Discrete Fourier Transform.rtf

7/30/2019 Discrete Fourier Transform.rtf

1/15

Discrete Fourier Transform of

Real Sequence

Motivation:

Consider a finite duration

signal )(tg of duration T

sampled at a uniform

rate st

such that

sNtT =

where N is an integer 0>NThen the Fourier transform of signal is given by

=T

ftj dtetgfG0

2)()(

If we now evaluate the above integral by trapezoidalrule of integration after padding two zeros at the

extremity on either side [signal is zero there infact], we

obtain the following expressions.

=

=1

0

2)()(

N

n

fntj

sssentgtfG

(1)

The inverse DFT (IDFT) which is used to reconstruct the signal is given by:

Fig (1)

= dfefGtg ftj 2)()((2)

If, from equation (1) we could compute complete frequency spectrum i.e. ffG ),( then (2) would imply that we can obtain Tttg 0)( . The fallacy

in the above statement is quite obvious as we have only finite samples and the curveconnecting any 2-samples can be defined plausibly in infinitely many ways (see fig (2)).

This suggests that from (1), we should be able to derive only limited amount of frequency

domain information.

Since, we have N-data points [real] and)( fG a complex number contains both

magnitude and phase angle information in thefrequency domain (2-units of information), it is

reasonable to expect that we should be in a

position to predict atmost 2

N

transforms )( fG

for original signal.


2/15

Now, letN

f

NtTf s

s

===11

0

andN

mf

Nt

mmff s

s

=== 0(3)

then substituting (3) in (1), we getFig (2)3- different fits to the curve

=

=1

0

2

)()(N

n

ntNt

mj

sss

ssentgt

N

mfG

=

=1

0

2

)(N

n

N

mnj

ss entgt

Note that our choice of frequency is such that the exponential term in (1) is

independent of st . The intuition for choosing such f is that, basically we are attemptinga transform on discrete samples which may (or) may not have a corresponding analogparent signal. This suggests to us the following discrete version of Fourier transform for

a discrete sequence{ }110 ,......,, Nxxx

=

=1

0

2

)()(N

n

N

mnj

enxmX

(4)Our next job should be to come up with inverse transformation. Assuming for N-samples

110 .,........., Nxxx that (4) would be a transformation and if (2) defines IFT in continuous

domain, in the discrete domain, we can hypothesize following inverse transform.

=

=1

0

2)(1)(

N

m

Nmnj

emXK

nx

(5)

Where K is a suitable scaling factor.

Our next job is to verify that (4) and (5) indeed define a transformation pair

Substituting (4) in (5), we get following expression for right hand side of (5)

Right hand side

N

mnjN

m

N

k

N

mkj

eekxK

21

0

1

0

2

)(1

=

=

=

(6)

[Note the use of dummy subscript k]

Let us work this expression out in a long hand fashion; for compactness we use notation)(nxxn =


3/15

2

0 1 1

2 2 2 2 ( 1) 2

0 1 1

2 ( 1) 2 ( 1) 2 ( 1) 2 ( 1) 2 ( 1)

0 1 1

0

11

1

N

j n j j n j N j n

N N N N NN

j N n j N j N n j N j N n

N N N N NN

x x x m

x e x e e x e e mRHS

K

x e x e e x e e m N

+ + + = + + + + ==+ +

+ + + + =

In the above expression, for the first row m is set to zero, for the second row it is set to

one and for the last row 1= Nm

Now, grouping terms column wise, we get

++++

+++++++=

)1(

)1()1(1

)1()1(2

)1(2

1

)1)(1(2)1(2

1

)1(22

0

NnN

NjNn

N

j

N

N

nNj

N

nj

N

nNj

N

nj

eex

eexeex

KRHS

=

=

=

1

0

1

0

)(2

1 N

k

N

m

knN

mj

k exK

Note that this jugglery shows that we can interchange the summation order. One order

indicates row wise and another column wise summation

i.e.

