EE 422G Notes: Chapter 9
Instructor: Zhang
Chapter 9 Analysis and Design of Digital Filter
9-1 Introduction
What designs have we done in this course?
What do we mean by filters here?
What do we mean by filters design?
Given specifications (requirements) => H(z)
Let’s see how we can implement a digital filter (processor) if its H(z) is given?
9-2 Structures of Digital Processors
1. Direct-Form Realization
r
Y ( z)
H ( z) 

X ( z)
L z
i 0
m
i
i
1 k j z j
j 1
r
m
i 0
j 1
 y (nT )   Li x(nT  iT )  k j y (nT  jT )
The function is realized!
What’s the issue here?
Count how many memory
elements we need!
Page 9-1
EE 422G Notes: Chapter 9
Instructor: Zhang
Can we reduce this number?
If we can, what is the concern?
r
H ( z) 
 Li z  i
i 0
m
1   k j z i
j 1
 r

1
   Li z  i 
m

i 0
1   k j z  j
H1 ( z )
j 1


H 2 ( z)
 Y ( z)  H ( z) X ( z)  H1 ( z) H 2 ( z) X ( z)
Denote V ( z)  H 2 ( z) X ( z)  Y ( z)  H1 ( z)V ( z)
Implement H2(z) and then H1(z) ?
Why H2 is implemented?
(1)
V ( z)  X ( z)  k1z 1V ( z)    km z mV ( z)
(2)
(1  k1 z 1    k m z  m )V ( z )  X ( z )
V ( z) 
1
m
1  k jz
X ( z)
j
j 1
H2 is realized!
Can you tell why H1 is realized?
What can we see from this realization? Signals at A j and B j : always the same
 Direct Form II Realization
Page 9-2
EE 422G Notes: Chapter 9
Instructor: Zhang
1  0.3z 1  0.6 z 2  0.7 z 3
Example H ( z ) 
(1  0.2 z 1 )3
1  0.3z 1  0.6 z 2  0.7 z 3
Solution: H ( z ) 
1  0.6 z 1  0.12 z  2  0.008 z  3
Important: H(z)=B(z)/A(z) (1) A: 1+…. (2) Coefficients in A: in the feedback channel
Page 9-3
EE 422G Notes: Chapter 9
Instructor: Zhang
2. Cascade Realization
Factorize 1  0.3z 1  0.6 z 2  0.7 z 3  (1  a1 z 1 )(1  a 2 z 2 )(1  a3 z 3 )
H ( z) 
1  0.3 z 1  0.6 z 2  0.7 z 3 1  a1 z 1 1  a 2 z 1 1  a3 z 1



1
1
1
(1  0.2 z 1 ) 3
10
.2
z
10
.2
z
10
.2
z
H (1 ) ( z )
H (2) ( z)
H ( 3) ( z )
General Form
H ( z )  kz
M
i 1 (1  ai z 1 ) j 1 (1  b j z 1 )(1  b j z 1 )
D
D
(1  c z 1 ) l 1 (1  d l * z 1 )(1  d l * z 1 )

k 
1 k 




N1
N2
1
2
*
*

1 q i z 1

1 (b j  b j * ) z 1  b j b j * z 2
1 c k z 1
1 ( d l  d l * ) z 1  d l d l * z 2
Apply Direct II for each!
3. Parallel Realization (Simple Poles)
D1
D2
M
1  el z 1
1
H ( z )   Ai z i   Bk

C

l
1  C z 1 l 1 (1  d l z 1 )(1  d l * z 1 )
i 0
k 1


 
k 



If r  m
realize the
real
poles
realize
the
complex
congugate poles
Example 9-1
(1  z 1 )3
H ( z) 
1
1
(1  z 1 )(1  z 1 )
2
8
cascade and parallel realization!
Solution:
Page 9-4
EE 422G Notes: Chapter 9
Instructor: Zhang
(1) Cascade:
1  z 1
1  z 1
H ( z) 
(1  z 1 )
1
1
(1  z 1 ) (1  z 1 )
2
8
(2) Parallel
(1  z 1 )3
( z  1)3
H ( z) 

1 1
1 1
1
1
(1  z )(1  z ) z ( z  )( z  )
2
8
2
8
In order to make deg(num)<deg(den)
H ( z)
( z  1) 3

1
1
z
z 2 ( z  )( z  )
2
8
A B
C
D
 2 

z z 1/ 2 z 1/ 8
z
A  lim zH ( z ) 
z 0
Partial-Fraction
Expansion for s
z 2 ( z  1)3
(1)3

 16
z 2 ( z  1 / 2)( z  1 / 8) z  0 ( 1 )( 1 )
2 8
Page 9-5
EE 422G Notes: Chapter 9
Instructor: Zhang
C  lim ( z  1 / 2)
H ( z)
( z  1)3
4
 lim 2

z 1 / 2 z ( z  1 / 8)
z
3
D  lim ( z  1 / 8)
H ( z)
( z  1)3
343
 lim 2

z 1 / 8 z ( z  1 / 2)
z
3
z 1 / 2
z 1 / 8
z = anything other than 0, ½, 1/8, 1 B = -112
For example, z=2
H ( z)
13
4


3 15 45
z
4 
2 8
A  16

 4
4
z2
C
 4/3

 8 / 9
1
3
/
2
z
2
D
 343 / 3 343 8


1
15
/
8
3 15
z
8
A
lim H ( z )  lim(  B )
z 0
z 0 z
lim zH ( z )  lim( A  Bz )
z 0
z 0
d (H ( z) z)
B
z 0
dz


