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Transcript of UNIT - 4 ANALYSIS OF DISCRETE TIME SIGNALS. Sampling Frequency Harry Nyquist, working at Bell Labs...
UNIT - 4
ANALYSIS OF DISCRETE TIME SIGNALS
Sampling Frequency
• Harry Nyquist, working at Bell Labs developed what has become known as the Nyquist Sampling Theorem:– In order to be ‘perfectly’ represented by its samples, a signal must be
sampled at a sampling rate (also called sampling frequency) equal to at least twice its highest frequency component
– Or: fs = 2f
– Note that fs here is frequency of sampling, not the frequency of the sample
How often do you sample? The sampling rate depends on the signal’s highest frequency (for baseband)
Sampling Rate Examples• Take Concert A: 440 Hz
– What would be the minimum sampling rate needed to accurately capture this signal?
– fs = 2 x 440 Hz = 880 Hz
• Take your telephone used for voice, mostly– Highest voice component is: 3000 Hz– Minimum sampling rate: fs = 2 x 3000 Hz = 6000 Hz– Real telephone digitization is done at 8000 Hz sampling rate (supporting a 4
kHz bandwidth). Why? Remember that Nyquist said “equal to at least twice…”
Undersampling and Oversampling• Undersampling
– Sampling at an inadequate frequency rate– Aliased into new form - Aliasing– Loses information in the original signal
• Oversampling – Sampling at a rate higher than minimum rate – More values to digitize and process– Increases the amount of storage and transmission– COST $$
Effects of Undersampling
Original waveform
Reconstructed waveform
DISCRETE TIME FOURIER TRANSFROM (DTFT)
• Definition of DTFT:– The Fourier transform (FT) of discrete – time signals is called Discrete Time
Fourier Transform (i.e DTFT).
Let x(n) = Discrete time signal
X(ej) = Fourier transform of x(n)
The Fourier transform of a finite energy discrete time signal, x(n) is defined as,
X(ej)= F{x(n)} =
FREQUENCY SPECTRUM• The FT X(ej) of a signal x(n) represents the frequency content of x(n).
By taking FT, the signal x(n) is decomposed into its frequency components. Hence X(ej) is also called frequency spectrum of discrete time signal or signal spectrum.
• Magnitude and Phase Spectrum• The X(ej is a complex value function of and so it can be expressed in
rectangular form as,
X(ej) = Xr(ej) + jXi(ej)
Where Xr(ej) = Real part of X(ej)
Xi(ej) = Imaginary part of X(ej)
• The polar form of X(ej) is,• X(ej) = |X(ej)|∟X(ej)
• Where, |X(ej)| = Magnitude spectrum
∟X(ej) = Phase spectrum
• The magnitude spectrum is defined as,• |X(ej)| = X(ej)X*(ej)
Derivation of the Discrete-time Fourier Transform
Recall DTFS pair
where
The limit of integration is over any interval of 2p in w
Periodic in w with period 2p
Thus,
DTFT Pair
Conditions for Convergence
Examples
0
,1
1
,1
1
a
a
IDTFT
6) Complex Exponentials
DTFT of Periodic Signals
Recall the following DTFT pair:
Represent periodic signal x[n] in terms of DTFS:
Example: A discrete-time Sine Function
Example: A discrete-time Periodic Impulse Train
The DTFS coefficients for this signal are:
ck
Properties of DTFT
Properties of DTFT
Convolution Property
Multiplication Property
PROPERTIES OF DTFT
The z-Transform
Content
• Introduction• z-Transform• Zeros and Poles• Region of Convergence• Important z-Transform Pairs• Inverse z-Transform• z-Transform Theorems and Properties• System Function
The z-Transform
Introduction
Why z-Transform?• A generalization of Fourier transform• Why generalize it?
– FT does not converge on all sequence– Notation good for analysis– Bring the power of complex variable theory deal
with the discrete-time signals and systems
The z-Transform
z-Transform
Definition
• The z-transform of sequence x(n) is defined by
n
nznxzX )()(
Let z = ej.
