Discrete -Time Fourier Transform Discrete Fourier ...tania/teaching/dsp/Lectures 3-4 DTFT DFT... ·...

239
Discrete Discrete - - Time Fourier Transform Time Fourier Transform Discrete Fourier Transform Discrete Fourier Transform z z - - Transform Transform Tania Stathaki 811b [email protected]

Transcript of Discrete -Time Fourier Transform Discrete Fourier ...tania/teaching/dsp/Lectures 3-4 DTFT DFT... ·...

Page 1: Discrete -Time Fourier Transform Discrete Fourier ...tania/teaching/dsp/Lectures 3-4 DTFT DFT... · Discrete -Time Fourier Transform Discrete Fourier Transform z-Transform Tania Stathaki

DiscreteDiscrete--Time Fourier TransformTime Fourier Transform

Discrete Fourier TransformDiscrete Fourier Transform

zz--TransformTransform

Tania Stathaki

811b

[email protected]

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Joseph Fourier (1768-1830)

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DiscreteDiscrete--Time Fourier TransformTime Fourier Transform

• Definition - The Discrete-Time Fourier

Transform (DTFT) of a sequence

x[n] is given by

• In general, is a complex function

of the real variable ω and can be written as

)( ωjeX

)( ωjeX

∑∞

−∞=

−=n

njj enxeX ωω ][)(

)()()( imreωωω jjj eXjeXeX +=

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform

• and are, respectively,

the real and imaginary parts of , and

are real functions of ω• can alternately be expressed as

where

)( ωjeX

)( ωre

jeX )( ωim

jeX

)( ωjeX)()()( ωθωω = jjj eeXeX

)}(arg{)( ω=ωθ jeX

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform

• is called the magnitude function

• is called the phase function

• Both quantities are again real functions of ω• In many applications, the DTFT is called

the Fourier spectrum

• Likewise, and are called the

magnitude and phase spectra

)( ωjeX

)(ωθ

)( ωjeX )(ωθ

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)(

)()ω(tan

)()()(

)ω(sin)()(

)ω(cos)()(

)()()(

ω

re

ω

im

ω2

im

ω2

re

ωω

im

ωω

re

ωω2

ω

j

j

jjj

jj

jj

jjj

eX

eX

eXeXeX

eXeX

eXeX

eXeXeX

=

+=

=

=

= ∗

θ

θ

θ

DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform

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DiscreteDiscrete--Time Fourier TransformTime Fourier Transform

• For a real sequence x[n], and are

even functions of ω, whereas, and

are odd functions of ω (Prove using previous slide relationships)• Note:

for any integer k

• The phase function θ(ω) cannot be uniquely

specified for any DTFT

)( ωjeX

)(ωθ)(re

ωjeX

)(im

ωjeX

)2()()( kjjj eeXeX π+ωθωω =)()( ωθω= jj eeX

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform

• Unless otherwise stated, we shall assume

that the phase function θ(ω) is restricted to

the following range of values:

called the principal value

π<ωθ≤π− )(

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform

• The DTFTs of some sequences exhibit

discontinuities of 2π in their phase

responses

• An alternate type of phase function that is a

continuous function of ω is often used

• It is derived from the original phase

function by removing the discontinuities of

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform

• The process of removing the discontinuities

is called “unwrapping”

• The continuous phase function generated by

unwrapping is denoted as

• In some cases, discontinuities of π may be

present after unwrapping

)(ωθc

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform

• Example - The DTFT of the unit sample

sequence δ[n] is given by

• Example - Consider the causal sequence

1]0[][)( =δ=∑δ=ω∆ ω−∞

−∞=

nj

n

en

=<=otherwise0

01][,1],[][

nnnnx n µαµα

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform

• Its DTFT is given by

as

∑α=∑ µα=∞

=

ω−∞

−∞=

ω−ω

0

][)(n

njn

n

njnj eeneX

ω−α−

=

ω− =∑ α=jen

nje1

1

0

)(

1<α=α ω− je

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform

• The magnitude and phase of the DTFT

are shown below)5.01/(1)( ω−ω −= jj eeX

-3 -2 -1 0 1 2 30.5

1

1.5

2

ω/π

Mag

nitude

-3 -2 -1 0 1 2 3

-0.4

-0.2

0

0.2

0.4

0.6

ω/π

Phas

e in

rad

ians

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform

• The DTFT of a sequence x[n] is a

continuous function of ω

• It is also a periodic function of ω with a

period 2π:

)( ωjeX

∑=∞

−∞=

π+ω−π+ω

n

nkjkj oo enxeX)2()2(

][)(

nkj

n

njeenx o π−∞

−∞=

ω−∑= 2][ )(][ oo j

n

njeXenx

ω∞

−∞=

ω− =∑=

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform

• Inverse Discrete-Time Fourier Transform:

• Proof:

∫ ωπ

π−

ωω deeXnx njj )(2

1][

∫ ω

π=

π

π−

ω∞

−∞=

ω− deexnx njj

l

ll][

2

1][

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DiscreteDiscrete--Time Fourier Time Fourier TransformTransform

• The order of integration and summation can

be interchanged if the summation inside the

brackets converges uniformly, i.e.,

exists

• Then ∫ ω

π

π

π−

ω∞

−∞=

ω− deex njj

l

ll][

2

1

)( ωjeX

)(

)(sin][ω

2

1][ )(

l

lll

l

l

l −−

=

= ∑∫∑

−∞=−

−∞

−∞= n

nxdex nj

ππ

π

π

π

ω

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform

• Now

• Hence

l

l

l

l

==

−π−π

n

n

n

n

,0

,1

)(

)(sin

][ l−δ= n

][][][)(

)(sin][ nxnx

n

nx =∑ −δ=

−π−π

∑∞

−∞=

−∞= ll

lll

ll

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform

• Convergence Condition - An infinite

series of the form

may or may not converge

• Consider the following approximation

∑=∞

−∞=

ω−ω

n

njj enxeX ][)(

∑=−=

ω−ω K

Kn

njjK enxeX ][)(

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DiscreteDiscrete--Time Fourier TransformTime Fourier Transform

• Then for uniform convergence of ,

• If x[n] is an absolutely summable sequence, i.e., if

for all values of ω• Thus, the absolute summability of x[n] is a sufficient

condition for the existence of the DTFT

)( ωjeX

0)()(lim =− ωω

∞→

jK

j

KeXeX

∞<∑∞

−∞=nnx ][

∞<∑≤∑=∞

−∞=

−∞=

ω−ω

nn

njj nxenxeX ][][)(

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform

• Example - The sequence for

is absolutely summable as

and therefore its DTFT converges

to uniformly

][][ nnx nµα=1<α

∞<α−

=∑ α=∑ µα∞

=

−∞= 1

1][

0n

n

n

n n

)( ωjeX

)1/(1 ω−α− je

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform

• Since

an absolutely summable sequence has

always a finite energy

• However, a finite-energy sequence is not

necessarily absolutely summable

,][][

22

∑≤∑∞

−∞=

−∞= nn

nxnx

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• Example - The sequence

has a finite energy equal to

• However, x[n] is not absolutely summable since the summation

does not converge.

DiscreteDiscrete--Time Fourier TransformTime Fourier Transform

=][nx

0011

≤≥

nnn

,,/

6

1 2

1

2 π=∑

=

=nx

nE

∑∑∞

=

=

=11

11

nn nn

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DiscreteDiscrete--Time Fourier TransformTime Fourier Transform

• To represent a finite energy sequence that is not

absolutely summable by a DTFT, it is necessary to

consider a mean-square convergence of

where

)( ωjeX

0)()(lim2

=ω∫ −π

π−

ωω

∞→deXeX j

Kj

K

∑=−=

ω−ω K

Kn

njjK enxeX ][)(

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform

• Here, the total energy of the error

must approach zero at each value of ω as K

goes to

• In such a case, the absolute value of the

error may not go to

zero as K goes to and the DTFT is no

longer bounded

)()( ωω − jK

j eXeX

)()( ωω − jK

j eXeX

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform

• Example - Consider the DTFT

shown below

π≤ω<ω

ω≤ω≤=ω

c

cjLP eH

,0

0,1)(

)( ωjLP eH

cω π0

1

π− cω−ω

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DiscreteDiscrete--Time Fourier TransformTime Fourier Transform

• The inverse DTFT of is given by

• The energy of is given by (See slide 46 for proof. Parseval’s Theorem stated in slide 37 is used).

• is a finite-energy sequence,

but it is not absolutely summable

)( ωjLP eH

][nhLP

,sin

2

1

n

n

jn

e

jn

e cnjnj cc

πω

=

π=

ω−ω

ω∫π

ω−

ω denhc

c

njLP

2

1][

∞<<∞− n

πω /c

][nhLP

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform

• As a result

does not uniformly converge to

for all values of ω, but converges to

in the mean-square sense

)( ωjLP eH

)( ωjLP eH

∑ πω

=∑−=

ω−

−=

ω− K

Kn

njcK

Kn

njLP e

nn

enhsin

][

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform

• The mean-square convergence property of

the sequence can be further

illustrated by examining the plot of the

function

for various values of K as shown next

∑ πω

=−=

ω−ω K

Kn

njcjKLP e

nn

eHsin

)(,

][nhLP

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DiscreteDiscrete--Time Fourier Time Fourier TransformTransform

0 0.2 0.4 0.6 0.8 1

0

0.2

0.4

0.6

0.8

1

ω/π

Am

plitu

de

N = 20

0 0.2 0.4 0.6 0.8 1

0

0.2

0.4

0.6

0.8

1

ω/π

Am

plitu

de

N = 30

0 0.2 0.4 0.6 0.8 1

0

0.2

0.4

0.6

0.8

1

ω/π

Am

plitu

de

N = 10

0 0.2 0.4 0.6 0.8 1

0

0.2

0.4

0.6

0.8

1

ω/π

Am

plitu

de

N = 40

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform

• As can be seen from these plots, independent

of the value of K there are ripples in the plot

of around both sides of the

point

• The number of ripples increases as K

increases with the height of the largest ripple

remaining the same for all values of K

)(,ωj

KLP eH

cω=ω

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform• As K goes to infinity, the condition

holds indicating the convergence of to

• The oscillatory behavior of approximating in the mean-square sense at a point of discontinuity is known as the Gibbs phenomenon

