DFT : The Discrete Fourier Transform

Chapter 8

2

Roots of Unity

Definition: An Nth root of unity is a complex number x such that xN = 1

The Nth roots of unity are: ω0, ω1, …, ωN-1 where ω = e 2π j/N.Proof: (ωk)N = (e 2π j k / N) N = (e π j ) 2k = (-1) 2k = 1.

In DSP, we often take the principal root of unity as

ω0 = 1

ω1

ω2 = i

ω3

ω4 = -1

ω5

ω6 = -i

ω7

N = 8

2 /j NNW e π−=

8W

58W

3

Properties of N-th Root of Unity

There are exactly N N-th roots of unity: WN0, WN

1 , WN2 , …, WN

N-1

WN is called the principal N-th root of unityThe inverse of WN : WN

-1 = WNN-1

Proof: WN WNN-1 = WN

N =1WdN

dk = WNk

Proof: : WdNdk = = = WN

K

WNN/2 = W2 = -1 Proof: WN

N/2 = W(N/2)*2N/2 = (by property 3) W2 = = -1

If N>0 and N is even, then (WNk )2 = WN/2

k , k = 0, 1, …, N–1

2 /j NNW e π−=

2 /( )j dN dke π− 2 /( )j N ke π−

2 / 2je π−

1

0

1

0

0( )

( ) 1 ( ) 1Proof : ( ) 01 1

Nk n

Nn

k N N kNk n N N

N k kn N N

k rNW

N k rN

W WWW W

−

=

−

=

≠⎧= ⎨ =⎩

− −= = =

− −

∑

∑

r = an integer

4

Circular Shift of a Sequence

Consider length-N sequences defined for

Values of such sequences are not defined for n < 0 and If x[n] is such a sequence, then for any arbitrary integer , the shifted sequenceis no longer defined for the rangeWe thus need to define another type of a shift that will always keep the shifted sequence in the range

Nn ≥

10 −≤≤ Nn

on][][ onnxnx −=1

10 −≤≤ Nn

10 −≤≤ Nn

5

The desired shift, called the circular shift, is defined using a modulo operation:

For (right circular shift), the above equation implies

[ ] [(( )) ]c o Nx n x n n= −

0>on

⎩⎨⎧

+−−= ],[

],[][ nnNxnnxnxo

oc

oo

nnNnn<≤

−≤≤0for

1for

0[ ] [ ] [ ] [ ]cx n x n x n n x n−periodicextension

shift take oneperiod

6

Illustration of the concept of a circular shift

][nx6[(( 1)) ]x n −

6[(( 5)) ]x n= + 6[(( 2)) ]x n= +

6[(( 4)) ]x n −

7

As can be seen from the previous figure, a right circular shift by is equivalent to a left circular shift by sample periods

A circular shift by an integer number greater than N is equivalent to a circular shift by

ononN −

on(( ))o Nn

8

Circular Time Shift

Points shifted out to the right don’t disappear – they come in from the left

Like a ‘barrel shifter’:

5-pt sequence

‘delay’ by 2g[n]

1 2 3 4 n

g[((n-2))5]

1 2 3 4 n

origin pointer

9

Example: Circular Shift

10

Circular Time Reversal

Time reversal is tricky in ‘modulo-N ’ indexing - not reversing the sequence:

Zero point stays fixed; remainder flips

1 2 3 4 n5 6 7 8 9 10 11-7 -6 -5 -4 -3 -2 -1

[ ]x n

5-pt sequence made periodic

Time-reversedperiodic sequence

1 2 3 4 n5 6 7 8 9 10 11-7 -6 -5 -4 -3 -2 -1

[ ](( ))Nx n−

[ ] [ ] [ ] [ ]rx n x n x n x n−periodicextension

reversal take oneperiod

Discrete-Time Fourier Transform (DTFT)

The DTFT is useful for the theoretical analysis of signals and systems.But, the definition of DTFT is

The numerical computation of DTFT has several problems:Problem 1: The summation over n is infiniteProblem 2: The independent variable ω is continuousDTFT and z-transform are not numerically computable transforms.

ApproachesProblem 1 requirement 1 : If the sequence had only a finite number of non-zero terms!

This might be a good assumption since in many cases, we concern only a finite duration of the signal

• The signal itself might be of a finite duration• Only a segment is of interest at a time• Signal is periodic and thus has only finite number of

unique valuesProblem 2 Requirement 2 : If we needed to compute the spectra only at a finite number of frequencies!

This can be achieved by sampling the DTFT in the frequency domain or the z-transform on the unit circle.The remaining issue is that what will happen for the spectra at other frequencies. Aliasing?

DTFS for a periodic sequence satisfy both requirements

Analysis equation

Synthesis equation

But, DTFS is defined for only periodic signals!!!

Way to get there:Analyze periodic sequences on the basis that a periodic sequence can always be represented by a linear combination of harmonically related complex exponentials Discrete Fourier Series (DFS).Extend the DFS to finite-duration sequences Discrete Fourier Transform (DFT), the solution to the two problems!

[ ]1

0

1[ ] , ,1, 10 ,knN

N

kNx n X k W n

N−

=

−

= −= ∑

[ ]0

1

[ ] , 0,1, , 1N

knN

nNX k x n W k

=

−

= = −∑

TerminologyDFS = DTFS

Discrete Fourier Transform (DFT)

For finite duration sequences, an alternative Fourier representation is DFT (Discrete Fourier Transform)

The summation over n is finiteDFT itself is a sequence, rather than a function of a continuous variableTherefore, DFT is computable and important for the implementation of DSP systemsDFT corresponds to the samples of the Fourier transform

DFT is very important in DSP systems since there is FFT (Fast Fourier Transform), an extremely efficient and fast way of computing DFT.

8.1 Discrete Fourier Series (DFS)

A discrete-time signal is periodic if

Fundamental period: the smallest positive integer N

Fundamental frequency:

The DFS can represent a periodic discrete signal by a linear combination of harmonically related complex exponentials.

The Fourier series representation of continuous-time periodic signals require infinite number of complex exponentials.

