curs dsp complet
TRANSCRIPT
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Difference equation representation of discrete time signals
)()(0 0
knxbknyaN
k
M
k
kk = = =
, 0n (1)
where kk ba , are known
)()(0 0
knxbknyaN
k
M
k
kk +=+ = =
0n (2)
NM for stable states
Method I for finding the solution for the system:
)()([1
)0(0
)()([1
)(
)()()(
0 10
0 10
1 0
0
kyakxba
yn
knyaknxba
ny
knxbknyanya
M
k
N
k
kk
M
k
N
k
kk
N
k
M
k
kk
==
=
=+
= =
= =
= =
where y(-k), n1,k= represents the initial condition for the differential equation and
are assumed known.
n=1 )1()1([1
)1(0 10
kyakxba
yM
k
N
k
kk = = =
where y(0) was
earlier determined
n=2 y(2)=
DTSx(n) y(n)
(H)
Input signal
(excitation)Output signal
(response)
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Solution:
The characteristic equation is: 016
1
2
1
4
51
321 =+
016
1
2
1
4
5 23 =+ =>2
121 == ,
4
13 =
nnn
h AAAny 332211)( ++= =nnn
AAA
+
+
4
1
3
1
2
1321
n= -1 , y(-1)=6
321
1
3
1
2
1
1 4224
1
3
1)1(
2
1)1( AAAAAAyh +=
+
+
=
n= -2 , y(-2)=6
321
2
3
2
2
2
1
16844
1
3
1)2(
2
1)2( AAAAAAy
h
+=
+
+
=
n= -3 , y(-3)= -2
321
3
3
3
2
3
1 642484
1
3
1)3(
2
1)3( AAAAAAyh +=
+
+
=
The system6422 321 =+ AAA 2/91 =A61684
321
=+ AAAhas the following solutions:
4/52 =
A
264248 321 =+ AAA 8/13 =A
=)(ny hnnn
+
4
1
8
1
3
1
4
5
2
1
2
9
b). finding the particular solution of the differential equation (1)
The particular solution )(ny p is obtained first by determining )(ny the particular
solution of the equation )()(0
nxknyaN
k
k=
=
Using of the principle of Superposition Theorem we can write the final particular
solution:
)()(1
knybnyM
k
kp = =
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If )()()( nyctnyctnx p== is a linear combination of the version x(n) and its
delayed version x(n-1), x(n-2), ..
)()()( nyknynx pnn == has the same form
)(sin)(sin)( 00 knknxnnx == =knkn 0000 sincoscossin = nBnA 00 cossin +=
==> nBnAny p 00 cossin)( +=
nnx 0cos)( = => )(cos)( 0 knknx = =nBnAknkn 000000 sincossinsincoscos +=+=
==> nBnAny p 00 sincos)( +=
Example:
Consider the differential equation:
2sin2)2(
8
1)1(
4
3)(
nnynyny =+ , with initial conditions:
y(-1)=2 , y(-2)=4)()()( nynyny hp +=
Solution:
Finding the particular solution
2
cos2
sin)( n
Bn
Anyp
+=
n= -1, y(-1)= 2 and )1()1( = yy p = - A
n= -2 , y(-2)= 4 and )2()2( = yy p = 4
2cos
2sin)( nBnAny p +=
2sin
2cos
2
)1(cos
2
)1(sin)1(
nB
nA
nB
nAny p +=
+
=
2cos
2sin
2
)2(cos
2
)2(sin)2(
nB
nA
nB
nAny p =
+
=
2sin2
]2
sin2
sin[8
1)
2sin
2cos(
4
3
2cos
2sin
n
nB
nA
nB
nA
nB
nA
=
=+++
2sin2
2cos)
8
1
4
3(2
sin)8
1
4
3( nnBABnABA =+
The system
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08
1
4
3
28
1
4
3
=
=
BAB
ABA
has the following solutions
85
96
85
112
=
=
B
A
2cos85
96
2sin85
112
)(
nn
nyp =
Find the homogeneous solution:
08
1
4
31 21 =+ => 0
8
1
4
32 =+ => 4/11 = , 2/12 =
=>nnnn
h DCDCny )2
1()
4
1()( 21 +=+=
)()()( nynyny hp += =>2
cos85
96
2sin
85
112)
2
1()
4
1()(
nnDCny
nn ++=
n= -1 => y(-1)=2=2
cos
85
96
2
sin
85
112
2
1
4
111
++
DC
n= -2 => y(-2)=4 = )cos(85
96)sin(
85
112
2
1
4
122
++
DC
The system has the following solutions:
15
13D4
85
964D16C
17
8-C2
85
1122D4C
==++
==++
2cos
85
96
2sin
85
112)
2
1(
15
13)
4
1(
17
8)(
nnny nn ++=
Example II
Consider the differential equation
)1(2
1)()2(
8
1)1(
4
3)( +=+ nxnxnynyny with
2sin2)(
nnx = . Find the particular solution of the differential equation
)1(2
1)()( += nynyny ppp
2cos
85
96
2sin
85
112)(
nnnyp =
2)1(cos
8596
2)1(sin
85112)1( = nnnyp
2
)1(cos
85
96
2
1
2
)1(sin
85
112
2
1
2cos
85
96
2sin
85
112)(
+=
nnnnnyp
Finding the response of the system when )()( nnx =
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)()(0 0
knbknyaN
k
M
k
kk = = =
(3)
0;1)( == kn knkn == ;1)(
0;0)( = kn knkn = ;0)(With y(-1)=0
y(-2)=0 0n.
For Mn > the right side of the eq.(3) is 0 so we have the homogeneous equationThe N initial conditions are y(M);y(M-1);y(M-N+1)
Since MN for a causal system we have to determine only y(0); y(1); ; y(n).But M1,n = and y(k)=0 if 0 M , =
=M
k
hk nyknya0
)(0)(
Example
)1(2
1)()2(
8
1)1(
4
3)( +=+ nnnynyny
Solution:
N=2; M=1
2n homogeneous equation: 0)2(8
1)1(
4
3)( =+ nynyny
08
1
4
31 21 =+ =>2
11
= ,3
12 = ==>
nn
h AAny )4
1()
2
1()( 21 +=
[ ]01
0 0
aa
a =
)1(
)0(
y
y
1
0
b
b=> [ ]
04/3
01
=
)1(
)0(
y
y
1
0
b
b=>
011
0
012
01
0
. . . . .
. . . . . . .. . . . . . . . . .
. .. . . . . . . . . .. . . . . . . . . .
0. . . . . . .1. . . .
0. . . . . . .1. . . . . . .
0. . . . . . .1. . . . . . . .0
aaaa
aa
aaa
aa
a
MM
M
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=
)1(
)0(
y
y
2
1
1
;
=
=
4
5)1(
1)0(
y
y
=>
=+
=+
4|4
5
4
1
2
1
1
21
21
AA
AA
=+
=+
52
1|1
21
21
AA
AA
=
=
4
3
2
1
A
A
nn
h ny )4
1(3)
2
1(4)( =
Proof
)1(2
1)0()2(
8
1)1(
4
3)0(0n +=+= yyy
)0(2
1)1()1(
8
1)0(
4
3)1(1n +=+= yyy
Response at the )()( nunx = of the discrete
time system )()(0
knhxnyk
k =
=,
=
=
==00
)()()()()(k
kkkn
k
kknukuknukuny
=
=0
1 )()()(k
kn kuny , with 0)0( =y for 0
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0n , = => =
+==n
k
kn nny0
)1()1()(
0n , =>
=
=
==
++
=
11
01
11
)(1
)(1)()(
nnn
n
k
nnknny
= 1
=+
1
1)(
1nnny
If = )()( nunx =
=N
k
khns0
)()(
Example )(3
1)( nunh
n
=
=>=+=
)10(
)6()(2)]10()6([)]6()([2)(
nu
nununununununx
106 )3
1()
3
1()
3
1(2)( = nnnns 50 n
=
=n
k
kny0
)3
1(2)(
=
==n
k
nk
0
)3
1(3)
3
1(2
106 n ==n
kny
0....................)(
=> 10n =
=n
k
ny0
....................)(
Discrete Time System (DTS)
Definition:DTS is a device or an algorithm that operates on a discrete time signal called the
input (excitation) according to some well defined rules to produce another discrete
time signal called the output (response)of the system.
