discrete -time signals and systems z-transform & iir filters 1
TRANSCRIPT
Yogananda Isukapalli
Discrete - Time Signals and Systems
Z-Transform & IIR Filters 1
{ } { }
1 0
1 1
1
1 0 1
1 0 1
[ ] [ 1] [ ] [ 1]
[ ]
( ) ( ) ( ) ( )
[ 1] [ ] [ 1
)
]
(
st
Consider the difference equation of a general
1 order IIR system
Apply the Z transform on both sides
Z y n Z a y n b x n b x n
y n a y n b x n b x
Y z a z Y z b X z b z X z
Y
n
z
- -
-
= - + + -
Þ
- +
+
= +
+
-
=1 1
1 0 1(1 ) ( )( )a z X z b b z- -- = +
Transfer function: 1st Order IIR Filters
{ }
10 1
11
,
( )(1 )
( )...
( ) ( )( ) =
( ) ( )
k
Unlike FIR IIR filter transfer function is a ratio
of polyn
b b za
B z numerator polynomial is defined by the
weight
Y z B zH z
X
ing coefficie
omia
nts b that mult
l
z A z
s
z
-
-= =
+
-
[ ]
( )...
iply x n
and its delayed versions
A z denominator polynomial is defined by
feedback coefficients
An implementation view
10 1
11
1 2
1 2
0
2
1
1
1
( )
( )(1 )
( ) ( ) ( )1( ) ,
( ) ( )(
[ ] [ 1] [ ] [ 1]
( )
( ) ( )( )
) =( ) ( )
b b za z
using the cascading propertyH z H z H z
y n a
Y z B zH z
H z H z
y n b x n b x n
H z H
B zA z
X z A z
z
-
-
= - + + -
=
=
=
=
=
=+-
Direct form I
Direct form II
Example 1
( )
[ ] 2 [ 1] [ ] 3 4 [ 1]
stFind the transfer function of the 1 order IIR system
y n y n x n x n= - - + -
{ } ( ){ }
( )
( )
( )1
1 1
1 1
1( ) (
( ) 2 ( ) ( ) 3 4 ( )
[ ] 2 [ 1] [ ]
1 3 4 )( )
( ) (1 2
( )(1 2 ) ( )( 1 3 4
4 1]
)
3
)
[
Apply the Z transform on both sides
Z y n Z y n x n x
Y z z Y z X z z X z
Y z z
Y z zH z
X z
X z z
n
z
- -
- -
-
-
= - +
-
= - - + -
Þ
- =
-=
-
- +
+=
Example 2
[ ] [ 1] [ ]
'
[ 1]
',
st
Directly pick the numerator and denominator coefficients
coefficien
Find the tran
ts related with
sfer function of the 1 order IIR syste
numerator
coeffici
m
y n ay
ents
n bx
x
related wi
n cx n= - + + -
' ' ,th denominay rator
1
1
( ) ( )( )
( ) (1 )
Y z b czH z
X z az
-
-
+= =
+
Transfer function: 2nd order IIR Filters
{ }
1 2 0 1 2
1 2 0 1
2
[ ] [ 1] [ 2] [ ] [ 1] [ 2]
[ 1] [ 2] [ ] [ 1]
[ ]
ndConsider the difference equation of a general order IIR system
Apply the Z transform on both sides
Z
y n a y n a y n b x n b x n b x n
a y n a y n by n
x n b xZ
n
= - + - + + - + -
- + - -
-
=+ +
1 2 1 2
1 2 0 1
2
2
1 2 1 21 2 0 1 2
( ) ( ) ( )
( ) ( ) ( )
[ 2]
( )(1 ) ( )( )
Y z a z Y z a z Y z b X z b z X z b
b x n
z X z
Y z a z a z X z b b z b z
- - - -
- - - -
=
+
+ + +Þ
-
+
- - = + +
ì üí ýî þ
1
0 1
1
1
22
22
( )( )( )
( ) (1 )
b b zY zH z
X z a
b zz a z
-
-
-
-
+=
-
+=
-
Example
{ }1 2 1 2
( ) 0.5 [ 1] 0.3 [ 2] [ ] 3 [ 1]
2
[ ] 0
2 [ 2]
( ) 0.5 ( ) 0.3 ( ) ( ) 3 ( ) 2 ( )
.5 [ 1] 0.3 [ 2
(
] [ ]
3 [ 1] 2 [ 2]
ndFind th
Y z Z y n y n x n x n x n
Y z
e transfer function of a order IIR system
y n y n y n x n x n x n
z Y z z Y z X z z X z z X z
Y z
- - - -
= - + - - + -
= - + - - + - - -
Þ -
-
=
-
+ + -
1 2
1
1 2
2 1 2)(1 0.5 0.3 ) ( )( 1 3 2 )
( ) ( 1 3 2 )( )
( ) (1 0.5 0.3 )
Y z z zH z
X z
z
z z
z z X z z- -
-
- -
-
- -
- + -=
- - = - +
-
-
=-
1 0
0
0
1
1
(
)( )
( ) 1
[ ] [ ] [ ]
( ) ( ) ( )
N M
l kl
Mk
kk
Nl
ll
l
k
N Ml k
l kl k
k
b zY z
H zX z
feed
y n a y n l
a coefficients
b coefficient
back
a
feed forwa
b x n k
Y
z
z a z Y z b z X z
srd
= =
- -
= =
-
=
-
=
= - +
=
+
-
=
=
-
-
å
å å
å
å
å
!