But for discrete-time periodic signals, we have

Due to the periodicity of the complex exponentials, we only need Nexponentials for discrete time Fourier series

][~][~ Nnxnx +=

N/20 πω =

( )( ) ( ) ( ) ( )2 / 2 / 2 2 /j N k mN n j N kn j mn j N kne e e eπ π π π+ = =

DFS analysis and synthesis pair is expressed as follows:

The Fourier series coefficients can be obtained via

For convenience we may use the notation

Analysis equation

Synthesis equation

[ ] ( ) [ ] ( )1

2 / 2 /

0

1 1[ ]N

j N kn j N kn

k N kx n X k e X k e

N Nπ π

−

=< > =

= =∑ ∑

[ ] ( )1

2 /

0[ ]

Nj N kn

nX k x n e π

−−

=

=∑

( )2 /j NNW e π−=

[ ]1

0

1[ ]N

knN

kx n X k W

N

−−

=

= ∑

[ ]1

0[ ]

Nkn

Nn

X k x n W−

=

=∑

Example 8.1 : DFS of a periodic impulse train

Since the period of the signal is N

We can represent the signal with the DFS coefficients as

[ ] 1[ ]

0r

n rNx n n rN

elseδ

∞

=−∞

=⎧= − =⎨

⎩∑

[ ] ( ) ( ) ( )1 1

2 / 2 / 2 / 0

0 0[ ] [ ] 1

N Nj N kn j N kn j N k

n nX k x n e n e eπ π πδ

− −− − −

= =

= = = =∑ ∑

[ ] ( )1

2 /

0

1[ ]N

j N kn

r kx n n rN e

Nπδ

∞ −

=−∞ =

= − =∑ ∑

Example 8.2 : Here let the Fourier series coefficients be the periodic impulse train Y[k] , and given by this equation:

Substituting Y[k] into DFS equation gives

Comparing this result with the results for Example 8.1, we see that Y[k]=Nx[k] and y[n]=X[n].

[ ] [ ]r

Y k N k rNδ+∞

=−∞

= −∑

10

0

1[ ] [ ] 1N

knN N

ky n N k W W

Nδ

−−

=

= = =∑

Example 8.3 : DFS of an periodic rectangular pulse train

The DFS coefficients

[ ] ( )( )

( )( ) ( )

( )2 /10 54

2 /10 4 /102 /10

0

sin / 21sin /101

j kj kn j k

j kn

keX k e eke

ππ π

π

ππ

−− −

−=

−= = =

−∑

8.2 Properties of DFS

Linearity (all signals have the same period)

Shift of a Sequence

Duality

[ ] [ ][ ] [ ]

[ ] [ ] [ ] [ ]

1 1

2 2

1 2 1 2

DFS

DFS

DFS

x n X kx n X k

ax n bx n aX k bX k

←⎯⎯→←⎯⎯→

+ ←⎯⎯→ +

[ ] [ ][ ] [ ]

[ ] [ ]2 /

2 /

DFS

DFS j km N

DFSj nm N

x n X kx n m e X k

e x n X k m

π

π

−

←⎯⎯→− ←⎯⎯→

←⎯⎯→ −

[ ] [ ][ ] [ ]

DFS

DFS

x n X kX n Nx k

←⎯⎯→←⎯⎯→ −

0 1m N≤ ≤ −

Periodic ConvolutionTake two periodic sequences

Let’s form the product

The periodic sequence with given DFS can be written as a periodic convolution

Periodic convolution is commutative

[ ] [ ][ ] [ ]

1 1

2 2

DFS

DFS

x n X kx n X k

←⎯⎯→←⎯⎯→

[ ] [ ] [ ]3 1 2X k X k X k=

[ ] [ ] [ ]1

3 1 20

N

mx n x m x n m

−

=

= −∑

[ ] [ ] [ ]1

3 2 10

N

mx n x m x n m

−

=

= −∑

ProofSubstitute periodic convolution into the DFS equation

Interchange summations

The inner sum is the DFS of shifted sequence

Substituting

[ ]1 1 1

3 3 1 20 0 0

[ ] [ ] [ ]N N N

kn knN N

n n mX k x n W x m x n m W

− − −

= = =

⎛ ⎞= = −⎜ ⎟

⎝ ⎠∑ ∑ ∑

[ ]1 1

3 1 20 0

[ ] [ ]N N

knN

m nX k x m x n m W

− −

= =

⎛ ⎞= −⎜ ⎟

⎝ ⎠∑ ∑

[ ]1

2 20

[ ]N

kn kmN N

nx n m W W X k

−

=

− =∑

[ ] [ ] [ ] [ ]1 1 1

3 1 2 1 2 1 20 0 0

[ ] [ ] [ ]N N N

kn kmN N

m n mX k x m x n m W x m W X k X k X k

− − −

= = =

⎛ ⎞= − = =⎜ ⎟

⎝ ⎠∑ ∑ ∑

Example 8.4

Properties of DFS

8.3 Fourier Transform of Periodic Signals

Periodic sequences are not absolute or square summableHence they don’t have a Fourier Transform per se

We can represent them as sums of complex exponentials: DFSWe can combine DFS and Fourier transform(Generalized) Fourier transform of periodic sequences

Periodic impulse train with values proportional to DFS coefficients

This is periodic with 2π since DFS is periodicThe inverse transform can be written as

( ) [ ]2 2j

k

kX e X kN N

ω π πδ ω∞

=−∞

⎛ ⎞= −⎜ ⎟⎝ ⎠

∑

( ) [ ]

[ ] [ ]

2 2

0 0

212

00

1 1 2 22 2

1 2 1

j j n j n

k

kN j nj n N

k k

kX e e d X k e dN N

kX k e d X k eN N N

π ε π εω ω ω

ε ε

ππ ε ω

ε

π πω δ ω ωπ π

πδ ω ω

∞− −

− −=−∞

∞ −−

−=−∞ =

⎛ ⎞= −⎜ ⎟⎝ ⎠

⎛ ⎞− =⎜ ⎟⎝ ⎠

∑∫ ∫

∑ ∑∫

Example 8.5Consider the periodic impulse train

The DFS was calculated previously to be

Therefore the Fourier transform is

[ ][ ]r

p n n rNδ∞

=−∞

= −∑

[ ] 1 for all kP k =

( ) 2 2j

k

kP eN N

ω π πδ ω∞

=−∞

⎛ ⎞= −⎜ ⎟⎝ ⎠

∑

Relation between Finite-length and Periodic Signals

Consider finite length signal x[n] spanning from 0 to N-1Convolve with periodic impulse train