=

=

=

1

0

1

0

)(2

1 N

k

N

m

knN

mj

kex

KRHS

(7)

Our primary task now is to evaluate the expression.

=

1

0

)(2N

m

knN

mj

e

We now claim that

;

;

0

1

0

)(2

nkif

nkifNe

N

m

mknN

j

=

==

=

=

Proof: - for nk = meemj

mkn

N

j

==

1.0.

)(2

Hence, the first case is obvious.

Now, if nk , let1

knk = )1)((

21)(

20)(

21

0

)(2

=

+++=

NknN

jkn

N

jkn

N

jN

m

knN

mj

eeee


4/15

Now,

( )mkm

kN

jmk

N

j

aee1

11 22

=

=

whereN

j

ea

2

=1

11

1 1

22 2 21 1( )

2 20 0

1 10

1 1

j k Nmj m j kNN Nn k j k

N Nj j

k km m N N

e ee e

e e

= =

= = = =

if1k is integer, then 1

12 =kje

Note that we have used the following geometric series expression

r

raarara

nn

=+++ 1

)1(........ 1

Thus, RHS in (6) is equal to

nx

K

N

We see that equation (6) defines the inverse transformation if we choose K N= ;

Thus, N-point DFT and IDFT for samples[ ]110 ,....., Nxxx are defined as follows.

N

mnjN

m

N

n

N

nmj

n

emXN

nx

exmX

21

0

1

0

2

)(1

)(

)(

=

=

=

=

Note that in general DFT and inverse DFT can be defined in many ways, each only

differing in choice of constant 1C and 2CDFT IDFT

i.e.

=

=1

0

2

1)(N

n

N

nmj

nexCmX

=

=1

0

2

2 )(N

m

N

nmj

n emXCx

The constraint in choosing the constraints is that product NCC 121 =

For Example, when

NCC

1,1 21 ==

2

1221 == CNC

NC

NC

1121 ==

Choice of NC

21 =

is commonly used in relaying because it simplifies phasor estimation.

Phasor estimation will be discussed later.


5/15

We now discuss some important properties of DFT.

Important Properties of DFT (Lecture 7)

(1) Linearity: Let{ }

110

,......,N

xxxand

{ }110

,......,N

yyybe two sets of discrete

samples with corresponding DFTs given by )(mX and )(mY . Then DFT of

sample set{ }111100 ..,,........., +++ NN yxyxyx is given by )()( mYmX +

Proof:

=

=1

0

2

)(N

n

N

nmj

n exmX

;

=

=1

0

2

)(N

n

N

nmj

neymY

=

+=+1

0

2

)()()(N

n

N

mnj

nn eyxmYmX

(2) Periodicity : We have evaluated DFT at 1,........,1,0 = Nm . There after,

( )m N it shows periodicity. For example

)()2()()2()()( mkNXmNXmNXmNXmNXmX +=+=+=+=+=

Where k is an integer

Proof:

=

+

=+

1

0

)(2

)(

N

n

mkN

N

nj

nexmkNX

=

=1

0

2

2N

n

knjN

nmj

n eex

(8)

Both k and n are integers. Hence 12 = knje ; Therefore from (8) we set

)()(1

0

2

mXexmkNxN

n

N

nmj

n ==+

=

(3) DFT symmetry: If the samples nx are real, then extracting in frequency domain

)1()........0( NXX seems counter intuitive; because, from N bits of information

in one domain (time), we are deriving 2N bits of information in frequency

domain. This suggests that there is some redundancy in computation of


6/15

)1(....).........0( NXX . As per DFT symmetry property, following relationship

holds.