!
Transfer function: General expression
Transfer function: Impulse Response
1 0 11
0 1 1 1
1
,
[ ] [ 1] [ ] [ 1]
[ ] [ ] [ 1]1[ ]
, ( ) [ ]
1
n
n
n
zn
n
From previous lecture Impulse response of
Transfer function is also Z trans
y n a y n b x n b x n
h n b a u n b
form of impulse
response H z h
a u n
a
n
u nz
s
z
i
a
¥
-
-
=-¥
-
-
=
= - + + -
= + -
¬¾®-
å
!
10 11 1
1 1
10 1
11
1 1(
,
)1 1
( )
,
1
In summary except for polynomial functions
on both numerator and denominator gener
H z b b za z a z
b
al
properti
b zH z sam
es of
e resul
Z transforms are same for II
ta z
-
- -
-
-
= +- -
+=
-
-
æ ö æ öç ÷ ç ÷è ø è ø
!
R
and FIR filters
Example 1: Application of inverse Z-transform
1
1 1
1
1
1
1
1
1
1
11 3
1 2 1 2
1[ ] & ( ) [ 1]
1
1 3
1 3( )
1 2
( 2) [ ] & 3( 2
( )1 2 1 2
) [ 1]
[ ] ( 2) [ ] 3( 2)
z zn
z zn n
n
z
z z
z
zH z
z
a u n z X z x naz
u n u
H zz z
n
Find the impulse response of the feedback system
h n u n
-
- -
-
-
-
-
-
-
-
-
+ +
-=
--
-
=
-
+
+
-
+
¬¾® ¬¾®
¬¾® ¬
= - -
¾®
-
-
!
1 [ 1]n u n- -
Example 2
1 2
10.2 0.2
2
2
( 2)( 0.2)( 0.6)
2( )
( 0.4 .12)
( 0.2) ( 0.6)
( 0.2) 2
(
) 2.75( 0.6)z z
Find the impulse response
z zc X z
of the feedback system
zH z
z z
z z
z zz zc z c zz z
= =
+=
- +
= +-
- += = =
+
+
+=
+ -
( )
( ) ( )
2
0.6 0.6
2.75 1.75
( 0.6) 2( ) 1.75
( )0.2 0.6
[ ]
2.75(0.2) [ ][
0.2
] 1.75( 0.6) [ ]
zn
n
z
n
z
z zH z
z z
za u
z zc X
nz a
u nh n
n
zz z
u
=- =-
= -- +
-
=- -
+ += = = -
-
¬¾®
ì üï ïí ýï ïî þ
!
Poles & Zeros
0. ,
,
( )
,
FIR filters had poles only at z However IIR filters can
have poles anywhere because the poles are nothi
Consid
ng but the
roots of t
er a fir
he de
st order
nominator polynomial A z
transfer function
=
1 1
1
0 1
1
1
0 1
0
1
1 1
1
1
... 0
1
( )1
0
bZero b b z z
b
b
Pole a
b zH z
a z
z p a
-
-
-
-
+ = Þ =
=-
- = Þ
+
-
=!
Poles & Zeros — 2nd order IIR filter
2 1
1 2
0 1 2
1 2
0 1 2
2
1
1
1 2
2
( )( )(
' '
( )( )
)( ) )
(
1
)
(
Convert the function as a function of z
b z b z bH z
z
A polynomial of degree 'N' has 'N' roots
b b z b zY zH z
X z a
order 2 po
z a
lynomial will have 2 zeros
a z
z
an
a
d 2
- -
- -
+ +
+ += =
-
\
-
=- -
poles
[ ]
( )
2
1 1 0 2
1 2
0
2 2
1 1 0 2 1 1 0
2
0 1
2 2
1 0 2 1
2
0 2
1 2
1 2
0
1
2
0
*2
4 0 4
4,
2
4
,
4,
0
;
2 2
if b b b o
Zeros from numerator
b
r b b b
Then the z and z will
b b b
be a complex conjugate pai
bz z
b
b b b b b b b bz z
b
z
b
z
r
z b b
z
- ± -=
- + -
+ + =
-
- - -=
<
=
<
=
[ ]
( )
21 1 2
1 2
2 21 1 2 1 1 2
1
21
2
2
4,
2
4 4,
0
2
,
2
221 2
1
2 1
2
*1 2
if < 0 or a < -
Then th
a a ap p
a a a
Poles from denominator;
z a
e p and p will be a complex conjugate p
a
air
p = p
a ap p
a +4a 4
z
a
a-
± +=
+ + - +=
- =
=
Example 1
1
...2 0
[ ] 0.5 [ 1]
2 2( )
1 0.
