Post on 11-Sep-2021
DiscreteDiscrete--Time Fourier TransformTime Fourier Transform
Discrete Fourier TransformDiscrete Fourier Transform
zz--TransformTransform
Tania Stathaki
811b
t.stathaki@imperial.ac.uk
Joseph Fourier (1768-1830)
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 +=
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
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
)(
)()ω(tan
)()()(
)ω(sin)()(
)ω(cos)()(
)()()(
ω
re
ω
im
ω2
im
ω2
re
2ω
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
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
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− )(
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
2π
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
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
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
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
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− =∑=
DiscreteDiscrete--Time Fourier Time Fourier
TransformTransform
• Inverse Discrete-Time Fourier Transform:
• Proof:
∫ ωπ
=π
π−
ωω deeXnx njj )(2
1][
∫ ω
∑
π=
π
π−
ω∞
−∞=
ω− deexnx njj
l
ll][
2
1][
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
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
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 ][)(
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 ][][)(
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
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
• 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
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 ][)(
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
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
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−
ω denhc
c
njLP
2
1][
∞<<∞− n
πω /c
][nhLP
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
][
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
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
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
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
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
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
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
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
DiscreteDiscrete--Time Fourier Time Fourier
TransformTransform
• Thus
where we have used the sampling property
of the impulse function )(ωδ
∫ ∑ ωπ+ω−ωπδπ
=π
π−
∞
−∞=
ω
k
njo deknx )2(2
2
1][
njnjo
oedeω
π
π−
ω =ω∫ ω−ωδ= )(
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(],[
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
Table: Table: General Properties of DTFTGeneral Properties of DTFT
Table:Table: Symmetry relations of the Symmetry relations of the DTFT of a complex sequenceDTFT of a complex sequence
x[n]: A complex sequence
Table:Table: Symmetry relations of Symmetry relations of the DTFT of a real sequencethe DTFT of a real sequence
x[n]: A real sequence
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
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
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
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)(
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
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
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)(
Energy Density SpectrumEnergy Density Spectrum
• Therefore
• Hence, is a finite-energy sequence
∞<πω
=∫ ωπ
=∑ω
ω−
∞
−∞=
c
nLP
c
c
dnh2
1][2
][nhLP
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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 =−
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
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
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
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
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
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
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
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
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
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
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
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
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
Sampling the DTFTSampling the DTFT
,
• i.e.
{x[n]} cannot be recovered from {y[n]}
][][][][ 44 −+++= nxnxnxny 30 ≤≤ n
}{]}[{ 3264=ny↑
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
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
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
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
Table:Table: General Properties of DFTGeneral Properties of DFT
Table:Table: DFT Properties: DFT Properties: Symmetry RelationsSymmetry Relations
x[n] is a complex sequence
Table:Table: DFT Properties: DFT Properties: Symmetry RelationsSymmetry Relations
x[n] is a real sequence
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
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
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
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
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 ⟩⟨
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
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
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
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
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
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 =×+×+×+×= )()()()(
Circular ConvolutionCircular Convolution
• Likewise ∑ ⟩−⟨==
3
04]1[][]1[
mC mhmgy
][][][][][][][][ 23320110 hghghghg +++=
711102221 =×+×+×+×= )()()()(
∑ ⟩−⟨==
3
04]2[][]2[
mC mhmgy
][][][][][][][][ 33021120 hghghghg +++=
611202211 =×+×+×+×= )()()()(
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
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
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
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][
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
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
][][
][][
][][
][][
][
][
][
][
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
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 ,],[
][
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
Circular ConvolutionCircular Convolution
• Continuing the process we arrive at
,)()(][][][][][ 6222101101 =×+×=+= hghgy
][][][][][][][ 0211202 hghghgy ++=,)()()( 5202211 =×+×+×=][][][][][][][][][ 031221303 hghghghgy +++=
,)()()()( 521201211 =×+×+×+×=][][][][][][][ 1322314 hghghgy ++=
,)()()( 4211012 =×+×+×=
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
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
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
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 +=
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 ⟩−⟨=⟩⟨−
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↑
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
NN--Point Point DFTsDFTs of Two Lengthof Two Length--NN
Real SequencesReal Sequences
• Therefore
verifying the results derived earlier
}{]}[{ jjkG +−−= 1214
}{]}[{ jjkH +−= 1016
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 ],[][],[][
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 ⟩⟨−−=
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 ,][][
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↑
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
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 ][][][
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
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
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 ,],[
][
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
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 *
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
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
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
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 = *
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
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*
OverlapOverlap--Add MethodAdd Method
OverlapOverlap--Add MethodAdd MethodAdd
Add
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
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
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]
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
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
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
OverlapOverlap--Save MethodSave Method
][][][ 000 xhyL =][][][][][ 01101 xhxhyL +=
][][][][][][][ 0211202 xhxhxhyL ++=
][][][][][][][ 1221303 xhxhxhyL ++=
][][][][][ 22314 xhxhyL +=
][][][ 325 xhyL =
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
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 =
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
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
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
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
OverlapOverlap--Save MethodSave Method
][][][][][][][ 62514002 xhxhxhw ++= ←Reject
][][][][][][][ 72415012 xhxhxhw ++= ←Reject
][][][][][][][][ 642516022 yxhxhxhw =++= ←Save
][][][][][][][][ 752617032 yxhxhxhw =++= ←Save
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 == −−
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
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
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
OverlapOverlap--Save MethodSave Method
• Process is illustrated next
OverlapOverlap--Save MethodSave Method
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
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 ][)(
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 ][)(
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 −
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 ][
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 ][
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
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
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
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
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
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
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
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)(
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
Table:Table: Commonly UsedCommonly Used zz--Transform PairsTransform Pairs
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
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
....
....)( )(
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
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
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
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
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
..
..)(
Rational Rational zz--TransformsTransforms
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 ±=
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
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
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)
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
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
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
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
][][)(
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
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
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
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
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 =
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
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 <<
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
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=][
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
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
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
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
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
ROC of a Rational ROC of a Rational zz--TransformTransform
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
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)
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
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)(
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)
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
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][
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][
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)(][
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
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
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
..
....)(
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
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
lρ
lll λ=
−λ−=ρz
zGz )()1( 1
lλ>z
][)( nnµλρ ll
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
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
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
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=
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
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
iγ
[ ] ,)()1()()()!(
1 1
1 ν=
−−−
−
− ν−ν−−
=γz
L
iL
iL
iLi zGzzd
d
iLLi ≤≤1
lρ
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
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
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−
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 },........{]}[{↑
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
Table:Table: zz--Transform PropertiesTransform Properties
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
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
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
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
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=
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
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