)()(* mXmNX = 1,.....,1,0 = Nm

Proof:

=

=

1

0

2

)(N

n

Nnmj

n exmX

=

=1

0

)(2

)(N

n

mNN

nj

nexmNX

=

njN

n

N

nmj

n eex

21

0

2

=

Since2

1,j n

e = integer follows that

=

=1

0

2

)(N

n

N

nmj

n exmNX

=

=1

0

*

2

)(N

n

N

nmj

nex

*1

0

2

=

=

N

n

N

nmj

nex

* *( ( )) ( )X m X m= =

If the samples nx

are real; then they contain atmost N bits of information. On

the other hand, )(mX is a complex number and hence contains 2 bits of

information. Thus, from sequence{ }110 ,.....,, Nxxx , if we derive

{ })1(),.......,1(),0( NXXX , it implies that from N-bit of information, we are

derivingN2

bits of information. This is counter intuitive. We should expect

some relationship in the sequence { })1(),.......,1(),0( NXXX

Thus, we conclude that

)()( mXmNX =[Symmetry]


7/15

and )()( mXmNX = [Anti-symmetry]

Thus, a typical DFT magnitude and phase plots appear as follows. ( )X m

Fig (3)

Fig (4)

DFT phase shifting:-

DFT shifting property states that, for a periodic sequence with periodicity N i.e.

)()( lNmxmx += , l an integer, an offset in sequence manifests itself as a phase shift in

the frequency domain. In other words, if we decide to sample x(n) starting at n equal to

some integer K, as opposed to n=0, the DFT of those time shifted samples. )(mXshifted

2

shifted ( ) ( )j Km

NX m e X m

=

Starting DFT with say

this sample will lead

to offset of 2

Fig (5)

+

=

=1 2

)()(nk

kn

N

nmj

shifted ekxmX


8/15

Proof: By periodicity of samples, we have

)0()( xNx =

)1()1( xNx =+

)1()1( =+ KxKNx

=

=1

0

2

)(N

n

mN

nj

n exmX

=

=

+=1 21

0

2 N

Kn

N

nmj

n

K

n

N

nmj

n exex

=

=

+ +=1 21

0

2 N

Kn

N

nmj

n

K

n

N

nmj

nN exex

=+=

=

=

+

+ 121

21

0

)(2

N

NmjN

Kn

N

nmj

n

K

n

N

nNmj

Nn eexex

=

+

=

+=1 21 2 N

Kn

N

nmj

n

KN

Nn

N

nmj

n exex

+

=

=1 2

)(KN

Kn

mN

nj

nexmX

(8)

Now to compute shiftedX , let us map the samples 11 ,........,, ++ KNKK xxx to

110 .........,, Nyyy .

Apply DFT to sequence y

=

=1

0

2

)(N

n

N

nmj

nshifted eymX

=

+=1

0

2N

n

mN

nj

nK ex

=

+

+=1

0

)(22 N

n

KnN

mj

nKN

mKj

exe

+

=

=1 22 KN

Kn

NN

nj

nN

nKj

exe


9/15

)(

2

mXe nN

nKj

=(from (8))

Review Questions:

1. Compute 8 pt DFT ( 0, 7)m = of the following sequence. (0) 0.35x =

(1) 0.33x = (2) 0.68x = (3) 1.07x = (4) 0.40x = (5) 1.12x = (6) 1.35x =

(7) 0.35x = Hence, illustrate the various DFT properties discussed in this

lecture.

2. By using inverse DFT, show that discrete samples can be recovered with

knowledge of (0), ........ (7)x x

3. Calculate the N pt DFT of rectangular function given by, 0 1 1.......... 1Nx x x = = = .

Verify the various DFT properties for this signal.