H)()( nynx
= 9n61
6n02)(nx
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Example
Determine the response of the following system to the input signal :
a). y(n) = x(n)
b). y(n) = x(n-1)
c). y(n) = x(n+1)
d). y(n) = 1/3 [x(n+1)+x(n)+x(n-1)]
e). y(n) = max[x(n+1), x(n), x(n-1)]
f). =
+++==n
k
nxnxnxkxny0
.....)2()1()()()(
V
n .. -5 -4 -3 -2 -1 0 1 2 3 4 5
x(n) 0 0 3 2 1 0 1 2 3 0 0
a 0 0 3 2 1 0 1 2 3 0 0
b 0 0 0 3 2 1 0 1 2 3 0
c 0 3 2 1 0 1 2 3 0 0 0
d 0 1 5/3 2 1 2/3 1 2 5/3 1 0
e 0 3 3 3 2 1 2 3 3 3 0
f 0 0 0 3 5 6 7 9 12 12 12
d). n=0 y(0)=1/3[x(1)+x(0)+x(-1)]= 1/3(1+0+1) => y(0)=2/3
n=1 y(1)=1/3[x(2)+x(1)+x(0)] = 1/3 (2+1+0) => y(1) = 1
e). max [x(x+1), x(n), x(n-1) ]
n=0 y(0)=max[1,0,1]=1n=1 y(1)=max[2,1,0]=2
f). =
=n
k
kxny0
)()(
=
o t h e0
3n3-)(
nnx
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n=0 y(0)=6
n=1 y(1)=7
=
=
+=+==n
k
n
k
nxnynxkxkxny )()1()()()()(1
Building blocks for implementation of DTS (discrete time system)
=
=n
k
nxny1
)()(
Multiplier with a constant of a system :
)(nx ____ ______)(ny )()( nxny =
For convenience
x(n) ____ _____y(n)
)(nx )(ny
Unit delay
)(nx _____ ____)1( nx
for convenience
)(nx _________
_________ )1( nx
Advance unit
)(nx ________ __________)1( nx
Figura ce include x1 ,xk si y
Unit
delay
1z
z
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Example:
Using the basic building blocks, sketch the block diagram representation of a
discrete time system described by a equation
)1(2
1)(
2
1)1(
4
1)( ++= nxnxnyny where )(nx is the input signal and )(ny is the
output signal .
)(nx _____DTS_____ )(ny
Solution:
METHOD 1: We write directly the equation:
)1(4
1)1(
2
1)(
2
1)( ++= nynxnxny
Figura1z 1/2
)(nx + +)(ny
1z
METHOD 2:
[ ] )1(4
1)1()(
2
1)( ++= nynxnxny
Figura 1z
1/2
)(nx + + )(ny
1/41
z
Clock frequency: nnx =)( or nn =)(
Function generator
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[ ])(),...,(),()( 21 nxnxnxfny k=
Exemplul 1
Figura
Exemplul 2
1
)1()()(
0=
+=
y
kykuky ; =)(ku 0 for k0
k=0: +=+= 11)1( 0yyk=1: 21 11)2( ++=+= yy..
k : =++++= kky ...1)( 2
+ +
1
1 1kwhen 1 and
1+k when 1=
)()1(1)1()( kukykyky +=+= )()(...)2()1()( 21 kunkybkybkybky n =++++
-differential equation of a DTS
Figura)(1 nx
f y(n)
)(nxk
Figura
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Structures of a DTS
-Recursive system
-in time domain
)()(0 0
knxbknyaN
k
M
k
kk =
= =NM for a causal system .It is difficult to work in the time domain, so we apply the Z transform
)(nx ____________ )(zX )( knx _______ )(zXz
k
)()(0 0
zXzbzYza kN
k
M
k
k
k
k
= =
= ;
)(zX ______ ________ )(z
=
=
=N
k
k
k
M
k
k
k
za
zb
zX
zYzH
0
0
)(
)()( -transfer function
-Non recursive system
10 =a : )()()(1 0
knxbknyany
N
k
M
k
kk =+ = = )()()(
1 0
knxbknyanyN
k
M
k
kk += = =
-by passing to Z transform we get :
)()()(1 0
zXzbzYzazY kN
k
M
k
k
k
k
= =
+=
=
=
+= N
k
k
k
M
k
k
k
za
zb
zX
zYzH
1
0
1)(
)()( ; where H(z) is the transfer function
)()()( 21 zHzHzH =
kM
k
kzbzH
==
0
1 )( gives all the 0s of the transfer function H(z)
)(zH
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=
+=
N
k
k
kza
zH
1
2
1
1)(
gives all the poles of the transfer function H(z)
Observation|:
)()()()()()( 21 zXzHzHzXzHzY ==
Implementation
-Direct Form 1 for implementation of DTS)()()( 1 zXzHzV =)()()( 2 zHzVzY =
-Direct Form 2 for implementation of DTS)()()( 2 zXzHzW =)()()( 1 zHzWzY =
Particular cases
-first order structures
)1()()1()( 101 ++= nxbnxbnyany => 11
1
10
1)(
++
=za
zbbzH
Direct form 1)1()()( 10 += nxbnxbnv
)()1()( 1 nvnyany +=-2 adders
-2 unit delays-3 multipliers
Direct form 2 (regular)
)1()()( 1 = nwanxnw)()1()( 01 nwbnwbny +=
-2 adders
-2 unit delays
-3 multipliers
-2 adders
-1 unit delay
-3 multipliers
Figura Direct form I
Figura Direct form II
=>Figura
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-second order structures
)2()1()()2()1()( 21021 +++= nxbnxbnxbnyanyany
=> 2211
2
2
1
10
1)(
++
++=
zaza
zbzbbzH
Direct form 1
)2()1()()( 210 ++= nxbnxbnxbnv
)()1()2()( 12 nvnyanyany +=
Direct form 2
)2()1()()( 21 = nwanwanxnw
)2()1()()(210 ++=
nwbnwbnxbny
Signal Graph of direct form 2
)()()( 0 nwnxbny +=
)1()1()( 111 = nyanwbnw)2()2()( 222 = nyanxbnw
Transpose graph (overturn the previous)
Figura
Implementation of transpose graph
Figura
Figura
Figura
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Signal graph (obtained by overturning once more)
)()()( 0 nxnwbny +=
)1()1()( 11 = nyanwbnw )2()2()( 221 = nyanxbnw
Transpose Direct Form 2 (for reducing the summation circuits)
Direct Form 1 Implementation
)()(0
knxbnvM
k
k=
=
)()()(1
knyanvnyN
k
k += =
Figura
Figura
Figura
Figura
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Direct Form 2 (regular)
Figura
)()()(1
knyanxnwN
k
k = =
)()(1
knwbnvM
k
k = =
grad M< grad N
Transpose Direct Form 2
Figura
Figura
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Homework:
Given the transfer function H(z), find its regular form and transpose
21
21
125.075.01
21)(
+++
=zz
zzzH
Hint! See the coefficients; make implementation
METHOD 3:
Cascade form
+=
Mk
k
Nk
k
za
zbzH
1
0
1
)( ; (the numerator has the degree M, the denominator- N)
If Rba kk , then all poles and zeros are complex conjugate
)1()1()1(
)1)(1()1(
)(1*
1
1
1
1
1
1*1
1
1
21
21
=
=
=
=
=
zdzdze
zbzbzg
zH
k
N
k
k
N
k
k
M
k
kk
M
k
k
Where 21 2MMM += 21 2NNN +=
ke -real poles
kg -real zeros*
, kk bb -complex zeros*, kk dd -complex poles
For a causal system 1
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=
=1
1
1
2 )1()(N
k
kzezH
),min( 22 MNNs =
Figura
Implementation of )(3 zH
Figura
Homework :
Implement the structure of the DTS described by the ecuation
)25.01)(5.01(
)1(
125.075.01
21)(
11
21
21
21
+=
+
++=zz
z
zz
zzzH ;
Solution)()()( 21 zHZHzH =
1
1
15.01
1)(
+
=z
zzH
1
1
225.01
1)(
+
=z
zzH
a)Direct Form 2
b)Direct Form 1 - subsections
Figura
Figura
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1
( ) ( )k
k
k
H z C H z =
= +
k-integer part of1
2
N+
1 2 1
1 1 *
(1 )
1 (1 )(10 1 1
( )p
k k k
k k k
N N N A B e z k
k C z d z d z k k k
H z C z
= = =
= + + 1 2
N N N = +
- if M N then p N M N =
- if ,k ka b then , , , ,k k k k k A B C c e and the system function can beinterpreted as representing a parallel combination of first and second order systems
with pN possible delay paths.
Second method:
Grasping the real poles in pairs, then 10 1
1 21 1 2
( )1
p sN N
k k kk
k o k k k
e e z H z C z
a z a z
= =
+= +
, where sN is integer part of1
2
N+
The general difference equations for the parallel form with second order direct form
II sections are:
1 2( ) ( 1) ( 2) ( );k k k k k W n a W n a W n x n= + + k=1, sN 0 1( ) ( ) ( 1)k k k k k y n e W n e W n= +
0 1
( ) ( )p sN N
n k k
k k
y C x n k y n= =
= + Example: Parallel form structure for a sixth order system with real and complex
poles grasped in pairs.
