

n 1
T0



 bn
n 1
T0
since

T0


n 1
n 1
 

 2
T0
 1 cos(n  5) 0 t dt

2
 1 sin(n 5) t  1 sin(n  5) t dt
0
0
 2

2


n 1
n 1
x (t )  a0   an cos n 0t  b n sin n0t
 0 for all n except n = 5
at n = 5 cos(0) = 1
 0 for all n


x(t )cos 50tdt   an 1 
cos(n  5) 0 tdt 
cos(n  5) 0 t dt
n 1
2

T0
 T0
 cos(0)dt  T
o

T0



To
 0 for all n
 0 for all n


  bn 1 
sin(n  5) 0 tdt 
sin(n  5) 0 t dt
n 1
2

T0
 T0


T0
x(t )cos 50tdt  a5

1 (T )
o
2

a5  2
To

a5 is the projection of x(t) in the direction of cos50t
T0
x(t )cos50tdt


n 1
n 1
x (t )  a0   an cos n 0t  b n sin n0t
a5  2
To

x(t )cos50tdt
a5 is the projection of x(t) in the direction of cos50t
T0
an  2
To
We can repete the same procedure for any n
We can repete the same procedure for b5 and we will get

T0
T0
b5  2
To
Then for any n we get
bn  2
To

x(t )cos n0tdt
x(t )cos n0tdt

T0
x(t )sin50tdt
Therefor


n 1
n 1
x (t )  a0   an cos n 0t  b n sin n0t
1
a0 
x (t )dt
T 0 T0

The average of x(t)
2
an 
x (t ) cos n 0tdt
T 0 T0

2
bn 
x (t ) sin n 0tdt
T 0 T0

n 0
Example 3-4
x (t )
A
0
T
T
0
A 2
The average value of x(t) = 0
an  2 x (t ) cos n 0tdt
T 0 T0

2
T0
T0 2

0
t
0
 a0  0
T0
A cos n0tdt  2 (A ) cos n0tdt
T 0 T0 2
 sin n0t
2
A
 
T 0  n0

T0 2
0
sin n0t

n0
 0

T0 2 

T0
Thus all the an coefficients are zero
Note : x (t ) odd
x (t )cos n0t is odd
an  0
bn  2 x (t ) sin n 0tdt
T 0 T0

2
T0
T0 2

0
T0
A sin n0tdt  2 (A ) sin n0tdt
T 0 T0 2

n odd  cos n 1
 2A 1cos n 
n
n even cos n  1
bn  2A 1cos n 
n
 2A
 n
 

 0
n odd
n even
1
1


x (t )  4A  sin 0t  sin 30t  sin 50t  
3
5
 

n =1
n =3
n = 21
0
0

sin (n  1)  ( 1)
2
n /2

sin (n  1)  (1)
2
n /2
Now for n =`1
1


2

n /2
 2(1)
an  
2

(
n
 1)


0


n 1
n even
n odd
e
e
 jn0t
jn0t
 cos n0t  j sin n0t
 cos n0t  j sin n0t
e
cos n0t 
jn0t
e
sin0t 
e
2
jn0t
 jn0t
e
2j
 jn0t


n 1
n 1
x (t )  a0   an cos n 0t  b n sin n0t

x (t )  a0   an
n 1

1
e
jn 0t
e
2
 jn 0t
x (t )  a0   (an  jbn )e
n 1 2

 b n
n 1
jn0t

1
e
jn 0t
e
2j
 jn 0t
  (an  jbn )e
n 1 2
 jn0t

1
x (t )  a0   (an  jb n )e
n 1 2
1
jn0t

1
  (an  jb n )e
n 1 2
 jn 0t
2
term 1 and term 2 are complex conjugate of each other
Then we can write x(t) as
x (t ) {  X 2 e
 j 2 t
0
 X 1 e  j 0t }  X 0 {X 1 e j 0t  X 2 e j 20t 
Where
X 0  a0
1
1
1
X 1  (a1  jb1) X 2  (a2  jb2)  X n  (an  jbn )
2
2
2
1 (a  jb )
1
1

X

n
n
n
X 1  (a1  jb1) X 2  (a2  jb2)
2
2
2
}

1
x (t )  a0   (an  jb n )e
n 1 2
x (t ) {  X 2 e
 X ne
0

x (t )   X ne
n 1
1
  (an  jb n )e
 jn 0t
2
 X 1 e  j 0t }  X 0  {X 1 e j 0t  X 2 e j 20t 
 jn0t
n 1

n 1 2
1
 j 2 t

jn0t

  X ne

jn 0t
 X ne
jn 0t
n 1
n 1
jn 0t
}

 X 0   X ne
n 1
jn 0t

  X ne
jn 0t
n 



n 1
n 1
n 
x (t )  a0   an cos n 0t  b n sin n0t  X ne
jn 0t



n 1
n 1
n 
x (t )  a0   an cos n 0t  b n sin n0t   X ne
jn 0t
How to find X n ?
Since
1
X n  (an  jbn )
2












