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TRANSCRIPT
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LECTURE NOTES ON
DIGITAL SIGNAL
PROCESSING
Prof. Dr. A. Salim KAYHAN
Hacettepe University
Department of Electrical and Electronics Engineering
ANKARA-2014
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ELE 407 Digital Signal Processing
Fall 2014
Place: E-? Time: Wednesday 13:00-15:50 Course Outline: W1-Sep.24 Review of Discrete-time signals, systems, Fourier, Z-tr. W2-Oct.01 Sampling, Decimation, Interpolation. W3-Oct.08 Frequency Response . W4-Oct.15 Relation Between Magnitude and Phase, Flow Graph
Realization. W5-Oct.22 Flow Graph Realization of FIR sys., Quantization. W6-Oct.29 No Class. (Holiday) W7-Nov.05 EXAM W8-Nov.12 Butterworth and Chebyshev Filter Design. W9-Nov.19 Butterworth and Chebyshev Filter, FIR Filter Design. W10-Nov26 Discrete-time Fourier Series, DFT, Properties of DFT. W11-Dec03 Convolution with DFT (Overlap-add/save). FFT. W12-Dec10 Fast Fourier Transform(FFT). W13-Dec17 EXAM. W14-Dec24 2-D Signal Processing. Textbook: Discrete-time Signal Processing, Oppenheim and Schafer, Prentice-Hall. Grading: 2Term Exams (50%), Final Exam (50%).
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1Digital Signal Processing A.S.Kayhan
DIGITAL SIGNAL
PROCESSING
Textbook: Discrete-Time Signal Processing, Oppenheim and Schafer, Prentice-Hall.
Digital Signal Processing A.S.Kayhan
Course Outline
Review of Discrete-time signals, systems, Fourier tr.
Review of Z-transform
Sampling, Decimation, Interpolation
Time and frequency response of systems
Flow Graph Realization
Filter Design
Discrete-time Fourier Series, Discrete Fourier Transform
Fast Fourier Transform
2-D Signal Processing
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SIGNALS:A signal is a function of time representing a physical variable, e.g. voltage, current, spring displacement, share market prices, number of students asleep in the class, cash in the bank account.
Continuous time signals : x(t), x(t1,t2,...)Speech signals, Image signals,Video signals, Seismic signals, Biomedical signals (ECG,EEG,...)
Discrete-time signals: x[n], x[n1,n2,...]Inherently discrete-time (Stock market)Discretized continuous-time(Most signals)
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SIGNALS:Deterministic or Random
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Speech data (8kHz):
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Discrete-time signal example:Stock Market daily closing values
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Annual max. temperature and precipitation for Ankara (data: MGM)
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Graphical Representation of Signals
Continuous-timeSignal
Discrete-timeSignal
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Reflection: x[-n]
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Time-scaling: x(at)
Time-scaling in discrete-time is not straightforward.Discussed in Decimation and Interpolation.
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Time-shift: x[n-no]
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Basic Discrete-time Signals:Unit Step Function :
0,1
0,0][
n
nnu
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Unit Impulse Function:
0,1
0,0][
n
nn
n
k
knu
nunun
][][
]1[][][
][]0[][][ nxnnx
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Real Exponential: nn aCeCnx ][
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Complex Exponential: nj oeAnx ][Sinusoidal Signal: )cos(][ nAnx o
}{][ )( nj oAenx Re
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Periodicity Properties:
njnj oo ee )2(
Therefore, consider only 20 o
njNnj oo ee )(If periodic, then
N
m
e Nj o
2
1
0
m and N integer.
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If periodic with fundamental period N, then fundamental frequency is N/2
If there are a few exponentials in the form:
nN
jk
k enx2
][
they are harmonically relatedThere are only N such distinct signals.
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SYSTEM:Any process that transforms signals
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Interconnection of Systems:Series(Cascade), Parallel, Series/Parallel.
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Properties:Memoryless if output depends only on input at the same time.
Example: Discrete-time systems with memory:
n
k
kxny ][][
]1[][ nxny
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Causality: Causal if output depends only on inputs at the present time and in the past.
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M
Mk
knxM
ny ][12
1][
]1[][][ nxnxny
Example: Causal system:
Example: Noncausal systems:
][]1[][ nxnxny
t
dxC
ty )(1
)(
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Stability: Stable if small inputs do not lead to diverging outputs.
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]1[][][ nxnxnyExample: Stable system:
Example: Unstable systems:
].[][ if ],[)1(][][ nunxnunkxnyn
k
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Time Invariance: Time-invariant if a time shift in input causes same time shift in output signal.
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Example:Time-varying: ][][ nnxny
Consider two signals
][][],[ 121 onnxnxnx
][][ 11 nnxny
][][
][][
12
22
onnnxny
nnxny
shift ][1 ny
][][)(][ 211 nynnxnnnny ooo
Digital Signal Processing A.S.Kayhan
Linearity: Linear if posses superposition property:Additivity and scaling (homogeneity):
1. Response to is 2. Response to is
][][ 21 nbxnax
][1 nax ][1 nay
Combining these two, we get:
][][ 21 nxnx
][][ 21 nbynay
][][ 21 nyny
Example: Linear: ][][ nnxny
Example: Nonlinear: 2])[(][ nxny
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Linear Time-Invariant Systems:LTI systems can be analyzed in great detail.Many physical processess can be modeled by LTI systems.Unit impulse function will be used as building block and response of LTI systems to a unit impulse will be used to characterize such systems.
Digital Signal Processing A.S.Kayhan
Input/output relationship for a LTI system is given as:
k
knhkxny ][][][
Convolution of x[n] and h[n]. h[n] is the impulse response.
][*][][ nhnxny
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Example:
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Properties:Commutative:
][*][][*][ nxnhnhnx
r
knr
k
rhrnxknhkx ][][][][
Use of this property:If it is easier; reflect and shift x[k] to obtain x[n-k] first, then multiply with h[k] to find convolution result at time n.
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Associative:
][*])[*][(])[*][(*][ 2121 nhnhnxnhnhnx
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Distributive:
][*][][*][])[][(*][ 2121 nhnxnhnxnhnhnx
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Memory: System is memoryless if output depends only on input at same time.Then, system is memoryless if:
][][ nKnh then ].[][ nKxny
Similarly for continuous-time, memoryless if
).()( tKxty
)()( tKth hence
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Causal: Consider convolution sum :
k
knxkhny ][][][
For a LTI system to be casual y[n] must not depend on input x[k] for k > n.If system is causal :
.0for ,0][ nnh
Then,
n
kk
knhkxknxkhny ][][][][][0
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Stability: Stable if bounded inputs do not lead to diverging outputs. Assume
. allfor ,][ nBnx
Then,
k
knxkhny ][][][
k
knxkhny ][][][
with, Bknx ][
k
khBny ][][
System is stable iff,
k
kh ][
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Example: A system with impulse response h[n]=[n-no] shifts the input
n n
onnnh 1][][
Example: Consider the fallowing accumulator
n
k
kxny ][][
Unstable][][0
n n
nunu
this system has unit step as the impulse response
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Discrete-Time Fourier Transform (DTFT):For a discrete-time signal x[n], DTFT is defined as
n
njj enxeX ][)(
the inverse DTFT (synthesis equation) is defined as
2
)(2
1][ deeXnx njj
Differences between continuous-time FT and DTFT:X(ej) is periodic with 2. Integral in synthesis eq. is over 2 interval (0 to 2 or - to ).
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Example:Consider 1],[][ anuanx n
n
njnj nueaeX ][)(
jn
njj
aeaeeX
1
1)()(
0
)cos(21
1)(
2
2
aaeX j
)cos(1
)sin(tan)( 1
a
aeX j
)sin())cos(1(
1)(
jaaeX j
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Common DTFT Pairs:
)2(2)( allfor ,1][ leXnnx j
1)(][][ jeXnnx
)cos(][ nnx o
)]2()2([)( lleX ol
oj
nj oenx ][
l
oj leX )2(2)(
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Convergence: Analysis equation will converge if x[n] is absolutely summable or it has finite energy
n
nx ][
n
nx2
][
Example:Consider
Nn
Nnnx
,0
,1][
)2/sin(
))2/1(sin(
1
1)(
)12(
N
e
eeeeX
j
NjNj
N
Nn
njj
Digital Signal Processing A.S.Kayhan
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Properties of DTFT:Linearity: DTFT is linear operator:
)()(][][
)(][
)(][
jj
j
j
eYbeXanbynax
eYny
eXnx
Time and frequency shifting:
)(][
)(][
)(][
)( oo
o
jnj
jnjo
j
eXnxe
eXennx
eXnx
Digital Signal Processing A.S.Kayhan
Symmetry: If x[n] is real, then
)()( * jj eXeX
Real part is even, imaginary part is odd function of .Similarly, magnitude is even, phase is odd funtion.
Time and frequency scaling: Time scaling is not straightforward for discrete-time signals. Consider special case; x[an], a = -1
)(][ jeXnx
)()(
k of multiplenot isn if 0,
k of multiple isn if],/[][
1
1
jkj eXeX
knxnx
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Parsevals relation: Energy is equal on both time and frequency domains:
2
22
)(2
1][ deXnx j
n
Since the right hand term is the total energy in frequency domain, is called energy spectral density function (or spectrum).
2)( jeX
Convolution: Convolution in time corresponds to multiplication in frequency domain.