3

d  ( z  1)
 B  lim 
1
1 
z  0 dz
 ( z  )( z  ) 
2
8 

 112
lim
4
8 343 8
4 
 ]  112
45
9
3 15
Example 9-2: System having a complex conjugate pole pair at z  ae  j
B  2[
Transfer function
z2
1
H ( z) 

( z  ae  j )( z  ae  j ) (1  ae j z 1 )(1  ae  j z 1 )
z2
1
 2

z  2a(cos  ) z  a 2 1  2a(cos  ) z 1  a 2 z  2
H (e j 2r ) 
1
1  2a(cos )e
 j 2r
 a 2e  j 4r
How do we calculate the amplitude response
| H (e j 2r ) | and H (e j 2r ) ?
Page 9-6
EE 422G Notes: Chapter 9
Instructor: Zhang
How the distance between the pole and the unit circle influence
|H| and H ?
How the distance between the
pole and the unit circle influence
H ?
1 a
How the pole angle
H and H ?

influence
See Fig. 9-7
Page 9-7
EE 422G Notes: Chapter 9
Instructor: Zhang
Page 9-8
EE 422G Notes: Chapter 9
Instructor: Zhang
HW: 9-2(a), 9-11, 9-13, 9-14
Page 9-9
EE 422G Notes: Chapter 9
Instructor: Zhang
9-3 Discrete-Time Integration
A method of Discrete-time system Design: Approximate continuous-time system
t
y (t )   x( )d  a simple system
Integrator

Output
0
 system input
Discrete-time approximation of this system: discrete-time Integrator
1. Rectangular Integration
t
t0 
t
t
0
0
t0
t0
y (t )   x( )d   x( )d   x( )d  y (t 0 )   x( )d
change of y from t0 to t : y(t0 , t )
t = nT,
t0 = nT- T
 y (nT )  y (nT  T )  
nT
nT T
x( )d
 y (nT  T )  y (nT  T , nT )
Constant (  [nT  T , nT ])
T small enough => x( )  x(nT  T )

nT
nT T
y(nT) : System output
x(nT) : System input , to be
integrated
x( )d  
nT
nT T
x(nT  T )d
nT
 x(nT  T )  d
Constant
nT T
 Tx(nT  T )
y (nT )  y (nT  T )  Tx (nT  T )
A discrete-time integrator: rectangular integrator
 Y ( z )  z 1Y ( z )  z 1TX ( z )
Y ( z)
z 1T
 H ( z) 

X ( z ) 1  z 1
Page 9-10
EE 422G Notes: Chapter 9
Instructor: Zhang
2. Trapezoidal Integration
Constants

nT
nT T

x( )d  
nT
nT T
x(nT  T )  x(nT )
d
2
T
( x(nT  T )  x(nT ))
2
y (nT )  y (nT  T ) 
T
T
x(nT )  x(nT  T )
2
2
T
or y (nT )  y (nT  T )  Tx(nT  T )  2 [ x(nT )  x(nT  T )]

difference
between
two
dicretetime
int egrators
 Y ( z )  z 1Y ( z ) 
T
T
X ( z )  z 1 X ( z )
2
2
T 1  z 1
H ( z) 
2 1  z 1
3. Frequency Characteristics
3.1 Rectangular Integrator
Page 9-11
EE 422G Notes: Chapter 9
Instructor: Zhang
z 1T
H r ( z) 
1  z 1
Frequency Response H r (e
Or
H r (e
jT
j 2r
Te  jT
Te  jT / 2
Te  jT / 2
)


1  e  jT e jT / 2  e  jT / 2 2 j sin T / 2
Te jr
)
2 j sin r
1 

  s  2

T 

 /  s  r 


  T  2r 


Amplitude Response
Ar (r ) 
T
2 sin r
0r
Phase Response
 r (r )  e  jr  j  r 

1
(sin r  0)
2
0r
2
1
2
3.2 Trapezoidal Integrator
T 1  z 1
H t ( z) 
2 1  z 1
Frequency Response
T 1  e  j 2r T e jr  e  jr
H t (e ) 

2 1  e  j 2r 2 e jr  e  jr
T 2 cos r T cos r


2 2 j sin r 2 j sin r
T cos r
1
Amplitude: At (r ) 
0r
2 sin r
2
j 2r
Phase:
 t (r )  

2
0r
1
2
sin r  0 

 

cos

r

0


3.3 Versus Ideal Integrator
Ideal (continuous-time ) Integrator
Page 9-12
EE 422G Notes: Chapter 9
Instructor: Zhang
H ( j ) 
1
j
  2f  2rf s
H (r ) 
A(r ) 
1
j 2rf s
1
2rf s
 (r )  

2
when T=1 second (Different plots and relationships will result if T is different.)
(2r  1)
 Low Frequency Range T  1
(Frequency of the input is much lower than the sampling frequency:
It should be!)
Page 9-13
EE 422G Notes: Chapter 9
Instructor: Zhang
cos r 1
T 1
1
  At (r ) 

sin r r
2 r 2rf s
1
1
T
1
  Ar (r ) 

sin r r
2r 2rf s
 High Frequency: Large error (should be)
Example 9-4 Differential equation (system)
dy
 x(t )  y (t )
dt
Determine a digital equivalent.
Solution
(1) Block Diagram of the original system
(2) An equivalent
(3) Transfer Function Derivation
T 1  z 1
Y ( z) 
( X ( z )  Y ( z ))
2 1  z 1
 T 1  z 1 
T 1  z 1
1  
Y ( z ) 
X ( z)
1 
1

