# Fourier Transform Pair: Frequency and Time

```Numerical Methods
Fourier Transform Pair
Part: Frequency and Time
Domain
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Lecture # 5
Chapter 11.03: Fourier Transform
Pair: Frequency and Time Domain
Major: All Engineering Majors
Authors: Duc Nguyen
http://numericalmethods.eng.usf.edu
4/10/2015
http://numericalmethods.eng.usf.edu
5
Example 1
 t for 0  t  
f (t )  
 for   t  2  (  T )
f 1 ( t )  a 0  a1Cos ( t )  b1 Sin ( t )
f 2 ( t )  a 0  a1Cos ( t )  b1 Sin ( t )  a 2 Cos ( 2 t )  b 2 Sin ( 2 t )
6
f 4 ( t )  a 0  a1Cos ( t )  b1 Sin ( t )  a 2 Cos ( 2 t )  b 2 Sin ( 2 t )
 a 3 Cos ( 3 t )  b 3 Sin ( 3 t )  a 4 Cos ( 4 t )  b 4 Sin ( 4 t )
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Frequency and Time Domain
The amplitude (vertical axis) of a given periodic function
can be plotted versus time (horizontal axis), but it can
also be plotted in the frequency domain as shown in
Figure 2.
7
Figure 2 Periodic function (see Example 1 in Chapter 11.02
Continuous Fourier Series) in frequency domain.
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Frequency and Time Domain cont.
Figures 2(a) and 2(b) can be described with the
following equations from chapter 11.02,

~ ikw 0 t
f (t )   C k e
(39, repeated)
k  
where

~
1 T
 ikw 0 t
C k     f (t )  e
dt
T  0
8

(41, repeated)
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Frequency and Time Domain cont.
For the periodic function shown in Example 1 of
Chapter 11.02 (Figure 1), one has:
w 0  2 f 

2
T

2
2
1
2
~
1 
 ikt
 ikt
C k     t  e dt     e dt

T  0
9

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Frequency and Time Domain cont.
Define:


A  t  e
0
 ikt

   1   ikt 
 1
dt   t  
e    
  ik 
 0 0  ik
  ikt
 e dt

or
  
A  
  ik
  ik    1   ik 
  1
e
 k2  e


 
1 
    i   ik   1   ik 
   e
  2 e
 2 
k 
k 
  k 
10
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Frequency and Time Domain cont.
Also,
2
B   e

 ikt
  ikt   
dt   e 
 ik

2



     ik 2 
  i   ik 2 
 ik 
 ik 



B 
e

e

e

e

 
 ik 
 k 
11

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Frequency and Time Domain cont.
~
 1
Ck  
 2
~
 1    ik    i
Ck  
e  
 2  
 k
Thus:

 A  B 

1
 i  1   i   ik 2  
   2   e

2
k
k  k
 k 

Using the following Euler identities
 ik 
e
 cos(  k  )  i sin(  k  )
 cos( k  )  i sin( k  )
e
12
 ik 2 
 cos( k  )
 cos( k 2  )  i sin( k 2  )  cos( k 2  )
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Frequency and Time Domain cont.
Noting that cos( k 2 )  1 for any integer k
~
 1
Ck  
 2
13
1


 1 
 i 
  Cos ( k  )   2   2    Cos ( k 2  ) 
1 

k  k
 k 
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Frequency and Time Domain cont.
Also,
  1 for k  odd number (  1, 3 , 5 , 7 ,...)
cos( k  )  
  1 for k  even number (  2 , 4 , 6 ,8 ,...)
Thus,
~
 1
Ck  
 2
1
i 
    1
 2  

2
k
k 
 k
k


~
k
 1 
 1 
  1  1    i
Ck  
2 
 2 k 
 2k 
14
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Frequency and Time Domain cont.
From Equation (36, Ch. 11.02), one has
a k  ib k
~
(36, repeated)
Ck 
2
Hence; upon comparing the previous 2 equations,
one concludes:
 1 
k
(  1)  1
ak  
2 
 k 
15
 1
bk  

 k 
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Frequency and Time Domain cont.
For k  1, 2 , 3 , 4 ... 8 ; the values for a k and b k
(based on the previous 2 formulas) are exactly
identical as the ones presented earlier in Example
1 of Chapter 11.02.
16
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Frequency and Time Domain cont.
Thus:
2
a 1  ib1
~
C1 
 
2
 i (  1)

2
1


1
i
2
 1
0  i  
a 2  ib 2
1
~
 2
C2 

0 i
2
2
4
17
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Frequency and Time Domain cont.
  2   1

  i

a 3  ib 3  9    3    1  1
~
C3 


 i
2
2
 9  6
 1
0  i

a

ib
1
~
 4 
4
4
C4 

0 i
2
2
8
18
  2   1

  i

a 5  ib 5  25    5    1  1
~
C5 


i

2
2
 25   10
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Frequency and Time Domain cont.
 1
0  i

a 6  ib 6
1
~
 6 
C6 

0
i
2
2
12
  2   1

  i

a 7  ib 7  49    7    1  1
~
C7 


i

2
2
 49   14
19
 1
0  i

a 8  ib 8
1
~
 8 
C8 

0
i
2
2
16
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Frequency and Time Domain cont.