)(,ωj

KLP eH)( ωj

LP eH

0)()(lim2

, =ω∫ −π

π−

ωω

∞→deHeH j

KLPj

LPK

)(,ωj

KLP eH)( ωj

LP eH

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform• The DTFT can also be defined for a certain

class of sequences which are neither

absolutely summable nor square summable

• Examples of such sequences are the unit

step sequence µ[n], the sinusoidal sequence

and the exponential sequence

• For this type of sequences, a DTFT

representation is possible using the Dirac

delta function δ(ω)

)cos( φ+ω nonAα

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform• A Dirac delta function δ(ω) is a function of

ω with infinite height, zero width, and unit

area

• It is the limiting form of a unit area pulse

function as ∆ goes to zero, satisfying)(ω∆p

∫ ωωδ=∫ ωω∞

∞−

∞−∆→∆

ddp )()(lim0

ω2∆−

2∆0

∆1

)(ω∆p

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform

• Example - Consider the complex exponential

sequence

• Its DTFT is given by

where is an impulse function of ω and

nj oenxω=][

∑ π+ω−ωπδ=∞

−∞=

ω

ko

j keX )2(2)(

)(ωδπ≤ω≤π− o

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform

• The function

is a periodic function of ω with a period 2πand is called a periodic impulse train

• To verify that given above is

indeed the DTFT of we

compute the inverse DTFT of

∑ π+ω−ωπδ=∞

−∞=

ω

ko

j keX )2(2)(

)( ωjeXnj oenx

ω=][

)( ωjeX

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DiscreteDiscrete--Time Fourier Time Fourier

TransformTransform

• Thus

where we have used the sampling property

of the impulse function )(ωδ

∫ ∑ ωπ+ω−ωπδπ

π−

−∞=

ω

k

njo deknx )2(2

2

1][

njnjo

oedeω

π

π−

ω =ω∫ ω−ωδ= )(

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Commonly Used DTFT PairsCommonly Used DTFT Pairs

Sequence DTFT

1][ ↔δ n

∑ π+ωπδ↔∞

−∞=kk)2(21

∑ π+ω−ωπδ↔∞

−∞=

ω

ko

njke o )2(2

∑ π+ωδπ+−

↔µ∞

−∞=ω−

kj

ke

n )2(1

1][

ω−α−↔<αµ

jen

1

1)1(],[

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DTFT PropertiesDTFT Properties

• There are a number of important properties

of the DTFT that are useful in signal

processing applications

• These are listed here without proof

• Their proofs are quite straightforward

• We illustrate the applications of some of the

DTFT properties

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Table: Table: General Properties of DTFTGeneral Properties of DTFT

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Table:Table: Symmetry relations of the Symmetry relations of the DTFT of a complex sequenceDTFT of a complex sequence

x[n]: A complex sequence

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Table:Table: Symmetry relations of Symmetry relations of the DTFT of a real sequencethe DTFT of a real sequence

x[n]: A real sequence

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DTFT PropertiesDTFT Properties

• Example - Determine the DTFT of

• Let

• We can therefore write

• From Tables above, the DTFT of x[n] is

given by

1],[)1(][ <αµα+= nnny n

1],[][ <αµα= nnx n

][][][ nxnxnny +=

ω−ω

α−=

j

j

eeX

1

1)(

)( ωjeY

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DTFT PropertiesDTFT Properties

• Using the differentiation property of the

DTFT given in Table above, we observe

that the DTFT of is given by

• Next using the linearity property of the

DTFT given in Table above we arrive at

][nxn

2)1(1

1)(ω−

ω−

ω−

ω

α−

α=

α−ω=

ω j

j

j

j

e

e

ed

dj

d

edXj

22 )1(

1

1

1

)1()( ω−ω−ω−

ω−ω

α−=

α−+

α−

α=

jjj

jj

eee

eeY

Page 44: Discrete -Time Fourier Transform Discrete Fourier ...tania/teaching/dsp/Lectures 3-4 DTFT DFT... · Discrete -Time Fourier Transform Discrete Fourier Transform z-Transform Tania Stathaki

DTFT PropertiesDTFT Properties

• Example - Determine the DTFT of

the sequence v[n] defined by

• The DTFT of is 1

• Using the time-shifting property of the

DTFT given in Table above we observe that

the DTFT of is and the DTFT

of is][ 1−nv

]1[][]1[][ 1010 −δ+δ=−+ npnpnvdnvd

][nδ

]1[ −δ n ω− je

)( ωjeV

)( ωω− jj eVe

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DTFT PropertiesDTFT Properties

• Using the linearity property of we then

obtain the frequency-domain representation

of

as

• Solving the above equation we get

]1[][]1[][ 1010 −δ+δ=−+ npnpnvdnvd

ω−ωω−ω +=+ jjjj eppeVedeVd 1010 )()(

ω−

ω−ω

+

+=

j

jj

edd

eppeV

10

10)(

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Energy Density SpectrumEnergy Density Spectrum

• The total energy of a finite-energy sequence

g[n] is given by

• From Parseval’s relation given above we

observe that

∑=∞

−∞=ng ng

2][E

∫ ωπ

=∑=π

π−

ω∞

−∞=deGng j

ng

22

2

1)(][E

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Energy Density SpectrumEnergy Density Spectrum

• The quantity

is called the energy density spectrum

• Therefore, the area under this curve in the

range divided by 2π is the

energy of the sequence

π≤ω≤π−

2)()( ω=ω j

gg eGS

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Energy Density SpectrumEnergy Density Spectrum

• Example - Compute the energy of the

sequence

• Here

where

∞<<∞−πω

= nnn

nh cLP ,

sin][

∫ ωπ

=∑π

π−

ω∞

−∞=deHnh j

LPn

LP

22)(

2

1][

π≤ω<ω

ω≤ω≤=ω

c

cjLP eH

,0

0,1)(

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Energy Density SpectrumEnergy Density Spectrum

• Therefore

• Hence, is a finite-energy sequence

∞<πω

=∫ ωπ

=∑ω

ω−

−∞=

c

nLP

c

c

dnh2

1][2

][nhLP

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DTFT Computation Using DTFT Computation Using

MATLABMATLAB

• The function freqz can be used to

compute the values of the DTFT of a

sequence, described as a rational function in

the form of

at a prescribed set of discrete frequency

points

NjN

j

MjM

jj

ededd

epeppeX ω−ω−

ω−ω−ω

+++

+++=

....

....)(

10

10

lω=ω

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DTFT Computation Using MATLABDTFT Computation Using MATLAB

• For example, the statement

H = freqz(num,den,w)

returns the frequency response values as a vector H of a DTFT defined in terms of the

vectors num and den containing the

coefficients and , respectively at a

prescribed set of frequencies between 0 and

2π given by the vector w

• There are several other forms of the functionfreqz

}{ ip }{ id

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DTFT Computation Using MATLABDTFT Computation Using MATLAB

• Example – We illustrate the magnitude and phase of

the following DTFT

ωωωω

ωωωωω

432

432

41.06.17.237.21

008.0033.005.0033.0008.0)(

jjjj

jjjjj

eeee

eeeeeX

−−−−

−−−−

+++++−+−

=

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1Magnitude Spectrum

ω/π

Mag

nitude

0 0.2 0.4 0.6 0.8 1-4

-2

0

2

4Phase Spectrum

ω/π

Phas

e, rad

ians

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DTFT Computation Using DTFT Computation Using

MATLABMATLAB• Note: The phase spectrum displays a

discontinuity of 2π at ω = 0.72

• This discontinuity can be removed using the function unwrap as indicated below

0 0.2 0.4 0.6 0.8 1-7

-6

-5

-4

-3

-2

-1

0Unwrapped Phase Spectrum

ω/π

Phas

e, rad

ians

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Linear Convolution Using Linear Convolution Using

DTFTDTFT

• An important property of the DTFT is given

by the convolution theorem

• It states that if y[n] = x[n] h[n], then the

DTFT of y[n] is given by

• An implication of this result is that the

linear convolution y[n] of the sequences

x[n] and h[n] can be performed as follows:

*

)( ωjeY

)()()( ωωω = jjj eHeXeY

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Linear Convolution Using Linear Convolution Using

DTFTDTFT

• 1) Compute the DTFTs and

of the sequences x[n] and h[n], respectively

• 2) Form the DTFT

• 3) Compute the IDTFT y[n] of

)()()( ωωω = jjj eHeXeY

)( ωjeX )( ωjeH

)( ωjeY

×x[n]

h[n]

y[n]DTFT

DTFT

IDTFT)( ωjeY

)( ωjeX

)( ωjeH

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Discrete Fourier TransformDiscrete Fourier Transform

• Definition - For a length-N sequence x[n],

defined for only N samples of its

DTFT are required, which are obtained by

uniformly sampling on the ω-axis

between at ,

• From the definition of the DTFT we thus have

10 −≤≤ Nn

10 −≤≤ Nk)( ωjeX

π≤ω≤ 20 Nkk /2π=ω

,][)(][1

0

/2

/2∑==−

=

π−

π=ω

ω N

k

Nkj

Nk

j enxeXkX

10 −≤≤ Nk

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Discrete Fourier TransformDiscrete Fourier Transform

• Note: X[k] is also a length-N sequence in

the frequency domain

• The sequence X[k] is called the Discrete

Fourier Transform (DFT) of the sequence

x[n]

• Using the notation the

DFT is usually expressed as:

NjN eW /2π−=

10,][][1

0

−≤≤∑=−

=NkWnxkX

N

n

nkN

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Discrete Fourier TransformDiscrete 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−= Nn

nNWl

10,][1

][1

0

−≤≤∑=−

=

− NnWkXN

nxN

k

knN

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Discrete Fourier TransformDiscrete Fourier Transform

resulting in

∑ ∑∑−

=

=

−−

=

=

1

0

1

0

1

0

1N

n

nN

N

k

knN

N

n

nN WWkX

NWnx ll ][][

∑ ∑−

=

=

−−=1

0

1

0

1 N

n

N

k

nkNWkX

N

)(][ l

∑ ∑−

=

=

−−=1

0

1

0

1 N

k

N

n

nkNWkX

N

)(][ l

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Discrete Fourier TransformDiscrete Fourier Transform