The Fourier transform of the periodic sequence (called periodic extension) is

This implies that

DFS coefficients of a periodic expansion signal can be thought as equally spaced samples of the Fourier transform of one period (= the original finite duration sequence)

[ ] [ ][ ] [ ] [ ] [ ]r r

x n x n p n x n n rN x n rNδ∞ ∞

=−∞ =−∞

= ∗ = ∗ − = −∑ ∑

( ) ( ) ( ) ( )

( )2

2 2

2 2

j j j j

k

kjj N

k

kX e X e P e X eN N

kX e X eN N

ω ω ω ω

πω

π πδ ω

π πδ ω

∞

=−∞

∞

=−∞

⎛ ⎞= = −⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

∑

∑

[ ] ( )2

2

kj jNk

N

X k X e X eπ

ωπω=

⎛ ⎞= =⎜ ⎟

⎝ ⎠

Example 8.6

Consider the following sequence

The Fourier transform

The DFS coefficients

1 0 4[ ]

0n

x nelse≤ ≤⎧

= ⎨⎩

( ) ( )( )

2 sin 5 / 2sin / 2

j jX e eω ω ωω

−=

[ ] ( ) ( )( )

4 /10 sin / 2sin /10

j k kX k e

kπ π

π−=

8.4 Sampling the Fourier Transform

Consider an aperiodic sequence with its Fourier transform

Assume that a sequence is obtained by sampling the DTFT

Since the DTFT is periodic, resulting sequence is also periodicWe can also write it in terms of the z-transform

The sampling points are shown in figurecould be the DFS of a sequence

Write the corresponding sequence

[ ] ( )( )

( )( )2 /

2 /

j N kj

N kX k X e X e πω

ω π== =

( )[ ] DTFT jx n X e ω←⎯⎯→

[ ] ( ) ( )( )( )2 /

2 /N k

j N k

z eX k X z X eπ

π

== =

[ ] ( )1

2 /

0

1[ ]N

j N kn

kx n X k e

Nπ

−

=

= ∑

[ ]X k

The only assumption made on the sequence is that DTFT exist

Combine equation to get

Term in the parenthesis is

So we get the periodic extension of x[n]

( ) [ ]j j m

mX e x m eω ω

∞−

=−∞

= ∑ [ ] ( )1

2 /

0

1[ ]N

j N kn

kx n X k e

Nπ

−

=

= ∑[ ] ( )( )2 /j N kX k X e π=

[ ] ( ) ( )

[ ] ( ) ( ) [ ] [ ]

12 / 2 /

0

12 /

0

1[ ]

1

Nj N km j N kn

k m

Nj N k n m

m k m

x n x m e eN

x m e x m p n mN

π π

π

− ∞−

= =−∞

∞ − ∞−

=−∞ = =−∞

⎡ ⎤= ⎢ ⎥

⎣ ⎦⎡ ⎤

= = −⎢ ⎥⎣ ⎦

∑ ∑

∑ ∑ ∑

[ ] ( ) ( ) [ ]1

2 /

0

1 Nj N k n m

k rp n m e n m rN

Nπ δ

− ∞−

= =−∞

− = = − −∑ ∑

[ ] [ ] [ ][ ]r r

x n x n n rN x n rNδ∞ ∞

=−∞ =−∞

= ∗ − = −∑ ∑

In this case, the Fourier series coefficients for aperiodic sequence are samples of the Fouriertransform of one period

In this case, still the Fourier series coefficients for are samples of the Fourier transform of x[n]. But,one period of is no longer identical to x[n].This is just sampling in the frequency domain as compared in the time domain discussed before.

[ ]x n

[ ]x n

Samples of the DTFT of an aperiodic sequence can be thought of as DFS coefficients of a periodic sequence obtained through summing periodic replicas of original sequence. periodic expansionIf the original sequence is of finite length and we take sufficient number of samples of its DTFT the original sequence can be recovered by

It is not necessary to know the DTFT at all frequencies to recover the discrete-time sequence in time domainDiscrete Fourier Transform representing a finite length sequence by samples of DTFT

[ ] [ ] 0 10

x n n Nx n

else⎧ ≤ ≤ −

= ⎨⎩

Time-domain aliasingThe relationship between x[n] and one period of in the undersampled case is considered a form of time domain aliasing.Time domain aliasing can be avoided only if x[n] has finite length, just as frequency domain aliasing can be avoided only for signals being bandlimited.If x[n] has finite length and we take a sufficient number of equally spaced samples of its Fourier transform (specifically, a number greater than or equal to the length of ), then the Fourier transform is recoverable from these samples, equivalently is recoverable from .

[ ]x n

36

Time-Domain Aliasing vs. Frequency-Domain Aliasing

To avoid frequency-domain aliasingSignal is bandlimitedSampling rate in time-domain is high enough

To avoid time-domain aliasingSequence is finiteSampling interval (2π/N) in frequency-domain is small enough

8.5 The Discrete Fourier Transform

Consider a finite length sequence x[n] of length N

For given length-N sequence associate a periodic sequence (a periodic extension)

The DFS coefficients of the periodic sequence are samples of the DTFT of x[n]Since x[n] is of length N there is no overlap between terms of x[n-rN] and we can write the periodic sequence as

To maintain duality between time and frequencyWe choose one period of as the Fourier transform of x[n]