)()()(][*][][ jjj eHeXeYnhnxny
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nj
n
j enheH
][)(
where, is the frequency response defined as)( jeH
Example:Consider a LTI sytem withThen the frequency response is
][][ onnnh
onjj eeH )(
For any input x[n] with Fourier transform Fourier transform of output
)( jeX
)()( jnjj eXeeY o
The output is equal to the input with a time shift
].[][ onnxny
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Digital Signal Processing A.S.Kayhan
Example:Consider a LTI sytem with |a|,|b|
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Z- Transform :For a discrete-time signal x[n], z-transform is defined as
n
nznxzX ][)(
z is a general complex variable shown in polar form as jerz
We get Fourier transform as a special case when r=1
Digital Signal Processing A.S.Kayhan
Convergence: Z-transform exists if
n
nrnx |][|
Poles-Zeros: consider X(z) as a rational function
)(
)()(
zQ
zPzX
where P(z) and Q(z) are polynomials in z. Values of z making X(z)=0 are called zeros of X(z) (roots of P(z)).Values of z making X(z)= are called poles of X(z) (roots of Q(z)).
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||||,1
1)( 1 azaz
z
azzX
n
n
az || 1For convergence
0
1 )(][)(n
n
n
nn azznuazX
][][ nuanx nConsiderExample:
Digital Signal Processing A.S.Kayhan
||||,1
11)( 1 azaz
z
zazX
n
n
za || 1For convergence
0
1 )(1)(n
nzazX
]1[][ nuanx nConsiderExample:
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||||||,)( bzabz
z
az
zzX
Then
abnubnuanx nn ],1[][][
ConsiderExample:
||||,][ azaz
znua n
||||,]1[ bzbz
znub n
Digital Signal Processing A.S.Kayhan
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Properties of convergence:ROC is a ring or disk centered at origin.DTFourier transform exists if ROC includes Unit CircleROC can not contain any polesIf x[n] has finite duration ROC is entire plane.If x[n] is right-sided, ROC is outward from outermost poleIf x[n] is left-sided, ROC is inward from innermost poleIf x[n] is two-sided, ROC is a ring not containing any pole.
Digital Signal Processing A.S.Kayhan
Example:
],1[)2(][)4.0(][ nununx nn
|2||||4.0|,24.0
)(
zz
z
z
zzX
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Digital Signal Processing A.S.Kayhan
Transfer Function: Z-transform of impulse response h[n]
n
nznhzH ][)(
If impulse response h[n] is causal then ROC will be outward.If the system is causal and stable, all the poles will be inside the unit circle and ROC will include the unit circle.
Digital Signal Processing A.S.Kayhan
Inverse Z-Transform:Power Series Expansion: Consider
12
2
11
2
1)( zzzzX
remember
1012 ]1[]0[]1[]2[
][)(
zxzxzxzx
znxzXn
n
therefore]1[
2
1][]1[
2
1]2[][ nnnnnx
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Digital Signal Processing A.S.Kayhan
Example (Long Division): |z| > |a|
33221
11
1
1)( zazaaz
azzX
therefore ][][ nuanx n
Example (Long Division): |z| < |a|
33221)( zazazaza
zzX
therefore ]1[][ nuanx n
Digital Signal Processing A.S.Kayhan
Partial fraction expansion: Consider
kdzkk
N
k k
k
k
N
k
k
M
k
o
oN
k
kk
M
k
kk
zXzdAzd
A
zd
zc
a
b
za
zbzX
|)()1(,1
)1(
)1()(
1
11
1
1
1
1
0
0
then
N
k
nk nudAnx k
1
][][
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Example : Consider
.2
1||,
)21
1)(41
1(
1)(
11
zzz
zX
)21
1()41
1()(
1
2
1
1
z
A
z
AzX
2|)()2
11(
1|)()4
11(
2/11
2
4/11
1
z
z
zXzA
zXzA
][4
1][
2
12][ nununx
nn
Digital Signal Processing A.S.Kayhan
Properties of Z-Transform:Linearity:
)()(][][
)(][
)(][
zYbzXanbynax
zYny
zXnx
Time shift:
x
n
RROC
zXzzYnnxny
zXnxo
)()(][][
)(][
0
yx RRROC
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Digital Signal Processing A.S.Kayhan
Multiplication by Exponential:
xo
ono
RzROC
zzXnxz
zXnx
||
)/(][
)(][
Conjugation:
xRROC
zXnx
)(][ ***
Digital Signal Processing A.S.Kayhan
Time reversal:
xRROC
zXnx
/1
)/1(][ ***
xRROC
zXnx
/1
)/1(][
Convolution:
yx RRROC
zYzXnxny
)()(][*][
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Digital Signal Processing A.S.Kayhan
Example: ][][ nuanx n ][][ nunh
||||,)( azaz
zzX
|1|||,
1)(
z
z
zzH
][*][][ nhnxny
11
2
11
1
1
1)(
|1|||,)1)((
)()()(
az
a
zazY
zzaz
zzHzXzY
y[n] is obtained as:
])[][(1
1][ 1 nuanu
any n
Digital Signal Processing A.S.Kayhan
Sampling: Consider a signal x[n] which is obtained by taking samples of a continuous time signal xc(t):
)(][ nTxnx cwhere T is the sampling period, its reciprocal, is the sampling frequency.
Tf s1
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where has the quantized samples and xB[n] has the coded samples (such as 2s complement).
])[(][ nxQnx
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Frequency Domain Representation of Sampling: Consider the sampled signal xs(t) obtained from xc(t) by multiplying with periodic impulse train:
n
nTtts )()(
then
n
ccs nTttxtstxtx )()()()()(
The discrete time signal is x[n]= xc(nT). The Fourier transform of s(t) is
TkTS s
ks
2,)(
2)(
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Then, Fourier transform of xs(t) is
k
scs kXTSXX )(
1)(*)(
2
1)(
Ns 2
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Ns 2
We observe that when the sampling frequency is not chosen high enough, copies of spectrum overlap; this is called aliasing (distortion). The minimum sampling frequency (rate) to avoid aliasing is 2N (Nyquist rate) :
Ns T 2
2
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If sampling rate is greater than Nyquist rate, then the original signal xc(t) may be obtained without distortion using a low-pass filter.
Digital Signal Processing A.S.Kayhan
Fourier Tr. of xs(t) may be written as
n
njj enxeX ][)(
n
Tnjcs enTxX )()(
On the other hand DTFT of x[n] is
Comparing these equations we observe that )(|)()( TjT
js eXeXX
k
cj
T
k
TX
TeX )
2(
1)(
n
cs nTttxtx )()()(
k
scTj kX
TeX )(
1)(
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Example:Consider
ttx oc cos)(
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)cos()( ttx or
))cos(()( ttx osr
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Example: Consider
4000,cos)( ooc ttxwith sampling period T=1/6000, we obtain
nnTnx3
2cos4000cos][
Nyquist sampling condition is satisfied , there is no aliasing.
80002120002
os T
Digital Signal Processing A.S.Kayhan
Example: Consider
16000,cos)( ooc ttxwith sampling period T=1/6000, we obtain
nnn
nnTnx
3
2cos6000/40002cos
6000/16000cos16000cos][
160002120002
os T
Nyquist sampling condition is not satisfied , there is aliasing.
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Reconstruction from Samples:Use ideal LPF to recover original signal
Digital Signal Processing A.S.Kayhan
With
n
r nTthnxtx )(][)(
Ttth sinc)(
n
r T
nTtnxtx sinc][)(
n
ncs
nTtnx
nTtnTxtx
)(][
)()()(
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Discrete-time Processing of Cont.-time Signals:
)()(
)(][
THeH
nThTnh
cj
c
with h(t) is impulse response of continuous-time system
][ nh
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Changing the Sampling Rate:Downsampling and Decimation:Sampling rate can be reduced by further sampling in discrete-time:
)(][][ nMTxnMxnx cd
Let MTT '
Then )'(][ nTxnx cd
Digital Signal Processing A.S.Kayhan
If NcX for ,0)( NMTT 2
2
'
2 and
orNM
22
(N is BW of x[n])
Then xc(t) can be recovered without distortion.
Remember that
k
cj
T
k
TX
TeX )
2(
1)(
Similarly
k
cj
d T
k
TX
TeX )
'
2
'(
'
1)(
k
cj
d MT
k
MTX
MTeX )
2(
1)(
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Digital Signal Processing A.S.Kayhan
Let
otherwise,0
,2,,0],[][1
MMnnxnx
neM
nxnxM
l
Mnlj ,1
][][1
0
/21
( [ ]=Discrete Fourier Series rep. of impulse train with period M )
][][][ 1 MnxMnxnx d
nj
n
nj
nd
jd eMnxenxeX
][][)( 1
MknMnk /
Mkj
k
jd ekxeX
/1 ][)(
Digital Signal Processing A.S.Kayhan
1
0
//2
/1
0
/2
][1
1][)(
M
l k
MkjMlkj
Mkj
k
M
l
Mlkjjd
eekxM
eeM
nxeX
1
0
)2
()(
1)(
M
l
M
l
Mjj
d eXMeX
Combining the exponential terms
First obtain then shift by increments to get)( Mj
eX
2
)( )/2'( MljeX
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IfNM
22
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A general system for downsampling called decimator is
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Upsampling and Interpolation:Sampling rate can be increased by obtaining intermediate values:
2,/'),'(][ LLTTnTxnx cigiven ).(][ nTxnx c
,2,,0),/(]/[][ LLnLnTxLnxnx ci Consider the following system
Digital Signal Processing A.S.Kayhan
The block with inserts (L-1) zeros between samples:L
otherwise,0
2,,0,/][
LLnLnxnx e
or
.][ kLnkxnxk
e
In frequency domain:
n
nj
k
je ekLnkxeX
)(
)()( Ljk
kLjje eXekxeX
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L=2
Digital Signal Processing A.S.Kayhan
k
i LkLn
LkLnkxnx
/)(
/)(sin][][
Ln
Lnnh
/
/sin
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Changing the sampling rate by noninteger factor:Following system may be used
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End of Part 1
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1Digital Signal Processing A.S.Kayhan
DIGITAL SIGNAL
PROCESSING
Part 2
Digital Signal Processing A.S.Kayhan
Frequency Analysis of LTI Systems:the Input/Output (I/O) relation of a LTI system is given by convolution:
k
knxkhnhnxny ][][][*][][
In z-domain)()()( zHzXzY
jez
jjj eHeXeY
)()()(
On the unit circle
where is the frequency response of the system.