(N=M=6)
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E.g. H(z)=1 2 1
1 2 1 2
1 2 7 88
1 0.75 0.125 1 0.75 0.125
z z z
z z z z
+ + += +
+ +
a) parallel structure using 2nd order system
b) H(z)= 1 118 25
81 0.05 1 0.25z z
+ c)
Parallel structure using first order system
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Example: Determine the cascade and parallel structure for the system described by
the transfer function
1 1 1
1 1 1 1
1 210(1 )(1 )(1 2 )
2 3( )3 1 1 1
(1 )(1 ) 1 ( ) 1 ( )4 8 2 2 2 2
z z z
H z j j
z z z z
+=
+
a) Cascade implementation
1 2( ) 10 ( ) ( )H z H z H z =
1
11 2
21
3( )7 3
18 32
z
H z
z z
=
+
1 2
21 2
31
2( )1
12
z zH z
z z
+ =
+
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b) parallel implementation
3 31 2
1 1 1 1
( )3 1 1 1
1 1 1 14 8 2 2 2 2
A AA AH z
j jz z z z
= + + + +
1 1
1 2 1 2
14.75 12.91 24.5 26.82( )
7 3 11 1
8 32 2
z zH z
z z z z
+= +
+ +
1A =29.3; 3A =12.25-14.57j
2A =-17.68; 3 12.25 14.57A j = +
Exam problems:
1) Obtain the direct form I, direct form II, cascade, parallel and lattice structures
for the following systems
a)3 1 1
( ) ( 1) ( 2) ( ) ( 1)4 8 3
y n y n y n x n x n= + +
b) ( ) 0.1 ( 1) 0.72 ( 2) 0.7 ( ) 0.252 ( 2)y n y n y n x n x n= + +
c) ( ) 0.1 ( 1) 0.2 ( 2) 3 ( ) 3.6 ( 1) 0.6 ( 2)y n y n y n x n x n x n= + + + +
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d)1 1
( ) ( 1) ( 2) ( ) ( 1)2 4
y n y n y n x n x n= + + +
e)1
( ) ( 1) ( 2) ( ) ( 1) ( 2)2
y n y n y n x n x n x n= + +
f)1 1 2
1 1 2
2(1 )(1 2( )
(1 0.5 )(1 0.9 0.81 )
z z z H z
z z z
+ +=
+ +
2) For1 2 3 4( ) 1 2.88 3.404 1.74 0.4H z z z z z = + + + + sketch the direct form and lattice
structure and find the corresponding input output implementations.
3) Determine a direct form of implementation for the systems
a) { }( ) 1, 2,3, 4, 3, 2,1h n =b) { }( ) 1, 2,3,3, 2,1h n =
4) Determine the transfer function and the impulse response of the systems
drawn below:
a)
b)
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Lattice structure
1
1 1( )
( )1 ( )
Nk N
N
k
H za z
a k z
=
= =+
1
( ) ( ) ( ) ( )N
N
k
g n a k y n k x n=
= +
B
1
0
( ) ( )M
k
k
y n b x n k
=
= ( ) ( ) ( )x z h n y nuuuur z z
1
0
( ) ( )M
k
k
k
Y z b z x z
=
= ( ) ( ) ( )x z H z Y z uuuur1
0
( )( )
( )
Mk
k
k
Y z H z b z
X z
=
= = the response of the system for the unitsample is identical to the coefficients kb , i.e.:
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{ }( ) , 0 1nh n b n M =
0 0
( ) 20 log 20 log
k
n B H j k
= =
2arg ( ) j k
n H j e
=
a) Direct form structure (transversal structure)
=
=1
0
)()()(n
k
knxnhny
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n-1 - memory locations for storing the M-1 previous inputs
M - multipliers
M-1 - adders of M-1 past values of the input and weighted current value of the input
if h(n)=h(M-n-1) the system is SYMMETRICh(n)=-h(M-n-1) the system is ANTISYMMETRIC
=
=
+
+==
1
0
12
0 22)()()()()(
M
k
M
k
Mnx
Mhknxkhknxkhny
+=
+1
12
)()(M
Mk
knxkh
2)()()(
MnxhMhknxkh
[ ]
+++=
= 22)()()()(
12
0
Mnx
MhkMnxknxkhny
M
k
2
Mmultiplications
b) Cascade form structure
=
=k
k
k zHzH1
)()(
Example:
Fourth order section in a cascade implementation of a symmetric system:
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( )( )43
1
2
2
1
10*
111*1
0 1111)(
++
++=
=
zzc
zczccz
z
z
zzzzzczH
k
kkk
k
kkkk
c) Lattice structure1,0);()( == MmzAmH mm
The unit sample response of the mth system is:
=
+===
=m
k
mmm knxknxmymkkkh
h
1
0
)()()()(;,1);()(
1)0(
Direct forms structure
a)
m(1) m(2) m(m-1) m(m)
b)
-m(1) -m(2) -m(m-1) -m(m)
Analog Filters (Continuous in time)
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An ideal frequency selective passes certain frequencies without any change while it
stops the other frequencies completely.
Types of Ideal Filters:
LPF (Low Pass Filter)
HPF (High Pass Filter)
PBF (Pass Band Filter)
SBF (Stop Band Filter)
Types of Real Filters:
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LPF 1-1 H(j) 1+1
1-1 || p
2 |H(j) | 1
|s|
1,2 - riple
Bocle Diagrams:
=
=
=
n
k
l
h
m
k
l
k
ph
hz
ps
zs
csH
1
1
)(
)(
)(
s=j
=
=
=
m
k
l
k
m
k
l
h
pk
zk
pj
zj
cjH
1
1
)(
)(
)(
+=
==
= =
m
k
n
k
kpkkzk
d
dB
pjlzjlc
jHjH
1 1
||log||loglog20
|)(|log20|)(|
dB
jHjHjHjH dBdB
32log102log202log20
|)(|log20|)(|log20|)(|2
1|)(|
==+
+=
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=
0
)(
jHjH
d
n
octave 2=b
a
; decade 10=
b
a
=
00
log20
jjH
dB
n
decadedBjHjH
jHjHjHjH
a
a
aadBbdBa
b
a
2010log2010
log20|)(|log20
|)(|log20|)(|log20|)(||)(|10
==
=
===
octavedBjHjH dBbdBab
a 62log20|)(||)(|2 ==
Example:
1) H (j)=k
( ) ( ) kjHkjHjH nn ==
=
0
( )
=
0,
0,0a r g
k
kjHn
( )dBn
jH ( )jHnarg
( ) dBkjHdBn
log20=
normalized transfer function
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kkj
kkk
k
ejjjH
js
ssH
**)()(
)(
2===
==2)
2
00
)(
jkk
n ejHjH
=
=
Study of the digital filters
1. Butterworth filters
- we study the transfer function
nssH
+=
1
1)(
; substitution: js =
nj
jH)(1
1)(
+=
-transfer function;
Figura
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n
jH2
1
1)(
+=
-magnitude transfer function;
njH
2
2
1
1)(
+=
-squared magnitude transfer function;
Figura grafic Figura grafic
Maximally flat approximation
00
)(=
=
k
k
d
jHd
; 1,1 = nk
nnH 422
8
3
2
11)( +=
+
dBH 32log202
1log20)(log20
1
2 ====
nn
j
sj
sjs
sHsHH22
2
1
11
1)()()(
+
==+
==
=
Poles of
2)(H
are :
n
k
nn
j
se
j
s
j
s2
)12(
22
101
==
=
+
; 12,......,1,0 = nk
n
nkj
k es2
12 +
=; 12,0 = nk -poles of squared magnitude transfer function;
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n
kj
n
k
n
kj
n
k
n
nkj
n
nksk
2
12cos
2
12sin
22
12sin
22
12cos
2
12sin
2
12cos
+
=
=
+
+
+
=
++
+=
kkk js += , where
n
k
nk
k
k
2
12cos
212sin2
=
=
Figure
Example:
n
nkj
k
n
es
H
n
2
12
2
2
11)(
3
+
=
+=
=
;
6
312
5,0
+
=
=k
j
k es
k
0=k 2
3
2
13
0 jesj
+==
;
1=k 2
3
2
13
2
1 jesj
+==
;
2=k 12 ==j
es;
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3=k 2
3
2
13
4
31jes
j
==
;
4=k 2
3
2
13
5
41 jes
j
+==
;
5=k 125 ==
jes;
The poles from the left plane:
( )
( ) ( )111
12
3
2
1
2
3
2
1
1
))()((
1)(
2
321 +++=
+
+
=
=sss
sjsjs
sssssssH
TABLE 1
n Butterworth polynomials(factored form)
1 1+s2 122 ++ ss3 ( ) ( )11 2 +++ sss4 184776.1176536.0 22 ++++ ssss5 ( )119318.116180.0 22 +++++ sssss6 ( )( ( )19318.11215176.0 222 ++++++ ssssss7 ( )( )( )( )118022.112456.114450.0 222 +++++++ sssssss8 19622.116663.11111.113986.0
2222
++++++++ ssssssss
1.......)( 1
1
1 ++++=
sasasasn
n
n
n ;)(
1)(
ssH
n=
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TABLE 2
Coefficients of butterworth polynomials
n 1
a 2
a 3
a 4
a 5
a 6
a 7
a 8
a
1 1
2
2 1
3 2 2 1
4 2.613 3.414 2.613 1
5 3.236 5.236 5.236 3.236 1
6 3.864 7.464 9.141 7.464 3.864 1
7 4.494 10.10
3
14.60
6
14.60
6
10.10
3
4.494 1
8 5.126 13.12
8
21.12
8
25.69
1
21.12
8
13.12
8
5.126 1
Exercise:Design of a low-pass Butterworth filter
11)( H ; p
2)(
11)( H ; p
js = ; )()( jHsH
2)(
n
nnn jssHjHsH
=
= )()()(
n
c
c
H 2
2
1
1
+
=
;
12
1
1
1
=
+
n
c
p
(a)
Figura grafic
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n
c
p
pnH2
1
1)(
+
=
22
1
1
=
+
n
c
s
(b)
n
c
s
snH2
1
1)(
+
=
;
( ) 21
2
1
11
=
+
n
c
p
( )
11
12
1
2
=
n
c
p
=> =>
2
2
2
11
=
+
n
c
s
11
2
2
2
=
n
c
s
=>
( )( )
2
2
2
2
2
1
2
12
1
111
=
n
s
p
=>
( )( ) ( )
s
p
n
log
1111log
2
12
2
2
1
2
2
2
1
=
(c)
n - must be integer , choose nearest integer value
If c
is determined from equation (a), the PB specifications are met exactly,
whereas the SB are exceeded.