[ ] 0 outside of 0 1x n n N= ≤ ≤ −

[ ] [ ]r

x n x n rN∞

=−∞

= −∑

[ ] ( ) ( )( ) mod NN

X k X k X k⎡ ⎤⎡ ⎤= =⎣ ⎦ ⎣ ⎦

[ ]X k

[ ] [ ] 0 10

X k k NX k

else⎧ ≤ ≤ −

= ⎨⎩

[ ] ( ) ( )( ) mod NN

x n x n x n⎡ ⎤⎡ ⎤= =⎣ ⎦ ⎣ ⎦

The DFS pair

The equations involve only one period so we can write

[ ] ( )1

2 /

0[ ]

Nj N kn

nX k x n e π

−−

=

=∑

[ ]( )

12 /

0[ ] 0 1

0

Nj N kn

nx n e k N

X kelse

π−

−

=

⎧≤ ≤ −⎪= ⎨

⎪⎩

∑

[ ] ( )1

2 /

0

1 0 1[ ]

0

Nj N kn

kX k e n N

x n Nelse

π−

=

⎧≤ ≤ −⎪= ⎨

⎪⎩

∑

The DFT pair can also be written as

The Discrete Fourier Transform

[ ] [ ]DFTX k x n←⎯⎯→

[ ]( )

12 /

0[ ] 0 1

0

Nj N kn

nx n e k N

X kelse

π−

−

=

⎧≤ ≤ −⎪= ⎨

⎪⎩

∑

[ ] ( )1

2 /

0

1 0 1[ ]

0

Nj N kn

kX k e n N

x n Nelse

π−

=

⎧≤ ≤ −⎪= ⎨

⎪⎩

∑

Discrete Fourier Transform (DFT)

The Discrete Fourier Transform (DFT) of a finite length sequence

The Inverse Discrete Fourier Transform (IDFT) is defined by

A DFT and IDFT pair is shown as

{ } { }1,...,0for ][][DFT][1

0

2

−∈== ∑−

=

−NkenxnxkX

N

n

Nnkj

N

π

{ } { }1,...,0for ][1][IDFT][1

0

2

−∈== ∑−

=

NnekXN

kXnxN

k

Nnkj

N

π

][][ DFT kXnx ⎯⎯ →←

41

Proof of Inverse Discrete Fourier Transform

The inverse discrete Fourier transform (IDFT) is given by

To verify the above expression we multiply both sides of the above equation by and sum the result from n = 0 to 1−= Nnn

NW

10,][1][1

0−≤≤∑=

−

=

− NnWkXN

nxN

k

knN

∑ ∑∑−

=

−

=

−−

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

1

0

1

0

1

0

1N

n

nN

N

k

knN

N

n

nN WWkX

NWnx ][][

∑ ∑−

=

−

=

−−=1

0

1

0

1 N

n

N

k

nkNWkX

N)(][

∑ ∑−

=

−

=

−−=1

0

1

0

1 N

k

N

n

nkNWkX

N)(][

42

Making use of the identity

we observe that the RHS of the last equation is

but, for k=0,1,…,N-1, there is only one non-zero term at k=l , and RHS =

Hence

⎩⎨⎧=∑

−

=

−−1

0

N

n

nkNW )(

,,

0N ,rNk =−for

otherwiser an integer

][X

][][ XWnxN

n

nN =∑

−

=

1

0

1

0

[ ] [ ]N

k

RHS X k k l rNδ−

=

= − −∑

43

12 /

0[ ] [ ] , 0 1

Nj kn N

nX k x n e k Nπ

−−

=

= ≤ ≤ −∑

12 /

0

1[ ] [ ] , 0 1N

j kn N

kx n X k e n N

Nπ

−

=

= ≤ ≤ −∑

[ ]X k[ ]x n

DFT: analysis equation

IDFT: synthesis equation

time domain

frequency domain

44

Periodic Extension of DFT Sequence

We can show that the definitions of DFT and IDFT induces periodicity

12 /

0[ ] [ ] , 0 1

Nj kn N

nX k x n e k Nπ

−−

=

= ≤ ≤ −∑1

2 ( ) /

01

2 / 2

01

2 /

0

[ ] [ ]

[ ]

[ ] [ ]

Nj k N n N

nN

j kn N j n

nN

j kn N

n

X k N x n e

x n e e

x n e X k

π

π π

π

−− +

=−

− −

=−

−

=

+ =

=

= =

∑

∑

∑

45

Similarly

12 ( ) /

01

2 / 2

01

2 /

0

1[ ] [ ]

1 [ ]

1 [ ] [ ]

Nj n N k N

kN

j kn N j k

kN

j kn N

k

x n N X k eN

X k e eN

X k e x nN

π

π π

π

−+

=−

=−

=

+ =

=

= =

∑

∑

∑

12 /

0

1[ ] [ ] , 0 1N

j kn N

kx n X k e n N

Nπ

−

=

= ≤ ≤ −∑

46

DTFS and DFT

X[k] is the Fourier series coefficient (or spectral component) at the frequency Thus except the scale factor, we have the following relationship

Differences between the DFT and the DTFSthe DFT assumes signal periodicity while the DTFS requires periodicitythe DFT scales the synthesis equation by 1/N whereas the DTFS scales the analysis equation.

0kω

ˆ[ ] [ ]

[ ] [ ]

x n x n

X k X k

periodic extension

take one period

periodic extension

take one period

DFT

IDFT

DTFS

Example 8.7The DFT of a rectangular pulsex[n] is of length 5We can consider x[n] of any length greater than 5Let’s pick N=5Calculate the DFS of the periodic form of x[n]

[ ] ( )

( )