)( jeH
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2Digital Signal Processing A.S.Kayhan
)()()( jjj eHeXeY
Magnitude is
where is the magnitude response of the system.
)( jeH
Phase is )()()( jjj eHeXeY
where is the phase response of the system.
)( jeH
Digital Signal Processing A.S.Kayhan
Discrete-time ideal filters: LPF, HPF, BPF:
These are not realizable. Why?
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Phase distortion and delay:Assume that ][][ did nnnh
thendnjj
id eeH )(
ord
jid
jid neHeH
)(,1)(
Linear phase distortion causes simple delay.
Group delay: It measures linearity of the phase response.Consider nnsnx ocos][][ where is a narrowband (slowly varying) signal.][ ns
Digital Signal Processing A.S.Kayhan
Assume around o
dojj neHeH )(,1)(
Then doood nnnnsny cos][][
Thus, group delay of a system is
)(arg)]([
jj eHd
deHgrd
where is the continuous phase. )(arg jeH
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Example: Consider the following filter
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Input is the following signal
Digital Signal Processing A.S.Kayhan
Pulses are at 85.0,5.0,25.0
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Systems with Difference Equation Models:Assume that the Input/Output (I/O) relation of a system is given by a constant coefficient difference equation:
M
kk
N
kk knxbknya
00
][][
Applying z-transform, using linearity and shifting properties, we obtain
M
k
kk
N
k
kk zXzbzYza
00
)()(
)()(00
zXzbzYzaM
k
kk
N
k
kk
Digital Signal Processing A.S.Kayhan
Then, the transfer function (or system function) is
N
k
kk
M
k
kk
za
zb
zX
zYzH
0
0
)(
)()(
We can write the transfer function in terms of its poles, dk, and zeros, ck, as
N
kk
M
kk
o
o
zd
zc
a
bzH
1
1
1
1
)1(
)1()(
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7Digital Signal Processing A.S.Kayhan
Example: Consider system function
)4
31)(
2
11(
)1()(
11
21
zz
zzH
)(
)(
83
41
1
21)(
21
21
zX
zY
zz
zzzH
then
)()8
3
4
11()()21( 2121 zYzzzXzz
]2[8
3]1[
4
1][
]2[]1[2][
nynyny
nxnxnx
Digital Signal Processing A.S.Kayhan
Stability and Causality: A system is causal if ROC for the transfer function H(z) is outward.A sytem is stable if ROC for the transfer function H(z) includes the unit circle.
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8Digital Signal Processing A.S.Kayhan
Inverse Systems :Let be the inverse system of , then )( zH)( zH i
1)()()( zHzHzG i
then
)(
1)(,
)(
1)(
j
jii eH
eHzH
zH
Not all systems have an inverse. Ideal LPF does not have an inverse, we can not recover high frequency components.Now, consider
N
kk
M
kk
o
o
zd
zc
a
bzH
1
1
1
1
)1(
)1()(
nnhnhng i *Eq.(1)
Digital Signal Processing A.S.Kayhan
Then the inverse system has
M
kk
N
kk
o
oi
zc
zd
b
azH
1
1
1
1
)1(
)1()(
Poles become zeros of the inverse system, zeros become poles. For Eq.(1) to hold, ROC of and must overlap. If is causal, ROC is
)( zH i )( zH)( zH
kdz max
Some part of ROC of must overlap with this.)( zH i
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9Digital Signal Processing A.S.Kayhan
Example: Suppose
9.0,9.01
5.0)(
1
1
zz
zzH
The inverse system is
1
1
1
1
21
8.12
5.0
9.01)(
z
z
z
zzH i
Two possible ROC:
2z nununh nni 11 28.1121which is stable but noncausal.
2z 128.12 112
nununh nni
unstable but causal.
Digital Signal Processing A.S.Kayhan
Example: Suppose
9.0,9.01
5.01)(
1
1
zz
zzH
1
1
5.01
9.01)(
z
zzH i
The inverse system is
The only choice for ROC is overlaps with is 9.0z
5.0z
then
15.09.05.0 1 nununh nni
which is both stable and causal.
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10
Digital Signal Processing A.S.Kayhan
Impulse Response for Rational System:Assume that a stable LTI system has a rational transfer function.
N
kk
M
kk
o
o
zd
zc
a
bzH
1
1
1
1
)1(
)1()(
N
k k
kNM
NMifr
rr zd
AzBzH
11
00 1
)(
then
N
k
nkk
NM
rr nudArnBnh
10
1
Impulse response is of infinite length, called Infinite Impulse Response (IIR) system.
Digital Signal Processing A.S.Kayhan
If the system has no poles, then
M
k
kk zbzH
0
)(
M
kk knbnh
1
Impulse response is of finite length, called Finite Impulse Response (FIR) system.
Example: Consider a causal system with][]1[][ nxnayny
If then stable and the impulse response is
1a
.nuanh n
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11
Digital Signal Processing A.S.Kayhan
Example: Consider .
otherwise,0
0,
Mna
nhn
Then
1
11
0 1
1)(
az
zazazH
MMM
n
nn
Zeros at
.,,1,0,12
Mkaez Mk
j
k
Difference equation is
However, we can also write
1][]1[][ 1 Mnxanxnayny M
M
k
k knxany1
Digital Signal Processing A.S.Kayhan
Frequency Response for Rational System Functions:Assume that a stable LTI system has a rational transfer function. Then frequency response is obtained by evaluating it on the unit circle:
N
k
kjk
M
k
kjk
j
ea
eb
eH
0
0)(
N
k
jk
M
k
jk
o
oj
ed
ec
a
beH
1
1
)1(
)1()(
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12
Digital Signal Processing A.S.Kayhan
The magnitude response of the system is
N
k
jk
M
k
jk
o
oj
ed
ec
a
beH
1
1
1
1)(
The magnitude-squared response of the system is
)()()( *2 jjj eHeHeH
N
k
jk
jk
M
k
jk
jk
o
oj
eded
ecec
a
beH
1
*
1
*2
2
11
11)(
Digital Signal Processing A.S.Kayhan
Log magnitude or Gain in decibels(dB) :
N
k
jk
M
k
jk
o
oj
ed
eca
beH
110
1101010
1log20
1log20log20)(log20
Attenuation in dB = - Gain in dB
Note that :
)(log20
)(log20)(log20
10
1010
j
jj
eX
eHeY
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13
Digital Signal Processing A.S.Kayhan
Phase response for rational system function:
N
k
jk
M
k
jk
o
oj
ed
eca
beH
1
1
1
1)(
Group delay for rational system function:
N
k
jk
M
k
jk
j
edd
d
ecd
deH
1
1
]1arg[
]1arg[)(grd
Digital Signal Processing A.S.Kayhan
Frequency Response of A single Zero:Consider transfer function of a system as
11)( azzH
then with jrea jjeHjjj ereeeHeH
j 1)()( )(
then magnitude is
and magnitude in dB is
)cos(21)( 22
rreH j
)]cos(21[Log20)(Log20 21010 rreH j
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14
Digital Signal Processing A.S.Kayhan
then phase is
)cos(1
)sin(tan)( 1
r
reH j
group delay is derivative of the (unwrapped) phase function
d
eHdeHgrd
jj ))(()]([
Example: Consider two cases : r=0.9, =0 and r=0.9, =:
Digital Signal Processing A.S.Kayhan
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15
Digital Signal Processing A.S.Kayhan
magnitudes of vectors give the magnitude response
31
3)( vv
v
e
reeeH
j
jjj
jezz
azzH
)(
Digital Signal Processing A.S.Kayhan
phases of vectors give the phase response
31313
)(
vv
e
reeeH
j
jjj
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16
Digital Signal Processing A.S.Kayhan
Example: Consider a second order system with
11
1
11
1)(
zrezre
zezH
jj
j
21
3)(vv
veH j
Digital Signal Processing A.S.Kayhan
jez
j zHeH
)()(
:
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17
Digital Signal Processing A.S.Kayhan
Example: Consider a third order system with
211
211
7957.04461.11683.01
0166.11105634.0)(
zzz
zzzzH
Digital Signal Processing A.S.Kayhan
Relation Between Magnitude and Phase:In general, knowledge about magnitude does not provide information about phase, and vice versa.But, for rational system functions, with some additional information such as number of poles and zeros, magnitude and phase responses provide information about each other.Consider
jez
jjj
zHzH
eHeHeH
|)/1()(
)()()(
**
*2
with
N
kk
M
kk
o
o
zd
zc
a
bzH
1
1
1
1
)1(
)1()(
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18
Digital Signal Processing A.S.Kayhan
N
kk
M
kk
o
o
zd
zc
a
bzH
1
*
1
*
**
)1(
)1()/1(
Then
N
kkk
M
kkk
o
o
zdzd
zczc
a
bzHzHzC
1
*1
1
*12
**
)1)(1(
)1)(1()/1()()(
Notice that poles and zeros of C(z) occur in conjugate reciprocal pairs. If one pole/zero is inside the unit circle there is another outside.