2
2
1

z
1

z


Y ( z)
T (1  z 1 )
H ( z) 

Z ( z ) ( T  2)  ( T  2) z 1
Design: 9-4 Find Equivalence of a given analog filter (IIR):
Including methods in Time Domain and Frequency Domain.
9-5 No analog prototype, from the desired frequency response: FIR
9-6 Computer-Aided Design
Page 9-14
EE 422G Notes: Chapter 9
Instructor: Zhang
9-4 Infinite Impulse Response (IIR) Filter Design
(Given H(s)  Hd(z) )
9-4A Synthesis in the Time-Domain: Invariant Design
1. Impulse – Invariant Design
(1) Design Principle
(2) Illustration of Design Mechanism (Not General Case)
Assume:
(1) Given analog filter (Transfer Function)
m
Ki
i 1 s  si
H a (s)  
(a special case)
(2) Sampling Period T (sample ha(t) to generate ha(nT))
Derivation:
(1) Impulse Response of analog filter
m
ha (t )  L ( H a ( s))   k i e  sit
1
i 1
(2) ha(nT): sampled impulse response of analog filter
m
m
i 1
i 1
ha (nT )   ki e  si nT   ki (e  si T ) n
Page 9-15
EE 422G Notes: Chapter 9
Instructor: Zhang
(3) z-transform of ha(nT)
Sampled impulse response of analog filter


m
Z (ha (nT ))   ha (nT ) z  n   k i (e  siT ) n z  n
n 0
n  0 i 1
m

m
i 1
n 0
i 1
  k i  ( e  si T z )  n   k i
1
1  e  siT z 1
m

i 1
ki
1  e  siT z 1
(4) Impulse-Invariant Design Principle
h(nT )  ha (nT )  Z (h(nT ))  Z (ha (nT ))

Digital filter is so designed that its impulse response h(nT)
equals the sampled impulse response of the analog filter ha(nT)
Hence, digital filter must be designed such that
m
Z (h(nT ))  Z (ha (nT ))  
z-transfer function
H (z ) of the digital
filter. Of course,
the z-transfer
function of its
impulse response.
ki
i 1 1  e
m
ki
 H ( z)  
i 1 1  e
 si T
 si T
z 1
z 1
T 0
(5) TZ (h(nT ))  L(ha (t ))
z  e jT
m
=> H ( z )  T 
i 1 1  e
ki
 si T
(scaling)
z 1
(3) Characteristics
(1) H ( z ) z  e jT  H a ( j )
when T 0

frequency response of digital filter
(2) T  0
H ( z ) z  e jT  H a ( j )
(3) Design: Optimized for T = 0
Not for T  0 (practical case) (due to the design principle)
Page 9-16
EE 422G Notes: Chapter 9
Instructor: Zhang
(4) Realization: Parallel
(5) Design Example H a ( s ) 
1
s 1
T  2s
Solution: m  1, K1  1, s1  1
m
Ki
1
2

2

 siT 1
1  e 2 z 1 1  e 2 z 1
z
i 1 1  e
H ( z)  T 
2. General Time – Invariant Synthesis
(1) Design Principle
Page 9-17
EE 422G Notes: Chapter 9
Instructor: Zhang
(2) Derivation
Given: Ha(s)
Xa(s)
T
Find H(z)
transfer function of analog filter
Lapalce transform of input signal of analog Filter
sampling period
z-transfer function of digital filter
(1) Response of analog filter xa(t)
ya (t )  L1[ H a (s) X a (s)]
(2) ya(nT) sampled signal of analog filter output
ya (nT )  [ L1[ H a (s) X a (s)] t  nT
(3) z-transform of ya(nT)
Z[ ya (nT )]  Z{[L1[ H a (s) X a (s)] t  nT }
(4) Time – invariant Design Principle
y(nT )  ya (nT )  Z ( y(nT ))  Z[ ya (nT )]

Digital filter is so designed
that its output equals the sampled
output of the analog filter
Incorporate the scaling :
Z [ y (nT )]  G
 Z [ ya (nT )]


H (z) X (z)
G T

z-transfer function of digital filter
T
(5) Design Equation
H ( z) 
G
Z {[ L1 ( H a ( s ) X a ( s ))] t  nT }
X ( z)
special case X(z)=1, Xa(s) = 1 (impulse)
1
=> H ( z )  GZ{[ L ( H a (s))] t nT }
(6) Design procedure
A: Find L1[ H a (s) X a (s)]  ya (t )
(output of analog filter)
B: Find ya (nT )  ya (t ) t  nT
C: Find Z ( ya (nT ))
D: H ( z )  GZ ( ya (nT ))
Page 9-18
EE 422G Notes: Chapter 9
Example 9-5 H a ( s ) 
Instructor: Zhang
0.5( s  4)
( s  1)( s  2)
Find digital filter H(z) by impulse - invariance.
Solution of design:
(1) Find L1[ H a (s) X a (s)]  ya (t )
X a ( s)  1
H a ( s) 
0.5( s  4)
1.5
1