( ) ( )j j n
n
X e x n e
Fourier Transform
z-Plane
Re
Im
z = ej
n
nznxzX )()(
( ) ( )j j n
n
X e x n e
Fourier Transform is to evaluate z-transform on a unit circle.
Fourier Transform is to evaluate z-transform on a unit circle.
z-Plane
Re
Im
X(z)
Re
Im
z = ej
Periodic Property of FT
Re
Im
X(z)
X(ej)
Can you say why Fourier Transform is a periodic function with period 2?Can you say why Fourier Transform is a periodic function with period 2?
The z-TransformZeros and Poles
Definition
• Give a sequence, the set of values of z for which the z-transform converges, i.e., |X(z)|<, is called the region of convergence.
n
n
n
n znxznxzX |||)(|)(|)(|
ROC is centered on origin and consists of a set of rings.
ROC is centered on origin and consists of a set of rings.
Example: Region of Convergence
Re
Im
n
n
n
n znxznxzX |||)(|)(|)(|
ROC is an annual ring centered on the origin.
ROC is an annual ring centered on the origin.
xx RzR ||r
}|{ xx
j RrRrezROC
Stable Systems
• A stable system requires that its Fourier transform is uniformly convergent.
Re
Im
1
Fact: Fourier transform is to evaluate z-transform on a unit circle.
A stable system requires the ROC of z-transform to include the unit circle.
Example: A right sided Sequence
)()( nuanx n
1 2 3 4 5 6 7 8 9 10-1-2-3-4-5-6-7-8
n
x(n)
. . .
Example: A right sided Sequence
)()( nuanx n
n
n
n znuazX
)()(
0n
nn za
0
1)(n
naz
For convergence of X(z), we require that
0
1 ||n
az 1|| 1 az
|||| az
az
z
azazzX
n
n
10
1
1
1)()(
|||| az
aa
Example: A right sided Sequence ROC for x(n)=anu(n)
|||| ,)( azaz
zzX
Re
Im
1 aaRe
Im
1
Which one is stable?Which one is stable?
Example: A left sided Sequence
)1()( nuanx n
1 2 3 4 5 6 7 8 9 10-1-2-3-4-5-6-7-8n
x(n)
. . .
Example: A left sided Sequence
)1()( nuanx n
n
n
n znuazX
)1()(
For convergence of X(z), we require that
0
1 ||n
za 1|| 1 za
|||| az
az
z
zazazX
n
n
10
1
1
11)(1)(
|||| az
n
n
n za
1
n
n
n za
1
n
n
n za
0
1
aa
Example: A left sided Sequence ROC for x(n)=anu( n1)
|||| ,)( azaz
zzX
Re
Im
1 aaRe
Im
1
Which one is stable?Which one is stable?
The z-TransformRegion of
Convergence
Represent z-transform as a Rational Function
)(
)()(
zQ
zPzX where P(z) and Q(z) are
polynomials in z.
Zeros: The values of z’s such that X(z) = 0
Poles: The values of z’s such that X(z) =
Example: A right sided Sequence
)()( nuanx n |||| ,)( azaz
zzX
Re
Im
a
ROC is bounded by the pole and is the exterior of a circle.
Example: A left sided Sequence
)1()( nuanx n|||| ,)( az
az
zzX
Re
Im
a
ROC is bounded by the pole and is the interior of a circle.
Example: Sum of Two Right Sided Sequences
)()()()()( 31
21 nununx nn
31
21
)(
z
z
z
zzX
Re
Im
1/2
))((
)(2
31
21
121
zz
zz
1/31/12
ROC is bounded by poles and is the exterior of a circle.
ROC does not include any pole.
Example: A Two Sided Sequence
)1()()()()( 21
31 nununx nn
21
31
)(
z
z
z
zzX
Re
Im
1/2
))((
)(2
21
31
121
zz
zz
1/31/12
ROC is bounded by poles and is a ring.
ROC does not include any pole.
