# PPT - Math For College

```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
http://numericalmethods.eng.usf.edu
4/13/2015
http://numericalmethods.eng.usf.edu
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
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
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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
http://numericalmethods.eng.usf.edu
For more details on this topic

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Click on Keyword
Click on Discrete Fourier Transform
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
to Share – to copy, distribute, display and
perform the work
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


Attribution — You must attribute the work in
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(but not in any way that suggests that they
endorse you or your use of the work).
Noncommercial — You may not use this work
for commercial purposes.
Share Alike — If you alter, transform, or build
upon this work, you may distribute the resulting
work only under the same or similar license to
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
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
Discrete Fourier Transform
Part: Aliasing Phenomenon
Nyquist Samples, Nyquist rate
http://numericalmethods.eng.usf.edu
For more details on this topic
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Go to http://numericalmethods.eng.usf.edu
Click on Keyword
Click on Discrete Fourier Transform
You are free
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to Share – to copy, distribute, display and
perform the work
to Remix – to make derivative works
Under the following conditions



Attribution — You must attribute the work in
the manner specified by the author or licensor
(but not in any way that suggests that they
endorse you or your use of the work).
Noncommercial — You may not use this work
for commercial purposes.
Share Alike — If you alter, transform, or build
upon this work, you may distribute the resulting
work only under the same or similar license to
this one.
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
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