2
0
0.5 0 0.5
[
5 0.5
]
Ze
Find the poles and zeros of the feedback system
y n y n x n
Always convert the transfer
ro
function as a functi
z z
Pole z
on of z
z
z
zH z
z -=
= Þ =
- = Þ =
- +
- -
=
=
!
Fig 14.5
Example 2
1
1
0 , .
... lim ( ) 0
3 3( )
1 0.5
[
0.5 0 0.5
] 0.
0
5 [ 1 3 [ ]
.
] 1
5
z
if all poles and zeros at
z or are also counted no of pole
Zero H z z
Pole z z
Find the poles and zeros of the feedback system
y n y n x n
zH z
z z
-
-
®¥
= ¥
=
= Þ = ¥
- = Þ
-
=-
-
-
=
+
=
!
. s no of zeros=Fig 14.6
Example 3
1
1 2
1 2 2
1 2 2 2
2
1 2
2 2( )
1 ( )
,
2 2 2
... 0;
2( )
1 1
2 2 2
0,
( 1) 0
1
zH z
z zConvert H z into a function o
Find the poles and zeros of the feedback system
f z
z z z zH z
z z z z z
zZero z zs
z
z
z
-
- -
-
- -
=
Þ
+=
- +
+ += =
- +
-
+
+ =
= =
-
+
æ öç ÷è ø
2
31
32
3 3
1 0
1 1 4 1 32 2 2
1 1 4 1 32 2 2
2 ( 1)( )
( )(
. .
)
.
j
j
j j
The system function can now be written in factored form also
z z
p j e
p j e
z zH z
z e z
p l
e
o es
p
p
p p
-
-
- + =
+ -= = + =
- -= = - =
+=
- -
PolesZeros
3 3
2 ( 1)( )
( )( )j j
z zH z
z e z ep p-
+=
- -
Example 4
1 2 2 2
1 2 2 2
1 2
1 2
( ) (1 2
[ ] [ 1] [ 2] [
)( )
( ) (1 ) ( ) ,
1 2 2
] [ 1] [ 2
( )1
2 ]
z z z
Find the poles and zeros of the feedb
Y z z zH z
X z z zConvert H z into a fu
ack syste
nction
z zH z
z z z
of z
m
y n y n y n n n x n
z
x x- -
- -
- -
- -
+ += =
=
-
-
+ + + += =
- -
+ - + + -
-
+ -
-
æ öç ÷è ø
2 2... 0
1
z zr
z
Ze os + +
-
=
2
1
2
1
2
1 1 8 1 7
2 2 2
1
... 1 0
1 1 4 1 5
2 2
1 1 4 1 5
2 2
1 8 1 7
2 2 2
Poles z z
p
p
jz
jz
- + -= = - +
-
- - =
+ + += =
- + -= =
- -= = - -
Fig 14.8
Example 5
1
1 2
1 2
2 2 2
1 2 2
( ) (3 2 )( )
( ) (1 2 3 )
[ ] 2 [ 1] 3 [ 2] 3 [
( )
3 2 3
,
2(
] 2 [ 1] [ 2
)1 2 3
]Find the poles an
Y z z zH z
X z z zConvert H z into a f
d zeros of the IIR filt
z
unction
er
o
y n y n y n x n x n x n
z z z zH z
z z z
f z-
-
-
-
-
-
-
-
=
+ +=
- - -
+ +
+ + -
=-
+=
-
-
=
+
æ
+
ö+ç ÷+è ø
2
2
3 2 1.