( s  1)( s  2) s  1 s  2
ya (t )  L1[ H a ( s) X a ( s)]  1.5et  e2t
(2) Find ya (nT )  ya (t ) t  nT
ya (nT )  1.5(eT )n  (e2T )n
(3) Find Z ( ya (nT ))
Z ( ya (nT )) 
1.5
1

1  e T z 1 1  e  2T z 1
(4) Find z-transfer function of the digital filter
H ( z )  GZ ( ya ( nT ))
1
 1.5

 G

 T 1
 2T 1 
1 e z 
1  e z
use G = T
H ( z) 
1.5T
T

T 1
 2T 1
1 e z
1 e z
(5) Implementation
Page 9-19
EE 422G Notes: Chapter 9
Instructor: Zhang
Characteristics
(1) Frequency Response equations: analog and digital
0.5( j  4)
(1  j )(2  j )
1.5T
T
Digital : H (e jT ) 

1  eT e  jT 1  e  2T e  jT
Analog : H a ( j ) 
(2) dc response comparison (   0 )
0.5  4
1
1 2
1.5T
T
Digital: H (e j 0 )  H (1) 

1  e  T 1  e  2T
Analog: H a (0) 
Varying with T (should be)
eT  1  T , e2T  1  2T
1.5T
T
H (1) 

1
1  (1  T ) 1  (1  2T )
2
T  0 : for example T 
 0.31416s
( f s  20)
20
eT  0.7304 , e2T  0.5335
H (1)  1.0745 good enough
2
 0.31416 sec ond
(3) | H a ( j ) | versus | H (e jT ) | : T 
T  0:
20
Using normalized frequency r  f / f s   / s
| H (e j 2r ) |

0.25488  0.06983 cos 2r
10 2.74925  3.51275 cos 2r  0.77932 cos 4r
16   2
| H a ( j )  0.5
4  5 2   4
  2f  2f s r 
2
r
T
 H a (r )
Page 9-20
EE 422G Notes: Chapter 9
Instructor: Zhang
(4) H versus H a
(5) Gain adjustment when T  0
T  0 => frequency response inequality
adjust G => H (e jT )  H a ( j ) at a special 
for example   0
If G = T = 0.3142 => H (e jT ) | 0  1.0745
If selecting G = T/1.0745 => H (e jT ) | 0  1
Page 9-21
EE 422G Notes: Chapter 9
Instructor: Zhang
3. Step – invariance synthesis
1
1
X ( z) 
s
1  z 1
G
H ( z) 
Z {[L1[ H a ( s ) X a ( s )]] |t  nT }
X ( z)
1
 H ( z )  G (1  z 1 ) Z {[ L1[ H a ( s )]] |t  nT }
s
0.5( s  4)
Example 9-6 H a ( s ) 
. Find its step-invariant equivalent.
( s  1)( s  2)
X a ( s) 
Solution of Design
1
0.5( s  4)
1 1.5
0.5
H a ( s) 
 

s
s( s  1)( s  2) s s  1 s  2
1
ya ( nT )  ya (t ) |t  nT  L1[ H a ( s )] |t  nT  1  1.5e  nT  0.5e  2 nT
s
1
H ( z )  G (1  z ) Z [ ya ( nT )]
1
1.5
0.5
 G (1  z 1 )(


1  z 1 1  e T z 1 1  e  2T z 1
Comparison with impulse-invariant equivalent.
Page 9-22
EE 422G Notes: Chapter 9
Instructor: Zhang
9-4B Design in the Frequency Domain --- The Bilinear z-transform
1. Motivation (problem in Time Domain Design)
x (t )

X( f )
Xs( f ) 
xs ( nT )


 Cn X ( f
n  
 nf s )
 Cn X ( f )  [C1 X ( f  f s )  [C1 X ( f  f s )]

Introduced by sampling, undesired!
x(t) bandlimited ( X ( f  f s / 2)  0)
X ( f )  Xs( f )
for | f |
fs
2
X ( f )  X s ( f  fs )
f
for | f | s
2
Consider digital equivalent of an analogy filter Ha(f):
H a ( j )  H a ( j 2f ) )
Ha(f): bandlimited => can find a Hd(z)
Ha(f): not bandlimited => can not find a Hd(z) Such that H d (e jT )  H a ( )
2. Proposal: from  axis to 1 axis
    1  0.5s (  s : given sampling frequency)
(s plane to s1 plane)
Page 9-23
EE 422G Notes: Chapter 9
Instructor: Zhang
Observations: (1) Good accuracy in low frequencies
(2) Poor accuracy in high frequencies
*
(3) 100% Accuracy at   1  
a given specific number such that 0.01  s
Is it okay to have poor accuracy in high frequencies? Yes! Input is bandlimited!
What do we mean by good, poor and 100% accuracy?
Assume (1) H a ( ) 
1
j  1
(originally given analogy filter)
(2) The transform is   0.915244 tan 1
Then,
1
~
1
 H a ( 1 ) .
is a function of 1 . Denote
j 0.915 tan  1  1
j 0.915 tan  1  1
Good accuracy:
H a (  0.3) 
1
1
1
~


 H a (1  0.3)
j 0.3  1 j 0.915 tan( 0.3)  1 j 0.28  1
Poor Accuracy
H a (  1.5) 
1
1
1
~