Digital Signal Processing A.S.Kayhan
If H(z) is causal and stable, then all poles must be inside the unit circle, with this we can identify the poles. But zeros of H(z) can not be uniquely identified from zeros of C(z) with this constraint alone.
Example: Consider two stable systems with
14/14/
11
1 8.018.01
5.0112)(
zeze
zzzH
jj
and
14/14/
11
2 8.018.01
211)(
zeze
zzzH
jj
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19
Digital Signal Processing A.S.Kayhan
Pole/zero plots are
Digital Signal Processing A.S.Kayhan
zezezeze
zzzz
zHzHzC
jjjj 4/4/14/14/
11
**111
8.018.018.018.01
5.01125.0112
)/1()()(
zezezeze
zzzz
zHzHzC
jjjj 4/4/14/14/
11
**222
8.018.018.018.01
211211
)/1()()(
Observe that (with )
zzzz 21215.015.014 11
then
)()( 21 zCzC
1224 zz
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20
Digital Signal Processing A.S.Kayhan
Example: Consider pole/zero plot of C(z) for a system, determine H(z).
For a causal and stable system, poles of H(z) are: p1, p2,p3.For real ak, bk, zeros/poles occur in complex conjugate pairs.
Digital Signal Processing A.S.Kayhan
All-Pass Systems:Consider following stable system function
1
*1
1)(
az
azzH ap
j
jj
j
jj
ap ae
aee
ae
aeeH
1
)1(
1)(
**
then constant )(but ,1)( jap
jap eHeH
This is called an all-pass system .A general all-pass system has the following form
cr M
k kk
kkM
k k
kap zeze
ezez
zd
dzAzH
11*1
1*1
11
1
)1)(1(
))((
1)(
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21
Digital Signal Processing A.S.Kayhan
Example: Consider pole/zero plot of a typical all-pass system
Digital Signal Processing A.S.Kayhan
Example: First order all-pass system with a real pole: z=0.9 (z=-0.9)
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22
Digital Signal Processing A.S.Kayhan
Example: Second order all-pass system with poles:4/9.0 jez
Digital Signal Processing A.S.Kayhan
Notice that group delay for causal all-pass systems are positive (unwrapped/continuous phase is nonpositive).All-pass systems may be used as phase compensators.They are also useful in transforming lowpass filters into other frequency-selective forms.
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23
Digital Signal Processing A.S.Kayhan
Block-Diagram Representation:LTI systems with difference equation represetation (rational system function) may be imlemented by converting to an algorithm or structure that can be realized in desired technology. These structures consists of basic operations of addition, multiplication by a constant and delay.In block diagram representation:
Digital Signal Processing A.S.Kayhan
Example: Consider the second order system :][]2[]1[][ 21 nxbnyanyany o
with
22
111
)(
zaza
bzH o
Block diagram representation is
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24
Digital Signal Processing A.S.Kayhan
For a system with higher order difference equation
M
kk
N
kk knxbknyany
01
][][][
with system function
N
k
kk
M
k
kk
za
zbzH
1
0
1)(
Rewriting the equation as
][][
][][][
1
01
nvknya
knxbknyany
N
kk
M
kk
N
kk
where
M
kk knxbnv
0
][][
Digital Signal Processing A.S.Kayhan
N+MDelayelement
Direct Form I:
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25
Digital Signal Processing A.S.Kayhan
Previous diagram is implementation of
M
k
kkN
k
kk
zbza
zHzHzH0
1
21
1
1)()()(
or
)()()()(0
1 zXzbzXzHzVM
k
kk
)(1
1)()()(
1
2 zVza
zVzHzY N
k
kk
Digital Signal Processing A.S.Kayhan
We can rearrange the system function
)(1
1)()()(
1
2 zXza
zXzHzW N
k
kk
)()()()(0
1 zWzbzWzHzYM
k
kk
In the time domain
M
kk knwbny
0
][][
][][][1
nxknwanwN
kk
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26
Digital Signal Processing A.S.Kayhan
M = N
Digital Signal Processing A.S.Kayhan
Direct Form II:
Max(N,M)Delay elements
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27
Digital Signal Processing A.S.Kayhan
Example:Consider the system with
21
1
9.05.11
21)(
zz
zzH
Digital Signal Processing A.S.Kayhan
Flow Graph Representation:Similar to the block diagram representation:
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28
Digital Signal Processing A.S.Kayhan
Example:Consider the system with
)()()( 41 zXzWzW )()( 12 zWzW )()()( 23 zXzWzW
)()( 31
4 zWzzW
)()()( 42 zWzWzY
Digital Signal Processing A.S.Kayhan
1
1
1)(
z
zzH
)(1
)(1
1
zXz
zzY
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29
Digital Signal Processing A.S.Kayhan
Structures for IIR Systems:Some considerations:Computaional complexity(no. of multiplication, delay )Finite precision, Ease of implementation, etc.
Direct Forms:We have already seen direct forms.
Cascade Form:We factor numerator and denominator polynomials of H(z)
21
21
1
1*1
1
1
1
1*1
1
1
)1)(1()1(
)1)(1()1()( N
kkk
N
kk
M
kkk
M
kk
zdzdzc
zgzgzfAzH
Digital Signal Processing A.S.Kayhan
We have cascade of first or second order subsystems
s
ss
N
k kk
kk
o
N
k kk
kkokN
kk
zaza
zbzbb
zaza
zbzbbzHzH
12
21
1
22
~1
1
~
12
21
1
22
11
1
1
1
1)()(
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30
Digital Signal Processing A.S.Kayhan
Example:Consider the system with
11
11
21
21
25.015.01
11
125.075.01
21)(
zz
zz
zz
zzzH
Digital Signal Processing A.S.Kayhan
Parallel Form:H(z) may be written as sum of subsystems
crf N
k kk
kkN
kk
k
kN
kk zdzd
zeB
zc
AzCzH
01*1
1
00
1
)1)(1(
)1(
1)(
N
k kk
kokN
kk zaza
zeezCzH
f
02
21
1
11
0
1
1)(
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31
Digital Signal Processing A.S.Kayhan
Digital Signal Processing A.S.Kayhan
Example:Consider the system with
112121
25.01
25
5.01
188
125.075.01
21)(
zzzz
zzzH
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32
Digital Signal Processing A.S.Kayhan
Transposed Forms :Reverse the directions of all branches while keeping the values as they were and reverse roles of input and output. Transposed forms may be useful in finite precision implementation.
Example:Consider the second order system with
Digital Signal Processing A.S.Kayhan
Structures for FIR Systems:Consider an FIR system with following input-output relation
][][0
knxbnyM
kk
Observe that the impulse response of this system is
otherwise,0
0,][
Mnbnh n
Direct form (or tapped delay line or transversal filter) :
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33
Digital Signal Processing A.S.Kayhan
Transposed direct form structure:
Digital Signal Processing A.S.Kayhan
ss N
kkkok
N
kk zbzbbzHzH
1
22
11
1
)()()(
Cascade form:
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34
Digital Signal Processing A.S.Kayhan
Linear-Phase FIR Systems:Finite impulse response (FIR) systems with linear-phase have symmetry properties such as
MnnMhnh 0],[][
MnnMhnh 0],[][or
For M even and odd. Thus there are four types of linear phase FIR filters.