In general, one has
 1  1 
 k 2    2 k  i for k  1, 3 , 5 , 7 ,..  odd number



~
Ck  
 1 


 i for k  2 , 4 , 6 ,8 ,..  even number

 2k 
20
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THE END
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This material is based upon work supported by the National
Science Foundation under Grant # 0717624. Any opinions,
findings, and conclusions or recommendations expressed in
this material are those of the author(s) and do not necessarily
reflect the views of the National Science Foundation.
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Numerical Methods
Fourier Transform Pair
Part: Complex Number in Polar
Coordinates
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Lecture # 6
Chapter 11.03: Complex number in
polar coordinates (Contd.)
In Cartesian (Rectangular) Coordinates, a
~
complex number C k can be expressed as:
~
C k  R k   I k i
~
In Polar Coordinates, a complex number C k can
be expressed as:
~
i
C k  Ae  A cos(  )  i sin(  )   A cos(  )   A sin(  )i
29
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Complex number in polar coordinates cont.
Thus, one obtains the following relations between
the Cartesian and polar coordinate systems:
R k  A cos 

I k  A sin 

This is represented graphically in Figure 3.
30
Figure 3. Graphical representation of the complex number
system in polar coordinates.
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Complex number in polar coordinates cont.
Hence
R k  I k  A cos    A sin
2
2
2
2
2
2
   A 2 cos 2 ( )  sin 2 ( ) 
 Rk 
cos(  ) 
   cos 

A
 A 
Rk
sin(  ) 
Ik
A
31
1
   sin
1
and
 Ik 
 
 A
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Complex number in polar coordinates cont.
~
Based on the above 3 formulas, the complex numbers C
can be expressed as:
1  1 
~
i ( 2 . 13770783
C1 
   i  ( 0 . 59272353 ) e

2
32
)
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k
Complex number in polar coordinates cont.
Notes:
~
(a) The amplitude and angle C 1 are 0.59 and
2.14 respectively (also see Figures 2a, and
2b in chapter 11.03).
Im

(b) The angle  (in radian) obtained from
Cos ( ) 
Rk
A
However based on Sin ( ) 
Ik
A
33
o).
Then  = 1.004 radians (=57.52http://numericalmethods.eng.usf.edu
Re
Complex number in polar coordinates cont.
Since the Real and Imaginary components of  are
negative and positive, respectively, the proper selection
for  should be 2.1377 radians.
1
~
C 2  0  i  ( 0 . 25 ) e
4
 
i

 2 
 ( 0 . 25 ) e
i ( 1 . 57079633
~
 1 1
i ( 1 . 77990097
C3  

i

(
0
.
17037798
)
e

 9  6
34
)
)
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Complex number in polar coordinates cont.
1
~
C 4  0  i  ( 0 . 125 ) e
8
 
i

 2 
 ( 0 . 125 ) e
i ( 1 . 57079633
)
1
~
 1 
i ( 1 . 69743886
C5  

i

(
0
.
100807311
)
e

 25   10
1
~
C6  0 
i  ( 0 . 08333333 ) e
12
35
 
i 
 2 
 ( 0 . 08333333 ) e
)
i ( 1 . 57079633
http://numericalmethods.eng.usf.edu
)
Complex number in polar coordinates cont.
1
~
 1 
i ( 1 . 66149251
C7  
i  ( 0 . 07172336 ) e

 49   14
1
~
C8  0 
i  ( 0 . 0625 ) e
16
36
)
 
i 
 2 
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THE END
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Acknowledgement
This instructional power point brought to you by
http://numericalmethods.eng.usf.edu
Committed to bringing numerical methods to the
For instructional videos on other topics, go to
http://numericalmethods.eng.usf.edu/videos/
This material is based upon work supported by the National
Science Foundation under Grant # 0717624. Any opinions,
findings, and conclusions or recommendations expressed in
this material are those of the author(s) and do not necessarily
reflect the views of the National Science Foundation.
The End - Really
Numerical Methods
Fourier Transform Pair
Part: Non-Periodic Functions
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Lecture # 7
Chapter 11. 03: Non-Periodic
Functions (Contd.)
Recall

~ ikw 0 t
f (t )   C k e
(39, repeated)
k  

~
1 T
 ikw t
C k     f ( t )  e 0 dt
T  0

(41, repeated)
Define
T
2
 ikw 0 t
ˆ
F ( ikw 0 )   f ( t ) e
dt

(1)
T
2
45
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Non-Periodic Functions
Then, Equation (41) can be written as
~
1
Ck  
T
 ˆ
  F ( ikw 0 )

And Equation (39) becomes
1
f (t )   
k  
T

From above equation
f np ( t )  lim
or
46
T 
or  f  0
 ˆ
ikw 0 t
  F ( ikw 0 ) e


ikw t
f ( t )  lim  (  f )  Fˆ ( ikw 0 ) e 0
 f  0 k  

ik 2   ft
f np ( t )  lim  (  f )  Fˆ ( ik 2   f ) e
 f  0 k  
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Non-Periodic Functions cont.
From Figure 4,
k f  f
i 2  ft
ˆ
f np ( t )   df  F ( i 2  f ) e
i 2  ft
f np ( t )   Fˆ ( i 2  f ) e df
Figure 4. Frequency are discretized.
47
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Non-Periodic Functions cont.
Multiplying and dividing the right-hand-side of the
equation by 2  , one obtains
 1  ˆ
iw 0 t
f np ( t )  
  F ( iw 0 ) e d ( w 0 ) ; inverse Fourier
 2   
transform
Also, using the definition stated in Equation (1), one gets

 iw 0 t
ˆ
F ( iw 0 )   f np ( t ) e
d ( t ) ; Fourier transform

48
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THE END
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