Example: A Finite Sequence
10 ,)( Nnanx n
nN
n
nN
n
n zazazX )()( 11
0
1
0
Re
ImROC: 0 < z < ROC does not include any pole.
1
1
1
)(1
az
az N
az
az
z
NN
N
1
1
N-1 poles
N-1 zeros
Always StableAlways Stable
Properties of ROC• A ring or disk in the z-plane centered at the origin.• The Fourier Transform of x(n) is converge absolutely iff the ROC includes
the unit circle.• The ROC cannot include any poles• Finite Duration Sequences: The ROC is the entire z-plane except possibly
z=0 or z=. • Right sided sequences: The ROC extends outward from the outermost
finite pole in X(z) to z=. • Left sided sequences: The ROC extends inward from the innermost
nonzero pole in X(z) to z=0.
More on Rational z-Transform
Re
Im
a b c
Consider the rational z-transform with the pole pattern:
Find the possible ROC’s
Find the possible ROC’s
More on Rational z-Transform
Re
Im
a b c
Consider the rational z-transform with the pole pattern:
Case 1: A right sided Sequence.
More on Rational z-Transform
Re
Im
a b c
Consider the rational z-transform with the pole pattern:
Case 2: A left sided Sequence.
More on Rational z-Transform
Re
Im
a b c
Consider the rational z-transform with the pole pattern:
Case 3: A two sided Sequence.
More on Rational z-Transform
Re
Im
a b c
Consider the rational z-transform with the pole pattern:
Case 4: Another two sided Sequence.
The z-TransformImportant
z-Transform Pairs
Z-Transform Pairs
Sequence z-Transform ROC)(n 1 All z
)( mn mz All z except 0 (if m>0)or (if m<0)
)(nu 11
1 z
1|| z
)1( nu 11
1 z
1|| z
)(nuan 11
1 az
|||| az
)1( nuan 11
1 az
|||| az
Z-Transform Pairs
Sequence z-Transform ROC)(][cos 0 nun 21
0
10
]cos2[1
][cos1
zz
z1|| z
)(][sin 0 nun 210
10
]cos2[1
][sin
zz
z1|| z
)(]cos[ 0 nunr n 2210
10
]cos2[1
]cos[1
zrzr
zrrz ||
)(]sin[ 0 nunr n 2210
10
]cos2[1
]sin[
zrzr
zrrz ||
otherwise0
10 Nnan
11
1
az
za NN
0|| z
The z-TransformInverse z-Transform
The Inverse Z-Transform• Formal inverse z-transform is based on a Cauchy
integral• Less formal ways sufficient most of the time
– Inspection method– Partial fraction expansion– Power series expansion
• Inspection Method– Make use of known z-transform pairs such as
• Example: The inverse z-transform of
az az11
nua 1Zn
nu21
nx 21
z z
21
1
1zX
n
1
Inverse Z-Transform by Partial Fraction Expansion• Assume that a given z-transform can be expressed as
• Apply partial fractional expansion
• First term exist only if M>N– Br is obtained by long division
• Second term represents all first order poles• Third term represents an order s pole
– There will be a similar term for every high-order pole
• Each term can be inverse transformed by inspection
N
0k
kk
M
0k
kk
za
zbzX
s
1mm1
i
mN
ik,1k1
k
kNM
0r
rr
zd1
Czd1
AzBzX
Partial Fractional Expression
• Coefficients are given as
s
1mm1
i
mN
ik,1k1
k
kNM
0r
rr
zd1
Czd1
AzBzX
kdz
1kk zXzd1A
1idw
1sims
ms
msi
m wXwd1dwd
d!ms
1C
Example: 2nd Order Z-Transform