Digital Signal Processing A.S.Kayhan
)2/sin(
)2/5sin()( 2
jj eeH
otherwise,0
40,1][
nnh
Example: Consider
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35
Digital Signal Processing A.S.Kayhan
Example: Consider
]2[]1[2][][ nnnnh
)].cos(1[2
2
21)(
1
111
21
j
jjj
jjj
e
eee
eeeH
Digital Signal Processing A.S.Kayhan
Assume M is even
M
Mk
M
k
M
k
knxkh
MnxMhknxkh
knxkhny
12/
12/
0
0
][][
]2/[]2/[][][
][][][
With, in the last term, k=M-l, l=0M/2-1
12/
0
12/
0
][][
]2/[]2/[][][][
M
l
M
k
lMnxlMh
MnxMhknxkhny
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36
Digital Signal Processing A.S.Kayhan
][][ nMhnh If
]2/[]2/[
])[][]([][12/
0
MnxMh
kMnxknxkhnyM
k
If ][][ nMhnh
])[][]([][12/
0
kMnxknxkhnyM
k
Digital Signal Processing A.S.Kayhan
Finite-Precision Numerical Effects:Input Quantization:We have seen earlier that continuous-time signals are sampled, quantized and coded first
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37
Digital Signal Processing A.S.Kayhan
Example:
Digital Signal Processing A.S.Kayhan
In twos complement binary system leftmost bit is the sign bit. If we have (B+1)-bit twos complement fraction of the following form
Baaaa 210 Value is B
Baaaa 2222 22
11
00
Example: Binary Code Numeric value
110 4/3
101 2/1
011 4/1
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38
Digital Signal Processing A.S.Kayhan
Quantizer step size is12
2
B
mX
With][][
^^
nxXnx Bmwhere
)complement s(two'1][1^
nx B
Analysis of Quantization Errors:We observe that the quantized samples are in general different from the true values. The difference is the quantization error
].[][][^
nxnxne
and2/][2/ ne
Digital Signal Processing A.S.Kayhan
A simplified model of quantizer is
Assumptions about e[n]:e[n] is stationarye[n] is uncorrelated with x[n]e[n] is white noisee[n] is uniformly distributed
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39
Digital Signal Processing A.S.Kayhan
Example:
3bit quantizer
Error for 3bit
Error for 8bit
Sinusoidal signal
Digital Signal Processing A.S.Kayhan
Mean value of e[n] is zero
012/
2/
deee
Variance (or power) of e[n] is
12
1 22/
2/
22
deee
For (B+1)-bit quantizer with full-scale value Xm
12
2 222 mXB
e
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40
Digital Signal Processing A.S.Kayhan
Signal-to-Noise Ratio (SNR) is defined as the ratio of signal variance (power) to noise variance. Expressed in dB
x
m
m
xB
e
x
XB
X
10
2
22
102
2
10
log208.1002.6
212log10log10SNR
If
6dB-SNRSNR,2/
dB 6SNRSNR,1
xx
BB
Digital Signal Processing A.S.Kayhan
Coefficient Quantization in IIR Systems:Consider
N
k
kk
M
k
kk
za
zbzH
1
0
1)(
If the coefficients are quantized, we get
N
k
kk
M
k
kk
za
zbzH
1
^
0
^
^
1)(
wherekkkkkk aaabbb
^^
,
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41
Digital Signal Processing A.S.Kayhan
Each pole or zero will be affected by all the errors in the coefficient quantization. If the poles (or zeros) are close to each other (clustered), then quantization of coefficients may cause large shifts of poles (or zeros).Direct form structures are more sensitive to coefficient quantization than the other forms (parallel, cascade,...)
Digital Signal Processing A.S.Kayhan
Example: Consider a bandpass IIR elliptic filter of order 12 implemented in cascade form of 2nd order subsystems and direct form.
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42
Digital Signal Processing A.S.Kayhan
Passband cascade unquantized
Passband cascade 16-bit
Digital Signal Processing A.S.Kayhan
Passband parallel 16-bit
Direct form 16-bit
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43
Digital Signal Processing A.S.Kayhan
Poles of Quantized 2nd order subsystems :Because of robustness, parallel and cascade forms are used more than direct forms.We can further improve the robustness, by improving implementation of the 2nd order subsystems. Consider the following implementation in direct form:
Digital Signal Processing A.S.Kayhan
When coefficients are quantized, a finite number of pole possitions possible.
4bits 7bits
Circles correspond to r2, vertical lines to 2rcos
-
44
Digital Signal Processing A.S.Kayhan
Consider the following coupled form
Digital Signal Processing A.S.Kayhan
4bits 7bits
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45
Digital Signal Processing A.S.Kayhan
Coefficient Quantization in FIR Systems:Consider FIR system with transfer function
M
n
nznhzH0
][)(
If][][][
^
nhnhnh
)()(][)(0
^
zHzHznhzHM
n
n
then
where
M
n
nznhzH0
][)(and
)()()(^
jjj eHeHeH
Digital Signal Processing A.S.Kayhan
Example: Consider FIR filter of order 27
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46
Digital Signal Processing A.S.Kayhan
Digital Signal Processing A.S.Kayhan
Effects of Round-off Noise:Analysis of Direct Form IIR Structures:Consider Nth-order difference equation for Direct Form I
M
kk
N
kk knxbknyany
01
][][][
Assume that all signal values and coefficients are represented by (B+1)-bit fixed point binary numbers. Therefore, each multiplication is followed by a quantizer, then
M
kk
N
kk knxbQknyaQny
01
^^
][][][
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47
Digital Signal Processing A.S.Kayhan
Digital Signal Processing A.S.Kayhan
An alternative representation is as following
-
48
Digital Signal Processing A.S.Kayhan
Rounding or truncation of a product bx[n] is represented by a noise source of the form
][][][ nbxnbxQne
Assumptions about e[n]s:Each e[n] is stationaryEach e[n] is uncorrelated with x[n] and other e[n]sEach e[n] is uniformly distributed
For (B+1)-bit rounding
2/2][2/2 BB ne
Digital Signal Processing A.S.Kayhan
For (B+1)-bit truncation
0][2 neB
Mean and variance for rounding are
0e 122 22
B
e
Mean and variance for truncation are
2
2 Be
12
2 22B
e
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49
Digital Signal Processing A.S.Kayhan
Now, lets try to determine effect of quantization noise on the output of the system. We can redraw the system as
][][][][][][ 43210 nenenenenene
Digital Signal Processing A.S.Kayhan
Since all the noise sources are
12
255
22222222
043210
B
eeeeeee
For the general Direct Form I case
12
21
22
B
e NM
Now, we observe that
][][][1
neknfanfN
kk
For rounding mean of the output noise is zero.The variance for rounding or truncation is
22
2 ][12
21
n
ef
B
f nhNM
is impulse response for ][ nh ef )(/1)( zAzH ef
-
50
Digital Signal Processing A.S.Kayhan
Example: Consider the following system
.1,1
)(1
aazb
zH
][][ nuanh nef
Then the noise variance(power) at the output is
2222
2
1
1
12
22
12
22
aa
Bn
n
B
f
Digital Signal Processing A.S.Kayhan
Analysis of Direct Form FIR systems:Consider
M
k
knxkhny0
][][][
12
21,][][][
22
0
B
f
M
kk Mnenenf
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51
Digital Signal Processing A.S.Kayhan
End of Part 2
-
1Digital Signal Processing A.S.Kayhan
DIGITAL SIGNAL
PROCESSING
Part 3
Digital Signal Processing A.S.Kayhan
IIR Filters:Infinite impulse response (IIR) filters have rational transfer functions as
N
k
kk
M
k
kk
za
zbzH
0
0)(
Some IIR filter types are: Butterworth, Chebyshev, Elliptical. IIR filters have to be stable. IIR filters may not have linear-phase.
-
2Digital Signal Processing A.S.Kayhan
Filter Specifications (Low-pass):
)(log20 10
Design analog filter transform to Digital filter
Digital Signal Processing A.S.Kayhan
Analog Butterworth Filters:Approximate ideal LPF with magnitude-squared response
c
cKH,0
,)(
2
by the following rational function:
0,11
)(22
1
221
212
nn
n
nn b
bb
aaKH
|H()|2 must be even function of , and denominator degree must be higher than degree of numerator (lowpass).
-
3Digital Signal Processing A.S.Kayhan
.1,,2,1, niba ii
.1)( 22
nnb
KH
Then, we have
For maximal flatness at the origin, =0, The first 2n-1 derivatives of |H()|2 must be zero. This requires
Maximal flatness also at , =, requires
0,11
)(22
1
221
212
nn
n
nn b
bb
bbKH
.1,,2,1,0 niba ii
Digital Signal Processing A.S.Kayhan
.
1
1)( 2
2
n
o
H
Then, we have the Butterworth response
Let |H(=0)|2 = 1. Then
At the half-power frequency
n
onn
on
bb
H
2
2
2 1
1
1
2
1)(
.11
)(2
2
nnb
H
-
4Digital Signal Processing A.S.Kayhan
Consider the filter specifications:
.dB)(log20 10 H
.dB1log10
2
10
n
o
n
o
2
10/ 110 no 2/110/ 110
Using max and pn
o
p
2
10/ 110 max
Using min and sn
o
s
2
10/ 110 min
To find n:
Digital Signal Processing A.S.Kayhan
Dividing these equations
n
p
s
2
10/
10/
110
110max
min
Taking logarithm, we get the filter order as:
p
s
nlog2
110
110log 10/
10/
max
min
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5Digital Signal Processing A.S.Kayhan
then, set the denominator to zero
We can normalize the frequency so that o = 1. Then
To find the filter poles, we let =s/j,
nnn sjssHsH
22 )1(1
1
)/(1
1)()(
.11
)(2
2
nH
1)1(0)1(1 22 nnnn ss
If n is even, then
.12,...,1,0,
.12,...,1,0,1
)2
21(
)2(2
nkes
nkes
n
kj
k
kjn
Digital Signal Processing A.S.Kayhan
If n is odd, then
.12,...,1,0,
.12,...,1,0,1 22
nkes
nkes
n
kj
k
kjn
Note that all the poles lie on a circle with radius 1, because the frequency is normalized. (If we want to design analog filter we need to correct this). Also, choose the poles in the left-half plane to get a stable filter.
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6Digital Signal Processing A.S.Kayhan
Example: Design specs. for a Butterworth LP filter:At f = 2000 Hz, max= 3dBAt f = 3000 Hz, min= 10dBThen, the filter order n isn 2.7154 n = 3.Poles are at
.5,...,1,0,3 kesk
j
k
.
