Numerical Methods
Discrete Fourier Transform
Part: Discrete Fourier Transform
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Lecture # 8
Chapter 11.04 : Discrete Fourier
Transform (DFT)
Major: All Engineering Majors
Authors: Duc Nguyen
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Numerical Methods for STEM undergraduates
4/13/2015
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5
Discrete Fourier Transform
Recalled the exponential form of Fourier series
(see Eqs. 39, 41 in Ch. 11.02), one gets:
~ ikw0t
f (t )   Ck e

(39, repeated)
k  

~ 1 T
Ck     f (t )  e ikw0t dt
T  0
6

(41, repeated)
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Discrete Fourier Transform
If time “ t ” is discretized at
t1  t , t2  2t , t3  3t ,.......,tn  nt ,
then Eq. (39) becomes:
~ ikw0tn
f (t n )   Ck e
N 1
k 0
7
(1)
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Discrete Fourier Transform cont.
To simplify the notation, define:
tn  n
(2)
Then, Eq. (1) can be written as:
~ ikw0n
f ( n)   Ck e
N 1
(3)
k 0
 ilw0 n
Multiplying both sides of Eq. (3) by e
, and performing
the summation on “n”, one obtains (note: l = integer
number)
8
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Discrete Fourier Transform cont.
N 1
 f ( n)  e
n 0
ilw0 n
~ ikw0n ilw0n
   Ck e  e
N 1N 1
n 0 k 0
~
   Ck e
N 1N 1
n 0 k 0
9
(4)
i ( k l )
2
n
N
(5)
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Discrete Fourier Transform cont.
Switching the order of summations on the
right-hand-side of Eq.(5), one obtains:
N 1
 f ( n)  e
 2
 il 
 N

n

n 0
Define:
N 1
 2
N 1 i ( k l ) 
 N
k 0
n 0
~
  Ck  e

n

 2 
N 1 i ( k  l ) 
n
 N 
(6)
(7)
A  e
n 0
There are 2 possibilities for ( k  l ) to be
considered in Eq. (7)
10
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Discrete Fourier Transform—Case 1
Case(1): ( k  l ) is a multiple integer of N,
such as: (k  l )  mN ; or k  l  mN where
m  0,1,2,......
Thus, Eq. (7) becomes:
N 1
A  e
n 0
im 2n
N 1
  cos(mn2 )  i sin( mn2 )
n 0
(8)
Hence:
A N
11
(9)
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Discrete Fourier Transform—Case 2
Case(2): ( k  l ) is NOT a multiple integer of N .
In this case, from Eq. (7) one has:

A   e
n 0

N 1
 2 
i ( k l ) 

 N 



n
(10)
Define:
ae
12
i ( k l )
2
N
2 
2 


 cos(k  l ) )  i sin (k  l ) )
N 
N 


(11)
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Discrete Fourier Transform—Case 2
a  1;because (k  l ) is “NOT” a multiple integer of N
Then, Eq. (10) can be expressed as:
A   a
N 1
n 0
13
n
(12)
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Discrete Fourier Transform—Case 2
From mathematical handbooks, the right side of
Eq. (12) represents the “geometric series”, and can
be expressed as:
A   a  N ; if
N 1
n
n 0
a 1
1 a

; if a  1
1 a
(13)
N
14
(14)
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Discrete Fourier Transform—Case 2
Because of Eq. (11), hence Eq. (14) should be used
to compute A . Thus:
1  a N 1  e i ( k l ) 2
A

1 a
1 a
e
15
i ( k l ) 2 
(See Eq. (10))
 cos(k  l )2   i sin(k  l )2   1
(15)
(16)
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Discrete Fourier Transform—Case 2
Substituting Eq. (16) into Eq. (15), one gets
A0
(17)
Thus, combining the results of case 1 and case 2,
we get
A N 0  N
16
(18)
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THE END
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Acknowledgement
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Committed to bringing numerical methods to the
undergraduate
For instructional videos on other topics, go to
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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.
The End - Really
Numerical Methods
Discrete Fourier Transform
Part: Discrete Fourier Transform
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this one.
Lecture # 9
Chapter 11.04: Discrete Fourier
Transform (DFT)
Substituting Eq.(18) into Eq.(7), and then referring
to Eq.(6), one gets:
N 1
 f ( n)e
n 0
ilw0 n
~
  Ck  N
N 1
(18A)
k 0
Recall k  l  mN (where l, m are integer numbers),
And since k must be in the range 0  N  1, m  0.
Thus:
25
k  l  mN becomes k  l
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Discrete Fourier Transform—Case 2
Eq. (18A) can, therefore, be simplified to
N 1
~
 ilw0 n
 Cl  N
 f ( n )e
n 0
~ ~  1  N 1
ikw0 n
Cl  Ck     f (n)e
 N  n 0
 1  N 1
    f (n)cos(kw0 n)  i sin( kw0 n)
 N  n 0
where n  t n and
(18B)
Thus:
(19)
f (n)   Ck eikw0n   Ck cos(kw0 n)  i sin( kw0 n)
26
N 1 ~
N 1 ~
k 0
k 0
(1, repeated)
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Discrete Fourier Transform cont.
Equations (19) and (1) can be rewritten as
~
C n   f ( k )e
N 1
2

 ik  w0 
N


n

k 0
1 ~
f (k )     Cn e
 N  n 0
N 1
27
2 

ik  w0 
n
N 

(20)
(21)
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Discrete Fourier Transform cont.
To avoid computation with “complex numbers”,
Equation (20) can be expressed as
~ R ~ I N 1 R
I