• Making use of the identity

we observe that the RHS of the last

equation is equal to

• Hence

=∑

=

−−1

0

N

n

nkNW

)( l

,,

0N ,rNk =− lfor

otherwiser an integer

][lX

][][ ll XWnx

N

n

nN =∑

=

1

0

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Discrete Fourier TransformDiscrete Fourier Transform

• Example - Consider the length-N sequence

• Its N-point DFT is given by

11001−≤≤

=Nn

n,,

=][nx

10 01

0

=== ∑−

=N

N

n

knN WxWnxkX ][][][

10 −≤≤ Nk

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Discrete Fourier TransformDiscrete Fourier Transform

• Example - Consider the length-N sequence

• Its N-point DFT is given by

=][ny

111001

−≤≤+−≤≤=

Nnmmnmn

,,,

kmN

kmN

N

n

knN WWmyWnykY === ∑

=][][][

1

0

10 −≤≤ Nk

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Discrete Fourier TransformDiscrete Fourier Transform

• Example - Consider the length-N sequence

defined for

• Using a trigonometric identity we can write

10),/2cos(][ −≤≤π= NrNrnng

( )NrnjNrnj eeng /2/2

2

1][ π−π +=

( )rn

N

rn

N WW += −

2

1

10 −≤≤ Nn

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Discrete Fourier TransformDiscrete Fourier Transform

• The N-point DFT of g[n] is thus given by

∑−

==

1

0

N

n

knNWngkG ][][

,2

1 1

0

)(1

0

)(

∑+∑=−

=

+−

=

−− N

n

nkrN

N

n

nkrN WW

1,0 −≤≤ Nrk

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Discrete Fourier TransformDiscrete Fourier Transform

• Making use of the identity

we get

−==

=otherwise0

for2

for2

,

,/

,/

][ rNkN

rkN

kG

=∑

=

−−1

0

N

n

nkNW

)( l

,,

0N ,rNk =− lfor

otherwiser an integer

1,0 −≤≤ Nrk

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Matrix RelationsMatrix Relations

• The DFT samples defined by

can be expressed in matrix form as

where

10,][][1

0

−≤≤∑=−

=NkWnxkX

N

n

knN

xDX N=

[ ]TNXXX ]1[]1[]0[ −= KX

[ ]TNxxx ]1[]1[]0[ −= Kx

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Matrix RelationsMatrix Relations

and is the DFT matrix given byND NN ×

=

−−−

21121

1242

121

1

1

1

1111

)()()(

)(

)(

NN

NN

NN

NNNN

NNNN

N

WWW

WWW

WWW

L

MOMMM

L

L

L

D

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Matrix RelationsMatrix Relations

• Likewise, the IDFT relation given by

can be expressed in matrix form as

where is the IDFT matrix

10,][][1

0

−≤≤∑=−

=

− NnWkXnxN

k

nkN

NN ×1−ND

XDx1−= N

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Matrix RelationsMatrix Relations

where

• Note:

=

−−−−−−

−−−−

−−−−

21121

1242

121

1

1

1

1

1111

)()()(

)(

)(

NN

NN

NN

NNNN

NNNN

N

WWW

WWW

WWW

L

MOMMM

L

L

L

D

*NN

NDD

11 =−

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DFT Computation Using DFT Computation Using

MATLABMATLAB

• The functions to compute the DFT and the IDFT are fft and ifft

• These functions make use of FFT

algorithms which are computationally

highly efficient compared to the direct

computation

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DFT Computation Using DFT Computation Using

MATLABMATLAB• Example - The DFT and the DTFT of the

sequence

are shown below

150),16/6cos(][ ≤≤π= nnnx

0 0.2 0.4 0.6 0.8 10

2

4

6

8

10

Normalized angular frequency

Mag

nitude

indicates DFT samples

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DTFT from DFT by DTFT from DFT by

Interpolation Interpolation

• The N-point DFT X[k] of a length-N

sequence x[n] is simply the frequency

samples of its DTFT evaluated at N

uniformly spaced frequency points

• Given the N-point DFT X[k] of a length-N

sequence x[n], its DTFT can be

uniquely determined from X[k] !

)( ωjeX

)( ωjeX

10,/2 −≤≤π=ω=ω NkNkk

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DTFT from DFT by DTFT from DFT by

InterpolationInterpolation

• Thus

∑=−

=

ω−ω 1

0

][)(N

n

njj enxeX

njN

n

N

k

nkN eWkX

N

ω−−

=

=

−∑

∑=1

0

1

0

][1

∑ ∑=−

=

=

π−ω−1

0

1

0

)/2(][1 N

k

N

n

nNkjekXN

S

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DTFT from DFT by DTFT from DFT by

InterpolationInterpolation

• To develop a compact expression for the

sum S, let

• Then

• From the above

1111 −+=−+∑= −=

NNNn

n rSrr

)/2( Nkjer π−ω−=

∑= −=

10S N

nnr

11S 111 −+∑+=∑= −==

NNn

nNn

n rrrr

1111 −+=−+∑= −=

NNNn

n rSrr

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DTFT from DFT by DTFT from DFT by InterpolationInterpolation

• Or, equivalently,

• Hence

Nrrr −=−=− 1S)1(SS

)]/2([

)2(

1

1

1

1S

Nkj

kNjN

e

e

r

rπ−ω−

π−ω−

−=

−−

=

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

2

2sin

2

2sin

−π−ω−⋅

π−ω

π−ω

= NNkje

N

kN

kN

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DTFT from DFT by DTFT from DFT by

InterpolationInterpolation

• Therefore

∑ ⋅

π−ω

π−ω

=−

=

−π−ω−1

0

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

2

2sin

2

2sin

][1 N

k

NNkje

N

kN

kN

kXN

)( ωjeX

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Sampling the DTFTSampling the DTFT

• Consider a sequence x[n] with a DTFT

• We sample at N equally spaced points

, developing the N

frequency samples

• These N frequency samples can be

considered as an N-point DFT Y[k] whose N-

point IDFT is a length-N sequence y[n]

)( ωjeX

)( ωjeX

Nkk /2π=ω 10 −≤≤ Nk

)}({ kjeXω

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Sampling the DTFTSampling the DTFT

• Now

• Thus

• An IDFT of Y[k] yields

∑=∞

−∞=

ω−ω

l

ll

jj exeX ][)(

)()(][ /2 NkjjeXeXkY k πω ==

∑=∑=∞

−∞=

−∞=

π−

l

l

l

lll

kN

Nkj Wxex ][][ /2

∑=−

=

−1

0

][1

][N

k

nkNWkY

Nny

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Sampling the DTFTSampling the DTFT

• i.e.

• Making use of the identity

∑ ∑−

=

−∞

−∞==

1

0

1 N

k

nkN

kN WWx

Nny l

l

l][][

= ∑∑

=

−−∞

−∞=

1

0

1 N

k

nkNWN

x )(][ l

l

l

=∑

=

−−1

0

1 N

n

rnkNWN

)(

,,

01 mNnr +=for

otherwise

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Sampling the DTFTSampling the DTFT

we arrive at the desired relation

• Thus y[n] is obtained from x[n] by adding

an infinite number of shifted replicas of

x[n], with each replica shifted by an integer

multiple of N sampling instants, and

observing the sum only for the interval

10 −≤≤+= ∑∞

−∞=NnmNnxny

m

],[][

10 −≤≤ Nn

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Sampling the DTFTSampling the DTFT

• To apply

to finite-length sequences, we assume that

the samples outside the specified range are

zeros

• Thus if x[n] is a length-M sequence with

, then y[n] = x[n] for

10 −≤≤+= ∑∞

−∞=NnmNnxny

m

],[][

NM ≤ 10 −≤≤ Nn

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Sampling the DTFTSampling the DTFT

• If M > N, there is a time-domain aliasing of

samples of x[n] in generating y[n], and x[n]

cannot be recovered from y[n]

• Example - Let

• By sampling its DTFT at ,

and then applying a 4-point IDFT to

these samples, we arrive at the sequence y[n]

given by

}{]}[{ 543210=nx↑

)( ωjeX 4/2 kk π=ω30 ≤≤ k

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Sampling the DTFTSampling the DTFT

,

• i.e.

{x[n]} cannot be recovered from {y[n]}

][][][][ 44 −+++= nxnxnxny 30 ≤≤ n

}{]}[{ 3264=ny↑

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Numerical Computation of the Numerical Computation of the

DTFT Using the DFTDTFT Using the DFT

• A practical approach to the numerical

computation of the DTFT of a finite-length

sequence

• Let be the DTFT of a length-N

sequence x[n]

• We wish to evaluate at a dense grid

of frequencies , ,

where M >> N:

)( ωjeX

)( ωjeX

Mkk /2π=ω 10 −≤≤ Mk

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Numerical Computation of the Numerical Computation of the

DTFT Using the DFTDTFT Using the DFT

• Define a new sequence

• Then

∑=∑=−

=

π−−

=

ω−ω 1

0

/21

0

][][)(N

n

MknjN

n

njjenxenxeX kk

−≤≤

−≤≤=

1,0

10],[][

MnN

Nnnxnxe

∑=−

=

π−ω 1

0

/2][)(M

n

MknjjenxeX k

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Numerical Computation of the Numerical Computation of the DTFT Using the DFTDTFT Using the DFT

• Thus is essentially an M-point DFT

of the length-M sequence

• The DFT can be computed very

efficiently using the FFT algorithm ifM is

an integer power of 2

• The function freqz employs this approach

to evaluate the frequency response at a

prescribed set of frequencies of a DTFT

expressed as a rational function in

)( kjeXω

][kXe ][nxe

][kXe

ω− je

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DFT PropertiesDFT Properties

• Like the DTFT, the DFT also satisfies a

number of properties that are useful in

signal processing applications

• Some of these properties are essentially

identical to those of the DTFT, while some

others are somewhat different

• A summary of the DFT properties are given

in Tables in the following slides

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Table:Table: General Properties of DFTGeneral Properties of DFT