If c
is determined from equation (b), the SB specifications are met exactly,
whereas the PB are exceeded.
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Steps in finding H(s) :
1. Determine n from equation (c) using the values of sp ,21 ,, and round up
to the nearest greater integer.
2. Determine c from the equation (a) or (b).
3. For the value of n determined at step 1. , determine the nominator of H(s) usingtable 1, table 2 or H(s).
4. Find the )(sHn by replacing s withc
s
.
Example:
Design a BF (Butterworth filter) to have attenuation no more than 1dB for 1000rad/s and least 10 dB for 5000 rad/s.
( )
= = 10lo g2011lo g20
2
1
=> == 31623.0
108749.02
1
=
=
5000
1000
s
p
=> 012.1=n . Round 2=n => 12
1)(
2 ++=
sssH
From equation (b) we have:
n=2 => 75.28862 = rad/s
22
2
275.288648.4082
75.2886
175.2886
2
75.2886
1)(++=
++
= ssss
sHc
n
DESIGN METHOD
Given: - the maximum pass band attenuation ( pA )
- the minimum stop band attenuation ( sA )
- the pass band edge frequency ( pf )
- the stop band edge frequency ( sf )
Find the transfer function H(s) of a Butterworth filter.
p
n
c
pA=
+
2
1log10
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s
n
c
s A=
+
2
1log10
=>
s
p
n
c
s
n
c
p
A
A=
+
+
2
2
1log
1log
pA
n
c
p 1.0
2
101 =
+
110 1.0
2
=
pA
n
c
p
sA
n
c
s 1.0
2
101 =
+
110
1.0
2
=
pA
n
c
s
110
1101.0
1.0
=s
p
A
A
s
p
s
p
A
A
s
p
A
A
s
p
s
p
n
log
110
110log
log
110
110log
2
1
2
1
1.0
1.0
1.0
1.0
=
=
Notations:
s
p
s
p
s
p
f
f
f
fk ===
2
2
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Given :dBA
dBA
s
p
20
1
=
=
kHzf
kHzf
s
p
5
1
=
=, find : H(s)=?
Solution :
110
110
110
110
2.010*5
10*1
20*1.0
1*1.0
1.0
1.0
3
3
=
=
===
s
p
A
A
s
p
d
f
fk
2
10*6099.1
=d
== 565.2log
log
k
dn choose n=3
Poles:
2
12cos
2
12sin
n
kj
n
ksk
+
= ; 5,0=k
Impose the stability condition for H(s)
k=1,2,3n Chebyshev polynomials ( )nC
0 1
1
2 12 2 3 34 3 4 188 24 + 5 52016 35 +6 1184832 246 ++
7 75611264 357 +8 132160256128 2468 ++
2. CHEBYSHEV FILTERS
Chebyshev polynomials
( ) ( )( )
>=
1;c o s hc o s h
10;c o sc o s1
1
n
nCn
1cos =u( ) nuCn cos=
Figura grafic
Figura graphic cosh x
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n=0 : ( ) 10cos0 ==Cn=1 : ( ) ( ) === 11 coscoscosuCn=2 : ( ) 121coscos21cos22cos 21222 ==== uuCn=3 : ( ) ( )[ ] ( )[ ] 34cos3coscos4cos3cos43cos 313133 ==== osuuuC
( ) ( )[ ]( )( ) ( )[ ]( ) ( ) ( ) uCunuCC
unuunuunC
nuC
unuunuunC
nnn
n
n
n
coscoscos2
sinsincoscos1cos
cos
sinsincoscos1cos
11
1
1
==+=+=
=+==
+
+
Recursive formula for Chebyshev polynomials:( ) ( ) ( ) 11 2 + = nnn CCC ; n=1,2,3;
( )( )
==
1
0 1
C
C
Properties of ( )nC :1) ( ) 10;10 nC ; ( ) 1;1 >> nC 2) ( )nC increases monotonically for 1>3) ( )nC is: -even polynomial for n-even
-odd polynomial for n-odd
4) ( )0C =
e vn
o dn
,1
,0
( )( )
22
2
1
1
nCH
+= squared magnitude transfer function
Properties:
1) n the zeros of ( )nC are located in the interval 1 2) -for ( ) ( )
21;1 HCn oscillates about unity such that the maximum
value is 1 and minimum value is 21
1
+-for ( ) nC;1> increases rapidly,as becomes large , then ( ) 2H
approaches zero rapidly.
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Figura grafic
n- even
Figura grafic
n-odd
-for ( ) 1,1 1 == C for all n ( ) 2
2
1
11
+=H
-for ( ) ( )
22
2
1
11
n
n
C
H
+=
==>==
S=j
2
12)(cos 1
n
kjs
j
s ==>=
2
12cos
2
12cos
n
kjs
n
kjs kk
==>
=
)(
11
)(1
11
)(1
)(
)(222222
22
r
n
r
n
r
n
r
nr
CCC
C
H ++
=+
=
2
)(log10)(
rHA = [dB]
+=)(
11log10)(
22
r
nC
A [dB]
+=)(
11log10)(
22
p
rn
p
C
A
[dB]
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)1
1log(10
)(
11log10)(
222
+=
+=
r
r
n
r
C
A [dB]
110
1
1.0 =
rA
E
k
dn1cosh
1cosh
1
1
Elliptic filters
For LPF only)(1
1)(
22
2
RH
+= where R()-Tchebyshev rational functions.
The roots of R () are related to the Jacobi elliptic sine function.
Properties of R ():
1) R() are even functions for even n and odd for odd n;
2) Zeros of R() are in the range 1 ;3) R() oscillates between values -1 and +1 in PB;
4) R()=1 for =1;
5) R() oscillates between d+ and in the SB (d= discriminator factor)
6)
=
=
=
2
1
122
22
2
1
022
22
,1
,1
)(n
i i
i
n
i i
i
e v en
o dn
R
Normalized to center: frequency 10 =
7) )(
1)
1(
RR =
The poles and zeros of R() are reciprocal and exhibit geometric symmetry
.center frequency 0.
1) 110)1log(10 1.02 ==>+= AppA
2)110
)1log(101.02
2
== >+=
rAr d
dA ;
r
pk
= ; rp =0
3)
( ) 41
2
4
1
2
0
11
)1(1
2
1
k
kq
+
=
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4)13
00
5
00150152 qqqqq +++=
5)
q
dn1
log
10log
2
=
6)
=
=
+
+
+=
1
0
)1(
1
2cosh)1(21
)12sinh()1(2
2
m
mm
m
mmmn
mq
mqq
a
11
11ln
2
1
2
2
++=
n
7)
=
=
+
+
+
=
1
0
)1(
1
2cos)1(21
12sin)1(2
2
m
mm
m
mmmn
l
l
n
mq
ln
mqq
il
il
=
=2
1
oddnn
i
evennn
i
=
=
,2
1,1
,2
,1
8) )1()1(V2
2
ik
k ii
=
9) 21
i
ia
=
10) 221
2
i
ii
ia
vab
+=
11) )1)(1(2
2
k
akaU ++=
12) 222
22
)1(
)()(
i
ii
i a
UaVc
+
+=
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13)
+
=
=
=
2
1
12
2
1
1
0
,1
1
,
n
i i
i
n
i i
i
e v ena
c
o d dna
ca
H
14)
++
+
+
++
+
=
=
=
2
12
2
0
2
12
2
0
,
,
n
i ii
i
n
i ii
i
s
o dncsbs
as
as
H
e v encsbs
asH
H
Example: Design a filter with no more than 2 dB ripple in the PB up to an edge
frequency of 3000 Hz. The filter is to attenuate the out of hand signals beyond
4000 Hz by at least 60 Hz.
Find H(s).
dBAp 2 13000*22 == radsfpp
dBAr 60
14000*22
== radsfrr
75.0===r
p
r
p
f
fk
( ) 41.0
1.0
10*6478.7
110
110 =
=
r
p
A
A
d
4
1
2
4
1
2
0
)1(1
)1(1
2
1
k
kq
+
= ; 130
9
000 15052 qqqqq +++=
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7.51log
16log
2
=
q
dn . We choose n=6.