42 /5

02

2 /5

11

5 0, 5, 10,...0

j k n

nj k

j k

X k e

ee

kelse

π

π

π

−

=

−

−

=

−=

−= ± ±⎧

= ⎨⎩

∑

If we consider x[n] of length 10

We get a different set of DFT coefficients

Still samples of the DTFT but in different places

8.6 Properties of DFT

][][ and ][][ kNXkXkNXkX −−∠=∠−=

50

Linearity

n

)(1 nx

0 N1−1

Duration N1

n)(2 nx

0 N2−1

Duration N2

),max( 21 NNN =

)()( 11 kXnx ⎯⎯ →←DFT

)()( 22 kXnx ⎯⎯ →←DFT

)()()()( 2121 kbXkaXnbxnax +⎯⎯ →←+ DFT

51

Review: Circular Shift of a Sequence

0n

)(nx

N

n

)(~ nx

0 N

n

)(~)(~1 mnxnx −=

0 N

⎩⎨⎧ −≤≤−=

=otherwise

Nnmnxnxnx N

010))(()(~

)( 11

52

DFT of Circular Shifted Sequence

⎩⎨⎧ −≤≤−=

=otherwise

Nnmnxnxnx N

010))(()(~

)( 11

)()( kXnx ⎯⎯ →←DFT

(2 / )(( )) , 0 1 ( ) ( )j N km kmN Nx n m n N e X k W X kπ−− ≤ ≤ − ←⎯⎯→ =DFT

53

Duality

)()( kXnx ⎯⎯ →←DFT

10 ,))(()( −≤≤−⎯⎯ →← NkkNxnX NDFT

54

Example: Duality

Choose N=10

Re[X(k)]

Im[X(k)]

Re[x1(n)]= Re[X(n)]

Im[x1(n)]= Im[X(n)]

X1(k) = 10x((−k))10

55

Notations

Notation : Circular reversal

Properties of DFT

(( )) (( ) )Nx n x n

( ) (( ))y n x n= −

(0) (0)y x=( ) (( )) ( )y n x n x N n= − = −

* * *

(( )) [ ] [ ]( ) [ ] [ ]

x n X k X N kx n X k X N k

− ↔ − = −

↔ − = −

56

Review: Even and Odd Components

Any real-valued signal can be expressed as the sum of an odd signal and an even signal.

( ) ( ) ( ) [ ] [ ] [ ]where

1 1( ) { ( ) ( )} [n]= { [ ] [ ]} anevenfunction2 2

and1 1( ) { ( ) ( )} [ ]= { [ ] [ ]} an odd function2 2

e o e o

e e

o o

x t x t x t x n x n x n

x t x t x t x x n x n

x t x t x t x n x n x n

= + = +

= + − + −

= − − − −

57

Even and Odd Components of a Finite Sequence

For a real finite duration sequence

( ) ( )

( ) ( )

1 1( ) ( ) (( )) ( ) ( )2 21 1( ) ( ) (( )) ( ) ( )2 2

e

o

x n x n x n x n x N n

x n x n x n x n x N n

= + − = + −

= − − = − −

58

Review: Even and Odd Components

If x(t) is a complex valued function:x(t)=xR(t) + j xI(t)x(t) is even iff both the real and imaginary parts of x(t) are even.Similarly x(t) is odd iff both the real and imaginary components of x(t) are odd.

We can also define conjugate symmetry, or Hermitian symmetry: real part of x(t) is even but the imaginary part of x(t) is odd

xR(t)= xR(-t) and xI(t)= -xI(-t) i.e. x*(t)= x*(-t)

59

Symmetry Properties

Real-Imaginary Decomposition

Symmetry Properties (Proakis p.415)

( ) ( ) ( )

[ ( ) ( )] [ ( ) ( )]( ) ( ) ( )

[ ( ) ( )] [ ( ) ( )]

R Ie o e oR R I I

R Ie o e oR R I I

x n x n jx n

x n x n j x n x nX k X k jX k

X k X k j X k X k

= +

= + + += +

= + + +

( ) [ ( ) ( )] [ ( ) ( )]

( ) [ ( ) ( )] [ ( ) ( )]

e o e oR R I I

e o e oR R I I

x n x n x n jx n jx n

X k X k X k jX k jX k

= + + +

= + + +

60

[Re{ ( )}] [ ( ) ( )]

[ ] [ ]

[ Im{ ( )}] [ ( ) ( )]

[ ] [ ]

e oR R

e oR I

e oI I

o eR I

DFT x n DFT x n x n

X k jX k

DFT j x n DFT jx n jx n

X k jX k

= +

= +

= +

= +

Hermitian Symmetry

Hermitian anti-Symmetry

61

The Fourier transform of real signals exhibits special symmetry which is important to us in many cases. Basically, The transform of a real signal x[n] is therefore conjugate Symmetric(Hermitian symmetric)

Real part Symmetric (even)

Imaginary part Antisymmetric (skew-symmetric, odd)

Magnitude Symmetric (even)

Phase Antisymmetric (odd)

Symmetry Relations for DFT of Real Signal

62

Example

)(nx

10 ),(~ =Nnx

|)(| kX

)(kX∠

63

Linear Convolution (Review)

)(*)()()()( 2113 nxnxmnxmxnxm

=−= ∑∞

−∞=

)()( 11ω⎯→← jeXnx FT )()( 22

ω⎯→← jeXnx FT

)()()()(*)()( 213213ωωω =⎯→←= jjj eXeXeXnxnxnx FT

Circular Convolution (Cyclic Convolution)

Circular convolution of two finite length sequences

In linear convolution, one sequence is multiplied by a time –reversed and linearly shifted version of the other. For convolution here, the second sequence is circularly time reversed and circularly shifted.So it is called an N-point circular convolution

[ ] [ ] [ ]

[ ] ( )( )

[ ] ( )( )

1

3 1 201

1 201

2 10

, 0 1

N

mN

NmN

Nm

x n x m x n m

x m x n m

x m x n m n N

−

=

−

=

−

=

= −

⎡ ⎤= −⎣ ⎦

⎡ ⎤= − ≤ ≤ −⎣ ⎦

∑

∑

∑

The cyclic convolution is not the same as the linear

convolution of linear system theory. It is a by-

product of the periodicity of DFS/DFT.

When the DFT X[k] is used, the periodic interpretation of the signal

x[n] is implicit: if

then for any integer m:

Thus, just as the DFT X[k] is implicitly period-N, the inverse DFT is

also implicitly period-N — the periodic extension of x[n].