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Table:Table: DFT Properties: DFT Properties: Symmetry RelationsSymmetry Relations

x[n] is a complex sequence

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Table:Table: DFT Properties: DFT Properties: Symmetry RelationsSymmetry Relations

x[n] is a real sequence

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Circular Shift of a SequenceCircular Shift of a Sequence

• This property is analogous to the time-

shifting property of the DTFT, but with a

subtle difference

• Consider length-N sequences defined for

• Sample values of such sequences are equal

to zero for values of n < 0 and Nn ≥

10 −≤≤ Nn

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Circular Shift of a SequenceCircular Shift of a Sequence

• If x[n] is such a sequence, then for any

arbitrary integer , the shifted sequence

is no longer defined for the range

• We thus need to define another type of a

shift that will always keep the shifted

sequence in the range

][][ onnxnx −=1

on

10 −≤≤ Nn

10 −≤≤ Nn

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Circular Shift of a SequenceCircular Shift of a Sequence

• The desired shift, called the circular shift,

is defined using a modulo operation:

• For (right circular shift), the above

equation implies

][][ Noc nnxnx ⟩−⟨=

0>on

+−−

=],[

],[][

nnNx

nnxnx

o

oc

o

o

nn

Nnn

<≤−≤≤

0for

1for

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Circular Shift of a SequenceCircular Shift of a Sequence

• Illustration of the concept of a circular shift

][nx ]1[ 6⟩−⟨nx

]5[ 6⟩+⟨= nx ]2[ 6⟩+⟨= nx

]4[ 6⟩−⟨nx

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Circular Shift of a SequenceCircular Shift of a Sequence

• 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

on

onN −

on

Non ⟩⟨

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Circular ConvolutionCircular Convolution

• This operation is analogous to linear

convolution, but with a subtle difference

• Consider two length-N sequences, g[n] and

h[n], respectively

• Their linear convolution results in a length-

sequence given by)( 12 −N

2201

0

−≤≤−= ∑−

=Nnmnhmgny

N

mL ],[][][

][nyL

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Circular ConvolutionCircular Convolution

• In computing we have assumed that

both length-N sequences have been zero-

padded to extend their lengths to

• The longer form of results from the

time-reversal of the sequence h[n] and its

linear shift to the right

• The first nonzero value of is

, and the last nonzero value

is

12 −N

][nyL

][nyL

][nyL][][][ 000 hgyL =

][][][ 1122 −−=− NhNgNyL

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Circular ConvolutionCircular Convolution

• To develop a convolution-like operation

resulting in a length-N sequence , we

need to define a circular time-reversal, and

then apply a circular time-shift

• Resulting operation, called a circular

convolution, is defined by

10],[][][1

0

−≤≤∑ ⟩−⟨=−

=Nnmnhmgny

N

mNC

][nyC

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Circular ConvolutionCircular Convolution

• Since the operation defined involves two

length-N sequences, it is often referred to as

an N-point circular convolution, denoted as

y[n] = g[n] h[n]

• The circular convolution is commutative,

i.e.

g[n] h[n] = h[n] g[n]

N

N N

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Circular ConvolutionCircular Convolution

• Example - Determine the 4-point circular

convolution of the two length-4 sequences:

as sketched below

},{]}[{ 1021=ng }{]}[{ 1122=nh

↑ ↑

n3210

1

2 ][ng

n3210

1

2 ][nh

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Circular ConvolutionCircular Convolution

• The result is a length-4 sequence given by

• From the above we observe

][nyC

,][][][][][3

04∑ ⟩−⟨==

=mC mnhmgnhngny 4

30 ≤≤ n

∑ ⟩⟨−==

3

04][][]0[

mC mhmgy

][][][][][][][][ 13223100 gghghghg +++=

621101221 =×+×+×+×= )()()()(

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Circular ConvolutionCircular Convolution

• Likewise ∑ ⟩−⟨==

3

04]1[][]1[

mC mhmgy

][][][][][][][][ 23320110 hghghghg +++=

711102221 =×+×+×+×= )()()()(

∑ ⟩−⟨==

3

04]2[][]2[

mC mhmgy

][][][][][][][][ 33021120 hghghghg +++=

611202211 =×+×+×+×= )()()()(

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Circular ConvolutionCircular Convolution

and

• The circular convolution can also be computed using a DFT-based approach as indicated in previous Table

521201211 =×+×+×+×= )()()()(

][][][][][][][][ 03122130 hghghghg +++=

∑ ⟩−⟨==

3

04]3[][]3[

mC mhmgy

][nyC

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Circular ConvolutionCircular Convolution• Example - Consider the two length-4

sequences repeated below for convenience:

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

n3210

1

2 ][ng

n3210

1

2 ][nh

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

3021 232 ≤≤++= −− kee kjkj ,// ππ

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Circular ConvolutionCircular Convolution

• Therefore

• Likewise,

,][ 21212 −=−−=G

,][ jjjG −=+−= 1211

jjjG +=−+= 1213][

,][ 41210 =++=G

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

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

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Circular ConvolutionCircular Convolution

• Hence,

• The two 4-point DFTs can also be

computed using the matrix relation given

earlier

,][ 611220 =+++=H

,][ 011222 =−+−=H

,][ jjjH −=+−−= 11221

jjjH +=−−+= 11223][

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Circular ConvolutionCircular Convolution

+−−=

−−−−

−−=

=

j

j

jj

jj

g

g

g

g

GGGG

12

14

1021

111111

111111

3

2

1

0

3210

4

][

][

][

][

][][][][

D

+

−=

−−−−

−−=

=

j

j

jj

jj

hhhh

HHHH

10

16

1122

111111

111111

3210

3210

4

][][][][

][][][][

D

is the 4-point DFT matrix4D

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Circular ConvolutionCircular Convolution

• If denotes the 4-point DFT of

then from Table above we observe

• Thus

30 ≤≤= kkHkGkYC ],[][][

][kYC ][nyC

−=

=

20

224

33

22

11

00

3

2

1

0

j

j

HG

HG

HG

HG

Y

Y

Y

Y

C

C

C

C

][][

][][

][][

][][

][

][

][

][

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Circular ConvolutionCircular Convolution

• A 4-point IDFT of yields][kYC

=

][

][

][

][

*

][

][

][

][

3

2

1

0

4

1

3

2

1

0

4

C

C

C

C

C

C

C

C

Y

Y

Y

Y

y

y

y

y

D

=

−−−−−−=

5676

20

224

111111

111111

4

1

j

j

jj

jj

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Circular ConvolutionCircular Convolution

• Example - Now let us extended the two

length-4 sequences to length 7 by

appending each with three zero-valued

samples, i.e.

≤≤≤≤=

640

30

n

nngnge ,

],[][

≤≤≤≤=

64030

nnnh

nhe ,],[

][

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Circular ConvolutionCircular Convolution

• We next determine the 7-point circular

convolution of and :

• From the above

60,][][][6

07 ≤≤∑ ⟩−⟨=

=nmnhmgny

mee

][nge ][nhe

][][][][][ 61000 eeee hghgy +=

][][][][][][][][ 16253443 eeeeeeee hghghghg ++++

22100 =×== ][][ hg

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Circular ConvolutionCircular Convolution

• Continuing the process we arrive at

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

][][][][][][][ 0211202 hghghgy ++=,)()()( 5202211 =×+×+×=][][][][][][][][][ 031221303 hghghghgy +++=

,)()()()( 521201211 =×+×+×+×=][][][][][][][ 1322314 hghghgy ++=

,)()()( 4211012 =×+×+×=

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Circular ConvolutionCircular Convolution

• 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]

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

111336 =×== )(][][][ hgy

][nyL

][nyL

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Circular ConvolutionCircular Convolution

• The N-point circular convolution can be

written in matrix form as

• Note: The elements of each diagonal of the

matrix are equal

• Such a matrix is called a circulant matrix

NN ×

−−−

−−−

=

]1[

]2[

]1[

]0[

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

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

]1[

]2[

]1[

]0[

Ng

g

g

g

hNhNhNh

hhhhhNhhhhNhNhh

Ny

y

y

y

C

C

C

C

M

L

MOMMM

L

L

L

M

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Computation of the DFT of Computation of the DFT of

Real SequencesReal Sequences

• In most practical applications, sequences of

interest are real

• In such cases, the symmetry properties of

the DFT can be exploited to make the DFT

computations more efficient

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NN--Point Point DFTsDFTs of Two Lengthof Two Length--NN

Real SequencesReal Sequences

• Let g[n] and h[n] be two length-N real

sequences with G[k] and H[k] denoting their

respective N-point DFTs

• These two N-point DFTs can be computed

efficiently using a single N-point DFT

• Define a complex length-N sequence

• Hence, g[n] = Re{x[n]} and h[n] = Im{x[n]}

][][][ nhjngnx +=

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NN--Point Point DFTsDFTs of Two Lengthof Two Length--NN

Real SequencesReal Sequences

• Let X[k] denote the N-point DFT of x[n]

• Then, DFT properties we arrive at

• Note that

]}[*][{][2

1NkXkXkG ⟩⟨−+=

]}[*][{][2

1Nj

kXkXkH ⟩⟨−−=

][*][* NN kNXkX ⟩−⟨=⟩⟨−

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NN--Point Point DFTsDFTs of Two Lengthof Two Length--NN

Real SequencesReal Sequences

• Example - We compute the 4-point DFTs of

the two real sequences g[n] and h[n] given

below

• Then is given by

}{]}[{},{]}[{ 11221021 == nhng↑ ↑

]}[{]}[{]}[{ nhjngnx +=

}{]}[{ jjjjnx +++= 12221↑

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NN--Point Point DFTsDFTs of Two Lengthof Two Length--NN

Real SequencesReal Sequences• Its DFT X[k] is

• From the above

• Hence

+

=

+

++

−−−−

−−=

22

2

64

1

22

21

111111

111111

3210

j

j

j

j

j

j

jj

jj

XXXX

][][][][

][][* 22264 jjkX −−−=

]22264[]4[* 4 −−−=⟩−⟨ jjkX

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NN--Point Point DFTsDFTs of Two Lengthof Two Length--NN