– Order of nominator is smaller than denominator (in terms of z-1)
No higher order pole
1
2
1
1
z21
1
A
z41
1
AzX
1
41
21
1
1zXz
41
1A1
41
z
11
2
21
41
1
1zXz
21
1A1
21
z
12
21
z :ROC z
21
1z41
1
1zX
11
Example Continued
• ROC extends to infinity – Indicates right sided sequence
21
z z
21
1
2
z41
1
1zX
11
nu41
-nu21
2nxnn
Example #2
• Long division to obtain Bo
1z z1z
21
1
z1
z21
z23
1
zz21zX
11
21
21
21
1z5
2z3z
21z2z1z
23
z21
1
12
1212
11
1
z1z21
1
z512zX
1
2
1
1
z1A
z21
1
A2zX
9zXz21
1A
21
z
11
8zXz1A1z
12
Example #2 Continued
• ROC extends to infinity– Indicates right-sides sequence
1z z1
8
z21
1
92zX 1
1
n8u-nu21
9n2nxn
Inverse Z-Transform by Power Series Expansion
• The z-transform is power series
• In expanded form
• Z-transforms of this form can generally be inversed easily
• Especially useful for finite-length series
n
nz nxzX
2112 z 2xz 1x 0xz 1xz 2xzX
12
1112
z21
1z21
z
z1z1z21
1z zX
• Example,
12
1112
z21
1z21
z
z1z1z21
1z zX
1n21
n1n21
2nnx
2n0
1n21
0n1
1n21
2n1
nx
Z-Transform Properties: Linearity• Notation
• Linearity
– Note that the ROC of combined sequence may be larger than either ROC– This would happen if some pole/zero cancellation occurs– Example:
• Both sequences are right-sided• Both sequences have a pole z=a• Both have a ROC defined as |z|>|a|• In the combined sequence the pole at z=a cancels with a zero at z=a• The combined ROC is the entire z plane except z=0
xZ RROC zXnx
21 xx21
Z21 RRROC zbXzaXnbxnax
N-nua-nuanx nn
Z-Transform Properties: Time Shifting
• Here no is an integer– If positive the sequence is shifted right– If negative the sequence is shifted left
• The ROC can change the new term may– Add or remove poles at z=0 or z=
• Example
xnZ
o RROCzXznnx o
41
z z
41
1
1z zX
1
1
1-nu41
nx1-n
Z-Transform Properties: Multiplication by Exponential• ROC is scaled by |zo|
• All pole/zero locations are scaled
• If zo is a positive real number: z-plane shrinks or expands
• If zo is a complex number with unit magnitude it rotates
• Example: We know the z-transform pair
• Let’s find the z-transform of
xooZn
o RzROC z/zXnxz
1z:ROC z-1
1nu 1-
Z
nure21
nure21
nuncosrnxnjnj
on oo
rz zre1
2/1
zre1
2/1zX
1j1j oo
Z-Transform Properties: Differentiation
• Example: We want the inverse z-transform of
• Let’s differentiate to obtain rational expression
• Making use of z-transform properties and ROCCopyright (C) 2005 Güner Arslan
351M Digital Signal Processing 86
x
Z RROC dz
zdXznnx
az az1logzX 1
1
11
2
az11
azdz
zdXz
az1az
dzzdX
1nuaannx 1n
1nuna
1nxn
1n
Z-Transform Properties: Conjugation
• Example
Copyright (C) 2005 Güner Arslan
351M Digital Signal Processing 87
x**Z* RROC zXnx
nxZz nxz nxzX
z nxz nxzX
z nxzX
n
n
n
n
n
n
n
n
n
n
Z-Transform Properties: Time Reversal
• ROC is inverted• Example:
• Time reversed version of
Copyright (C) 2005 Güner Arslan
351M Digital Signal Processing 88
x
Z
R1
ROC z/1Xnx