1555.1
1 == radsk
r
1
0 3464*2== radsrp
RsH 410*1374.7)( =
)7341.005386.0)(194425.0)(10903.035518.0(
)3955.1)(2153.2)(8451.13(222
222
++++++++
=sssss
sssR
Bessel Filters
a))(
1)(
sBsH
n
=
==
n
k kkn
sasB0
)(, n-th order Bessel ..
)!(!2
)!2(
knk
kna
knk =
b) )()()12()( 22
1 sBssBnsB nnn += with initial conditions ;1)0( =nB 1)1( +=sBn
Frequency transformations for analog filters
Type oftransformation
Transformation Band edge frequenciesof new filter
LPFss
p
p
1
HPF
ss
pp 1
SBF
lu
lu
ps
ss
+
2
)(
PBF
)(
2
lu
lu
ps
s
s
+
Transform Analysis of LTI Systems
In time domain
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0cT Tc
1
H(e )j
h(n) impulse response of LTI for )()( nnx =
y(n)=h(n)*x(n)=
=
k
knxkh )()(
Z transform)(arg)( jeH=
D.F.T.
)(*)()(
jjjez eXeHeYj
= =
)(*)()(
)()(
jj
j
jj eeH
eX
eYeH == , where )(arg)( jeH= - phase
We define arg )( jeH has principal value of phase: )(arg jeH
)( > ZrreArgHeH jj += )(),(2)()(
= )()(
jeArgH
d
dtime delay except pants,
where Arg )( jeH has discontinuities.
a) A linear phase a low-pass filter
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id
nd n
n
n
j
ee
ne
jnnde
nnh c
jnjnjnjn
LPF
c
c
c
c
sin
2
11
2
1
2
1)( ====
Comment: If n increases 0 LPFh , but not fast enoughb) Phase shift
d
j
i d
j
i d
j
i dnjj
i d
di d
n
d
eHd
neH
eHeeH
nnh
d
=
=
=
=
=
)4()3(
)5()2(
)6()1(
)7()0(
hh
hh
hh
hh
H(z) = h(0) (1 + h-7) + h(1) (z-1 + z-6) + h(2) (z-2 + z-5)+ h(3) (z-3 + z-4) =
= z 27
[h(0) (z 27
+ z 27
) + h(1) (z 25
+ z 25
) + h(2) (z 23
+ z 23
) h(3) (z 21
+ z 21
)]
H(ej) = H(z)|z=ej
= e 27
j [h(0) 2cos2
7 + h(1) 2cos
2
5+ h(2) 2cos
2
3+ h(3) 2cos
2
]
=
= e 27j H
~ () ej = H~
7 () e
+
2
7j
= 0 | H~ 7 (ej)| 0 = |H(ej)| < 0
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H~
7 () = 2h(0) cos2
7 + 2h(1) 2cos
2
5 + 2h(2) cos
2
3 + 2h(3) 2cos
2
1
=
+=
2
72
7
g r d
H(ej) = e
+
2
Nj H
~ ()
H~ () = 2
+
=
+2
1
1 2
1
N
n
nN cos
2
1n
In general:
-
=
=
2
2
Ng r d
N
Antisymmetric impulse response with odd length (type 3)
forN = even h(n) = -h(N-n)
N=8 H(z) = =
8
0
)(n
nh .z-n = h(0).z0 + h(1).z-1 + h(2).z-2 + h(3).z-3 + h(4).z-4 +
+ h(5).z-5 + h(6).z-6 + h(7).z-7 + h(8).z-8
h(n)= - h(8-n) = >
== >=
=
==
=
0)4()4()4(
)5()3(
)6()2()7()1(
)8()0(
hhh
hh
hhhh
hh
H(z) = h(0) (1+z-8) + h(1) (z-1 - z-7) + h(2) (z-2 - z-6) + h(3) (z-3 - z-5) =
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= z-4 [h(0) (z4 - z-4) + h(1) (z3 - z-3) + h(2) (z2 - z-2) + h(3) (z1 - z-1)]
z = ej zm + z-m = ejm- e-jm = 2jsin m
H(ej) = H(z)|z= ej=
= e-4j[h(0) 2jsin 4 + h(1) 2jsin 3 + h(2) 2jsin 2 + h(3) 2jsin] == e
+
24
j [h(0) 2sin4 + + h(3) sin= e
+
24
j H
~ () ej
= 0 H~ () 0 = H~ () < 0
),0(
==
++=
4
24
d
dg r d
In general:
- H(ej) = e
++
22
Nj H
~ ()
- H~ () = 2
=
2
1 2
N
n
nN sin n
Antisymmetric impulse response with even length (type 4)
forN = odd
h(n) = -h(N-n)
N=7 H(z) ==
7
0
)(n
nh z-n = h(0) + h(1) z-1 + h(2) z-2 + h(3) z-3 + h(4) z-4 +
+ h(5) z-5 + h(6) z-6 + h(7) z-7
h(n)= -h(7-n) =>
=
=
=
=
)4()3(
)5()2(
)6()1(
)7()0(
hh
hh
hh
hh
H(z) = h(0) (1 - h-7) + h(1) (z-1 - z-6) + h(2) (z-2 - z-5)+ h(3) (z-3 - z-4) =
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= z 27
[h(0) (z 27
- z 27
) + h(1) (z 25
- z 25
) + h(2) (z 23
- z 23
) h(3) (z 21
- z 21
)]
H(ej) = H(z)|z=ej = e 2
7j [h(0)( e 2
7j - e 2
7j ) + h(1)( e 2
5j - e 2
5j ) + h(2) ( e 2
3j
- e 23
j ) + h(3) ( e 21
j - e 21
j )] =
= 2j e 27
j [h(0) 2sin2
7 + h(1) 2sin
2
5 + h(2) 2sin
2
3 + h(3) 2sin
2
1] = =e
++
22
7j H
~ ()
H~
() = 2[h(0) sin2
7+ h(1) sin
2
5 + h(2) sin
2
3 + h(3) sin
2
1
=
+=
2
722
7
g r d
= 0 H~ () 0 = H~ () < 0In general
- H(ej) = e
++
22
Nj H
~ ()
- H~ () = 2
+
=
+21
1 2
1
N
n
nN sin
2
1n ;
=
++=
2
22
Ng r d
N
General form of frequency response
H(e
j
) = e 2N
j
e
j
H~
()
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() =
z = 1
is also zero of H(z)
If z = ej (complex number on a unit circle) => z = e-j is also a zero of H(z)
If z = rej is a zero of H(z) => z =je
r
1
is also a zero of H(z)
If zero at z = 1 is its own reciprocal, implying it can appear only single
A type 2 FIR filter must have a zero at z = -1, since H(-1) = (-1)N |H(-1)| = -H(-
1)
For type 3 and 4 FIR filter H(-1) = -(-1)N H(-1) = -H(-1) forcing H(-1) = 0
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Bounded real transfer function of FIR filter
H(ej) =)(
)(
j
j
eX
eY; |H(ej) | 1 - bounded real transfer function for causal filter
(BR)
|Y(ej)|2 |X(ej)|2 => y(n)2 x(n)2
y(n) = x(n) - lossless transmitting signal
Law Pass Digital Filter Design
Eg: The moving average filter
Ho(z) = ( )z
zz
1
2
11
2
1 1 +=+
=
=
p oz
z ez
p
z
0
1
Z = ej
H(ej) = H(z)|z = ej
=
j
j
e
e 1
2
1 +
Zz = -1 => z =
Zz = 0 => p = 0 ( ),0( )
H(ej) =
2
cos
2
12
222
j
j
jjj
e
e
eee
=
+
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=
=
=
2c o s)(
2
1
2
1
jeH
g r d
2
1)(2 =CjeH
20 log )( Cj
eH
= 20 log2
1H( )0
je = -20 log 2 = 3dB
)( Cj
eH
=2
1= cos
2c
=> cos2
cos4
c = => c =2
High Pass Filter
LPF zz HPF
=
=
0
1
p
z
z
z
H1(z) = ( ) zz
zz
1
2
111
2
1)(1
2
1 1 =
=+
H1(ej) =
2sin
2
11
2
1 22
222
=
=
+
j
j
jjj
j
j
ee
eee
e
e
Filter design steps
== >=
== >=
001
zp
zz
zz
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1. Specification of the filter requirementsa) signal characteristics
- type of signal source
- input output interface
- data rates and width- highest of frequency
b) characteristics of filter
b1) desired amplitude
} and their tolerances
b2) phase response
b1) desired amplitude is specific in frequency domain in tolerance scheme for
selective filters
b2) phase response - are used in equalize or compensate in phase response
- speed of operation
- modes of filtering in real time and not in real time
c) level of implementation
- high level language routines in a computer
- DSP processor based system
- choice of signal processor
- costs
2. Coefficients computation for FIR filters- random method
- frequency sampling method
- optimal method
Computation of coefficients for IIR filters
- impulse invariant method
- bilinear method
- pole zero placement method
3. -- Representation of the filter by a suitable structure (realization)4. -- Analysis of the effect of finite word length in filter tolerance
5. -- Implementation of filter in software and hardware mode
IIR Filters
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1) y(n) =
=
0
)()(k
knxnh
2) y(n) = ==
+N
k
N
k
k knynhknxa00
)()()(
3) H(z) =k
N
k
k
N
k
k
k
zb
za
=
=
+0
0
1
Reconstruction of a bandwidth signal from its samples.
