Numerical Methods Discrete Fourier Transform Part: Discrete Fourier Transform .

Numerical Methods

Discrete Fourier Transform Part: Discrete Fourier Transform

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Chapter 11.04 : Discrete Fourier Transform (DFT)

Major: All Engineering Majors

Authors: Duc Nguyen


Numerical Methods for STEM undergraduates04/19/23

http://numericalmethods.eng.usf.edu 5

Lecture # 8

Discrete Fourier TransformRecalled the exponential form of Fourier series (see Eqs. 39, 41 in Ch. 11.02), one gets:

k

tikw

keCtf 0~)(

Ttikw

k dtetfT

C0

0)(1~

(39, repeated)

(41, repeated)

http://numericalmethods.eng.usf.edu6


,,.......,3,2, 321 tnttttttt n

then Eq. (39) becomes:

1

0

0~)(

N

k

ntikw

kn eCtf (1)

If time “ ” is discretized at t

Discrete Fourier Transform

Discrete Fourier Transform cont.

To simplify the notation, define:

ntn (2)

Then, Eq. (1) can be written as:

1

0

0~)(

N

k

nikw

keCnf (3)

Multiplying both sides of Eq. (3) by nilwe 0

, and performing

the summation on “ ”, onen obtains (note: l = integernumber)



nilwN

n

N

k

nikw

k

N

n

nilw eeCenf 01

0

1

0

01

0

0 ~)(

1

0

1

0

2)(~N

n

N

k

nN

lki

keC

(4)

(5)



Switching the order of summations on the right-hand-side of Eq.(5), one obtains:

1

0

1

0

2)(1

0

2 ~)(

N

k

N

n

nN

lki

k

N

n

nN

il

eCenf

(6)

Define:

1

0

2)(N

n

nN

lki

eA

(7)

There are 2 possibilities for to be considered in Eq. (7)

)( lk


Discrete Fourier Transform—Case 1

Case(1): is a multiple integer of N, such as: ; or where

)( lk mNlk )( mNlk

,......2,1,0 m

Thus, Eq. (7) becomes:

1

0

1

0

2 )2sin()2cos(N

n

N

n

nim mnimneA (8)

Hence:

(9)NA



Case(2): is NOT a multiple integer of In this case, from Eq. (7) one has:

)( lk .N

1

0

2)(N

n

n

Nlki

eA

(10)

Define:

)

2)(sin)

2)(cos

2)(

Nlki

Nlkea N

lki

(11)



;1a because is “NOT” a multiple integer of )( lk NThen, Eq. (10) can be expressed as:

1

0

N

n

naA (12)



From mathematical handbooks, the right side of Eq. (12) represents the “geometric series”, and can be expressed as:

;1

0NaA

N

n

n

if 1a (13)

;1

1

a

aN

if 1a (14)



Because of Eq. (11), hence Eq. (14) should be usedto compute . Thus:A

a

e

a

aA

lkiN

1

1

1

1 2)(

(See Eq. (10)) (15)

12)(sin2)(cos2)( lkilke lki (16)



Substituting Eq. (16) into Eq. (15), one gets

0A (17)

Thus, combining the results of case 1 and case 2, we get

NNA 0 (18)


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Acknowledgement

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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

Discrete Fourier Transform Part: Discrete Fourier Transform


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Substituting Eq.(18) into Eq.(7), and then referring to Eq.(6), one gets:

1

0

1

0

0 ~)(

N

kk

N

n

nilw NCenf (18A)

Recall (where are integer numbers), And since must be in the range

mNlk ml,k .0,10 mN

mNlk lk becomes

Chapter 11.04: Discrete Fourier Transform (DFT)

Lecture # 9

Thus:


Eq. (18A) can, therefore, be simplified to

NCenf l

N

n

nilw

~)(

1

0

0

(18B)

Thus:

1

000

1

0

0

)sin()cos()(1

)(1~~

N

n

N

n

nikw

kl

nkwinkwnfN

enfN

CC

(19)

where ntn and

1

000

~1

0

0~

)sin()cos()(N

kk

N

k

nikw

k nkwinkwCeCnf(1, repeated)



Equations (19) and (1) can be rewritten as

1

0

20

)(~ N

k

nN

wik

n ekfC

(20)

1

0

20~1

)(N

n

nN

wik

neCNkf

(21)



To avoid computation with “complex numbers”, Equation (20) can be expressed as

1

0)sin()cos()()(

~~ N

k

IRI

n

R

n ikfikfCiC (20A)

where

nN

wk

2

0



)sin()()cos()(

)sin()()cos()(~~ 1

0

kfkfi

kfkfCiC

RI

N

k

IRI

n

R

n

(20B)

The above “complex number” equation is equivalent to the following 2 “real number” equations:

1

0)sin()()cos()(

~ N

k

IRR

n kfkfC

1

0)sin()()cos()(

~ N

k

RII

n kfkfC

(20C)

(20D)



Acknowledgement

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Numerical Methods

Discrete Fourier Transform Part: Aliasing Phenomenon Nyquist Samples, Nyquist rate


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Chapter 11.04: Aliasing Phenomenon, Nyquist samples,

Nyquist rate (Contd.)When a function ),(tf which may represent the signals from some real-life phenomenon (shown in Figure 1), is sampled, it basically converts that function into a sequence )(

~kf at discrete locations of .t


Lecture # 10

Figure 1: Function to be sampled and “Aliased” sample problem.

Aliasing Phenomenon, Nyquist samples, Nyquist rate cont.

)(~kf ,)( 0 tkttattf represents the value of Thus,

where 0tis the location of the first sample ).0( kat

In Figure 1, the samples have been taken with a fairly large Thus, these sequence of discrete data will not be able to recover the original signal function

.t).(tf


Aliasing Phenomenon, Nyquist samples, Nyquist rate cont.

These piecewise linear interpolation (or other interpolationschemes) will NOT produce a curve which closely resembles the original function . This is the case where the data has been “ALIASED”.

)(tf


For example, if all discrete values of were connected by piecewise linear fashion, then a nearly horizontal straight line will occur between through and through respectively (See Figure 1).

)(tf

1t 11t16t12t

“Windowing” phenomenon

Another potential difficulty in sampling the function is called “windowing” problem. As indicated in Figure 2, while is small enough so that a piecewise linear interpolation for connecting these discrete values will adequately resemble the original function , however, only a portion of the function has been sampled (from through ) rather than the entire one. In other words, one has placed a “window” over the function.

t

)(tf

0t 12t


“Windowing” phenomenon cont.

Figure 2. Function to be sampled and “windowing” sample problem.


“Nyquist samples, Nyquist rate”

Figure 3. Frequency of sampling rate versus maximum frequency content

)( Sw ).( maxw

In order to satisfy the frequency ( ) should be between points A and B of Figure 3.

max0)( wwforwF w


Hence:

maxmax wwww s

which implies:

max2wws

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).



Figure 4. Correctly reconstructed signal.



In Figure 4, a sinusoidal signal is sampled at the rate of 6 samples per 1 cycle (or ). Since this sampling rate does satisfy the sampling theorem requirement of , the reconstructed signal does correctly represent the original signal.

06wws

max2wws



Figure 5. Wrongly reconstructed signal.

In Figure 5 a sinusoidal signal is sampled at the rate of 6 samples per 4 cycles

04

6wwor s

Since this sampling rate does NOT satisfy the requirement the reconstructed signal was wrongly represent the original signal!

,2 maxwws




Acknowledgement

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Numerical Methods Discrete Fourier Transform Part: Discrete Fourier Transform .

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