nuanx n
nuan
111-
1-1
az za-1
za-az1
1zX
Z-Transform Properties: Convolution• Convolution in time domain is multiplication in z-domain• Example:Let’s calculate the convolution of
• Multiplications of z-transforms is
• ROC: if |a|<1 ROC is |z|>1 if |a|>1 ROC is |z|>|a|• Partial fractional expansion of Y(z)
Copyright (C) 2005 Güner Arslan
351M Digital Signal Processing 89
2x1x21
Z21 RR:ROC zXzXnxnx
nunx and nuanx 2n
1
az:ROC az11
zX 11
1z:ROC z1
1zX 12
1121 z1az11
zXzXzY
1z :ROC asume az11
z11
a11
zY 11
nuanua1
1ny 1n
The z-Transformz-Transform Theorems and Properties
Linearity
xRzzXnx ),()]([Z
yRzzYny ),()]([Z
yx RRzzbYzaXnbynax ),()()]()([Z
Overlay of the above two
ROC’s
Shift
xRzzXnx ),()]([Z
xn RzzXznnx )()]([ 0
0Z
Multiplication by an Exponential Sequence
xx- RzRzXnx || ),()]([Z
xn RazzaXnxa || )()]([ 1Z
Differentiation of X(z)
xRzzXnx ),()]([Z
xRzdz
zdXznnx
)()]([Z
Conjugation
xRzzXnx ),()]([Z
xRzzXnx *)(*)](*[Z
Reversal
xRzzXnx ),()]([Z
xRzzXnx /1 )()]([ 1 Z
Real and Imaginary Parts
xRzzXnx ),()]([Z
xRzzXzXnxe *)](*)([)]([ 21R
xj RzzXzXnx *)](*)([)]([ 21Im
Initial Value Theorem
0for ,0)( nnx
)(lim)0( zXxz
Convolution of Sequences
xRzzXnx ),()]([Z
yRzzYny ),()]([Z
yx RRzzYzXnynx )()()](*)([Z
Convolution of Sequences
k
knykxnynx )()()(*)(
n
n
k
zknykxnynx )()()](*)([Z
k
n
n
zknykx )()(
k
n
n
k znyzkx )()(
)()( zYzX
The z-TransformSystem Function
Shift-Invariant System
h(n)h(n)x(n) y(n)=x(n)*h(n)
X(z) Y(z)=X(z)H(z)H(z)
Shift-Invariant System
H(z)H(z)X(z) Y(z)
)(
)()(
zX
zYzH
Nth-Order Difference Equation
M
rr
N
kk rnxbknya
00
)()(
M
r
rr
N
k
kk zbzXzazY
00
)()(
N
k
kk
M
r
rr zazbzH
00)(
Representation in Factored Form
N
kr
M
rr
zd
zcAzH
1
1
1
1
)1(
)1()(
Contributes poles at 0 and zeros at cr
Contributes zeros at 0 and poles at dr
Stable and Causal Systems
N
kr
M
rr
zd
zcAzH
1
1
1
1
)1(
)1()( Re
ImCausal Systems : ROC extends outward from the outermost pole.
Stable and Causal Systems
N
kr
M
rr
zd
zcAzH
1
1
1
1
)1(
)1()( Re
ImStable Systems : ROC includes the unit circle.
1
Example
Consider the causal system characterized by
)()1()( nxnayny
11
1)(
azzH
Re
Im
1
a
)()( nuanh n
Determination of Frequency Response from pole-zero pattern
• A LTI system is completely characterized by its pole-zero pattern.
))(()(
21
1
pzpz
zzzH
Example:
))(()(
21
1
00
0
0
pepe
zeeH jj
jj
0je
Re
Im
z1
p1
p2
Determination of Frequency Response from pole-zero pattern
• A LTI system is completely characterized by its pole-zero pattern.
))(()(
21
1
pzpz
zzzH
Example:
))(()(
21
1
00
0
0
pepe
zeeH jj
jj
0je
Re
Im
z1
p1
p2
|H(ej)|=?|H(ej)|=? H(ej)=?H(ej)=?
Determination of Frequency Response from pole-zero pattern
• A LTI system is completely characterized by its pole-zero pattern.
Example:0je
Re
Im
z1
p1
p2
|H(ej)|=?|H(ej)|=? H(ej)=?H(ej)=?
|H(ej)| =| |
| | | | 1
2
3
H(ej) = 1(2+ 3 )
Example
11
1)(
azzH
Re
Im
a
0 2 4 6 8-10
0
10
20
0 2 4 6 8-2
-1
0
1
2
dB