 H a (1  1.5)
j1.5  1 j 0.915 tan(1.5)  1 j12.9  1
100% Accuracy (Equal)
H a (  0.5) 
1
1
1
~


 H a (1  0.5)
j 0.5  1 j 0.915 tan( 0.5)  1 j 0.5  1
Page 9-24
EE 422G Notes: Chapter 9
Instructor: Zhang
~
~
Is H ( 1 ) bandlimited? That is, can we find a 1 such that H ( 1 ) =0?
Yes, 1   / 2 .  We have no problem to find a digital equivalent
~
H d (e j T )  H a (1 ) without aliasing!
Let’s use  ( L ) as a number (for example 0.2) representing any low frequency,
~
Then, because H a ( ( L)  0.2) is a good approximation of H a ( ( L )  0.2) ,
1
H d (e j (
( L)
0.2 )T
)  H a ( ( L )  0.2) should be a good approximation.
A digital filter can thus be designed for an analogy filter H a ( ) which is not
bandlimted!
Two Step Design Procedure:
Given: analogy filter H a (s)
~
(1) Find an bandlimited analogy approximation ( H a (s1 ) ) for H a (s)
~
(2) Design a digital equivalent ( H d ( z )) for the bandlimited filter H a (s1 ) .
~
Because of the relationship between ( H a (1 ) ) for H a ( ) ,
H d (z ) is also digital equivalent of H a (s) .
The overlapping (aliasing) problem is avoided!
The designed digital filter can approximate H a ( ) (for 1 and  take the
same value) at low frequency.
3.  axis to 1 axis (s plane to s1 plane) transformation
Requirement :     1 
s
2
(  s is given sampling frequency.)
Proposed transformation :
Variable in 
domain


  
T
  C tan  1   C tan 1
1
2
2
 s 
2

Variable in 1
domain
Constant
Effect of C:
We want the transformation map
  r (for example, r  100rad / s ) to 1   r ( * )
=>
 r  C tan
 rT
2
 C   r cot
 rT
2
i.e. when the sampling period T is given, C is the only parameter
which determines what  will be mapped into 1 axis with the
same value.
Page 9-25
EE 422G Notes: Chapter 9
Instructor: Zhang
c2
Example: H a ( s )  2
s  2 c s   c 2
H a ( j ) 
c 2
not bandlimited
( c 2   2 )  j 2 c
If we want to map   2  100 to 1  2  100
 r  200
C  200 cot
200T
2
Hence, for any given T or  s  2
H a ( jC tan
1T
2
1
T
)  H a (( j 200 cot
T
200T
) tan 1 )
2
2
c 2

( c  C 2 tan 2
2
1T
2
)  j 2C tan
1T
is bandlimited as a function of 1 by | 1 |
T
~
H a (1 )  H a ( jC tan 1 )  H a ( j ) when
2

Further
2
s
2
 c


T
1   at low frequencies.
~
H a (1   r  200 )  H a ( j  j r  j 200 )
Exactly holds!
How to select  r or sampling frequency  s at which 1   ?
(1)
 rT
2
should be small?
 rT

1 2
 1  r   1 or  r   s ,

2
s
f s s
(The accuracy should be good at low frequencies)
why? T 
(2) When  r is given or determined by application,  s should be large
enough such that r  s to ensure the accuracy in the frequency
range including  r
Page 9-26
EE 422G Notes: Chapter 9
When
 rT
2
Instructor: Zhang
 1 :
C  r
2
2

 rT T
cot x 
since
1
x
for small x.
4. Design of Digital Filter using bilinear z-transform
~
 A procedure: (1) H a ( j )  H a (1 )


(not bandlimited,
original analog)
or
(bandlimited,
analog)
~
H a (s)  H a (s1 )
~
(2) H d ( z )  H a ( s1 ) |es1T  z


(Transfer
function of
digital filter)
replace
e s1T
by z
* Question: Can we directly obtain Hd(z) from Ha(s) ? Yes! (But how?)
 Bilinear z-transform
sin x
e jx  e  jx
Preparation : (1) tan x 
  j jx
cos x
e  e  jx
T
(2)   C tan 1
2
Hence,   C tan
1T
2
 C ( j)
T
j 1
e 2
T
j 1
e 2
T
j 1
e 2
T
j 1
e 2

Replace j by s , j1 by s1 (   s / j   js,
e s1T / 2  e  s1T / 2
 js   jC s T / 2
e 1  e  s1T / 2
e s1T / 2  e  s1T / 2
1  e  s1T
 s  C s T /2
C
e 1  e  s1T / 2
1  e  s1T
1   js1 )
Page 9-27
EE 422G Notes: Chapter 9
Instructor: Zhang
Replace es1T  z for digital filter
Bilinear z –
transformation
1  z 1
sC
 direct transformation from s to z (bypass s1)
1  z 1
Example
c2
H a ( s)  2
s  2 c s   c 2
Digital Filter
H d ( z) 
c2
(1  z 1 ) 2
1  z 1
2
C
 2 c
 c
1 2
1
(1  z )
1 z
2
 c 2 (1  z 1 ) 2
 2
2
C (1  z 1 ) 2  2 c (1  z  2 )   c (1  z 1 ) 2
C: only undetermined parameter in the digital filter.
To determine C: (1)
T
(s )
(2)
 r (related to the frequency range of interest)
Example 9-7
c2
H a ( s)  2
s  2 c s   c 2
c : break frequency
Take r  c
Consider f c  500Hz
f s  2000Hz
(c  2  500)
1
(T 
 0.0005sec)
fs
 C determined => Hd(z) determined

