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EXPERT SYSTEMS AND SOLUTIONS

Email: expertsyssol@gmail.com

expertsyssol@yahoo.com

Cell: 9952749533www.researchprojects.info

PAIYANOOR, OMR, CHENNAI

Call For Research Projects Final

year students of B.E in EEE, ECE, EI,

M.E (Power Systems), M.E (Applied

Electronics), M.E (Power Electronics)

Ph.D Electrical and Electronics.

Students can assemble their hardware in our

Research labs. Experts will be guiding theprojects.

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Discrete Fourier Transform

&Fast Fourier Transform

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Review To use Matlab, we have to truncate sequences and

then evaluate the expression at finitely many points.

The evaluation were obviously approximations to theexact calculations.

In other words, the DTFT and the z-transform are not

numerically computable transform.

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Introduction Therefore we turn our attention to a numerically computable

transform.

It is obtained bysamplingthe DTFT transform in the

frequency domain (or the z-transform on the unit circle).

We develop this transform by analyzingperiodic sequences.

From FT analysis we know that aperiodic function can always

be represented by a linear combination of harmonically related

complex exponentials (which is form of sampling).

This give us the Discrete Fourier Series representation.

We extend the DFS to finite-duration sequences, which leads

to a new transform, called the Discrete Fourier Transform.

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Introduction The DFT avoids the two problems mentioned above

and is a numerically computable transform that is

suitable for computer implementation. The numerical computation of the DFT for long

sequences isprohibitively time consuming.

Therefore several algorithms have been developed to

efficiently compute the DFT.

These are collectively calledfast Fourier transform

(or FFT) algorithms.

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The Discrete Fourier Series Definition: Periodic sequence

N: thefundamental periodof the sequences

From FT analysis we know that theperiodic functions can besynthesized as a linear combination of complex exponentialswhose frequencies are multiples (orharmonics) of the

fundamental frequency (2pi/N). From the frequency-domain periodicity of the DTFT, we

conclude that there are a finite number of harmonics; thefrequencies are {2pi/N*k,k=0,1,,N-1}.

knkNnxnx ,),(~)(~ !

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The Discrete Fourier Series A periodic sequence can be expressed as

.,1,0,)(1

)(1

0

2

s!!

!

nekXnxk

knj T

},1,0),(~

{ .s!kK are called the discrete Fourier seriescoefficients, which are given by

.,1,0,)(~)(~ 1

0

2

s!!

!

kenxk

N

n

knjNT

The discrete Fourier series representation of periodic sequences

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The Discrete Fourier Series X(k) is itself a (complex-valued)periodic

sequence with fundamental period equal to N.

!

!

!!

!!

!

1

0

1

0

)(1)]([I FS)(

)()]([DFS)(

Let2

N

k

nk

N

N

n

nk

N

j

N

WkXN

kXnx

WnxnxkX

eW NT

Example 5.1

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To perform frequency analysis on a discrete-time signal,x(n),

need to convert the time-domain to an equivalent frequency-domainrepresentation. In order to this, need to use a powerful computationaltool to perform frequency analysis called

Discrete Fourier Transform or DFT.

The continuous Fourier Transform is defined as below:

However, this integral equation of Fourier Transform is not suitable toperform frequency analysis due to this 2 reasons:

Continuous nature can be handled by Computer

The limits of integration cannot be from minus infinity to infinity.There should be a finite length sequences that can be handled by

computer.

DFT : DEFINATION

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DFT : PROPERTIES

The are 4 properties ofDFT:

1. Periodicity

If X(k) is the N-point DFT ofx(k),

x(n+N) =x(n), for all nX(k+N) = X(k), for all k

It shows that DFT is periodic with period N, also known

as Cyclic property of the DFT

2. Linearity

If X1(k) and X2(k) are the N-point DFTofx1(n) andx2(n),

ax1(n) + bx2(n)DFT aX1(k) + bX2(k)

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3. Circular ShiftingLetx(n) be a sequence of length N and X(k) is N

-point DFT, thus the sequence,x,

(n) obtained from

x(n) by shiftingx(n) cyclically by m units. Then,

x,(n) DFT X(k)e-j2Tkm/N

4. Parsevals Theorem

ifx(n)DFT

X(k) andy(n) DFT Y(k)

thus, y*(n) = 1/N Y*(k)

!

1

0

)(N

n

nx

!