Computation of Phasor from DFT: (Lecture 8)

Consider a sinusoidal input signal of frequency 0

, given by

)sin(2)( 0 += tXtx (9)

This signal is conveniently represented by a phasorX

(cos sin )jX Xe X j = = + (10)

Assuming that ( )tx is sampled N times per cycle s.t 0 sT Nt= of the 50Hz waveform,

0 50f = to produce the sample set { kx }, Then,

)

2sin(2

+= k

NXxk

(11)

Hence, An integer m, in the transform domain corresponds to frequency mF 0. Thus,

choice m=1 corresponds to extract the fundamental frequency component. The discrete

Fourier Transform of kx

contains a fundamental frequency component given by sF N

=

=

1

0

2

1

2 N

k

kN

j

kexN

X

(12)


10/15

=

=

=1

0

1

0

)2

sin(2

)2

cos(2 N

k

N

k

kk kN

xN

jkN

xN

scjXX =

(13)

where

==

1

0

)2cos(2N

k

kc kN

xN

X

(14)

=

=1

0

)2

sin(2 N

k

ks kN

xN

X

(15)

Substituting for kx

from (11) in (12) and (13) it can be shown that for a sinusoidal input

signal given by (9)

cos2sin2 XXXX sc == (16)

From equation (10), (13) and (14) it follows that

( )1

1 1

2 2s cX jX X jX= = +

(17)

When the input signal contains other frequency components as well, the phasor

calculated by equation (17) is a filtered fundamental frequency phasor. It is presumed

here that the input signal must be band-limited to satisfy the Nyquist Criterion, to avoid

errors due to aliasing affects.

Effect on phasor computation

In relaying applications during steady state we will be working with moving

window, such that each window having most recent N-samples. Letw

cX andw

sX indicate

cX and sX

component of DFT forthw window (window no. bery equal to 1st sample

no.). Then, by DFT define

1

0

2 2cosNw

c k w

k

X x kN N

+

=

= 1

0

2 2 22 sin cos

N

k

X k w kN N N

=

= + +


11/15

1

0

2 4 2 2sin sin

N

k

Xk w w

N N N N

=

= + + + +

2 20 sin

XN w

N N

= + +

22 sinX w

N

= +

Similarly,

1

0

2 2sin

Nw

k w

k

sX x kN N

+=

= 1

0

2 2 22 sin sin

N

k

X k w kN N N

=

= + +

1

0

2 2 4 2cos cos

N

k

Xw k w

N N N N

=

= + + +

2 2cos

XNw

N N

= +

22 cosX w

N

= +

Hence, DFT estimated atth

w window for (m = 1) is given by

2 2(1) 2 sin cos

w w w

c sX X jX X w j wN N

= = + +

Hence, phasor estimate atth

w window

( )1 2 2

cos sin2

w w w

s cX X jX X w j wN N

= + = + + +

2j w

NX e

+ =

2j w

j NX e e

=

Thus, we see that with moving window, the phasor estimate rotates at a speed of

2j

Ne

per

window. This rotate in phasor during computation can be directing derived from DFT phase shifting properties.

This suggests that

N

j

K ejXN

KX

2

1 )1(2

)(

=

phasor estimated forthn window rotates at

speed N2

per sample.


12/15

In the computation of phasor, we would not like this phasor shift to occur due to DFT

phase shifting. This can be avoided if we use equation (8) for computing DFT. Replace k

by w , the window number, we get

21

2( )

j kmk w

Nk c s

k w

X m x e X jXN

+

== =

Where

1 2( ) cos

N ww

c k

k w

kmX m x

N

+

=

= (9)

12 2( ) sin

N ww

s k

k w

kmX m x

N N

+

=

= (10)

Example: let )sin( tx =

With sampling rate of N-samples per cycle we have N

Tts

0=;

)2

sin()2

sin( 0

0 N

n

N

Tn

Txn

==

Substituting kx

in equation (9), we get

12 2 1 2 22sin cos

2

N ww

c

k w

X k kmX

N N N

+

=

= +

12 2 1 2 2sin (1 ) sin (1 )