H d ( z) 
0.292893  0.585786 z 1  0.292893 z 2
1  0.171573 z  2
 H d (e j 2r ) , | H d (e j 2r ) | , H d (e j 2r )
To compare the frequency response with the original analog filter Ha :
f s 4 f c
H a ( j )  H a ( j 2f )  H a ( j 2rf s )  H a ( j8rf c )
( replace s by j8rfc in H a ( s)
)
 | H a | , H a
Page 9-28
EE 422G Notes: Chapter 9
Instructor: Zhang
Too low fs => poor accuracy in fc.
Page 9-29
EE 422G Notes: Chapter 9
Instructor: Zhang
9-4C Bilinear z-Transform Bandpass Filter
1. Construction Mechanism
(1) From an analog low-pass filter Ha(s)
s 2  c 2
to analog bandpass filter H a (
)
sb
s 2  c 2
i.e., replace s by
to form a bandpass filter
sb
In the low pass filter
For example H a ( s ) 
1
s 1
1
s 2  c 2
1
sb
low-pass
band-pass
Why? Original low-pass
H a ( j )

Low => High Gain
High => Low Gain
After Replacement
  2  c 2
 2  c 2
Ha (
)  Ha ( j
)
jb
b
high
 2  c 2  2




=> high => low gain
b
b b
low
 2  c 2
 c 2 / b


=> high => low gain
b

(2) From analog to digital
s 2  c 2
1  z 1
) by C
Replace s in H a (
sb
1  z 1
 H d ( z 1 )
for example
Page 9-30
EE 422G Notes: Chapter 9
1
s 2  c 2
1
sb
Instructor: Zhang

1
C 2 (1  z 1 ) 2
 c 2
1 2
(1  z )
1
1  z 1
b
1  z 1
2. Bilinear z-transform equation
Analog Low-pass
s
digital bandpass filter

Bandpass (analog)
2
s2  c
s b

 1  z 1 
2
 c
C
1 
 1 z 
1  z 1 
C

1  b
1  z 
2
C 2 (1  z 1 ) 2   c (1  z 1 ) 2

C b (1  z  2 )
2
Direct Transformation
s (in low-pass)
C 2 (1  z 1 ) 2   c (1  z 1 ) 2
C b (1  z  2 )
2

C 2  c

C b
s (in low-pass)

A
2
 C 2   c 2  1
1  2 2
z  z 2
2 
 C   c 
1  z 2
1  Bz 1  z 2
1  z 2
with
C 2  c
A
Cb
2
C 2  c
2
C 2  c
2
B2
3. How to select ( C,c ,b ) for bandpass filter
Page 9-31
EE 422G Notes: Chapter 9
Instructor: Zhang
(design)
Important parameters of bandpass
(1) center frequency c
(2) u upper critical frequency
(3)  l low critical frequency
Selection of c for bandpass: c  ul
2
Design of ( C,c ,b )
We want  u  C tan
 uT
2
, Also want  l  C tan
one parameter C => impossible
2
 solution  c  C tan
2
 uT
tan
lT
2
lT
2
2
T
T
 bandwidth  b  C tan u  C tan l
2
2
Hence, A and B can be determined to perform the transform.
4. Convenient design equation
A
C 2  C 2 tan
C 2 (tan

1  tan
 uT
 uT
2
 uT
2
tan
2
tan
 tan
lT
lT
2
lT
2
)
2
T
T
tan u  tan l
2
2
why no C?
Page 9-32
EE 422G Notes: Chapter 9
Instructor: Zhang
C 2  C 2 tan
 uT
tan
 lT
2
2
 T
T
C 2  C 2 tan u tan l
2
2
 T
T
1  tan u tan l
2
2
2
 T
T
1  tan u tan l
2
2
B2
T
2

2
T
cos( u   l )
2
1  Bz 1  z 2
s A
1  z 2
1 tan x tan y cos( x  y )

1 tan x tan y cos( x  y )
In low-pass
1
Example : s   
cos( u   l )
1
1  z 2

1  Bz 1  z 2
A(1  Bz 1  z 2 )   (1  z 2 )
A

1  z 2
5. In the normalized frequency
Reference frequency: sampling frequency f s

f
ru  u  u
s
fs


=>
 uT

rl 
2f u
1
fs
 ru ,
l
f
 l
s fs
lT
2
2
2
 A  cot  (ru  rl )

=>  B  2 cos  (ru  rl )

cos  (ru  rl )

1  Bz 1  z 2
s => A
In low-pass
1  z 2
(s )
 rl
Page 9-33
EE 422G Notes: Chapter 9
Instructor: Zhang
Example 9-9 Lowpass H a ( s ) 
1
s 1
Transfer function of bandpass digital filter
H d ( z)
1
1  Bz 1  z  2
A
1
1  z 2
A and B? Determined by design requirements.
 f u  1000 Hz
sampling frequency fs = 5000Hz

f

500
Hz
 l
ru  f u / f s  0.2
 
rl  f l / f s  0.1
A  cot  (0.2  0.1)  3.0776835
cot  (0.2  0.1)
 1.2360680
cot  (0.2  0.1)
1  z 2
 H ( z) 
4.0776835  3.8042261z 1  2.0776835z  2
H (e j 2r ) ,
| H (e j 2r ) | ,
H (e j 2r )