Real SequencesReal Sequences

• Therefore

verifying the results derived earlier

}{]}[{ jjkG +−−= 1214

}{]}[{ jjkH +−= 1016

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22NN--Point DFT of a Real Point DFT of a Real

Sequence Using an Sequence Using an NN--point DFTpoint DFT

• Let v[n] be a length-2N real sequence with

an 2N-point DFT V[k]

• Define two length-N real sequences g[n]

and h[n] as follows:

• Let G[k] and H[k] denote their respective N-

point DFTs

Nnnvnhnvng ≤≤+== 0122 ],[][],[][

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22NN--Point DFT of a Real Point DFT of a Real

Sequence Using an Sequence Using an NN--point DFTpoint DFT

• Define a length-N complex sequence

with an N-point DFT X[k]

• Then as shown earlier

]}[{]}[{]}[{ nhjngnx +=

]}[*][{][2

1NkXkXkG ⟩⟨−+=

]}[*][{][2

1Nj

kXkXkH ⟩⟨−−=

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22NN--Point DFT of a Real Point DFT of a Real

Sequence Using an Sequence Using an NN--point DFTpoint DFT

• Now ∑−

==

12

02

N

n

nkNWnvkV ][][

∑ ∑−

=

=

+++=1

0

1

0

122

22 122

N

n

N

n

knN

nkN WnvWnv )(][][

∑ ∑−

=

=+=

1

0

1

02

N

n

N

n

kN

nkN

nkN WWnhWng ][][

∑ ∑−

=

=−≤≤+=

1

0

1

02 120

N

n

N

n

nkN

kN

nkN NkWnhWWng ,][][

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22NN--Point DFT of a Real Point DFT of a Real

Sequence Using an Sequence Using an NN--point DFTpoint DFT

• i.e.,

• Example - Let us determine the 8-point

DFT V[k] of the length-8 real sequence

• We form two length-4 real sequences as

follows

120],[][][ 2 −≤≤⟩⟨+⟩⟨= NkkHWkGkV NkNN

}{]}[{ 11102221=nv↑

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22NN--Point DFT of a Real Point DFT of a Real

Sequence Using an Sequence Using an NN--point DFTpoint DFT

• Now

• Substituting the values of the 4-point DFTs

G[k] and H[k] computed earlier we get

}{]}[{]}[{ 10212 == nvng↑

}{]}[{]}[{ 112212 =+= nvnh↑

70],[][][ 484 ≤≤⟩⟨+⟩⟨= kkHWkGkV k

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22NN--Point DFT of a Real Point DFT of a Real

Sequence Using an Sequence Using an NN--point DFTpoint DFT

1064000 =+=+= ][][][ HGV

][][][ 111 18HWGV +=

41422111 4 .)()( / jjej j −=−+−= − π

202222 228 −=⋅+−=+= − /][][][ πjeHWGV

][][][ 333 38 HWGV +=

41420111 43 .)()( / jjej j −=+++= − π

264004 48 −=⋅+=+= − πjeHWGV ][][][

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22NN--Point DFT of a Real Point DFT of a Real

Sequence Using an Sequence Using an NN--point DFTpoint DFT

][][][ 115 58 HWGV +=

41420111 45 .)()( / jjej j +=−+−= − π

202226 2368 −=⋅+−=+= − /][][][ πjeHWGV

][][][ 337 78 HWGV +=

41422111 47 .)()( / jjej j +=+++= − π

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Linear Convolution Using the Linear Convolution Using the

DFTDFT

• Linear convolution is a key operation in

many signal processing applications

• Since a DFT can be efficiently implemented

using FFT algorithms, it is of interest to

develop methods for the implementation of

linear convolution using the DFT

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Linear Convolution of Two Linear Convolution of Two

FiniteFinite--Length SequencesLength Sequences

• Let g[n] and h[n] be two finite-length

sequences of length N andM, respectively

• Denote

• Define two length-L sequences

1−+= MNL

−≤≤−≤≤=

10

10

LnN

Nnngnge ,

],[][

−≤≤−≤≤=

1010

LnMMnnh

nhe ,],[

][

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Linear Convolution of Two Linear Convolution of Two

FiniteFinite--Length SequencesLength Sequences

• Then

• The corresponding implementation scheme

is illustrated below

][][][][][][ nhngnynhngny CL === * L

Zero-paddingwith

zeros1)( −N point DFT

Zero-paddingwith

zeros1)( −M

−−+ )( 1MN

point DFT

×

g[n]

h[n]

Length-N

][nge

][nhe −−+ )( 1MN

−−+ )( 1MN

point IDFT

][nyL

Length-M Length- )( 1−+ MN

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Linear Convolution of a FiniteLinear Convolution of a Finite--Length Sequence with an Length Sequence with an InfiniteInfinite--Length SequenceLength Sequence

• We next consider the DFT-based

implementation of

where h[n] is a finite-length sequence of

lengthM and x[n] is an infinite length (or a

finite length sequence of length much

greater than M)

][][][][][ nxnhnxhnyM

∑−

==−=

1

0l

ll *

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OverlapOverlap--Add MethodAdd Method

• We first segment x[n], assumed to be a

causal sequence here without any loss of

generality, into a set of contiguous finite-

length subsequences of length N each:

where

][nxm

∑∞

=−=

0mm mNnxnx ][][

−≤≤+=

otherwise010

,],[

][NnmNnx

nxm

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OverlapOverlap--Add MethodAdd Method

• Thus we can write

where

• Since h[n] is of lengthM and is of

length N, the linear convolution

is of length

][nxm][][ nxnh m*

][][][ nxnhny mm = *

∑∞

=−==

0mm mNnynxnhny ][][][][ *

1−+MN

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OverlapOverlap--Add MethodAdd Method

• As a result, the desired linear convolution

has been broken up into a

sum of infinite number of short-length

linear convolutions of length

each:

• Each of these short convolutions can be

implemented using the DFT-based method

discussed earlier, where now the DFTs (and

the IDFT) are computed on the basis of

points

][][][ nxnhny = *

1−+MN

)( 1−+MN

][][][ nhnxny mm = L

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OverlapOverlap--Add MethodAdd Method

• There is one more subtlety to take care of

before we can implement

using the DFT-based approach

• Now the first convolution in the above sum,

, is of length

and is defined for

∑∞

=−=

0mm mNnyny ][][

1−+MN

20 −+≤≤ MNn

][][][ 00 nxnhny = *

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OverlapOverlap--Add MethodAdd Method

• The second short convolution

, is also of length

but is defined for

• There is an overlap of samples

between these two short linear convolutions

• Likewise, the third short convolution

, is also of length

but is defined for

1−+MN

22 −+≤≤ MNnN

20 −+≤≤ MNn

1−M

][][ nxnh 2*

][][ nxnh 1*

=][2 ny

=][1 ny

1−+MN

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OverlapOverlap--Add MethodAdd Method

• Thus there is an overlap of samples

between and

• In general, there will be an overlap of

samples between the samples of the short

convolutions and

for

• This process is illustrated in the figure on

the next slide for M = 5 and N = 7

][][ nxnh r 1−* ][][ nxnh r*

1−M

1−M

][][ nxnh 1* ][][ nxnh 2*

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OverlapOverlap--Add MethodAdd Method

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OverlapOverlap--Add MethodAdd MethodAdd

Add

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OverlapOverlap--Add MethodAdd Method

• Therefore, y[n] obtained by a linear

convolution of x[n] and h[n] is given by

],[][ nyny 0=],[][][ 710 −+= nynyny

],[][ 71 −= nyny

],[][][ 147 21 −+−= nynyny

],[][ 142 −= nyny•••

60 ≤≤ n107 ≤≤ n1311 ≤≤ n1714 ≤≤ n2018 ≤≤ n

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OverlapOverlap--Add MethodAdd Method

• The above procedure is called the overlap-

add method since the results of the short

linear convolutions overlap and the

overlapped portions are added to get the

correct final result

• The function fftfilt can be used to

implement the above method

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OverlapOverlap--Add MethodAdd Method

• We have created a program which uses fftfilt

for the filtering of a noise-corrupted signal y[n]

using a length-3 moving average filter. The

• The plots generated by running this program is

shown below

0 10 20 30 40 50 60-2

0

2

4

6

8

Time index n

Am

plitu

de

s[n]y[n]

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OverlapOverlap--Save MethodSave Method

• In implementing the overlap-add method

using the DFT, we need to compute two

-point DFTs and one -

point IDFT since the overall linear

convolution was expressed as a sum of

short-length linear convolutions of length

each

• It is possible to implement the overall linear

convolution by performing instead circular

convolution of length shorter than

)( 1−+MN )( 1−+MN

)( 1−+MN

)( 1−+MN

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OverlapOverlap--Save MethodSave Method

• To this end, it is necessary to segment x[n]

into overlapping blocks , keep the

terms of the circular convolution of h[n]

with that corresponds to the terms

obtained by a linear convolution of h[n] and

, and throw away the other parts of

the circular convolution

][nxm

][nxm

][nxm

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OverlapOverlap--Save MethodSave Method

• To understand the correspondence between

the linear and circular convolutions,

consider a length-4 sequence x[n] and a

length-3 sequence h[n]

• Let denote the result of a linear

convolution of x[n] with h[n]

• The six samples of are given by

][nyL

][nyL

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OverlapOverlap--Save MethodSave Method

][][][ 000 xhyL =][][][][][ 01101 xhxhyL +=

][][][][][][][ 0211202 xhxhxhyL ++=

][][][][][][][ 1221303 xhxhxhyL ++=

][][][][][ 22314 xhxhyL +=

][][][ 325 xhyL =

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OverlapOverlap--Save MethodSave Method

• If we append h[n] with a single zero-valued

sample and convert it into a length-4

sequence , the 4-point circular

convolution of and x[n] is given

by

][][][][][][][ 2231000 xhxhxhyC ++=

][][][][][][][ 3201101 xhxhxhyC ++=

]0[]2[]1[]1[]2[]0[]2[ xhxhxhyC ++=

][][][][][][][ 1221303 xhxhxhyC ++=

][nhe][nhe][nyC

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OverlapOverlap--Save MethodSave Method