Reconstruction scheme of a signal from its sampling signal
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=
=n
nTtnxtx )()()(*
=
=
=
===
00 0
00
~
)()()()()(
)()()()(*)()(
mm
T
mTxmTthdmTttxh
dttxhdtxhty
omTnTh = )( for mn
n
n m
n
on
zmTxmTnThznTyzy
=
=
= ==
0 0
~
)()()()(*
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)(*)(*
)()()()()(*00
)(
0 0
zXzH
zmTxzkThzmTxkThzy m
m
k
k
mk
n m
=
===
=
=
+
=
=
IIR filter design by impulse invariant method.
)()( ded nThTnh = , eh - impulse response of continuous. in time filterh(n) impulse of digital filter
=
+
=k
kTT
j
c
j ddeHeH )()()
2(
Frequency for continuous time filter is noted by .
Frequency for digital filter is noted by .Transfer function for CTF in band limited dT=
=
=
) ,()(
,0)(
dTj
c
j
d
c
eHeH
TjH
=
=N
k
z
k
kc
ss
AsH
1
)( zT
s ln1
=
1L
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system fct of a digital IIR filter)(sHa - transfer fct. Of an analog filter
This approximation is valid only in LPF or BPF ( low frequency filter ) and cannot
be used for high frequency.
==
+=
L
k
k
nTt T
kTnTykTnTya
dt
dy
1
)()(
Z
=
=L
k
Kk
k ZZaT
s1
)(1
jeZ=
( ) = =
==L
k
L
k
k
kjkj
k kaT
jeeaT
s1 1
sin21
=
==L
k
k kaT
js1
sin2
Is difficult to calculate { }ka for the poles that are inside the unit circle.
Ex 1:
Convert the analogue band pass filter with system function9)1.0(
1)(
2
++=
s
sHa into
a IIR digital filter by use of the backward difference for derivative.
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21
2
2
2
21
1
01.92.01
1
01.92.01
)1.01(21
01.92.01
91.01
1)()( 1
=
+++
+++
++=
=
+
+
==
ZTT
ZTT
T
TT
T
T
ZsHZH
T
Zsa
Choose T=0.1 =>5.16
2,1 949.0jep =
Ex2:
Convert the analogue band pass filter with9)1.0(
1)(
2 ++=
ssHa into a digital IIR filter by the
use of the mapping ( )11 = zzT
s
( ) 12.0)01.92(2.0......
91.01
1)(
2234
22
2
1++++
==
+
+
= TzzTTzz
TZ
zzT
zH
Bilinear transform
+
=
1
1
1
12
z
z
T
s
=
+=
jr eZ
js
++
+++
=
+=
cos21
sin2
cos21
12
1
1222
2
1
1
rr
rj
rr
r
Ter
er
Ts
j
j
=>
cos21
12
2
rrr++
=
cos21
sin22
rr
r
++=
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If r
r=1 0= => =js ,2
tan2
)cos1(2
sin22
TdTd=
+=
-the transformation is not bijective; is non linear.
Ex: Convert the analogue filter with the system fct.16)1.0(
1.0)(
2 +++
=s
ssHa into a
digital IIR filter by means of the bilinear transformation. The digital filter is to have
a resonant frequency:2
=r
4=r 2
=r
2
1
2tan
2 == TdrTd
r
1
1
1
14
+
=Z
Zs
21
21
2
1
1
1
1
1
14
975.00006.01
122.0006.0128.0
161.01
14
1.01
14
)()(1
1
+
=
+++
=
=
+
+
+
++
==
zz
zz
z
z
z
z
sHzHz
zsa
poles:2
2,1
2
987.0
0975.01
+=
=+
j
ez
z
095.101222.0006.0128.0 2,121 ==+ zzzzzeros
Usually, the design begins with the digital filter specifications, after that consider the
analogue filter which satisfies the imposed specifications.
Ex 1: Design a single pole low-pass filter with 3dBbandwidth of 0.2 using thebilinear transformation applied to the analogue filter.
c
c
ssH
+
=)( , where c is the
3dB bandwidth of analogue filter.
c=0.2
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c= Td2
tan2
c=T
2tan
2
0.2=
Td
65.0
Then the transfer function will be: H(s)=
T d
T d
s 6 5.0
6 5.0
+
H(z)=H(s)|s= Td2
(
1
1
1
1
+
z
z
)=Tdz
zTd
Td
65.0
1
12
65.0
)( 11 +
+
=1
1
509.01
)1(245.0
+
z
z
Z=ej
H(ej
)= j509.01
)1(245.0
+
e
e jw
=0 H(ej )=509.01
490.0
1
c=0.2 H(ej0.2
) = 707.0
509.01
)1(245.0j0.2-
-j0.2
=
+
e
e 3 dB attenuation
Ex. 2:
The specifications on the discrete-time filter are:
0.819125 |H( -j0.2e ) 1; 0 0.2| H( -j0.2e )| 0.17783; 0.3
For this specific filter, the analogue filter must have:
0.89125 | Hc( -je
)| 1; 0 Td
2tan
2
0.2
| H( -je )| 0.17783;Td
2tan
2
0.2
Td
2tan2
()
For convenience choose Td=1{ (| Hc 2jtan0.1e )|0.89125{ (| Hc j2tan0.15e )| 0.17783
|Hc(ej
)|2= Nc
2)(1
1
+ Butterworth filter
1+(c1.0tan2
)2N=89125.0
1
1+( c
15.0tan2
)
2N
= 17783.0
1
N=5.30466
Choose N=6 c=0.76622
We take the functions from the tabels for various degrees. We choose the poles
from the left in order to have stability.
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Hc(s)= )5781.0480.1)(5871.00836.1)(5871.03966.0(20238.0
222 ++++++ ssssssH(z)=
)2155.09044.01)(3583.00106.11)(7051.02686.11(
)1(0007378.0212121
1
++++
zzzzzz
z
Draw the signal flow-graph of implementation of the system:-as a cascade of 2nd order section
-in direct form I and direct form II
-in the parallel form using 2nd order section
IIR filters Form
Y(n)==
N
k
k knxa0
)( +=
N
k
x knyb1
)( (1)
Or
H(z)=
=
=
+N
k
k
k
N
ok
kk
zb
za
1
1
(2)
or
H(z)=))...()((
)(...))((
21
21
n
n
pzpzpz
zzzzzzk
(3)
Pole-zero placement method:
Ex.1: A bandpass filter is required to meet the following specifications:
-a complete rejection on d.c. and 250 Hz
-a narrow band centered at 125 Hz
-a 3 dB bandwidth of 10 Hz
Assuming a Fs=500Hz, obtain the transfer function of the filter by suitable placing
z-plane poles and zeros and its diference equation. Sketch the block-diagram of the
filter.
0o complete rejection
oo
180
500
250*360=
Passband must be centered at 125 Hz and we must have the poles
oo
90500
125*360=
r=1- 500
10=0.937
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H(z)= 2
2
2
2
077969.01
1
877969.0
1
)937.0)(937.0(
)1)(1(
22
+
=+
=
+z
z
z
z
ezez
zz
jj
y(n)=-0.877969y(n-2)+x(n)+x(n-2) the form we need
In this form the coefficients are:a0=1 ;a1=0; a2=-1;
b1=1; b2=0.877969.
Ex.2: A digital notch filter has the following specifications:
-notch frequency: 50 Hz (notch = cutting, very sharp stop-band)
-3 dB width of the notch: +/- 5Hz
-sampling frequency: Fs=500 Hz
To reject the comp. of 50 Hz we place a pair of zeroes at points in the corresponding
unit circle.o
o
36500
50*360=
-for poles: r=1- 937.0500
51 ==
Fs
bw
H(z)=
21
11
2
2
3636
3636
87.0516.11
618.11
87.05161.1
1618.1
)937.0)(937.0(
))((
++
=
=+
+=
zz
zz
zz
zz
ezez
ezezoo
oo
jj
jj
The differential equations for this filter:
The E of this notch filter is:
y(n)=x(n)-1.618x(n-1)+x(n-2)+516y(n-1)-0.878y(n-2)
In the coefficient form:
a0=1; a1=-1.618; a2=1;
b1=-1.5161; b2=0.878
Measure of information
X- space of discrete random variables with possible outcomes ix , ni ,1=
Y- space of discrete RV with possible outcomes jy , mj ,1=
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- If X,Y are statistically independent, then the measure of information they will
define should be zero (no information)
- If X,Y are mutual dependent (causal link), then the measure of information they
will define should be given by ixx = , when jyy = .