13
.. 0
2
z z
z z
Zeros + +
- +
=
1
1
2
2
2
-2 4 -12 2 2-1 -1 2
2 2
-2 - 4 -12 2 2-1- -1- 2
... 2 3 0
2 4 12 2 81 2
2 2
2 4 12 2 82
2
2
12
2
Poles
j
z z
j
z j
jz j
p j
jp j
+= = + = +
= =
- + =
+ - += = = +
- - -= = = -
=
Fig 14.9
Poles locations & Stability•All FIR filters are stable• The location of poles plays an important role inthe stability of IIR filters due to recursion
1
0 1 1 1
1
0 1
1
1
( )1
[ ] [ ] [ 1]
,
n n
Consider the s
h n b a u n b a u n
ystem functi
b b zH z
o
z
n
athe impulse response
-
-
-
= +
-
-
+=
( )( )
( )
1
0
10 1 1 1
1 1
0 1 1 1
10 1 1
[ ]=
0 0 0
1
lim 0
1,
1
n
n
n
n
h n
for nb for n
b b a a fo
Consider the part with n b b a a
le
r n
t b b a
ka a
k
if
-
-
-
®¥
³ +
+
ì <ïï =íï +
=
³ïî
® <
1
1
1
,
1
lim 1
n
n
It is desirable to have the impulse response die
down as n otherwise an unboun
a implies
d output result
that th
s
even from a few inpu
e pole wi
t samp
ll be inside the unit
ka if
les
a®¥
® ¥ ³
® ¥
<
Thus a causal LTI IIR system with initial res
circle
t conditions
is stable only if all the poles lie inside the unit
in z
c
plane
ircle
-
Unit Circle
|z|=1Re(z)
w
Im(z)
1
ka
stable system
<
1
ka
Unstable system
³
IIR stability: All poles must be inside unit circle
Location of Zeros has nothing to do with stability
Fig 14.10
Example 1
1
1
1
1
1 2( )
1 0.8
...1 2 0 2.
0.8 1..1 0.8 0 0.
,8
.
zH
Comment on the stability of the IIR filter with transfer functio
z
n
z
zero z z
polinside unit circle
The filter is sta
e z
b e
z
l
-
-
-
-
-=
-
- = Þ =
<
- = Þ =
Fig 14.11
Example 2
1
1
1
1
1 0.8( )
1
[ ] 2 [ 1]
2
[ ] 0.8 [ 1]
2 1, .
...1 0.8 0 0.8...1 2 0 2
Comment on the stability of the filter with diferrence equatio
zH z
z
n
y n y n x n x n
outside unit circle
The filter is
zero z z
pole z z
-
-
-
-
-=
-
= - + - -
>
- = Þ =
- = Þ =
unstable
Example 3
[ ]
2
2 2
2 2
2
2
1 2
2
[ ] [ 2] [
... 0 , [0,0]
...
1( )
11
1
1
]
1
0
C
zeros z z
pole
omment on the stability of the feedback system
y n y n x n
H zz
zz
z
zs
zz z
-
-
=+
= =+ +
+
= Þ =
=
= - - +
æ öç ÷è ø
[ ]1 2
1,
.
1 , [ , ] j on the unit circle
The filte
z p p
r is unsta
j
ble
jÞ = ± - =
=
= -
2nd orderzero
Fig 14.13
Example 4
1 2
1 2
1 2 2
1 2 2
2
2
[ ] [ 1] 2 [ 2] [ ] 2 [ 1] 3 [ 2
1 2 3( )
1 2
1 2 3
1 2
2 3
2
]
Comme
z zH z
z
nt on the stability of the feedback system
y n y n y n
z
z z z
x n x n
z z z
z z
z
x
z
n- -
- -
- -
- -
- +=
-
= - - - + - - + -
+
- +=
- +
- +=
- +
æ öç ÷è ø
2
2
1
2
1
2
... 0
2 4 121 2
2
2 4 121 2
2
... 0
1 1 8 1 7
2 2 2
1 1 8 1
2
7
2 2 2
3
2
zeros
z j
z j
poles
j
z z
z z
p
jp
=
+ -= = +
- -= = -
=
+ -= = +
- -=
+
= -
-
- +
Unstable systemFig 14.14
Example 5
1 2
1 2
1 2 2
1 2 2
2
2
1 5 6( )
4 2
1 5 6
4 [ ]
4 2
5 6
[ 1] 2 [ 2] [ ] 5 [ 1] 6 [ 2]
4 2
z zH z
z z
z
Comment on the stability of the IIR system
y n y n y n x n x n x n
z z
z z z
z z
z z
- -
- -
- -
- -
- +
= - - -
=-
- +=
-
- +=
- +
+ - - + -
+æ öç ÷+è ø
[ ]
2
1
2
2
1
2
... 5 6 0, ( 2)( 3) 0
, [2,3]
...4 2 0
1 1 32 1 318 8 8
1 1 32 1 318 0. 7
8 870
Pol
Magnitude of pole
es z
Zeros
z
p j
p j
z z z
z
s
z
z
- + =
+ -= = +
- -= = -
- + = - -
=
=
=
Þ
Stable systemFig 14.15
James H. McClellan, Ronald W. Schafer and Mark A. Yoder, “ 8.3, 8.4, and 8.9 “Signal Processing First”, Prentice Hall, 2003
Reference