• If we compare the expressions for the

samples of with the samples of ,

we observe that the first 2 terms of do

not correspond to the first 2 terms of ,

whereas the last 2 terms of are

precisely the same as the 3rd and 4th terms

of , i.e.,

][nyL

][nyL

][nyL

][nyC][nyC

][nyC

],[][ 00 CL yy ≠ ][][ 11 CL yy ≠

],[][ 22 CL yy = ][][ 33 CL yy =

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OverlapOverlap--Save MethodSave Method

• General case: N-point circular convolution

of a length-M sequence h[n] with a length-N

sequence x[n] with N > M

• First samples of the circular

convolution are incorrect and are rejected

• Remaining samples correspond

to the correct samples of the linear

convolution of h[n] with x[n]

1−M

1+−MN

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OverlapOverlap--Save MethodSave Method

• Now, consider an infinitely long or very

long sequence x[n]

• Break it up as a collection of smaller length

(length-4) overlapping sequences as

• Next, form

][nxm∞≤≤≤≤+= mnmnxnxm 0302 ,],[][

][][][ nxnhnw mm = 4

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OverlapOverlap--Save MethodSave Method

• Or, equivalently,

• Computing the above for m = 0, 1, 2, 3, . . . ,

and substituting the values of we

arrive at

][][][][][][][ 2231000 mmmm xhxhxhw ++=

][][][][][][][ 3201101 mmmm xhxhxhw ++=

][][][][][][][ 0211202 mmmm xhxhxhw ++=

][][][][][][][ 1221303 mmmm xhxhxhw ++=

][nxm

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OverlapOverlap--Save MethodSave Method

][][][][][][][ 22310000 xhxhxhw ++=

][][][][][][][ 32011010 xhxhxhw ++=][][][][][][][][ 202112020 yxhxhxhw =++=

][][][][][][][][ 312213030 yxhxhxhw =++=

←Reject

←Reject

←Save

←Save

][][][][][][][ 42512001 xhxhxhw ++=

][][][][][][][ 52213011 xhxhxhw ++=][][][][][][][][ 422314021 yxhxhxhw =++=

][][][][][][][][ 532415031 yxhxhxhw =++=

←Reject

←Reject

←Save

←Save

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OverlapOverlap--Save MethodSave Method

][][][][][][][ 62514002 xhxhxhw ++= ←Reject

][][][][][][][ 72415012 xhxhxhw ++= ←Reject

][][][][][][][][ 642516022 yxhxhxhw =++= ←Save

][][][][][][][][ 752617032 yxhxhxhw =++= ←Save

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OverlapOverlap--Save MethodSave Method

• It should be noted that to determine y[0] and

y[1], we need to form :

and compute for

reject and , and save

and

][nx 1−

,][,][ 0100 11 == −− xx

][][][ nxnhnw 11 −− = 4 30 ≤≤ n][01−w ][11−w ][][ 021 yw =−

][][ 131 yw =−

][][],[][ 1302 11 xxxx == −−

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OverlapOverlap--Save MethodSave Method

• General Case: Let h[n] be a length-N

sequence

• Let denote the m-th section of an

infinitely long sequence x[n] of length N

and defined by

withM < N

10)],1([][ −≤≤+−+= NnmNmnxnxm

][nxm

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OverlapOverlap--Save MethodSave Method

• Let

• Then, we reject the first samples of

and “abut” the remaining samples of

to form , the linear convolution of

h[n] and x[n]

• If denotes the saved portion of ,

i.e.

][nwm

][nwm

][][][ nxnhnw mm = N

1−M

1+−MN][nyL

][nym ][nwm

−≤≤−−≤≤=

21],[20,0

][NnMnw

Mnny

mm

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OverlapOverlap--Save MethodSave Method

• Then

• The approach is called overlap-save

method since the input is segmented into

overlapping sections and parts of the results

of the circular convolutions are saved and

abutted to determine the linear convolution

result

11],[)]1([ −≤≤−=+−+ NnMnyMNmny mL

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OverlapOverlap--Save MethodSave Method

• Process is illustrated next

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OverlapOverlap--Save MethodSave Method

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zz--TransformTransform

• The DTFT provides a frequency-domain

representation of discrete-time signals and

LTI discrete-time systems

• Because of the convergence condition, in

many cases, the DTFT of a sequence may

not exist

• As a result, it is not possible to make use of

such frequency-domain characterization in

these cases

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zz--TransformTransform• A generalization of the DTFT defined by

leads to the z-transform

• z-transform may exist for many sequences

for which the DTFT does not exist

• Moreover, use of z-transform techniques

permits simple algebraic manipulations

∑=∞

−∞=

ω−ω

n

njj enxeX ][)(

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zz--TransformTransform

• Consequently, z-transform has become an

important tool in the analysis and design of

digital filters

• For a given sequence g[n], its z-transform

G(z) is defined as

where z = Re(z) + jIm(z) is a complex

variable

∑∞

−∞=

−=n

nzngzG ][)(

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zz--TransformTransform

• If we let , then the z-transform

reduces to

• The above can be interpreted as the DTFT

of the modified sequence

• For r = 1 (i.e., |z| = 1), z-transform reduces

to its DTFT, provided the latter exists

ω= jerz

∑=∞

−∞=

ω−−ω

n

njnj erngerG ][)(

}][{ nrng −

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zz--TransformTransform• The contour |z| = 1 is a circle in the z-plane

of unity radius and is called the unit circle

• Like the DTFT, there are conditions on the

convergence of the infinite series

• For a given sequence, the set R of values of

z for which its z-transform converges is

called the region of convergence (ROC)

∑∞

−∞=

n

nzng ][

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zz--TransformTransform

• From our earlier discussion on the uniform

convergence of the DTFT, it follows that the

series

converges if is absolutely

summable, i.e., if

∑=∞

−∞=

ω−−ω

n

njnj erngerG ][)(

}][{ nrng −

∞<∑∞

−∞=

n

nrng ][

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zz--TransformTransform

• In general, the ROC R of a z-transform of a

sequence g[n] is an annular region of the z-

plane:

where

• Note: The z-transform is a form of a Laurent

series and is an analytic function at every

point in the ROC

+− << gg RzR

∞≤<≤ +− gg RR0

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zz--TransformTransform

• Example - Determine the z-transform X(z)

of the causal sequence and its

ROC

• Now

• The above power series converges to

• ROC is the annular region |z| > |α|

][][ nnx nµα=

∑α=∑ µα=∞

=

−∞

−∞=

0

][)(n

nn

n

nn zznzX

1for,1

1)( 1

1<α

α−= −

− zz

zX

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zz--TransformTransform

• Example - The z-transform µµµµ(z) of the unit

step sequence µ[n] can be obtained from

by setting α = 1:

• ROC is the annular region

1for,1

1)( 1

1<α

α−= −

− zz

zX

∞≤< z1

1for1

1 1

1<

−= −

−z

zz ,)(µµµµ

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zz--TransformTransform

• Note: The unit step sequence µ[n] is not

absolutely summable, and hence its DTFT

does not converge uniformly

• Example - Consider the anti-causal

sequence

]1[][ −−µα−= nny n

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zz--TransformTransform

• Its z-transform is given by

• ROC is the annular region

∑α−=∑ α−=∞

=

−−

−∞=

1

1)(

m

mm

n

nn zzzY

1for,1

1 1

1<α

α−= −

− zz

z

zzz

m

mm

1

1

0

1

1 −

−∞

=

−−

α−

α−=∑αα−=

α<z

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zz--TransformTransform

• Note: The z-transforms of the two

sequences and are

identical even though the two parent

sequences are different

• Only way a unique sequence can be

associated with a z-transform is by

specifying its ROC

]1[ −−µα− nn][nnµα

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zz--TransformTransform

• The DTFT of a sequence g[n]

converges uniformly if and only if the ROC

of the z-transform G(z) of g[n] includes the

unit circle

• The existence of the DTFT does not always

imply the existence of the z-transform

)( ωjeG

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zz--TransformTransform

• Example - The finite energy sequence

has a DTFT given by

which converges in the mean-square sense

∞<<∞−πω

= nnn

nh cLP ,

sin][

π≤ω<ω

ω≤ω≤=ω

c

cjLP eH

,0

0,1)(

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zz--TransformTransform

• However, does not have a z-transform

as it is not absolutely summable for any value

of r

• Some commonly used z-transform pairs are

listed on the next slide

][nhLP

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Table:Table: Commonly UsedCommonly Used zz--Transform PairsTransform Pairs

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Rational Rational zz--TransformsTransforms

• In the case of LTI discrete-time systems we

are concerned with in this course, all

pertinent z-transforms are rational functions

of

• That is, they are ratios of two polynomials

in :

1−z

1−z

NN

NN

MM

MM

zdzdzdd

zpzpzpp

zD

zPzG

−−−−

−−−−

++++

++++==

)(

)(

....

....

)(

)()(

11

110

11

110

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Rational Rational zz--TransformsTransforms

• The degree of the numerator polynomial

P(z) is M and the degree of the denominator

polynomial D(z) is N

• An alternate representation of a rational z-

transform is as a ratio of two polynomials in

z:

NNNN

MMMM

MN

dzdzdzd

pzpzpzpzzG

++++

++++=

−−

−−

11

10

11

10

....

....)( )(

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Rational Rational zz--TransformsTransforms

• A rational z-transform can be alternately

written in factored form as

∏∏

=−

=−

−=

N

M

zd

zpzG

11

0

11

0

1

1

l l

l l

)(

)()(

λ

ξ

∏∏

=

=−

−=

N

MMN

zd

zpz

10

10

l l

l l

)(

)()(

λ

ξ

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Rational Rational zz--TransformsTransforms

• At a root of the numerator polynomial

, and as a result, these values of z

are known as the zeros of G(z)

• At a root of the denominator

polynomial , and as a result,

these values of z are known as the poles of

G(z)

lξ=z

lλ=z∞→)( lλG

0=)( lξG

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Rational Rational zz--TransformsTransforms

• Consider

• Note G(z) has M finite zeros and N finite

poles

• If N > M there are additional zeros at

z = 0 (the origin in the z-plane)

• If N < M there are additional poles at

z = 0

MN −

NM −

∏∏

=

=−

−=

N

MMN

zd

zpzzG

10

10

l l

l l

)(

)()( )(

λ

ξ

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Rational Rational zz--TransformsTransforms• Example - The z-transform

has a zero at z = 0 and a pole at z = 1

1for1

11

>−

=−

zz

z ,)(µµµµ

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Rational Rational zz--TransformsTransforms

• A physical interpretation of the concepts of

poles and zeros can be given by plotting the

log-magnitude as shown on

next slide for

)(log zG1020

21

21

640801

882421−−

−−

+−

+−=

zz

zzzG

..