Cn  iCn    f (k )  i f (k )  cos( )  i sin( )
k 0

(20A)
where
2 

  k  w0  n
N

28
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Discrete Fourier Transform cont.
~ R ~ I N 1 R
Cn  iCn    f (k )  cos( )  f I (k ) sin( )
k 0
 i f I (k ) cos( )  f R (k ) sin( )
(20B)
The above “complex number” equation is equivalent to
the following 2 “real number” equations:
~ R N 1 R
Cn    f (k ) cos( )  f I (k ) sin( )
(20C)
~ I N 1 I
Cn    f (k ) cos( )  f R (k ) sin( )
(20D)
k 0
k 0
29
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THE END
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Acknowledgement
This instructional power point brought to you by
Numerical Methods for STEM undergraduate
http://numericalmethods.eng.usf.edu
Committed to bringing numerical methods to the
undergraduate
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
Discrete Fourier Transform
Part: Aliasing Phenomenon
Nyquist Samples, Nyquist rate
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Lecture # 10
Chapter 11.04: Aliasing Phenomenon,
Nyquist samples, Nyquist rate (Contd.)
When a function f (t ), which may represent the signals
from some real-life phenomenon (shown in Figure
1), is sampled, it basically converts that function
~
into a sequence f (k ) at discrete locations of
t.
Figure 1: Function to be sampled and
“Aliased” sample problem.
38
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Aliasing Phenomenon, Nyquist samples,
Nyquist rate cont.
~
Thus, f (k ) represents the value of f (t ) at t  t0  kt ,
where t0 is the location of the first sample (at k  0).
In Figure 1, the samples have been taken with a fairly
large t. Thus, these sequence of discrete data will
not be able to recover the original signal function f (t ).
39
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Aliasing Phenomenon, Nyquist samples,
Nyquist rate cont.
For example, if all discrete values of f (t ) were
connected by piecewise linear fashion, then a nearly
horizontal straight line will occur between t1 through t11
and t12 through t16 respectively (See Figure 1).
These piecewise linear interpolation (or other interpolation
schemes) will NOT produce a curve which closely
resembles the original function f (t ). This is the case
where the data has been “ALIASED”.
40
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“Windowing” phenomenon
Another potential difficulty in sampling the function is
called “windowing” problem. As indicated in Figure 2,
while t is small enough so that a piecewise linear
interpolation for connecting these discrete values will
adequately resemble the original function f (t ),
however, only a portion of the function has been
sampled (from t0 through t12) rather than the entire
one. In other words, one has placed a “window” over
the function.
41
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“Windowing” phenomenon cont.
Figure 2. Function to be sampled and
“windowing” sample problem.
42
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“Nyquist samples, Nyquist rate”
Figure 3. Frequency of sampling rate (wS ) versus ( wmax ).
maximum frequency content
In order to satisfy F (w)  0 for w  wmax the
frequency (w) should be between points A and B
of Figure 3.
43
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“Nyquist samples, Nyquist rate”
Hence:
wmax  w  ws  wmax
which implies:
ws  2wmax
Physically, the above equation states that one
must have at least 2 samples per cycle of the
highest frequency component present (Nyquist
samples, Nyquist rate).
44
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“Nyquist samples, Nyquist rate”
45
Figure 4. Correctly reconstructed
signal.
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“Nyquist samples, Nyquist rate”
In Figure 4, a sinusoidal signal is sampled at the
rate of 6 samples per 1 cycle (or ws  6w0).
Since this sampling rate does satisfy the
sampling theorem requirement
of ws  2wmax  , the reconstructed signal does
correctly represent the original signal.
46
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“Nyquist samples, Nyquist rate”
In Figure 5 a sinusoidal signal is sampled at the
rate of 6 samples per 4 cycles  or ws  6 w0 
Since this sampling rate
does NOT satisfy the
requirementws  2wmax ,
the reconstructed signal
was wrongly represent
the original signal!

4

Figure 5. Wrongly reconstructed signal.
47
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THE END
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Acknowledgement
This instructional power point brought to you by
Numerical Methods for STEM undergraduate
http://numericalmethods.eng.usf.edu
Committed to bringing numerical methods to the
undergraduate
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