digital filters- firsite.iugaza.edu.ps/masmar/files/dsp_disc_digital-filters_part3.pdf · 4 linear...
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1
3
FIR
22 1
2
:
1+Z+ZLet h(n)={1,1,1} , H(z)=Z Z 1
Z
no poles except at Z=0
Ex
4
LINEAR PHASE FIR FILTERS
Type 1 ,2,3 and 4
Filter Design: FIR
1. Windowed Impulse Response
2. Window Shapes
3. Design by Iterative Optimization
FIR filtersno poles (just zeros)
no precedent in analog filter design
5
The coefficients are symmetric
h(n) = h(N − n)
and the order N is even.
General form of a FIR filter type I
Ex: h(n)={h(0),h(1),h(2),h(3),h(4)}The order N=4Number of coefficient=N+1=5h(0)=h(4) , h(1)=h(3)
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7
Ex:
h(n)={h(0),h(1),h(2),h(3),h(4)}
Symmetric FIR filter
Prove that FIR filters have a linear phase.
1 2 3 4
2 2 1 2
2 2 2 1 1
2 2 2
( ) (0) (1) (2) (3) (4)
(0) (1) (2) (3) (4)
( ) ( ) 4 (3)
( ) (0) (1) (2)
( ) (0) (1) (2)j j j j j
H z h h Z h Z h Z h Z
Z h Z h Z h h Z h Z
h n h N n h(0) = h( ) , h(1) h
H z Z h Z Z h Z Z h
H e h e e h e e h
2
2
( ) 2 (0)cos(2 ) (1)cos(2 ) (2)
( ) *
j
j
H e h h h
H e real
8
2 ( )
Rel Rel
Rel
Rel
2
Rel
Rel
Rel
( ) * ( ) | ( ) |
: ( ) 0
( ) 2
: ( ) 0
( ) ( )
2 ( ) 0( )
2 ( ) 0
j j
j j
H e H H e
If H
If H
H e e H
H
H
9
The coefficients are symmetric
h(n) = h(N − n)
and the order N is odd.
Ex: h(n)={h(0),h(1),h(2),h(3),h(4),h(5)}The order N=5Number of coefficient=N+1=6h(0)=h(5) , h(1)=h(4) , h(2)=h(3)
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The coefficients are antisymmetric
h(n) = -h(N − n)
and the order N is even.
Ex: h(n)={h(0),h(1),h(2),h(3),h(4)}The order N=4Number of coefficient=N+1=5h(0)=-h(4) , h(1)=-h(3)
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The coefficients are antisymmetric
h(n) = -h(N − n)
and the order N is odd.
Ex: h(n)={h(0),h(1),h(2),h(3),h(4),h(5)}The order N=5Number of coefficient=N+1=6h(0)=-h(5) , h(1)=-h(4) , h(2)=-h(3)
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Properties of Linear Phase FIR Filters
For the four types of FIR filters discussed above:
linear phase (or equivalently constant group delay)
( ) ( ) ( ) ( )oj n
oY e X y n x n n
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1 3 2 3
2 3 2
1 1 1 1
* *
* * *
Z Z Z Z
Z Z Z
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54.1
1
54.1
2
128.
1 2 3 4 5
6
3
4
6
6 5 4 3 2 1
6
( ) 0.2 0.8 0.6 0.4
0.2 0.8 0.6 0.4( )
0.5861 0.8102 1
0.5861 0.8102 1
0.432 0.5396 0.6
:
9
o
o
o
j
j
j
H z 0.4 + 0.6Z 0.8Z Z Z Z Z
0.4Z + 0.6Z 0.8Z Z Z ZH z
Z
zeroe
Z j e
Z j e
Z j e
Z
s
type(1)
128.6
128.6
5
12
1 2 3
8.6
6
4 5 6
6
0.432 0.5396 0.69
0.904 1.1290 1.45
0.904 1.1290 1
:
6 poles at Z=0
( )( )( )( )( )( )( )
.45
o
o
o
j
j
j
j
Poles
Z Z Z Z Z Z Z Z Z Z Z ZH z
j
Z
e
Z e
Z j e
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54.1
1
54.1
2
128.6
3
128.6
4
128.6
5
128.6
6
1
1
0.69
0.69
1.45
1.45
o
o
o
o
o
o
j
j
j
j
j
j
Z e
Z e
Z e
Z e
Z e
Z e
128.6
6128.63
128.6
5128.64
1 11.45
0.69
1 11.45
0.69
o
o
o
o
j
j
j
j
e ZZ e
e ZZ e
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54.1
1
54.1
2
128.6
3
128.6
4
128.6
5
128.6
6
1
1
0.69
0.69
1.45
1.45
o
o
o
o
o
o
j
j
j
j
j
j
Z e
Z e
Z e
Z e
Z e
Z e
1 1 *Z Z
3 3 *Z Z
3 3
1 1
*Z Z
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Properties of Linear Phase FIR Filters
Type 4Type 3Type 2Type 1
have an odd
number of
zeros
have an odd
number of zeros
have an even
number of
zeros or no
zeros
have an even
number of
zeros or no
zeros
Zeros at
Z=1
have an even
or odd
number of
zeros
have an odd
number of zeros
have an odd
number of
zeros
have an even
number of
zeros or no
zeros
Zeros at
Z=-1
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1 2
For type(2) FIR filter,has the following zero locations:
1 1Z , Z
2 2
(a) Find the remaining zeros ???