][][][ nhnxnyN⊗=

][*][][ nhnxny =

{ } { }1,...,0for ][1][IDFT][1

0

2

−∈== ∑−

=

NnekXN

kXnxN

k

Nnkj

N

π

∑∑−

=

−

=

+

==1

0

221

0

)(2

][][1][1 N

k

kmjNknjN

k

NmNnkj

nxeekXN

ekXN

πππ

66

Circular Convolution Theorem

1

3 1 1 20

( ) ( ) (( )) ( ) ( )N

Nm

x n x m x n m x n x n−

=

= − =∑

)()( 11 kXnx ⎯⎯ →←DFT )()( 22 kXnx ⎯⎯ →←DFT

3 1 2 3 1 2( ) ( ) ( ) ( ) ( ) ( )x n x n x n X k X k X k= ←⎯⎯→ =DFT

both of length N

67

Proof: Circular Convolution Theorem

Suppose we have two finite duration sequences x1[n] and x2[n] of length N.Suppose we multiply the two DFTs together: X3[k] = X1[k]X2[k] What happens when I take the IDFT of X3[k] ?

1

3 301

1 20

1 1 1

1 20 0 0

1 1 1( )

1 20 0 0

1( ) [ ]

1 [ ]

1 [ ] [ ]

1 [ ] [ ]

Nkm

kN

km

k

N N Nkn kl km

k n l

N N Nk n l m

n l k

x m X k WN

X X k WN

x n W x l W WN

x n x l WN

−−

=

−−

=

− − −−

= = =

− − −+ −

= = =

=

=

⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎡ ⎤

= ⎢ ⎥⎣ ⎦

∑

∑

∑ ∑ ∑

∑ ∑ ∑

and we have , the circular convolution of x1 and x2

68

( )

1( )

0

1 ( : an integer), if (( ))

Thus, 0,

k n l m

NNk n l m

k

W iff n l m Np pN l m n NP m n

Wotherwise

+ −

−+ −

=

= + − =

= − + = −⎧= ⎨⎩

∑

1

3 1 20

( ) ( ) (( ))N

Nn

x m x n x m n−

=

= −∑

69

Circular Convolution Example

70

Example: Circular Convolution Theorem

⎩⎨⎧ −≤≤

==otherwise

Lnnxnx

0101

)()( 21

3 1 2( ) ( ) ( )x n x n x n=

L=N=6

⎩⎨⎧ =

=

== ∑−

=

otherwisekN

WkXkXN

n

knN

00

)()(1

021

⎩⎨⎧ =

==otherwisekN

kXkXkX0

0)()()(

2

213

0 L

)(1 nx

0 L

)(2 nx

0 L

)(3 nxN

71

3 1 2( ) ( ) ( )x n x n x n=

⎪⎩

⎪⎨⎧

−≤≤−

−=

==

+−

−

=∑

otherwise

LkW

W

WkXkX

kL

LkN

L

n

knL

0

1201

1

)()(

2

1)1(

1

0221

⎪⎩

⎪⎨

⎧−≤≤⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−==

+−

otherwise

LkW

WkXkXkX k

L

LkL

0

1201

1 )()()(

2

2

1)1(2

213

⎩⎨⎧ −≤≤

==otherwise

Lnnxnx

0101

)()( 21

L=6, N=12

0 L

)(2 nx

N

0 L

)(1 nx

N

0

)(3 nx

N

N

Why is cyclic convolution not true linear convolution?

Because a wraparound effect occurs at the “ends”:

The procedure of each pair are summed around the circle.

In a while, it will be seen that can be computed using

][*][][ nhnxny =

].[][][ nhnxnyN⊗=

0][][][ 101kn

NDFT WnXnnnx =⎯⎯ →←−= δ

][][][][ 22130 kXWkXkXkX kn

N==

]))[((

]))[((][

][][][

02

2

1

01

213

N

N

N

k

nnx

knxkx

nxnxnx

−=

−=

⊗=

∑−

=

Example 8.11

Circular convolution of two rectangular pulses L=N=6

DFT of each sequence

Multiplication of DFTs

And the inverse DFT

[ ] [ ]1 2

1 0 10

n Lx n x n

else≤ ≤ −⎧

= = ⎨⎩

[ ] [ ]21

1 20

00

N j knN

n

N kX k X k e

else

π− −

=

=⎧= = = ⎨

⎩∑

[ ] [ ] [ ]2

3 1 20

0N k

X k X k X kelse

⎧ == = ⎨

⎩

[ ]3

0 10N n N

x nelse

≤ ≤ −⎧= ⎨⎩

We can augment zeros to each sequence L=2N=12The DFT of each sequence

Multiplication of DFTs

[ ] [ ]2

1 2 21

1

LkjN

kjN

eX k X ke

π

π

−

−

−= =

−

[ ]

22

3 21

1

LkjN

kjN

eX ke

π

π

−

−

⎛ ⎞−⎜ ⎟= ⎜ ⎟⎜ ⎟−⎝ ⎠

77

Example : Consider the two length-4 sequences repeated below for convenience:

We will take DFT approach

n3210

1

2 ][ng

n3210

1

2 ][nh

78

The 4-point DFT G[k] of g[n] is given by

Therefore

,][ 21212 −=−−=G,][ jjjG −=+−= 1211

jjjG +=−+= 1213][

,][ 41210 =++=G

4210 /][][][ kjeggkG π−+=4644 32 // ][][ kjkj egeg ππ −− ++

3021 232 ≤≤++= −− kee kjkj ,// ππ

79

Likewise,

Hence, ,][ 611220 =+++=H

,][ 011222 =−+−=H,][ jjjH −=+−−= 11221

jjjH +=−−+= 11223][

4210 /][][][ kjehhkH π−+=4644 32 // ][][ kjkj eheh ππ −− ++

3022 232 ≤≤+++= −−− keee kjkjkj ,// πππ

80

4

[0] [0] 1 1 1 1 1 4[1] [1] 1 1 2 1

W[2] [2] 1 1 1 1 0 2[3] [3] 1 1 1 1

G gG g j j jG gG g j j j

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − +⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