))23
21
())(23
21
()(1(
1
)1)(1(
1)(
2
jsjss
ssssH
Digital Signal Processing A.S.Kayhan
Analog Chebyshev Filters:We can generalize Butterworth response as:
.)(11
)(1
1)(
22
2
nn F
H
Fn() is a function of . Now, consider:
))(coscos()cos(
)(cos)cos(1
1
xnny
xx
n
This is the Chebyshev polynomial of order n. When
))(coshcosh()(,1 1 xnxCx n
When x 1 , Cn(x) 0.
-
7Digital Signal Processing A.S.Kayhan
Cn(x) has zeros in -1< x < 1 :
-1 0 1-1
0
1
x
C1(x)
-1 0 1-1
0
1
x
C2(x)
-1 0 1-1
0
1
x
C3(x)
-1 0 1-1
0
1
x
C4(x)
-1 0 1-1
0
1
x
C5(x)
.)(,1)(
),()(2)(
1
11
xxCxC
xCxCxxC
o
nnn
Digital Signal Processing A.S.Kayhan
Use Cn(.) in :
)(1
1)(
22
2
nCH
))(coscos()(,1
))(coshcosh()(,11
1
nC
nC
n
n
2 1 , is known as the ripple factor.
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8Digital Signal Processing A.S.Kayhan
Behaviour at =0:
even isn if,)1(
1)0(,1)(
odd isn if,1)0(,0)(
2
22
22
HC
HC
n
n
Behaviour at =1:
)1(
1)1(
n allfor ,1)(
2
2
2
H
C n
Digital Signal Processing A.S.Kayhan
110)(1log10 10max/22 nCThe attenuation is
When .dB01.31)(22 nC
This defines the half-power frequency (3dB cut-off) hp.Then,
).1()).1
(cosh1
cosh(
)1
(cosh1
)(cosh
))(coshcosh(1
)(
1
11
1
hphp
hp
hphpn
n
n
nC
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9Digital Signal Processing A.S.Kayhan
The specifications for a Chebyshev filter are max, min, s(p=1). Bandpass is between 0 1 rad/s.To find n:
)(110
)(1log102210/
22min
minsn
sn
C
C
With 110 10max/2
110
110))(coshcosh(
10/
10/1
max
min
sn
Finally,
)(cosh
110110cosh
1
10/
10/1
max
min
s
n
Digital Signal Processing A.S.Kayhan
To find the filter poles, we let =s/j,
)/(1
1)()(
22 jsCsHsH
n
then
1
)/(0)/(1 22 jjsCjsC nn
Letjsjvuw /)cos()cos(
wjs )/(cos 1
With
)sinh()sin()cosh()cos()/(
)cos())/(coscos()/( 1
nvnujnvnujsC
nwjsnjsC
n
n
)cosh(2
)cos(,2
)cos( xee
jxee
xxxjxjx
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10
Digital Signal Processing A.S.Kayhan
With
1
)/( jjsC n then
1
)sinh()sin(
0)cosh()cos(
nvnu
nvnu
cosh(nv) can never be zero, therefore cos(nu) must be zero. It is possible for
.12,,1,0),12(2
,2
5,
2
3,
2
nkkn
u
nnnu
k
k
Digital Signal Processing A.S.Kayhan
For these values of u, sin(nu) = 1, then
.)1
(sinh1 1 an
v k
Remember that
jvuwjs )/(cos 1then
jakn
jwjs kk 122cos)cos(
The poles are :12,,1,0, nkjs kkk
)cosh()2
12cos(
)sinh()2
12sin(
an
k
an
k
k
k
Choose left half poles.
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11
Digital Signal Processing A.S.Kayhan
Discrete Time Filter Design(IIR):There are 3 approches:1-Sampling (Impulse invariance)2-Bilinear transformation3-Optimal procedures
Impulse invariance method:We can sample the impulse response hc(t) of an analog filter with desired specifications as (Td is sampling interval):
.][ dcd nThTnh then
).2
()( kTT
HHdk d
c
Digital Signal Processing A.S.Kayhan
If dc TH /,0)( then
.),()(
d
c THH
Analog and digital frequencies have a linear relation: . dT
Consider the transfer function of a system expressed as
,)(1
N
k k
kc ss
AsH
Then, the impulse response is
.0,)(1
teAthN
k
tskc
k
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12
Digital Signal Processing A.S.Kayhan
The impulse response of the discrete time filter is
.)(
)(
1
1
nueATnh
nueATnThTnh
N
k
nTskd
N
k
nTskddcd
dk
dk
Transfer (or system) function of the discrete time filter is
.1
)(1
1
N
kTs
kd
ze
ATzH
dk
Poles are .dk Tsk ezss
If Hc(s) is stable (k < 0), then H(z) is also stable (|z| < 1).
Digital Signal Processing A.S.Kayhan
Example: Transfer function of analog filter is
.
)23
21
(
12
121
)23
21
(
12
121
1
1
)1)(1(
1)(
2
js
j
js
j
s
ssssH c
then
.
1
)12
121
(
1
)12
121
(
1)(
)2
3
2
1(1)2
3
2
1(1
1
jT
d
jT
d
Td
dd
d
ez
jT
ez
jT
ez
TzH
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13
Digital Signal Processing A.S.Kayhan
Bilinear Transformation:Transformation needed to convert an analog filter to a discrete time filter must have following properties:1- j axis of the s-plane must be mapped onto the unit circle of the z-plane,2- stable analog filters must be tranformed into stable discrete time filters (left hand plane of the s-plane must be mapped into inside the unit circle of the z-plane).Following BT satisfies these conditions:
./1
/1
1
11
1
Ks
Ksz
z
zKs
Digital Signal Processing A.S.Kayhan
Let jrezjs ,
then
.)/()/1(
)/()/1(22
222
KK
KKr
Therefore
)u.c. theinside(1)planeLH(0 r
)u.c. the(1)axis (0 rj
)u.c. theoutside(1)planeRH(0 r
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14
Digital Signal Processing A.S.Kayhan
Digital Signal Processing A.S.Kayhan
Now, let jezjs ,
then.
)(
)(
1
12/2/2/
2/2/2/
jjj
jjj
j
j
eee
eeeK
e
eKj
).2/tan()2/cos(
)2/sin(
jKKjj
or).2/tan( K
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15
Digital Signal Processing A.S.Kayhan
Observations:1- According to Taylor series expansion of tan(), for small
.2/24/2/ 3 KK 2- For high frequencies, the relation is nonlinear causing a distortion called warping effect. Therefore, BT is usually used for the design of LPF to avoid this.
Digital Signal Processing A.S.Kayhan
Discrete Butterworth Filter Design:
We convert a frequency normalized analog Butterworth filterto a discrete filter using the BT.
Normalized half power frequecy HP=1 is mapped into the discrete half power frequency HP. To do that, we set K to
).2/cot()2/tan(/)1( HPHPHPbK
-
16
Digital Signal Processing A.S.Kayhan
Then, we use ).5.0tan(/)5.0tan()5.0tan( HPbK
in nnH 2
2
1
1)(
And get the discrete Butterworth Low Pass Filter as:
n
HP
nH 22
)5.0tan()5.0tan(
1
1)(
Using the design specifications, we find the filter order as
)5.0tan()5.0tan(log2
110110log 10/
10/
max
min
p
s
n
Digital Signal Processing A.S.Kayhan
The half-power freq. is
npHP 2/110/
1
110
)5.0tan(tan2
max
).5.0cot( HPbK
Example: Consider the second order analog filter
)12(
1)(
22
sssH
Applying the BT, we get (Kb = 1)
2929.0)1716.0(
)1()(
2
2
2
z
zzH
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17
Digital Signal Processing A.S.Kayhan
Remarks:1- Given the specs.:
a) find the filter order nb) obtain the analog filter H(s)(using LHP poles)c) calculate HP and Kbd) use biliear transformation to find H(z).
2- H(z) is BIBO stable, because H(s) is stable.3- Applying the BT to high order filters may be tedious. Therefore, first express H(s) as product or some of first and second order functions, then apply the BT.