2

N w

k w

X k km m

N N N

+

=

= + + + +

2 sinX =

12 2 1 2 22sin sin

2

N ww

k w

s

X k kmX

N N N

+

=

= +

12 2 1 2 2cos( (1 ) ) cos( (1 ) ) 2 cos

2

N w

k w

X k km m X

N N N

+

=

= + + + =

sin( )x t = +

( )1

2

ww w

s sX X jX X= + =

a


13/15

From equation (10) and (11), we have

12 2( ) cos

N ww

c k

k w

kmX m x

N N

+

=

= and

12 2( ) sin

k N ww

s k

k w

kmX m x

N N

= +

=

=

. The computation forw

cX can be visualized from the

following pair wise multiplication and add sequence show in red in the figure -

Now, to compute1( )wcX m

+

, we get one new sample. w Nx + and computation of

1( )WcX m+

is( )wcX m shown by pairing in green boundary. It is now obvious, that

( )12 2

( ) ( ) cos cosw w

c c w N wX m X m x N w m x wmN N

++

= + + . Since,

( )2 2 2

cos cos 2 cosN w m m wm wmN N N

+ = + = for m an integer, we get

( )12

( ) ( ) cosw w

c c w N wX m X m x x wmN

++

= + -------- (12)

Similarly, for sine component, we have

1 2

( ) ( ) [ ]sin

w w

w N ws sX m X m x x wmN

++

= + ----(13)

Eqns. 12 and 13 provide a recursive update for DFT computation. The advantage of

recursive form is that it reduces computation from 2-N multiply- add operation in normal

DFT into additions and 2 multiplication is respectively.

To begin with we will get( ) ( ) 0w ws sX m X m= = . When the first window is full populated

with N-samples, we will correct values of( )cX m and

( )mX m . Afterwards, for a

stationary phasor w N wx x+ = and hence, the DFT latches in the appropriate phasor.

wx

2 ( 2)cos

w m

N

+

2wx + 1w N

x +

2cos ( 1)N w m

N

+

w Nx +

2cos ( )N w m

N

+

1( )wcX m+

( )wcX m


14/15

Half Cycle DFT Form for Phasor Estimation

If our primary interest is to extract fundamental phasor component in the signal then, it

can be verified that, restricting moving window to half a cycle does not alter the end

result of eqn. (5) and (6) provided that N-now represents, number of samples per half-

cycle. Now, the sample kx

is given by

2 sinkx kN

= + . Thus, half cycle form of

DFT phasor estimation is given by the following eqns.

12cos

N ww

c k

k w

X x kN N

+

=

= ----(14)

1

2 sinN w

wk

k w

sX x kN N

+

==

----(15)

Advantage of half cycle algorithm is that the moving window latches on to the post fault

signal in1

2 a cycle. Thus, compared to full cycle version, it is twice as fast. A keen

observer would have noticed that DFT based on moving window phasor estimation

equations are identical to the full cycle and cycle fourier algorithms derived in lecture-x.

Thus, the frequency response of fourier algorithms developed in lecture x applies to the

DFT version. In particular, it is not surprising to see that harmonic rejection property of

half cycle algorithm is inferior to its full cycle avatar. This is consistent with the Speed

vs accuracy conflict, we have discussed earlier.

Half Cycle Recursive DFT

Recursive form of half cycle DFT can be derived in an analogous manner to full cycle

DFT. Realizing that

( )cos cosN w m m mwN N

+ = +

cos mod dmwN

=

( )sin sinN w m m mwN N

+ = +


15/15

sin mod dmwN

=

we get following recursive update forms for fundamental phasor computation

1 ( ) cosw wc c w N w

X X x x wN

+

+= +1 ( ) sinw w w N ws sX X x x w

N

++= +

Review Questions

1. Consider, the following signal

( ) 10sin(2 50 30 )tx t= +

a) Generate samples on this waveform using sampling frequency of samples

per cycle.b) Apply full cycle and half cycle algorithms to estimate phasor from

generalized DFT approach.

c) Repeat (b) using recursive forms.

2. Now consider the following signal.

10( ) 10sin(2 50 30 ) 5sin(4 50 17.5 ) cos(6 50 75 )

3x t t t t = + + + + +

Repeat (2) and comment on the accuracy of full cycle and half cycle estimation

methods.

Documents

Discrete Fourier Transform.rtf