B2
Page 9-34
EE 422G Notes: Chapter 9
Instructor: Zhang
9-5 Design of Finite-Duration Impulse Response (FIR) Digital Filter
Direct Design of Digital Filters with no analog prototype.
Can we also do this for IIR? Yes!
Using computer program in next section.
9-5A A few questions
1. How are the specifications given?
By given A( ) and  ( )
2. What is the form of FIR digital filter?
y (nT )   h(kT ) x(nT  kT )
Difference Equation
k
(What is T ? sampling period)
H ( z )   h(kT ) z  k
Transfer function
k
3. How to select T ?
4. After T is fixed, can we define the normalized frequency r and
A(r ) and  (r ) ? Yes!
Can we then find the desired frequency response
H (r )  A(r )e j ( r )
r is not
time
? Yes!
5. Why must H( r ) be a periodic function for
digital filter?
H( r ) = H ( n + r ) ? Why? What is its
period?
Page 9-35
EE 422G Notes: Chapter 9
Instructor: Zhang
6. Can H( r ) be expressed in Fourier Series ? Yes!
How?
 0 : sampling frequency? No!
See general formula :



 X ne jn t   X n e jn2f t   X ne
x (t ) 
0
0
n  
1
Xn 
T0
n  
jn
T0 : period of x(t )
2
t
T0
n  
1
 jn0t
x
(
t
)
e
dt

T0
T0
T0 / 2

T0 / 2
 jn
x(t )e
2
t
T0
dt
In our case for H(r):

x  H
H (r )   X n e jn2 r

n  
tr 
1/ 2
 jn 2 r
X


n
1 / 2 H (r )e dr
T0  1 
What does this mean? Every desired frequency response H(r) of digital
filter can be expressed into Fourier Series ! Further, the coefficients of
the Fourier series can be calculated using H(r)!
9-6B Design principle
H (r ) 

X
n  
Denote
n
e jn2 r
hd (nT )  X n


 jn 2 r
 H (r )   hd (nT )e
n  

h (nT )  X  1/ 2 H (r )e jn 2 r dr
n
 d
1/ 2

Consider a filter with transfer function
 hd (nT ) z  n
n  
What’s its frequency response ?

h
n  
d
(nT )(e
j 2 r
)
n


h
n  
d
(nT )e  j 2 rn  H (r )
given specification of digital filter’s frequency response!
Page 9-36
EE 422G Notes: Chapter 9
Instructor: Zhang
9-6 C Design Procedure
(1) Given H(r)
(2) Find H(r)’s
Fourier series
H (r ) 

 h (nT )e
n  
 j 2 rn
d
1/ 2
jn2 r
dr
where hd (nT )  1/ 2 H (r )e
(3) Designed filter’s transfer function

 hd (nT ) z  n
n  
What’s hd(nT) ? Impulse response!
1
2
Example 9-10: H (r )  (1  cos 2 r )
Solution :
(1) Given H (r ) : done
(2) Find H (r ) ’s Fourier series

1
H (r )  (1  cos 2r )   hd (nT )e  jn2 r
2
n  
where hd ( nT ) 
1 1/ 2
(1  cos 2r )e jn 2r dr

2 1 / 2
n=0
hd (0) 
n0
1 1/ 2
1 1/ 2
1 1/ 2
1
(
1

cos
2

r
)
dr

dr

cos 2rdr 



2 1 / 2
2 1 / 2
2 1 / 2
2
1 1 / 2 jn 2 r
1 1 / 2 e j 2 r  e  j 2 r j 2nr
hd (nT )   e
dr  
e
dr
2 1 / 2
2 1 / 2
2
1 1/ 2
1 1/ 2
  e j 2 ( n 1) r dr   e j 2 ( n 1) r dr
4 1 / 2
4 1 / 2
1 1/ 2
1
dr 

4 1 / 2
4
1 1/ 2
1
hd (T )   dr 
4 1 / 2
4
hd (nT )  0
n  0,1
 hd (T ) 


1
1 1

H
(
r
)

hd (nT )e  j 2n r  e  j 2 r   e  j 2 r


4
2 4
n  

1
h (0)  1 ,
hd (T )  , hd (nT )  0, | n | 2
 d
2
4
Page 9-37
EE 422G Notes: Chapter 9
Instructor: Zhang
(3) Digital Filter

h
n  
d
(nT ) z  n  hd (T ) z 1  hd (0)  hd (T ) z 1 
1
1 1
z   z 1
4
2 4
9-6D Practical Issues : Infinite number of terms and non-causal
(1) H nc ( z ) 

 hd (nT ) z  n
2M+1 terms
n  
Truncation => H nc ( z ) 
M
 h (nT ) z
n M
n
d
(

 w (n)h (nT ) z
n  
r
d
n
)
Rectangular window function
| n | M
| n | M
1
wr (n)  
0
Truncation  window
Effect of Truncation (windowing):
Time Domain: Multiplication ( h and w )
Frequency Domain: Convolution
M
sin  (2M  1)r
j 2 r
wr (e
)   e  j 2 n r 
sin  r
n M
(After Truncation: The desired frequency Hr
 H r wr frequency response of truncated filter )
The effect will be seen in examples!
(2) Causal Filters:
H c ( z)  z
M
k = n+M H c ( z ) 
M
h
n M
d
(nT ) z
n