Digital Signal Processing A.S.Kayhan
Example: Design specs. for a LP digital Butterworth filter:At f = 0 Hz, 1= 18dBAt f = 2250 Hz, 2= 21dBAt f = 2500 Hz, 3= 27dBSampling freq. fsmp= 9000 Hz(ej0)= 18dBmax = 2 1=3dBmin = 3 1=9dB smp
ss
smp
pp f
f
f
f 2,2
Filter order n 5.52,n = 6, Kb = 1 and a gain G for = 18dB
)02.0.0)(17.0)(59.0(
)1(0037.0)(
222
6
6
zzz
zzH
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18
Digital Signal Processing A.S.Kayhan
Discrete Chebyshev Filter Design:
We find the constant K by transforming (normalized passband freq.) P=1 into the discrete passband frequency P. We get
)5.0tan(/)5.0tan()5.0tan(/1 PPcK
And get the discrete Chebyshev Low Pass Filter as:
))5.0tan(/)5.0(tan(1
1)(
22
2
pnCH
Digital Signal Processing A.S.Kayhan
Using C1(1)=1, we let max, p
110
1log10
10/
2max
min
Also using min, sWe find the filter order
)]5.0tan(/)5.0[tan(cosh
110110cosh
1
10/
10/1
max
min
ps
n
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19
Digital Signal Processing A.S.Kayhan
Example: Design specs. for a LP digital Chebyshev filter:At f = 0 Hz, 1= 0dBAt f = 2250 Hz, max= 3dBAt f = 2500 Hz, min= 10dBSampling freq. fsmp= 9000 Hz
Filter order n = 3, Kc = 1
)54.0)(72.015.0(
)1(09.0)(
2
3
3
zzz
zzH
smp
ss
smp
pp f
f
f
f 2,2
Digital Signal Processing A.S.Kayhan
Example: Design specs. for a LP digital Chebyshev filter:At = 2/5 rad, max= 1dBAt = /2 rad, min= 9dB
Filter order n = 3
)472.0)(619.0505.0(
)1(736.0)(
2
3
3
zzz
zzH
Example: Design specs. for a LP digital Butterworth filter:At = /2 rad, max= 3dB = 0 rad, = 0dBAt = 5/9 rad, min= 10dB
Filter order n = 7, Kb = 1
)052.0)(232.0)(636.0(
)1(01656.0)(
222
7
7
zzzz
zzH
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20
Digital Signal Processing A.S.Kayhan
Frequency Transformations:
The objective is to obtain transfer functions of other types of dicrete filters from already available prototype Low Pass Filters using tranformation functions g(z) as
))(()( zgHzH LP
Example: Design specs. for a LP digital Chebyshev filter:At f = 0 Hz, = 0dBAt f = 2250 Hz, 2= 21dBAt f = 2500 Hz, 3= 27dBSampling freq. fsmp= 9000 Hz
Digital Signal Processing A.S.Kayhan
39.0802.0691.0
)1(09.0)(
23
3
zzz
zzH LPF
-
21
Digital Signal Processing A.S.Kayhan
Now, we want a HPF with cutt-off at 3.6kHz, we use following transformation
)5095.01(
)5095.0(1
11
z
zz
6884.0102.2361.2
)133(0066.0)(
23
23
zzz
zzzzH HPF
Digital Signal Processing A.S.Kayhan
Design of FIR Filters by Windowing:
Ideal frequency response and corresponding impulse response functions are .),( nheH djd
Generally hd[n] is infinitely long. To obtain a causal practical FIR filter, we can truncate it as
otherwise.,0
0, Mnnhnh d
In a general form we can write it as nwnhnh d
Where w[n] is window function.
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22
Digital Signal Processing A.S.Kayhan
In frequency domain, we have
.)()(2
1)(
deWeHeH jjdj
Digital Signal Processing A.S.Kayhan
Some commonly used window functions are Rectangular, Bartlett, Hamming, Hanning, Blackman, Kaiser.
Some important factors in choosing windows are: Main lobe width and Peak side lobe level. For rectangular:
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23
Digital Signal Processing A.S.Kayhan
Main lobe width and transition region; Peak side lobe level and oscillations of filter are related.
Digital Signal Processing A.S.Kayhan
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24
Digital Signal Processing A.S.Kayhan
Digital Signal Processing A.S.Kayhan
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25
Digital Signal Processing A.S.Kayhan
Discrete Time Fourier Series (DTFS):
Consider a periodic sequence with period N:
.][11
0
2~~
N
k
nkN
jekX
Nnx
Which can be represented by a Fourier series as:
.][21
0
~~ nkN
jN
n
enxkX
The DTFS coefficients are obtained as:
.~~
Nnxnx
nx~
Digital Signal Processing A.S.Kayhan
Let NjN eW
2
The DTFS analysis and synthesis equations are:
.][1
0
~~nk
N
N
n
WnxkX
.][11
0
~~
N
k
nkNWkXN
nx
Example: Consider the periodic impulse train:
r
rNnnx ][~
The DTFS
. allfor 1][1
0
~
kWnkX nkNN
n
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26
Digital Signal Processing A.S.Kayhan
Similarly, ][~~
lkXnxW nlN
Properties: 1- Linearity: It is linear.2- Shift of a sequence: If ][
~~
kXnx
then ][~~
kXWmnx kmN
3- Periodic convolution: Consider two periodic sequences with period N
][~
1
~
1 kXnx ][~
2
~
2 kXnx
Then,
][][][~
2
~
1
~
3 kXkXkX .][][1
0
~
2
~
13
~
N
m
mxmnxnx
Digital Signal Processing A.S.Kayhan
Example: Consider periodic convolution of two sequences:
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27
Digital Signal Processing A.S.Kayhan
Sampling the Fourier Transform:
Consider signal x[n] with DTFT X() and assume by sampling X() , we get
kN
XXkXk
N
2|)(][ 2
~
][~
kX could be the sequence of DTFS coefficients of a periodic signal which may be obtained as
.][11
0
~~
N
k
nkNWkXN
nx
Digital Signal Processing A.S.Kayhan
Substituting,
k
Nm
mj XkXemxX 2
~
|)(,
.*~
rr
rNnxrNnnxnx then,
-
28
Digital Signal Processing A.S.Kayhan
If N is too small, then aliasing occurs in the time domain.
If there is no aliasing, we can recover x[n].
Digital Signal Processing A.S.Kayhan
Discrete Fourier Tranform (DFT):
DFT is obtained by taking samples of the DTFT. Remember DTFT is defined as:
n
njj enxeX ][)(
We take samples of X(ej) at uniform intervals as:
.1,1,0,)(][ 2
NkeXkXk
N
j
We define: N
j
N eW2
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29
Digital Signal Processing A.S.Kayhan
Then the DFT is defined as (analysis equation):
N
n
knNWnxkX
0
][][
for k=0,1,...,N-1. The inverse DFT is defined as (synthesis equation):
N
n
knNWkXN
nx0
][1
][
for n=0,1,...,N-1.
Digital Signal Processing A.S.Kayhan
Example:N=5
-
30
Digital Signal Processing A.S.Kayhan
N=10
Digital Signal Processing A.S.Kayhan
Properties:1- Linearity: DFT is a linear operation.2- Circular shift:
].[10,2
kXeNnmnxkm
Nj
N
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31
Digital Signal Processing A.S.Kayhan
3- Circular convolution:
kXkXkX
mxmnxnxnxnxN
mN
213
1
021213 ][][
Example:
Digital Signal Processing A.S.Kayhan
Example: N=L
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32
Digital Signal Processing A.S.Kayhan
N=2L
Digital Signal Processing A.S.Kayhan
4- Multiplication(Modulation):
.213213 kXkXkXnxnxnx
Linear Convolution Using the DFT:
Since there are efficient algorithms to take DFT, like FFT, we can use it instead of direct convolution as:
1. Compute N point DFTs, 2. Multiply them to get 3. Compute the inverse DFT, to get:
. and 21 kXkX .213 kXkXkX
nxnxnx 213
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33
Digital Signal Processing A.S.Kayhan
Linear Convolution of Finite Length Signals:Consider two sequences x1[n] of length L and x2[n] of
length P, and x3[n] = x1[n]* x2[n].Observe that x3[n] will be of length (L+P-1).
Circular Convolution as Linear Convolution with Aliasing :
Consider again x1[n] of length L and x2[n] of length P, and x3[n] = x1[n]* x2[n].
)()()( 213 jjj eXeXeX
Digital Signal Processing A.S.Kayhan
Taking N samples .213 kXkXkX
Now, taking the inverse DFT, we have
otherwise.,0
10,33
NnrNnxnx
rp
nxnxnx p 213
This circular convolution is identical to the linear convolution corresponding to X1(ej) X2(ej), if N (L+P-1).
-
34
Digital Signal Processing A.S.Kayhan
Example:
Digital Signal Processing A.S.Kayhan
Example:
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35
Digital Signal Processing A.S.Kayhan
Digital Signal Processing A.S.Kayhan
LTI Systems Using the DFT :Consider x[n] of length L and h[n] of length P, and y[n] =
x[n]* h[n] will be of length (L+P-1).To obtain the result of linear convolution using the DFT,
we must use N ( (L+P-1))-point DFTs. For that x[n] and h[n] must be augmented with zeros (zero-padding).
Overlap-Add Method :Consider h[n] of length P, and length of x[n] is much
grater than P.We can write x[n] as combination of length L segments:
,otherwise,0
10,
LnrLnx
nx r
r
r rLnxnx
-
36
Digital Signal Processing A.S.Kayhan
Since the system is LTI then
nhnxnyrLnyny rrr
r *,
Each output segment yr[n] can be obtained using (L+P-1) point DFT.
Example:
Digital Signal Processing A.S.Kayhan
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37
Digital Signal Processing A.S.Kayhan
Overlap-Save Method :Consider again h[n] of length P, and length of x[n] is much
grater than P.In this method L-point circular convolution (or DFT) is
used. Since the first (P-1) points will be incorrect, input segments must overlap.