..)(

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Rational Rational zz--TransformsTransforms

Page 184: Discrete -Time Fourier Transform Discrete Fourier ...tania/teaching/dsp/Lectures 3-4 DTFT DFT... · Discrete -Time Fourier Transform Discrete Fourier Transform z-Transform Tania Stathaki

Rational Rational zz--TransformsTransforms

• Observe that the magnitude plot exhibits

very large peaks around the points

which are the poles of

G(z)

• It also exhibits very narrow and deep wells

around the location of the zeros at

6928040 .. jz ±=

2121 .. jz ±=

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ROC of a Rational ROC of a Rational

zz--TransformTransform

• ROC of a z-transform is an important

concept

• Without the knowledge of the ROC, there is

no unique relationship between a sequence

and its z-transform

• Hence, the z-transform must always be

specified with its ROC

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ROC of a Rational ROC of a Rational

zz--TransformTransform

• Moreover, if the ROC of a z-transform

includes the unit circle, the DTFT of the

sequence is obtained by simply evaluating

the z-transform on the unit circle

• There is a relationship between the ROC of

the z-transform of the impulse response of a

causal LTI discrete-time system and its

BIBO stability

Page 187: Discrete -Time Fourier Transform Discrete Fourier ...tania/teaching/dsp/Lectures 3-4 DTFT DFT... · Discrete -Time Fourier Transform Discrete Fourier Transform z-Transform Tania Stathaki

ROC of a Rational ROC of a Rational

zz--TransformTransform

• The ROC of a rational z-transform is

bounded by the locations of its poles

• To understand the relationship between the

poles and the ROC, it is instructive to

examine the pole-zero plot of a z-transform

• Consider again the pole-zero plot of the z-

transform µµµµ(z)

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ROC of a Rational ROC of a Rational zz--TransformTransform

• In this plot, the ROC, shown as the shaded

area, is the region of the z-plane just outside

the circle centered at the origin and going

through the pole at z = 1

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ROC of a Rational ROC of a Rational zz--TransformTransform

• Example - The z-transform H(z) of the

sequence is given by

• Here the ROC is just outside the circle going through the point 60.−=z

][)6.0(][ nnh nµ−=

,.

)(1601

1−+

=z

zH

60.>z

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ROC of a Rational ROC of a Rational

zz--TransformTransform

• A sequence can be one of the following

types: finite-length, right-sided, left-sided

and two-sided

• In general, the ROC depends on the type of

the sequence of interest

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ROC of a Rational ROC of a Rational

zz--TransformTransform

• Example - Consider a finite-length sequence

g[n] defined for , where M and

N are non-negative integers and

• Its z-transform is given by

NnM ≤≤−∞<][ng

N

MN nMNN

Mn

n

z

zMngzngzG

∑∑

+ −+

−=

− −== 0

][][)(

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ROC of a Rational ROC of a Rational

zz--TransformTransform

• Note: G(z) has M poles at and N poles

at z = 0 (explain why)

• As can be seen from the expression for

G(z), the z-transform of a finite-length

bounded sequence converges everywhere in

the z-plane except possibly at z = 0 and/or at

∞=z

∞=z

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ROC of a Rational ROC of a Rational

zz--TransformTransform

• Example - A right-sided sequence with

nonzero sample values for is

sometimes called a causal sequence

• Consider a causal sequence

• Its z-transform is given by

∑∞

=

−=0

11 ][)(n

nznuzU

0≥n

][1 nu

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ROC of a Rational ROC of a Rational zz--TransformTransform

• It can be shown that converges

exterior to a circle , including the

point

• On the other hand, a right-sided sequence

with nonzero sample values only for

with M nonnegative has a z-transform

with M poles at

• The ROC of is exterior to a circle

, excluding the point

)(1 zU

1Rz =

2Rz =

∞=z

∞=z

∞=z

)(2 zU

][2 nu

Mn −≥

)(2 zU

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ROC of a Rational ROC of a Rational

zz--TransformTransform

• Example - A left-sided sequence with

nonzero sample values for is

sometimes called a anticausal sequence

• Consider an anticausal sequence

• Its z-transform is given by

0≤n

][1 nv

∑−∞=

−=0

11 ][)(n

nznvzV

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ROC of a Rational ROC of a Rational zz--TransformTransform

• It can be shown that converges

interior to a circle , including the

point z = 0

• On the other hand, a left-sided sequence

with nonzero sample values only for

with N nonnegative has a z-transform

with N poles at z = 0

• The ROC of is interior to a circle

, excluding the point z = 0

Nn ≤

)(1 zV

3Rz =

)(2 zV

)(2 zV

4Rz =

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ROC of a Rational ROC of a Rational

zz--TransformTransform

• Example - The z-transform of a two-sided

sequence w[n] can be expressed as

• The first term on the RHS, ,

can be interpreted as the z-transform of a

right-sided sequence and it thus converges

exterior to the circle

∑∑∑−

−∞=

−∞

=

−∞

−∞=

− +==1

0

][][][)(n

n

n

n

n

n znwznwznwzW

5Rz =

∑∞=

−0

][n

nznw

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ROC of a Rational ROC of a Rational

zz--TransformTransform

• The second term on the RHS, ,

can be interpreted as the z-transform of a left-

sided sequence and it thus converges interior

to the circle

• If , there is an overlapping ROC

given by

• If , there is no overlap and the

z-transform does not exist

∑−−∞=

−1][

nnznw

6Rz =

65 RR <

65 RR >65 RzR <<

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ROC of a Rational ROC of a Rational zz--TransformTransform

• Example - Consider the two-sided sequence

where α can be either real or complex

• Its z-transform is given by

• The first term on the RHS converges for , whereas the second term converges

for

nnu α=][

∑∑∑−

−∞=

−∞

=

−∞

−∞=

− α+α=α=1

0

)(n

nn

n

nn

n

nn zzzzU

α>zα<z

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ROC of a Rational ROC of a Rational

zz--TransformTransform

• There is no overlap between these two

regions

• Hence, the z-transform of does

not exist

nnu α=][

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ROC of a Rational ROC of a Rational

zz--TransformTransform

• The ROC of a rational z-transform cannot

contain any poles and is bounded by the

poles

• To show that the z-transform is bounded by

the poles, assume that the z-transform X(z)

has simple poles at z = α and z = β• Assume that the corresponding sequence

x[n] is a right-sided sequence

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ROC of a Rational ROC of a Rational

zz--TransformTransform

• Then x[n] has the form

where is a positive or negative integer

• Now, the z-transform of the right-sided

sequence exists if

for some z

( ) β<α−µβ+α= ],[][ 21 onn Nnrrnx

oN

][ on Nn −µγ

∞<γ∑∞

=

oNn

nnz

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ROC of a Rational ROC of a Rational

zz--TransformTransform

• The condition

holds for but not for

• Therefore, the z-transform of

has an ROC defined by

∞<γ∑∞

=

oNn

nnz

γ>z γ≤z

( ) β<α−µβ+α= ],[][ 21 onn Nnrrnx

∞≤<β z

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ROC of a Rational ROC of a Rational

zz--TransformTransform

• Likewise, the z-transform of a left-sided

sequence

has an ROC defined by

• Finally, for a two-sided sequence, some of

the poles contribute to terms in the parent

sequence for n < 0 and the other poles

contribute to terms

( ) β<α−−µβ+α= ],[][ 21 onn Nnrrnx

α<≤ z0

0≥n

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ROC of a Rational ROC of a Rational

zz--TransformTransform

• The ROC is thus bounded on the outside by

the pole with the smallest magnitude that

contributes for n < 0 and on the inside by

the pole with the largest magnitude that

contributes for

• There are three possible ROCs of a rational

z-transform with poles at z = α and z = β( )

0≥n

β<α

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ROC of a Rational ROC of a Rational zz--TransformTransform

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ROC of a Rational ROC of a Rational

zz--TransformTransform

• In general, if the rational z-transform has N

poles with R distinct magnitudes, then it has

ROCs

• Thus, there are distinct sequences with

the same z-transform

• Hence, a rational z-transform with a

specified ROC has a unique sequence as its

inverse z-transform

1+R

1+R

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ROC of a Rational ROC of a Rational

zz--TransformTransform

• The ROC of a rational z-transform can be

easily determined using MATLAB

determines the zeros, poles, and the gain

constant of a rational z-transform with the

numerator coefficients specified by the vector num and the denominator coefficients

specified by the vector den

[z,p,k] = tf2zp(num,den)

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ROC of a Rational ROC of a Rational

zz--TransformTransform

• [num,den] = zp2tf(z,p,k)

implements the reverse process

• The factored form of the z-transform can be obtained using sos = zp2sos(z,p,k)

• The above statement computes the

coefficients of each second-order factor given as an matrix sos6×L

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ROC of a Rational ROC of a Rational

zz--TransformTransform

where

=

LLLLLLaaabbb

aaabbb

aaabbb

sos

210210

221202221202

121101211101

MMMMMM

∏=

−−

−−

++++

=L

k kkk

kkk

zazaa

zbzbbzG

12

21

10

22

110)(

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ROC of a Rational ROC of a Rational

zz--TransformTransform

• The pole-zero plot is determined using the function zplane

• The z-transform can be either described in terms of its zeros and poles:

zplane(zeros,poles)

• or, it can be described in terms of its

numerator and denominator coefficients:zplane(num,den)

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ROC of a Rational ROC of a Rational zz--TransformTransform

• Example - The pole-zero plot of

obtained using MATLAB is shown below

pole−×zeroo −

12181533

325644162)(

234

234

−+−+++++

=zzzz

zzzzzG

-4 -3 -2 -1 0 1

-2

-1

0

1

2

Real Part

Imag

inar

y P

art

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Inverse zInverse z--TransformTransform