1

0

)(N

k

k

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DFT : RELATIONSHIP WITH z-TRANSFORM

The z-transform of the sequence,x(n) is given by:

X(z) = z-n

, ROC include unit circle

by defining zk= ej2Tk/N, k = 0,1, 2,, N-1

X(k) = X(z)|zk = e

j2Tk/N , k = 0,1,2,, N-1

= e-j2Tnk/N

where k= 2Tk/N, k = 0,1,2,,N-1

g

g!n nx )(

g

g!n

nx )(

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DFT : EXAMPLES

EXAMPLE 1:EXAMPLE 1:

Find the DFT for the following finite length sequence,

x(n) = { , , }

Solution :Solution :

1. Determine the sequence length, NN = 3, k = 0,1,2

2. Use DFT formula to determine X(k)

X(k) = e-j2Tnk/N , k = 0,1, 2

X(0) = + + =

X(1) = + e-j2T/3 + e-j4T/3

= + [cos (2T/3) jsin(2T/3) + [cos (4T/3) jsin(4T/3)= + [-0.5 j0.866] + [ -0.5 + j0.866]

= + [-1] = 0

!

1

0

)(N

n

nx

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Continued from Examples 1:Examples 1:

X(2) = + e-j4T/3 + e-j8T/3

= + [cos (4T/3) jsin(4T/3)] + [cos (8T/3) jsin(8T/3)]

= + [-0.5 +j0.866] + [-0.5

j0.866]

= 0

Thus,

X(k) = { , 0, 0}

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Examples 2:Examples 2:

Given the following the finite length sequences,

x(n) = {1,1,2,2,3,3}

Perform DFT for this sequences.

Solution :Solution :1. Determine the sequence length, N = 6.

2. Use DFT formula to determine X(k).

X(k) = e-j2Tnk/N , k = 0,1,2,3,4,5

X(0) = 12, X(1) = -1.5 + j2.598

X(2) = -1.5 + j0.866, X(3) = 0

X(4) = -1.5 j0.866, X(5) = -1.5 j2.598

Thus,

X(k) = {12, -

1.5 + j2.598, -

1.5 + j0.866, 0, -

1.5 j0.866,

-1.5 j2.598}

!

1

0

)(N

n

nx

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Examples 3:Examples 3:

Find the DFT for the convolution of 2 sequences :

x1(n) = {2,1, 2,1} &x2(n) = {1, 2, 3, 4}

Solution :Solution :

1. Determine the sequence length for eachsequence, N = 4. Thus, k = 0,1,2,3

2. Perform DFT for each sequences,

(i) X1(0) = 6, X1(1) = 0, X1(2) = 2, X2(3) = 0

X1(k) = {6,0,2,0}

(ii) X2(0) = 10, X2(1) = -2+j2, X2(2) = -2, X2(3) = -2-j2X2(k) = {10,-2+j2,-2,-2-j2}

3. Perform Convolution by :

X3(k) = X1(k) X2(k)

= {60, 0, -4, 0}

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IDFT : DEFINATION

IDFT is the inverse Discrete Fourier Transform.

The finite length sequence can be obtained from the Discrete Fourier

Transform by performing IDFT.

The IDFT is defined as :

x(n) = 1/N e-j2Tnk/N,

where n = 0,1,, N-1

!

1

0

)(N

k

kX

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IDFT :EXAMPLES

EXAMPLES 5:EXAMPLES 5:

Obtain the finite length sequence,x(n) from the DFT sequence inExample 3.

Solution :

1. The sequence in Example 3 is :

X3(k) = {60, 0, -4, 0}

2. Use IDFT formula to obtainx(n):

x3(n) = 1/4 e-j2Tnk/4,

x3(0) = 14, x3(1) = 16, x3(2) = 14, x3(3) = 16

Thus the finite length sequences are :

x3

(k) = {14,16,14,16}

!

3

0

)(

k

kX

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DFT & IDFT : COMPLEXITY OF DFT

A Large number of multiplications and additions are required to compute

DFT.

To compute the 8-point ofDFT of the sequence,x(n),

The X(k) will be the summation ofx(0)e-j2T(0)k/8 untilx(7)e-j2T(7)k/8 For

the eight terms, there will be 64 multiplication (82) and 56 addition (8 x

(8-1))

Hence, for N-point DFT, there will be N2 multiplication and N(N-1)

addition.

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Thus, need one algorithm to reduce the number of calculation and

speeds up the computation ofDFT. The algorithm is called Fast

F

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