We can write each L sample segment of x[n] as :
10,)1()1( LnPPLrnxnx r
11,
,)1()1(
LnPnyny
PPLrnyny
rpr
rr
Digital Signal Processing A.S.Kayhan
Example:
-
38
Digital Signal Processing A.S.Kayhan
Efficient Computation of the DFT :Remember the definition of the DFT and inverse DFT
1
0
][][N
n
knNWkXnx
1
0
][][N
n
knNWnxkX
N2 comlex multiplications and N(N-1) additions or 4N2 real mult. and 4N(N-2) real additions necessary:
N
n
knN
knN
knN
knN
Wnx
Wnxj
Wnx
Wnx
kX0
}Re{}][Im{
}Im{}][Re{
}Im{}][Im{
}Re{}][Re{
][
Digital Signal Processing A.S.Kayhan
To improve the efficiency, we can use some properties as:
nNkN
knN
nNkN
knN
knN
nNkN
WWW
WWW)()(
*)(
)2
)()1
}.Re{}][Re{}][Re{}Re{}][Re{}Re{}][Re{ )(
knN
nNkN
knN
WnNxnx
WnNxWnx
Example:
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39
Digital Signal Processing A.S.Kayhan
Fast Fourier Transform(FFT):Fast Fourier Transform algorithms are used to implement DFT efficiently to reduce the number of multiplication and addition operations.Two main algorithms are:Decimation in time and Decimation in frequency. Both these algorithms require that N=2m.The number of operations using either of these will require (N/2)log2N complex multiplications and Nlog2Nadditions.Direct implementation of DFT requires N2 complex multiplications and additions. For N=1024= 210
N2=1048576 but Nlog2N=10240.
Digital Signal Processing A.S.Kayhan
Decimation in time FFT Algorithm:Assume N=2m
1,,1,0,][][1
0
NkWnxkXN
n
knN
Which can be written as
oddn
knN
evenn
knN WnxWnxkX ][][][
With n = 2r (even) and n = 2r+1 (odd)
.
]12[)(]2[
)(]12[)(]2[][
12/
02/
12/
02/
12/
0
212/
0
2
kHWkG
WrxWWrx
WrxWWrxkX
kN
N
rN
kN
N
rN
N
r
rkN
kN
N
r
rkN
rkrk
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40
Digital Signal Processing A.S.Kayhan
.][ kHWkGkX kN
Requires N+2(N/2)2 multiplications
Digital Signal Processing A.S.Kayhan
Since G[k] and H[k] requires 2(m-1)-point DFTs, each one can be similarly decomposed until we reach 2-point DFTs.
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41
Digital Signal Processing A.S.Kayhan
Digital Signal Processing A.S.Kayhan
Butterfly:(in place comp.)
Bit reversed order
-
42
Digital Signal Processing A.S.Kayhan
Decimation in frequency FFT Algorithm:Again assume N=2m
1,,1,0,][][1
0
NkWnxkXN
n
knN
then)12/(,,1,0,][]2[
1
0
2
NrWnxrXN
n
rnN
1
2/
212/
0
2 ][][]2[N
Nn
rnN
N
n
rnN WnxWnxrX
12/
0
)2/(212/
0
2 ]2/[][]2[N
n
NnrN
N
n
rnN WNnxWnxrX
12/
0
12/
02/2/ ][])2/[][(]2[
N
n
N
n
rnN
rnN WngWNnxnxrX
Digital Signal Processing A.S.Kayhan
Similarly for
.][
])2/[][(]12[
12/
02/
12/
02/
N
n
rnN
nN
N
n
rnN
nN
WWnh
WWNnxnxrX
)12/(0 Nr
-
43
Digital Signal Processing A.S.Kayhan
Procedure is repeated until 2-point DFTs
Digital Signal Processing A.S.Kayhan
Computation of Inverse DFT:
1,,1,0,][1
][0
NnWkXN
nxN
n
knN
1,,1,0,][][0
NkWnxkXN
n
knN
][DFT1][1][ *0
** kXN
WkXN
nxN
n
knN
** ][DFT1][ kXN
nx
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44
Digital Signal Processing A.S.Kayhan
End of Part 3
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1Digital Signal Processing A.S.Kayhan
DIGITAL SIGNAL
PROCESSING
Part 4
J.S. Lim, Two Dimensional Signal Processing, in Advanced Topics In Signal Processing, Prentice-Hall.
Digital Signal Processing A.S.Kayhan
Signals:Impulses: The 2-D impulse is,
otherwise,0
0,1, 2121
nnnn
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2Digital Signal Processing A.S.Kayhan
Any sequence may be represented as a linear combination of shifted impulses:
].,[],[, 221121211 2
knknkkxnnxk k
21, nnx
Digital Signal Processing A.S.Kayhan
Line impulses are:
otherwise,0
0,1, 1121
nnnnx T
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3Digital Signal Processing A.S.Kayhan
or
otherwise,0
0,1, 2221
nnnnx T
Digital Signal Processing A.S.Kayhan
or
otherwise,0
,1, 212121
nnnnnnx T
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4Digital Signal Processing A.S.Kayhan
Step sequence:
otherwise,0
0,,1, 2121
nnnnu
].,[,1
1
2
2
2121
n
k
n
k
kknnu
or
1,11,,1,, 2121212121 nnunnunnunnunn
Digital Signal Processing A.S.Kayhan
Separable sequences:A separable sequence can be written as
221121, nfnfnnx The impulse is separable
2121, nnnn Periodic sequences:
is periodic with period N1xN2 if 21,~ nnx 22121121 ,~,~,~ NnnxnNnxnnx
Example: Periodic with 2x4
2121 )2/(cos,~ nnnnx
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5Digital Signal Processing A.S.Kayhan
Linearity: A system is linear if
Systems:The relation between the input and the response of the system is given by
.,, 2121 nnxFnny
21 , nnx
.,,,, 212211212211 nnybnnyannxbnnxaF where .2,1,,, 2121 innxFnny ii
Time Invariance: A system is time-invariant if
.,, 22112211 mnmnymnmnxF
Digital Signal Processing A.S.Kayhan
Convolution:For a LSI system with impulse response, , the input/output relation is given by
Where * denotes 2-D convolution.
Convolution with a delayed impulse :
21 , nnh
].,[],[
,*,,
221121
212121
1 2
knknhkkx
nnhnnxnny
k k
].,[,*, 2211221121 mnmnxmnmnnnx
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6Digital Signal Processing A.S.Kayhan
Example:
Digital Signal Processing A.S.Kayhan
].,[],[, 221121211 2
knknhkkxnnyk k
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7Digital Signal Processing A.S.Kayhan
If the impulse response is separable
221121, nhnhnnh then
1
1 2
1 2
21111
22221111
2221112121
,][
][],[][
][][],[,
k
k k
k k
nkfknh
knhkkxknh
knhknhkkxnny
Digital Signal Processing A.S.Kayhan
Stability: A system is BIBO stable iff a bounded input leads to a bounded output. If
then the system is BIBO stable.
1 2
],[ 21n n
nnh
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8Digital Signal Processing A.S.Kayhan
Discrete Space Fourier Transform (2D F.T.):
.),()2(
1,
,),(
2121221
2121
2211
1 2
2211
1 2
ddeeXnnx
eennxX
njnj
njnj
n n
is periodic with 2 in both variables.),( 21 X
).2,(),2(),( 212121 XXX
2D Fourier Transform and its inverse are defined as
Digital Signal Processing A.S.Kayhan
Example:
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9Digital Signal Processing A.S.Kayhan
Frequency Response:
If the input to Linear Shift Invariant (LSI) system is
2211],[ 21njnj eennx
then 2211),(],[ 2121
njnj eeHnny
where
22111 2
2121 ,),(kjkj
k k
eekkhH
is the frequency response function of the system.
Digital Signal Processing A.S.Kayhan
Example:Given LPF
otherwise,0
and,1),( 2121
baH
This function can be written as
).()(),( 221121 HHH
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10
Digital Signal Processing A.S.Kayhan
.)sin()sin(
][][],[2
2
1
1221121
n
bn
n
annhnhnnh
then
Digital Signal Processing A.S.Kayhan
Example:Given impulse response and magnitude response of a 2D LPF as
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11
Digital Signal Processing A.S.Kayhan
Original LP-Filtered HP-Filtered
Digital Signal Processing A.S.Kayhan
Example:Given frequency response of a 2D LPF as
2122
21
22
21
21,and,0
,1),(
c
cH
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12
Digital Signal Processing A.S.Kayhan
where J1(x) is the Bessel function.
2221122
21
212
],[ nnJnn
nnh cc
then the impulse response sequence is
Digital Signal Processing A.S.Kayhan
2D Z-Transform:
211 2
212121 ,),(nn
n n
zznnxzzX
2D Z-Transform is defined as
where z1 and z2 are complex variables. The space represented by (z1, z2) is four-dimesional (4-D).
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13
Digital Signal Processing A.S.Kayhan
Example:Given 2D sequence 2121 ,],[ 21 nnubannx nn
then
bzazbzazz
b
z
a
zznnubazzX
n n
nn
nn
n n
nn
2112
110 0 21
212121
and,1
1
1
1
,),(
1 2
21
21
1 2
21
Digital Signal Processing A.S.Kayhan
Inverse Z-Transform:2-D polynomials, in general, can not be factored as a product of lower order polynomials, therefore the partial fraction expansion is not a general procedure for 2-D signals.
Linear, Constant Coefficient Difference Eq.:LCCDE is given in the following form
221121),(
221121),(
,,,,2121
knknxkkbknknykkakk Rkk R BA
where a and b are known, with boundary conditions.
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14
Digital Signal Processing A.S.Kayhan
Example:Given an IIR filter with a first quadrant impulse response as
12
11
21 5.01
1),(
zzzzH
then
12
1121
2121 5.01
1
),(
),(),(
zzzzX
zzYzzH
),(),(5.0),( 211
21
12121 zzXzzzzYzzY
],[]1,1[5.0],[ 212121 nnxnnynny
Digital Signal Processing A.S.Kayhan
End of Part 4