Postulates:
1) 0)( ixP (probability of event ix )
2) 1)( =XP (probability of the sample space X (certain event))
3) = ji yx , ,...2,1= ji
-these are called mutually exclusive events
==
=n
i
iii
xPxPU11
)()( - probability of the mutually exclusive events
Joint events and joint probability
Theorem:
If one experiment has the possible outcomes ix , ni ,1= and the second experiment
has the possible outcomes mjyj ,1, = , then the combined experiment has the
possible outcomes ),( ji yx , ni ,1= , mj ,1=The joint probability:
),( ii yxP of combined experiment satisfies the condition: 1),(0 ji yxP
Theorem:
If the outcomes jy ; mj ,1= are mutually exclusive, then =
=m
j
iji xPyxP1
)(),(
Similarly, if the outcomes nixi ,1, = are mutually exclusive events, then
= =n
i
jji yPyxP1
)(),(
If all the outcomes of the 2 experiments are mutually exclusive, then :
= =
=n
i
m
j
ji yxP1 1
1),(
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Conditional probabilities
)(
),()|(
j
ji
jiyP
yxPyxP = ; 0)( >jyP - conditional probability of the event ix given by
the accuracy of the event jy
)(
),()|(
i
ji
ijxP
yxPxyP = , )|()()|()(),(0)( ijijijjii xyPxPyxPyPyxPxP ==>
Notes
1. If ji yx ( ji yx & are mutually exclusive events) then 0)|( =ji yxP
2. If ix is a subset of jy ( iji xyx = ) then)(
)()|(
j
i
jiyP
xPyxP =
3. If jy is a subset of ix ( jji yyx = ) then 1
)(
)()|( ==
j
j
ji
yP
yPyxP
Bayes Theorem
If nixi ,1; = are mutually exclusive events such that ni
i Xx1=
= and Y is an arbitrary
event with 0)( >YP , then =
==
n
j
jj
iii
i
xPxYP
xPxYP
YP
YxPYxP
1
)()/(
)()/(
)(
),()|(
Statistical independence
If the occurrence of X doesnt depend on the occurrence of Y, then )()|( XPYXP =
and )()(),( YPXPYXP =
Example:
)()(),(
)()(),(
)()(),(
3131
3232
2121
xPxPxxP
xPxPxxP
xPxPxxP
===
Logarithmic measure of information
==)()(
),(log)(
)/(log),(ji
ji
b
i
ji
bjiyPxP
yxPxP
yxPyxI mutual information ji yx ,
if b=2 then:
- ),( ji yxI [bits]
if b=e then: b=e [nats]
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a
aae
ln44265.1log
log69315.0log
2
2=
Observations
1. If the random variables X & Y are statistically independent, then )()|( iji xPyxP = ,
then 0),( =ii yxI
2. If the random variable are fully dependent, then
==== )()(log)(
1log),( iib
i
bji xIxPxP
yxI self information.
Show that:
)()(
),(log
)(
)/(log),(
ji
ji
b
i
ji
bjiyPxP
yxP
xP
yxPyxI
==
)(log)( ibi xPxI =
Since:
),();(
)(
)/(
)()(
),(
)()(
)()/(
)(
)/(
ijji
j
ij
ji
ji
ji
jji
i
ji
xyIyxI
yP
xyP
yPxP
yxP
yPxP
yPyxP
xP
yxP
=
=
=
=
Examples:1) Suppose a discrete information source that emits a binary digit }1,0{=ix with
equal probability at every seconds. (2
1)( =ixP ) then the information content
of each output source is 12
1log)(log)( 22 === ii xPxI [bit]
2) If considered a block of K binary digits from the source, which occurs in a
time interval K then KixP2
1)( = then Kxi Ki ==
2
1log)(
2 [bits]
X- the set of signals on entry point
Y- the set of signals on exit point
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X & Y {0,1}
0P - Probability of error for input 0
1P - Probability of error for input 1
======
0
0
)0/1(
1)0/0(
PXYP
PXYP
======
1
1
)1/0(
1)1/1(
PXYP
PXYP
)1(2
1
2
1
2
1)1(
)1()1/0()0()0/0()0(
101 PPPP
XPXYPXPXYPYP
o +=+=
====+=====
)1(2
1
2
1)1(
2
1
)1()1/1()0()0/1()1(
1010 PPPP
XPXYPXPXYPYP
+=+=
====+=====
We know that = )(log)( ibi xPxI self information
==
)()(
),(log
)(
)/(log),( 2
ji
ji
b
i
ji
iiyPxP
yxP
xP
yxPyxI mutual information
Then:
)1(
2
1
1log
)0(
)0/0(log)0,0(
00
0
22
PP
P
YP
XYPI
+
=
===
=[bits]
Similarly:)1(
2
1
1log
)1(
)1/1(log)1,1(
01
122
PP
P
YP
XYPI
+
=
===
=
Special Cases
a) If
+=+
=
+=+
=
==
)1(l o g11
)1(2l o g)1,1(
1)1(l o g)1(
)1(2l o g)0,0(
22
22
10
PPP
PI
PPP
PI
PPP
Cases:
1: == 010 PP Errorless channel (without noise)
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=
=
1)1,1(
1)0,0(
I
Itotal information
2: ==== 0)1,1()0,0(2
110 IIPP dont transform information
3: 587,0)1,1(3log1)0,0(4
1210 ==+=== IIPP [bits]
Conditional self information
I(xi/yj)= logb ( )ji
yxP /
1= logbP(xi/yj) information about x= I after having
observed the event Y= jy
I(xi/yj)= I(xi)-I(xi/yj)
0)y/x(I0I ( x i )ii
I(xi/yj)>;0; average number of bits per symbol
JxHRxH J
1)()( +
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9 5.02 1.21 1.2)(
2 1.2)(
1 1.2)(l o g)()(
7
1
7
1
2
===
==
==
=
=
RxH
nxpR
b i t sxpxpxH
i
ii
i
ii
Code II:
H(x) = 2.11 bits ; R
= 2.66 ; ..8.066.2
11.2)(==
R
xHLetter p(xi) - log2p(xi) Code Length
x1
x2
0.45
0.35
1.156
1.52
1
0 1
1
2
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x3 0.2 2.33 0 0 2
9 7.0)(
5 5.1)(
5 1 8.1)(l o g)()(
3
1
3
1
2
==
==
==
=
=
R
xH
nxpR
b i t sxpxpxH
i
ii
i
ii
Letter pair p(xi ,yj) -log2p(xi, yj) Code Length ni
x1 x1 0.2025 1 0 2x1 x2 0.1575 0 0 0 3
x2 x1 0.1575 0 1 0 3
x2 x2 0.1225 0 1 1 3
x1 x3 0.09 1 1 0 3
x3 x1 0.09 0 0 1 0 4
x2 x3 0.07 0 0 1 1 4
x3 x2 0.07 1 1 1 0 4
x3 x3 0.04 1 1 1 1 4
9.0)(
/0 6 7.3
0 3 6.3)(==
=
=
R
xH
l e t t e r b i t sR
b i t sxH
Figure Code
Or code II figure
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%9
)(
/0 6 7.3
0 3 6.3)(
==
=
=
R
xH
l e t t e r b i t sR
b i t sxH
Coding for analog sources
An analog source emits a waveform ( )tx that is a sample function of a stochasticprocess ( )tX .
If ( )tX is a stationary stochastic process: autocorrelation function is ( )xx power density function is ( )fxx
If ( )tX is a stationary stochastic process with band limited, then ( ) 0= fxx .For the signals above, we ca use the Sampling Theorem:
( )
=
+
=
s
s
s
s
n s
f
ntf
f
ntf
f
nXtX
sin
where: max2 ffs = -Nyquist criterionmax
f -highest frequency of the signal
sf -sampling frequency
Type of encoding represent each sample (discrete amplitude level) by a
sequence of binary digits. Hence for L levels the number of binary digits is:
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R=[ ] k
k
forLL
forLL
21log
2log
2
2
+
=
If the levels are not equally probable and the probabilities of the output levels are
known ,then we use the Huffman coding(entropy coding).
If the levels are not equally probable and the probabilities are not known, we canestimate the encoding source.
Quantization means both compression of data and distortion of the waveform (loss
of signal fidelity).
Minimization of the distortion can be made with PCM, DPCM and DM.