M
h
n M
d
(nT ) z ( M  n )
2M
 hd (kT  MT ) z  k
k 0
Define Lk  hd (kT  Mt ) 
Relationship: H c ( z )  H nc ( z ) z
H c ( z) 
2M
 Lk z  k
k 0
M
 Ac (r )  Anc (r )
j 2r
j 2r
 j 2Mr
H
(
e
)

H
(
e
)
e


Frequency Response
c
nc
 c (r )   nc (r )  2Mr
Page 9-38
EE 422G Notes: Chapter 9
Instructor: Zhang
Design Examples
n

0.54  0.46 cos
M
Hamming window: wh (n)  
0
| n | M
| n | M
Example 9-11 Design a digital differentiator
Step1 : Assign H (r )
H (r ) should be the frequency response of the analog differentiator
H(s) = s
=> Desired H (r )  j | 2 f  j 2 f s r
f  rf s
Step2 : Calculate hd(nT)
1/ 2
hd (nT )  
1 / 2
1/ 2

1 / 2
1/ 2
H (r )e j 2 nr dr  
1 / 2
1/ 2
H (r )e j 2 nr dr  
1 / 2
b
 udv  uv
a
a
b
  vdu
u  r , v  e j 2 nr
a
( j 2 f s r )e j 2 nr dr
fs
n
f
 s
n
f
 s
n

f
 s
n
 j 2 nr
1
j 2 nr 
re

e


j 2n

 r 1 / 2

b
( j 2 f s r )e j 2 nr dr
1/ 2
1 / 2
1/ 2

j 2 nre j 2 nr dr
rde j 2 nr r
1 / 2
re j 2 nr

r 1 / 2
e j 2 nr d (r )

1 / 2
1/ 2
r  1 / 2

r 1 / 2
1
1  j n f s  1
1   j n

e



 2 j 2n 
 2 j 2n  e
n




f
fs
 s e j n  e  j n 
e j n  e  j n
2
2n
j 2 n
f
fs
2 j sin  n
 s 2 cos  n 
2n
j 2n 2

fs
n




Page 9-39
EE 422G Notes: Chapter 9
Instructor: Zhang
0
fs 2 j
fs
 fs
n
(

1
)

sin

n

(1) n
2
 n
n
j 2n
hd (nT )  
 f s  f s sin n  f s [n  sin n]
 n n 2
n 2
n0
d
[n  sin n ]
dn
 f s lim
n 0
d n 2
dn
f
   cos n
 s lim
2 n 0
n
f
d (   cos n ) / dn
 s lim
2 n 0
d (n ) / dn
fs
 2 sin n
 lim
0
2 n 0

 fs
( 1)n

i.e., hd (nT )   n
0
n0
n 1
n0
Step 3: Construct nc filter with hamming window (M=7)
n  1
n 7
fs
f
n
n
H ( z )   (1) wh (n) z   s (1) n wh (n) z  n
n  7 n
n 1 n
H c ( z )  z 7 H ( z )
Page 9-40
EE 422G Notes: Chapter 9
Instructor: Zhang
Example 9-12: Desired low-pass FIR digital filter characteristic
| r | 0.15
1
H (r )  
0.15 | r | 0.5
0
hd (nT )  
0.15
e j 2 n r dr 
0.15
1
1
(e j 0.3  e  j 0.3 ) 
sin 0.3 n
j 2n
n
d (sin 0.3 n) / d ( n)
0.3 cos 0.3 n
 lim
 0.3
 n0
 n0
d ( n) / d ( n)
1
hd (0T )  lim
hd (0)  0.3

 
sin 0.3n
hd (nT )  n
n0
Page 9-41
EE 422G Notes: Chapter 9
Instructor: Zhang
8
NC filter with 17 weight’s window: H NC ( z )   hd (nT ) wh (n) z n , HC  z 8 H NC ( z )
n  8
Page 9-42
EE 422G Notes: Chapter 9
Instructor: Zhang
Example 9-13 (90o phase shifter)
 j
H (r)  
j
hd (nT )  
0
1
 0.5  r  0
1/ 2
1 / 2

0  r  0.5
je j 2 n r dr  
0
1
(1  e  jn ) 
( j )e j 2 n r dr
(e jn  1)
2 n
2 n
1
1
cos  n
1

(e jn  e  jn ) 


2 n
n
n
n
1  cosn
sin n
 lim
0
n  0
n  0
n
1
n = 0 => hd (0)  lim
hd (0)  0

 
1
h
(
nT
)

[1  ( 1)n ]
d


n
n0
2

 n

0
2

=> hd ( nT )  n
0
Filter:
n  odd
n  0, n  even
n  odd
n  even
H NC ( z ) 
7
h
n  7
d
(n) wh (n) z  n
M 7
7
HC  z H NC ( z )
Page 9-43
EE 422G Notes: Chapter 9
Instructor: Zhang
Fig. 9-32 Amplitude response of digital 90 degree phase shifter
9-6
Computer-Aided Design of Digital Filters
9-6A Command yulewalk for IIR
Example 9-14
9-6B Command remez for FIR
Example 9-15
Page 9-44
EE 422G Notes: Chapter 9
Instructor: Zhang
Chapter 9 Homework
2, 4, 6, 11, 13, 14, 25, 26, 27, 29, 30, 31, 44, 48, 53, 61, 62, 64, 66, 68, 69
Page 9-45