(b) Find h(n) ???
j j
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1 2 1
3 4 3
1 1
90 90
5 6 5
For type(2) FIR filter,has the following zero locations:
1 1Z = Z =Z *=
2 2
1 1Z 1.414 Z =Z *= 1.414
Z Z *
Z 1 Z Z * 1
:
o o
o o
o o
j45 j45
j45 j45
j j
j 0.707e 0.707e
e e
j e e
have an even number
type(2)
of zer
7assume one Z 1
The order (N)=7 , the length(# of coeffi
the zeros:
cients)=N+1
a
, ,1.4
t
1
=
4
8
o oj45 j45 j450.707e 0.707e e
have
no zeros at z = 1
zero at z =
an odd number of zeros
os or no zeros at z
1
= 1
z = 1
90 90,1.414 ,1 ,1 , 1o o o oj45 j je e e
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1 2 3 4 5 6 7
7
90 90
7
7 6 5 4 3 2
7
( )( )( )( )( )( )( )( )
( )( )( 1.414 )( 1.414 )( 1 )( 1 )( ( 1))( )
2 2.5 0.5 0.5 2.5 2 1( )
( ) {1, 2,2.5,
o o o o o oj45 j45 j45 j45 j j
Z Z Z Z Z Z Z Z Z Z Z Z Z ZH z
Z
Z 0.707e Z 0.707e Z e Z e Z e Z e ZH z
Z
Z Z Z Z Z Z ZH z
Z
h n
0.5, 0.5,2.5, 2,1}
: (0) (7) 2 5 3 4note h h , h(1) = h(6) , h( ) = h( ) , h( ) = h( )
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WINDOWS Method
1 1 1( ) ( ) 1. |
2 2 2
sin( )
, , n 02
cc
cc
c c
c
j nj n j n
j n j n n
n
eh n H e d e d
jn
e en
j n
The ideal LPF has a frequency response characte
1 |
ris
|<( )
0 <
tics:
c
c
H
23Limit in freq. unlimited in time, Sol: Using windows
0
sin( )( ) , , n 0
1 1( ) ( ) | 1 , n=0
2 2
c
c
c
j n
n
c
nh n n
n
h n H e d d
For n = 0
24
,
( )sin( )
,
c
LP
c
n = 0
h nn
n 0n
25
26
( )HPh n
( )BPh n
( )BSh n
?prove
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( )LPh n
( )w n
( )wh n
( )h n
Coefficients of the FIR filter modified by a rectangular window function
,
( )sin( )
,
c
LP
c
n = 0
h nn
n 0n
( ) ( ) ( )LPh n h n w n
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( )LPh
( )w
( ) ( ) ( )LPh h w
Gibbs Phenomenon
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Rectangular window
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The truncation of the fourier series is known to
introduce ripples in the frequency response
characteristic H(w) due to the non uniform
convergence of the fourier series at a discontinuity .
The oscillatory behavior near the band edge of the
filter is called the Gibbs phenomenon
To reduce Gibbs phenomenon used
other different type of windows.
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Other windows
Note: the number of coefficients generated by the window functions is
33
Note: the number of coefficients generated by the window functions is
34
Shapes of several window functions
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3 tap FIR filter
Length 3
the order N 2
N 2M M 1
8000.1
8000
0.1
cc
s
c
Ff
F
36
,
( )sin( )
,
( 1) (1) 0.187
(0) 0.2
c
LP
c
n = 0
h nn
n 0n
h h
h
37
1
( 1) (1) 0.08
(0) 1
M
w w
w
38
( ) ( ) ( )
( 1) ( 1) ( 1) (0.187)(0.08) 0.01497
(0) (0) (0) (0.2)(1) 0.2
( ) {0.01497, ,0.014972
0.0149
}
by one sample(causal)
( ) { ,2,0.0 }7 1497
w
w
w
w
w
h n h n w n
h h w
h h w
h n
shifting
h n
39
40
41
Consider the pole-zero plot shown in figure
(a) Does it represent an FIR filter?
(b) Is it a linear-phase system?
60 60
2
60 60
3 4
5 6
7
4 4(b) ,
3 3
3 3 ,
4 4
4 3
3 4
1
o o
o o
j j
1
j j
Z e Z e
Z e Z e
Z Z
Z
( ) Yes, But it may be linear phase FIR or not.a
7 zeroes
42
2
3 4
5
6
7
*
1 1 ,
*
1
*
1
1
1 1
Z Z
Z ZZ Z
ZZ
Z
60 60
2
60 60
3 4
5 6
7
4 4 ,
3 3
3 3 ,
4 4
4 3
3 4
1
o o
o o
j j
1
j j
Z e Z e
Z e Z e
Z Z
Z
Linear phase
43
1 2 3 4 5 6 7
7
60 60 60 60
7
7 6 5 4 3 2
7
( )( )( )( )( )( )( )( )
4 4 3 3 4 3( )( )( )( )( ( ))( ( ))( 1)
3 3 4 4 3 4( )
0 2.8 2.8 0 1 1( )
( ) {1, 1,0,2.8, 2.8,0,1, 1}
: (0)
o o o oj j j j
Z Z Z Z Z Z Z Z Z Z Z Z Z ZH z
Z
Z e Z e Z e Z e Z Z Z
H zZ
Z Z Z Z Z Z ZH z
Z
h n
note h
(7) 2 5 3 4
the order N=7 , length=N+1=8
anti-symmetric
anti-symmetric (4)
h , h(1) = h(6) , h( ) = h( ) , h( ) = h( )
odd
odd type
60 60 60 60
2 3 4
5 6 7
4 4 3 3 , , ,
3 3 4 4
4 3 , , 1
3 4
o o o oj j j j
1Z e Z e Z e Z e
Z Z Z
44
3 ( ) 5 ( 1) 4 ( 2) 5 ( 3) 7 ( 4)
(b) FIR,No poles
(d) for linear phase we require a symmetric or anti-symmetric
but, h(0) h(4) the filter is not linear phase.
(a) h(n) = n n n n n