PCM-Pulse Code Modulation
( )tx -a sample function emitted by the source
( )nx
-samples taken at a sample ratemax2 ffs
In PCM, each sample of the signal is quantized to one of R2 amplitude levels and
the rate of source is sbitsfR s .
nnn qxx +=~
- mathematical model of quantization
-nx
~-quantized value of nx
- nq -noise
Uniform Quantizer
7 bit quantizerinput-output characteristic for a uniform quanitzer:
pdf- probability density function
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( ) =1qp
R= 2 - size of quantization
MSE mean square error -2q
( ) ( )
=
=
=
== 2
2
222
2
3
22
2
22
12
2
123
11R
qdqqdqqqpq
( ) ( ) dBRRqqR
8.10612log102log2012
2log10log10
2
2log
2 ====
ex: for R=7bits,
( ) dBdBq 8.522 =
uniform quantization non-uniform quantization
An uniform quantizer provides the same spacing between successive levels
throughout the entire dynamic range of a signal.
A better approach is to have more closely spaced levels at the large signal amplitude=>non uniform quantizer.
A classic non uniform quantizer is a logarithmic compressor (Javant 1974 for speech
processing)
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y - magnitude of output
x - magnitude of input
- compression factor
( ))1log(
1log
x
xy
+
+=
ex: =225; R=7 => ( ) dBq 772 =
A non quantizer is made from a non-linear device that compresses the signal and a
uniform quantizer.
DPCM Differential Pulse Code Modulation
In PCM each sample is encoded independently. However, most sources
sampled at Nyquist rate or faster exhibit significant correlation between successivesamples (the average change is successive sample and is small).
Exploiting the redundancy => lower rate for the output
Simple solution encoding the differences between successive samples.
Refinement: to predict the current sample based on previous p samples.
=
=p
i
inin xax1
^
;
{ }ia - coefficient of predictionn
x - current sample^
nx- predicted sample
MSE for computation of { }ia coefficients:
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( )
( ) ( ) ( )
=
==
=
+=
=
=
==
p
j
jninji
p
i
p
i
innin
p
i
ninnnnp
xxEaaxxEaxE
xaxExxEeE
111
2
2
1
2^
2
2
The source is stationary with ( )n -autocorrelation function.If the source is wide sense stationary, results:
( ) ( ) ( )
= = =+=
p
i
p
i
p
j
jiip ijaaia1 1 1
20
Minimization of p with respect to predicted coefficients { }ia results in a set oflinear equations:
( ) ( )jjiap
i
i ==1
; j=1,2,,p Yule-Walker equations
If ( )n is unknown apriory, it can be estimated by the relation:
( ) =
+=N
i
niixxN
n1
^ 1; n=0,1,2,,p
nx - sampled value ne - predicted error~
nx - quantized signal~
ne - quantized predicted error
^~
nx- predicted quantized signal nq - quantized error
Encoder:
Decoder:
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=
==p
i
ininnnn xaxxxe1
~^~
nnnnnnnnnnnxxxxexxeeeq =+=
==
~
^
~~
^
~~~
The quantized value~
nx differs from the input nx by the quantization error nq
independent of the predictors prediction. Therefore the quantization errors do not
accumulate.
Improvement in the quality of estimation => inclusion of the linearly filtered past
values of the quantized error.
= =
+=p
i
m
i
iniinin ebxax1 1
~~
Encoder:
Decoder
figure
Adaptive PCM and DPCM
Many real sources are quasistationary. The variance and autocorrelation functions of
the source output vary slowly with time. The PCM and DPCM encoders are
figure
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designed on the basis that the source output is stationary. The efficiency and
performance can be improved by having them adapted to the slowly time-variant
statistics of the source. The quantization error q n has a time-variant variance.
Reducing of the dynamic range of q n by using an adaptive quantifier
n+1 = n M(n), where :
n is the step size of the quantizer for processingM(n) is the multiplication factor dependent of the quantizer level for the sample x(n)
PCM DPCM
2 3 4 2 3 4
M(1) 0.60 0.83 0.8 0.8 0.9 0.9
M(2) 2.20 1.00 0.8 1.6 0.9 0.9
M(3) 1.00 0.8 1.25 0.9M(4) 1.50 0.8 1.70 0.9
M(5) 1.20 1.20
M(6) 1.60 1.60
M(7) 2.00 2.00
M(8) 2.40 2.40
In DPCM the predictor can also be made adaptive when the source output is
stationary. The predictor coefficients can be changed periodically to reflect the
changing signal statistics of the source. The short-term estimate of the
autocorrelation function of x n replaces the ensemble correlation function. Thepredictor coefficients are passed along the receiver with the quantization error en.
The source predictor is implemented at the receiver and a higher bit rate results. The
advantage of decreased bit rate produced by using a quantizer with fewer bits is lost.
An alternative is the using of a predictor at the receiver that may compute its own
predicted coefficients from en and xn .
Delta Modulation (DM)Adaptive Delta Modulation (ADM)
Delta Modulation is a simplified form of DPCM in which a 2-level (1 bit) quantizer
is used in conjunction with a fixed first-order prediction.
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x n = xn-1 en-1
x n = x n-1 +q n-1 + x n-1 x n-1 = x n-1 + q n-1q n = en en = en (x n x n)
The estimated value of x n is the previous sample x n-1 modified by the quantization
noise q n-1. The difference equation represents an integrator with an input e n.
An equivalent realization of the one-step predictor is an accumulator with an input
equal to the quantized error e n
In general the quantized error signal is scaled by some value, say 1, called stepsize.
The encoder shown in the figure approximates the waveform x(t) by a linear
staircase function (the waveform must change slowly relative to the sampling rate. Itresults that the sampling rate must be about 5 times greater or equal to the Nyquist
rate.
+
Sampler _ Quantizer to transmitter
xn= xn-1
Encoder Unit delay + z-1
en LPF output
Decoder
z-1
____________________________________________________________________
____________________________________________________________________
+ en = 1Sampler _ Quantizer to transmitter
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Encoder xnAccumulator +
1
en+ Accumulator LPF output
1 Decoder
Joyant (1970)e n e n-1 n = n-1 K , K > 1
en = 1Sampler Quantizer
z-1
Accumulator +
Continuous variable slope:
+=
+=
21
11
K
K
nn
nn
21
~;~;~ nnn eee have the same sign
K1 >>K2 > 0 ; 0
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Instead of sending the samples of the source waveform, the parameters of the linear
system are transmitted along with the appropriate excitation signal.
- The source output sampled at a rate > Nyquist rate
{ } 1,0 = Nnnx
- The sample sequence is assumed to have been generated by an all-pole filter with:
=
=
p
k
k
kza
GxH
1
1
)(
Excitation functions: - impulse- sequence of impulses
- white noise with unit variance
The difference equations for the filter
= =
+=p
k
p
k
knkknkn vbxaX1 1
or
=
=+=p
k
nknkn NnGvxaX1
,0;
In general, the observed source output does not satisfy the difference equation, only
its model does.
If the input is a white noise sequence or an impulse, we may for an estimate
(prediction) of Xn :
=
>=p
k
knkn nxaX1
0;
The error of the observed value Xn with respect to the predict value nX is
=
==p
k
knknnnnxaxxxe
1
The filter coefficients are chosen to minimize the mean square of the error
= = =
=
+=
==
p
k
p
k
p
m
mkkk
p
k
knknnp
mkaaa
xaxEeE
1 1 1
2
1
2
)(2)0(
])[()(
where )(m is auto correlating function of sequence Xn.
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p is identical to MSE for a predictor used in DPCM and results the Julle-Walker
equations.
To completely specify the filter H(z), the filter gain G has to be specified.
p
p
k
knknnn xaxEGvEGvGE ==== =
])(])[(]))([(2
1
2222
where p is the residual MSE obtained after substituting the optimum predictioncoefficients (Julle-Walker equations).
=
==p
k
kp kaG1
2)()0(
Usually the time auto correlation function of the source output is unknown, and we
use the estimate
=+ ==
nN
i
nii NnxxN
n1
1,0;1
)(
Julle-Walker equation =>)()(
,1);()(1
jbyreplacedjwith
pjjjiap
ki
= ==
Matriceal form of Julle-Walker equation:]]][ =a
where)(]
]
)(][
ielementswithvectorcolumnR
tscoefficienpredictorofvectorcolumnRa
jielementswithR
pxp
pxp
ji
pxp
=
Example : p=4;
)0()1()2()3(
)1()0()1()2(
)2()1()0()1(
)3()2()1()0(
= - Toeplitz matrix
Recursive algorithm for the inverse Toeplitz matrix is done by Durbin and Levinson
(1959):
) Start with 1st order predictor )0(0 =
2)
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)0()1(;
,1
1,1;)1(
)()(
)1(
,1,1
1
1
,1
==
=
=
=
=
=
iikiiiikiik
i
k
ki
ij
jiiji
aaaaa
pi
iki
kiai
a
a
3)
=
==
==p
k
kp
kpk
kaG
pkaa
1
2
,
)()0(
,1;
The recursive solutions give the predictive coefficients for all orders less than p.
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The residual MSE pii ,1; = form a monotone decreasing sequence:
1
... 011