• General Expression: Recall that, for ,

the z-transform G(z) given by

is merely the DTFT of the modified sequence

• Accordingly, the inverse DTFT is thus given

by

ω= jerz

∑=∑= ∞−∞=

ω−−∞−∞=

−n

njnn

n erngzngzG ][][)(

nrng −][

ω∫π

= ωππ−

ω− dereGrng njjn )(2

1][

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Inverse zInverse z--TransformTransform

• By making a change of variable ,

the previous equation can be converted into

a contour integral given by

where is a counterclockwise contour of

integration defined by |z| = r

ω= jerz

C′

dzzzGj

ngC

n∫

π=

−1)(2

1][

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Inverse zInverse z--TransformTransform• But the integral remains unchanged when

is replaced with any contour C encircling

the point z = 0 in the ROC of G(z)

• The contour integral can be evaluated using

the Cauchy’s residue theorem resulting in

• The above equation needs to be evaluated at

all values of n and is not pursued here

=

C

zzGngn

insidepolestheat

ofresidues 1)(][

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Inverse Transform by Inverse Transform by

PartialPartial--Fraction ExpansionFraction Expansion

• A rational z-transform G(z) with a causal

inverse transform g[n] has an ROC that is

exterior to a circle

• Here it is more convenient to express G(z)

in a partial-fraction expansion form and

then determine g[n] by summing the inverse

transform of the individual simpler terms in

the expansion

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Inverse Transform by Inverse Transform by PartialPartial--Fraction ExpansionFraction Expansion

• A rational G(z) can be expressed as

• If then G(z) can be re-expressed as

where the degree of is less than N

∑∑

=−

=−

==Ni

ii

Mi

ii

zD

zP

zd

zpzG

0

0)(

)()(

)(

)()(

zD

zPNM

zzG 1

0

∑−

=

− +=l

llη

NM ≥

)(zP1

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Inverse Transform by Inverse Transform by

PartialPartial--Fraction ExpansionFraction Expansion

• The rational function is called a

proper fraction

• Example - Consider

• By long division we arrive at

)(/)( zDzP1

21

321

20801

3050802−−

−−−

++

+++=

zz

zzzzG

..

...)(

21

11

20801

12555153

−−

−−

++

+++−=

zz

zzzG

..

....)(

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Inverse Transform by Inverse Transform by

PartialPartial--Fraction ExpansionFraction Expansion

• Simple Poles: In most practical cases, the

rational z-transform of interest G(z) is a

proper fraction with simple poles

• Let the poles of G(z) be at ,

• A partial-fraction expansion of G(z) is then

of the form

kz λ= Nk ≤≤1

λ−

ρ=

=−

N

zzG

111

)(l l

l

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Inverse Transform by Inverse Transform by

PartialPartial--Fraction ExpansionFraction Expansion

• The constants in the partial-fraction

expansion are called the residues and are

given by

• Each term of the sum in partial-fraction

expansion has an ROC given by

and, thus has an inverse transform of the

form

lll λ=

−λ−=ρz

zGz )()1( 1

lλ>z

][)( nnµλρ ll

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Inverse Transform by Inverse Transform by

PartialPartial--Fraction ExpansionFraction Expansion

• Therefore, the inverse transform g[n] of

G(z) is given by

• Note: The above approach with a slight

modification can also be used to determine

the inverse of a rational z-transform of a

noncausal sequence

∑ µλρ==

Nn nng

1

][)(][l

ll

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Inverse Transform by Inverse Transform by

PartialPartial--Fraction ExpansionFraction Expansion

• Example - Let the z-transform H(z) of a

causal sequence h[n] be given by

• A partial-fraction expansion of H(z) is then

of the form

).)(.().)(.(

)()(

11

1

601201

21

6020

2−−

+−

+=

+−+

=zz

z

zz

zzzH

12

11

6.012.01)( −− +

ρ+

ρ=

zzzH

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Inverse Transform by Inverse Transform by

PartialPartial--Fraction ExpansionFraction Expansion

• Now

and

75.26.01

21)()2.01(

2.01

1

2.01

1 =+

+=−=ρ

=−

=−

z

zz

zzHz

75.12.01

21)()6.01(

6.01

1

6.01

2 −=−

+=+=ρ

−=−

−=−

z

zz

zzHz

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Inverse Transform by Inverse Transform by

PartialPartial--Fraction ExpansionFraction Expansion

• Hence

• The inverse transform of the above is

therefore given by

11 601

751

201

752−− +

−−

=zz

zH.

.

.

.)(

][)6.0(75.1][)2.0(75.2][ nnnh nn µ−−µ=

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Inverse Transform by Inverse Transform by

PartialPartial--Fraction ExpansionFraction Expansion

• Multiple Poles: If G(z) has multiple poles,

the partial-fraction expansion is of slightly

different form

• Let the pole at z = ν be of multiplicity L and

the remaining poles be simple and at

,

LN −lλ=z LN −≤≤ l1

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Inverse Transform by Inverse Transform by PartialPartial--Fraction ExpansionFraction Expansion

• Then the partial-fraction expansion of G(z)

is of the form

where the constants are computed using

• The residues are calculated as before

∑ν−

γ+∑

λ−

ρ+∑ η=

=−

=−

=

− L

ii

iLNNM

zzzzG

11

11

0 )1(1)(

l l

l

l

ll

[ ] ,)()1()()()!(

1 1

1 ν=

−−−

− ν−ν−−

=γz

L

iL

iL

iLi zGzzd

d

iLLi ≤≤1

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PartialPartial--Fraction Expansion Fraction Expansion

Using MATLABUsing MATLAB

• [r,p,k]= residuez(num,den)

develops the partial-fraction expansion of

a rational z-transform with numerator and

denominator coefficients given by vectorsnum and den

• Vector r contains the residues

• Vector p contains the poles

• Vector k contains the constantslη

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PartialPartial--Fraction Expansion Fraction Expansion

Using MATLABUsing MATLAB

• [num,den]=residuez(r,p,k)

converts a z-transform expressed in a

partial-fraction expansion form to its

rational form

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Inverse zInverse z--Transform via Long Transform via Long

DivisionDivision

• The z-transform G(z) of a causal sequence

{g[n]} can be expanded in a power series in

• In the series expansion, the coefficient

multiplying the term is then the n-th

sample g[n]

• For a rational z-transform expressed as a

ratio of polynomials in , the power series

expansion can be obtained by long division

1−z

1−z

nz−

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Inverse zInverse z--Transform via Long Transform via Long

DivisionDivision

• Example - Consider

• Long division of the numerator by the

denominator yields

• As a result

21

1

12.04.01

21)( −−

−++

=zz

zzH

........)( +−+−+= −−−− 4321 2224040520611 zzzzzH

02224040520611 ≥−−= nnh },........{]}[{↑

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Inverse zInverse z--Transform Using Transform Using

MATLABMATLAB

• The function impz can be used to find the

inverse of a rational z-transform G(z)

• The function computes the coefficients of

the power series expansion of G(z)

• The number of coefficients can either be

user specified or determined automatically

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Table:Table: zz--Transform PropertiesTransform Properties

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zz--Transform PropertiesTransform Properties

• Example - Consider the two-sided sequence

• Let and

with X(z) and Y(z) denoting, respectively,

their z-transforms

• Now

and

]1[][][ −−µβ−µα= nnnv nn

][][ nnx nµα= ]1[][ −−µβ−= nny n

α>α−

= − zz

zX ,1

1)(

1

β<β−

= − zz

zY ,1

1)(

1

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zz--Transform PropertiesTransform Properties

• Using the linearity property we arrive at

• The ROC of V(z) is given by the overlap

regions of and

• If , then there is an overlap and the

ROC is an annular region

• If , then there is no overlap and V(z)

does not exist

11 1

1

1

1−− −

+−

=+=zz

zYzXzVβα

)()()(

α>z β<z

β<αβ<<α z

β>α

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zz--Transform PropertiesTransform Properties

• Example - Determine the z-transform and

its ROC of the causal sequence

• We can express x[n] = v[n] + v*[n] where

• The z-transform of v[n] is given by

][)(cos][ nnrnx on µω=

][][][2

1

2

1 nnernv nnjn o µα=µ= ω

rzzerz

zVoj

=α>−

⋅=α−

⋅=−ω− ,

1

1

1

1)(

12

112

1

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zz--Transform PropertiesTransform Properties• Using the conjugation property we obtain

the z-transform of v*[n] as

• Finally, using the linearity property we get

,1

1

*1

1*)(*

12

112

1−ω−− −

⋅=α−

⋅=zerz

zVoj

*)(*)()( zVzVzX +=

−+

−=

−ω−−ω 112

1

1

1

1

1

zerzer oo jj

α>z

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zz--Transform PropertiesTransform Properties

• or,

• Example - Determine the z-transform Y(z)

and the ROC of the sequence

• We can write where

rzzrzr

zrzX

o

o >+ω−

ω−= −−

−,

)cos2(1

)cos(1)(

221

1

][)1(][ nnny nµα+=

][][][ nxnxnny +=

][][ nnx nµα=

Page 238: Discrete -Time Fourier Transform Discrete Fourier ...tania/teaching/dsp/Lectures 3-4 DTFT DFT... · Discrete -Time Fourier Transform Discrete Fourier Transform z-Transform Tania Stathaki

zz--Transform PropertiesTransform Properties

• Now, the z-transform X(z) of

is given by

• Using the differentiation property, we arrive

at the z-transform of as

][][ nnx nµα=

][nxn

α>α−

= − zz

zX ,1

1)(

1

α>α−

α=− −

−z

z

z

dz

zXdz ,

)1(

)(1

1

Page 239: Discrete -Time Fourier Transform Discrete Fourier ...tania/teaching/dsp/Lectures 3-4 DTFT DFT... · Discrete -Time Fourier Transform Discrete Fourier Transform z-Transform Tania Stathaki

zz--Transform PropertiesTransform Properties

• Using the linearity property we finally

obtain

21

1

1 )1(1

1)( −

− α−α

+α−

=z

z

zzY

α>α−

= − zz

,)1(

121