2
1
2
Xn
x (t ) cos n0tdt  j
x (t ) sin n0tdt
T 0 T0
2 T 0 T0
1 x (t )e  jn0t dt
1


x (t ) cos n 0t  j sin n 0t dt
T 0 T0
T0

T0






n 1
n 1
n 
x (t )  a0   an cos n 0t  b n sin n0t   X ne
jn 0t
1 x (t )e  jn0t dt
1

X n  (an  jbn )
T 0 T0
2

Another method to find X n ?
 jm 0 t
Multiplying both side of x(t) by e

T0
x (t )e
 jm 0t
dt 

T0

 X ne
n 


  Xn e
n 
T0
jn 0t
and integrating over T0

e jm0t dt
j (n  m )0t
dt

x (t )   X ne
n 

x (t )e
 jm 0t
dt 
T0

 jn 0t
1
1
x (t )e
dt
X n  (an  jbn ) 
T 0 T0
2

jn 0t

 X ne
n 
T0


  Xn e
n 
T0

e
j (n  m )0t
T0

x (t )e
jn 0t
0
dt  
T
 0
 jnm 0t

j (n  m )0t
e jm0t dt
dt
if m  n
if m  n
dt  X m (T 0)
T0
Since it is true for all m then it is true for all n
 jm 0t
1
Xm
x (t )e
dt
T 0 T0
 jn0t
1
Xn
x (t )e
dt
T 0 T0


Example 3-6 Find the complex Fourier series coefficients for
A half-rectified sine wave
x (t )
A
T 0

T0
2
0

A sin 0t
x (t )  

0

T0
2
0 t 
T0
T0
2
T0
 t T0
2
t

A sin 0t
x (t )  

0

 jn0t
1
Xn
x (t )e
dt
T 0 T0

0 t 
T0
2
T0
 t T0
2
since sin0t  e
0  2 /T 0
n  1
j 0t
e
2j
 j 0t

A sin 0t
x (t )  

0

since
e
j (1 n )
e
j
e
 j 1n 
(1)
1
 0

X n  
 A 2
  (1 n )
 (1)
n
n
n odd
n even
n  1
0 t 
T0
2
T0
 t T0
2

A sin 0t
x (t )  

0

 0

X n  
 A 2
  (1 n )
X 1 A
2 jT 0
e


T0 / 2
0
Similarly
X 1   A
4j
n odd
n  1
n even
j 0t
e
 j 0t
e
 j 0t
0 t 
T0
2
T0
 t T0
2
X n  1 x (t )e
T 0 T0
A

dt
2 jT 0

1e
dt



T0 / 2
0
 j 20t
 jn0t
dt
A
4j

A sin 0t
x (t )  

0

First Entry in Table 3-1
0 t 
T0
2
T0
 t T0
2
Table 3-1(old Book)
Symmetry Properties of Fourier Series coefficients


n 1
n 1

x (t )  a0   an cos n 0t  b n sin n0t
a0  1 x (t )dt
T 0 T0
an  2 x (t ) cos n 0tdt
T 0 T0
bn  2 x (t ) sin n 0tdt
T 0 T0


  X ne
n 

 jn t
X n  1  x (t )e
T 0 
0
n 0

X 0 a0
X n  1 (an  jbn )
2
X n  1 (an  jbn )  X n
2
an  2Re[X n ]
bn 2Im[X n ]
an  X n  X
*
n
jn 0t
*
*
X
bn  n  X n
j
Line Spectra

x (t )   X ne
jn 0t
n 
{  X 2 e
 j 2 t
0
 X 1 e  j 0t }  X 0  {X 1 e j 0t  X 2 e j 20t 
where
X n | X n |  n
X ne
Therefore,
x(t)
jn0t
In general a complex number that can be represented as
a phasor
Is a rotating phasor of frequency
consists of a summation of rotating phasors
n 0
}
Ck  |Ck |  k
k
o
90
|Ck |
10
10
30o
2.5
2.0
3
2
  
1 0 1
3
2
30o
90o
k
1.5
2.5
1.5 2.0
3 2 1 0 1 2 3
k
since x(t ) is an odd function  Co  0
Fourier domain or Frequency domain
Time domain
Ck
one to one
Known

Let y(t) =Ax(t) + B
  Cky e
jk0t
k 
unknown
Question what are the unknown coefficients Cky interms of the known coefficients Ckx
Writing y(t) as
C
C
0y
 AC
0x
B
C
ky
 AC
kx
k 0
C
0y
ky
Let x(t) be as shown
one to one
Let y(t ) be as shown

y(t)   Cky e
jk0t
k 
one to one
unknown
what are C and C
oy
ky
Let y(t) =Ax(t) + B
We wish to find the Fourier series for the sawtooth signal y(t)
First, note that the total amplitude variation of x(t) is X o while the total variation of y(t ) is 4.
Also note that we invert x(t) to get y(t), yielding A=  4
Xo
 y(t ) =Ax(t ) + B   4 x(t ) + 1
Xo
one to one

y(t)   Cky e
jk0t
k 
one to one
unknown
what are C and C
oy
y(t ) =Ax(t ) + B   4 x(t ) + 1
Xo
ky