4

[0] [0] 1 1 1 1 2 6[1] [1] 1 1 2 1

W[2] [2] 1 1 1 1 1 0[3] [3] 1 1 1 1

H hH h j j jH hH h j j j

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − +⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

is the 4-point DFT matrix4W

81

If denotes the 4-point DFT of then

Thus

30 ≤≤= kkHkGkYC ],[][][

][kYC ][nyC

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

20

224

33221100

3210

j

j

HGHGHGHG

YYYY

CCCC

][][][][][][][][

][][][][

82

The 4-point IDFT of yields][kYC

4

[0] [0][1] [1]1 *W[2] [2]4[3] [3]

C C

C C

C C

C C

y Yy Yy Yy Y

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−−−=

5676

20

224

111111

111111

41

j

j

jj

jj

83

4 pt sequences: g[n]={1 2 0 1}

n1 2 3

h[n]={2 2 1 0}

n1 2 3

× 1

× 2

× 0

× 1

nh[((n – 0))4]1 2 3

nh[((n -1))4]1 2 3

nh[((n – 2))4]1 2 3

nh[((n – 3))4]1 2 3

n

g[n] h[n]={4 7 5 4}

1 2 3

4

check: g[n] h[n]={2 6 5 4 2 1 0}

*

[ ] [ ]1

0

(( ))N

Nm

g m h n m−

=

−∑

84

Example : Now let us extended the two length-4 sequences to length 7 by appending each with three zero-valued samples, i.e.

⎩⎨⎧

≤≤≤≤= 640

30nnngnge ,

],[][

⎩⎨⎧

≤≤≤≤= 640

30nnnhnhe ,

],[][

85

We next determine the 7-point circular convolution of and :

From the above

6

70

[ ] [ ] [(( )) ], 0 6e em

y n g m h n m n=

= − ≤ ≤∑

][nge][nhe

[0] [0] [0] [1] [6] [2] [5] [3] [4][4] [3] [5] [2] [6] [1]

[0] [0] 1 2 2

e e e e e e e e

e e e e e e

y g h g h g h g hg h g h g hg h

= + + ++ + +

= = × =

86

Continuing the process we arrive at

,)()(][][][][][ 6222101101 =×+×=+= hghgy

][][][][][][][ 0211202 hghghgy ++= ,)()()( 5202211 =×+×+×=

][][][][][][][][][ 031221303 hghghghgy +++=

,)()()()( 521201211 =×+×+×+×=

][][][][][][][ 1322314 hghghgy ++= ,)()()( 4211012 =×+×+×=

,)()(][][][][][ 1111023325 =×+×=+= hghgy

111336 =×== )(][][][ hgy

87

As can be seen from the above that y[n] is precisely the sequence obtained by a linear convolution of g[n] and h[n]

][nyL

][nyL

DFT and DFS

Periodic Extension: Given a finite-length sequence

define the periodic sequence by

does not have a z-transform or a convergent Fourier sum (why?). But it does have a DFS representation.

The DFS that is the true frequency representation of discrete-time periodic signals. The DFT is just one period of the DFS.

The length-N DFT of the length-N signal contains all the information about . It is convenient to work with.

Whenever the DFT is used, actually the DFS is being used – computations involving are affected by the true periodicity of the coefficients.

{ }1,...,0];[ −= Nnnx{ }+∞<<−∞ nnx ];[~

{ }] mod [][][~ Nnxnxnx N ==][~ nx

][kX ][nx][nx][kX ][~ kX

][kX][~ kX

DFT and z-Transform

The DFT samples the z-transform at evenly spaced samples of the unit circle over one revolution:

{ }1,...,0for |)(][ /2 −∈==

NkzXkX Nkjez π

DFT and DTFT

The DFT samples on period of the discrete-time Fourier transform (DTFT) at N evenly spaced frequencies

{ }1,...,0for |)(][ /2 −∈= = NkeXkX Nkj

πωω

91

DTFT from DFT by Interpolation

It can readily be shown that

This is the dual of interpolation of original time function from sampled values by an ideal low-pass filtering

∑ ⋅⎟⎠⎞

⎜⎝⎛ π−ω

⎟⎠⎞

⎜⎝⎛ π−ω

=−

=

−π−ω−1

0

]2/)1)][(/2[(

22sin

22sin

][1 N

k

NNkje

NkN

kN

kXN

( )jX e ω

[ ]

( ) [ ]

x n

X X kω

IDTFT

DTFT

DFT

IDFTsampling

interpolation

for a finite-durationsignal with no aliasing

92

Relations to the DTFT

[ ] [ ] [ ]

( ) [ ]

m

x n y n x n mN

X Y Kω

∞

=−∞

= +∑

DTFTIDTFT

Aliasing

X

N-point sampling

X

IDFT DFT

Example : A periodic sequence is constructed from the sequence

Discrete-Time Fourier Transform (DTFT)

Discrete Fourier Transform (DFT)

By comparing DTFT and DFT, we find

1 ],[][~ 1 ],[][ <+=→<= ∑∞

−∞=

arNnxnxanuanxr

n

ωωωω

jn

njn

n

njj

aeeaenxeX −

∞

=

−∞

−∞=

−

−=== ∑∑ 1

1][)(0

Nkj

N

n

nNj

aeenxkX /2

1

0

)/2(

11][~][~

ππ

−

−

=

−

−== ∑

{ }1,...,0for |)(][ /2 −∈= = NkeXkX Nkj

πωω

94

Matrix Representation

The DFT samples defined by

can be expressed in matrix form as

where and

10,][][1

0−≤≤∑=

−

=NkWnxkX

N

n

knN

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

=

]1[:

]1[]0[

Nx

xx

x

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

=

]1[:

]1[]0[

NX

XX

X

1

0

[ ] [ ]N

nkN N

n

X k x n W W−

=

= ⇒ =∑ X x

11

0

1[ ] [ ]N

nkN N

nx n X k W W

N

−− −

=

= ⇒ =∑ x X

95

and is the DFT matrix given byWN NN ×

2

1 2 ( 1)

2 4 2( 1)

( 1) 2( 1) ( 1)

1 1 1 111W

1

NN N N

NknN N NN N

N N NN N N

W W WW W WW

W W W

−

−

− − −

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎡ ⎤= =⎣ ⎦ ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

indexes column androw :,

)1)(1(,

jiW ji

Nji

N−−=W

2 /j NNW e π−=

96

Likewise, the IDFT relation given by

can be expressed in matrix form as

where is the IDFT matrix

1

0

1[ ] [ ] , 0 1N

k nN

kx n X k W n N

N

−−

=

= ≤ ≤ −∑

NN ×1WN−

1x W XN−=

2

1 2 ( 1)

1 2 4 2( 1)

( 1) 2( 1) ( 1)

1 1 1 1

11 1

1

1

NN N N

kn NN N N N N

N N NN N N

W W W

W W W WN N

W W W

− − − −

− − − − − −

− − − − − −

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎡ ⎤= =⎣ ⎦ ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

W

1 1 *W WN NN− =

97

Example: Matrix Representation

0 0 0 04 4 4 40 1 2 3

4 4 4 44 0 2 4 6

4 4 4 40 3 6 9

4 4 4 4

1 1 1 11 1

W1 1 1 11 1

W W W Wj jW W W W

W W W Wj jW W W W

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎣ ⎦⎣ ⎦

0 0 0 04 4 4 40 1 2 3

1 4 4 4 44 0 2 4 6

4 4 4 40 3 6 9

4 4 4 4

1 1 1 11 11W1 1 1 141 1

W W W Wj jW W W W

W W W Wj jW W W W

− − −−

− − −

− − −

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎣ ⎦⎣ ⎦

Example : The DFT matrices of dimension 2, 3, 4 are as follows:

Suppose

Where we observe that the real part of X[k] is even-symmetric, and the imaginary part is odd-symmetric – the DFT of the real signal.

⎥⎦

⎤⎢⎣

⎡−

=11

112W

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−−+−

+−−−=

231

2311

231

2311

111

3

jj

jjW

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−−

−−=

jj

jjW

111111

111111

4

,]2031[ T−=x

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8.7 Linier convolution using the DFT

Efficient algorithms are available for computing the DFT of finite-duration sequence, therefore it is computationally efficient to implement a convolution of two sequences by the following procedure:

Compute the N point discrete Fourier transforms X1[k] and X2[k] for the two sequences given.Compute the product X3[k]= X1[k]X2[k] Compute the inverse DFT of X3[k]

The multiplication of discrete Fourier transforms corresponds to a circular convolution. To obtain a linear convolution, we must ensure that circular convolution has the effect of linear convolution.

[ ] [ ] [ ]3 1 2m

x k x m x n m+∞

=−∞

= −∑

With aliasing

Without aliasing

Implementing LTI systems using the DFT

Let us consider an L point input sequence x[n] and a P point impulse response h[n]. The linear convolution has finite-duration with length L+P-1. Consequently for linear convolution and circular convolution to be identical, the circular convolution must have the length of at least L+P-1 points.

i.e. both x[n] and h[n] must be augmented with sequence amplitude with zero amplitude.

This process is often referred to as zero-padding.

106

Linear convolution is a key operation in many signal processing applicationsSince a DFT can be efficiently implemented using FFT algorithms, it is of interest to develop methods for the implementation of linear convolution using the DFTLet g[n] and h[n] be two finite-length sequences of length N and M, respectivelyPickDefine two length-L sequences

1L N M= + −

⎩⎨⎧

−≤≤−≤≤= 10

10LnN

Nnngnge ,],[][

⎩⎨⎧

−≤≤−≤≤= 10

10LnM

Mnnhnhe ,],[][

107

Linear Convolution of Two Finite-Length Sequences

Then

The corresponding implementation scheme is illustrated below

Zero-paddingwith

zeros1)( −N point DFT

Zero-paddingwith

zeros1)( −M−−+ )( 1MN

point DFT×

g[n]

h[n]

Length-N

][nge

][nhe −−+ )( 1MN

−−+ )( 1MNpoint IDFT

][nyL

Length-M Length- )( 1−+ MN

[ ] [ ] [ ] [ ] [[ ] ]CLy n g n h n y n g n h n== =∗

The size of DFT to be used can be any integer L≥N+M-1

108

Circular Convolution as Linear Convolution with Time Aliasing

L0

x1(n)n

P

x2(n)0

nL

0 Pn

L+P−1

x (n)

L

109

Terminology: Fast Convolution

110

Filtering Long Sequences

Sometimes we want to filter a sequence that is very longcould save up all the samples, then either

» do a really long time-domain convolution, or» use really big DFTs to do it in the frequency domain

but big DFTs may become impractical; besideswe get long latency: we have to wait a long time to get any output

Sometimes we want to filter a sequence of indefinite lengthand then even the methods above don’t work

111

Example: Filtering Long Sequences

Finite-length impulse response h[n] and indefinite-length signal x[n] to be filtered

Implementing Linear Time-Invariant Systems Using the DFTTheoretically, we store the entire samples and then implement the convolution procedure using a DFT for a large number points which is generally impractical to compute.No filter samples can be computed until all the input samples have been collected. Generally, we would like to avoid such a large delay in processing.

The solution of both problems is to use “block convoltuion”.

112

Block Convolution Techniques

All of samples will be segmented into section of appropriate length (L).Each section can then be convolved with the finite-length impulse response and the filtered sections fitted together in an appropriate way.

Overlap-Add MethodOverlap-Save Method

113

Using Linearity to Filter a Long Signal

114

115

Block Convolution - Overlap Add Method

116

Segmenting the Input in OLA

117

Putting the Output Pieces Together

118

Filtering the Segments

119

Block Convolution - Overlap Save Method

120

Linear Convolution Example Again

121

Aliased Convolution

122

123

OLS Method - Segmenting the Input

124

OLS Method - Extracting the Output

125

Overlap-add method is the procedure of decomposition of x[n] into nonoverlappingsections of length L and the result of convolving each section with h[n] which are overlapped and added to construct the output.

Comparison

126

Overlap-save method is the procedure of decomposition of x[n] into overlapping sections of length L and the result of convolving each section with h[n] which the portions of each filtered section to be discarded in forming the linear convolution.