Digital fiiter

EXPERT SYSTEMS AND SOLUTIONS
Email: expertsyssol@gmail.com
expertsyssol@yahoo.com
Cell: 9952749533
www.researchprojects.info
PAIYANOOR, OMR, CHENNAI
Call For Research Projects Final
year students of B.E in EEE, ECE,
EI, M.E (Power Systems), M.E
(Applied Electronics), M.E (Power
Electronics)
Ph.D Electrical and Electronics.
Students can assemble their hardware in our
Research labs. Experts will be guiding the
projects.

Digital Filter Structures

Content

Introduction
IIR Filter Structures
FIR Filter Structures

Introduction

 What is a digital filter
 A filter is a system that is designed to remove some
component or modify some characteristic of a
signal
 A digital filter is a discrete-time LTI system which
can process the discrete-time signal.
 There are various structures for the implementation
of digital filters
 The actual implementation of an LTI digital filter can
be either in software or hardware form, depending
on applications

Introduction

 Basic elements of digital filter structures
 Adder has two inputs and one output.
 Multiplier (gain) has single-input, single-output.
 Delay element delays the signal passing through it by
one sample. It is implemented by using a shift register.
a
z-1

a z-1

x (n) b0 y(n)
1 a2 2

z-1
z-1
a1
a1
5 3
z-1
x (n) y(n)
0 b a2 z-1
4

y( n) = b0 x ( n) + a1 y( n − 1) + a 2 y( n − 2)
w 2 ( n) = y( n)
w 3 ( n) = w 2 ( n − 1) = y( n − 1)
w 4 ( n) = w 3 ( n − 1) = y( n − 2)
w 5 ( n) = a1 w 3 ( n) + a 2 w4 ( n) = a1 y( n − 1) + a 2 y( n − 2)
w1 ( n) = b0 x( n) + w5 ( n) = b0 x ( n) + a1 y( n − 1) + a 2 y( n − 2)

Introduction
 The major factors that influence the choice of a
specific structure
 Computational complexity
refers to the number of arithmetic operations (multiplications,
divisions, and additions) required to compute an output value y(n)
for the system.
 Memory requirements
refers to the number of memory locations required to store the
system parameters, past inputs, past outputs, and any
intermediate computed values.
 Finite-word-length effects in the computations
refers to the quantization effects that are inherent in any digital
implementation of the system, either in hardware or in software.


 The characteristics of the IIR filter
 IIR filters have Infinite-duration Impulse Responses
 The system function H(z) has poles in 0 <| z |< ∞
 An IIR filter is a recursive system
M

Y (z) ∑ bk z − k
b0 + b1 z −1 +  + bM z − M
H (z) = = k =0
=
X (z) N
1 − (a1 z −1 +  + a N z − N )
1 − ∑ ak z − k
k =1

N M
y( n) = ∑ a k y( n − k ) + ∑ bk x( n − k )
k =1 k =0

The order of such an IIR filter is called N if aN≠0


 Direct form
In this form the difference equation is implemented
directly as given. There are two parts to this filter,
namely the moving average part and the recursive
part (or the numerator and denominator parts).
Therefore this implementation leads to two versions:
direct form I and direct form II structures

M
M N ∑ bk z − k
y( n) = ∑ bk x ( n − k ) + ∑ a k y( n − k ) H ( z ) = k =0
N
k =0 k =1
1 − ∑ ak z − k
k =1

y(n)
b 0
z-1 a1 x (n)
y( n − 1)
b1 z-1
z-1 a2 x ( n − 1)
y ( n − 2) z-1
b 2
x ( n − 2)
aN-1
y( n − N + 1) bM-1
x ( n − M + 1)
z-1 aN
b
z-1
y( n − N ) M
y2 ( n) y1 ( n) x( n − M )
k =1 k =0
y( n) = ∑ bk x ( n − k ) + ∑ a k y( n − k )
N M
Direct form I 

 Direct form II
For an LTI cascade system, we can change the order
of the systems without changing the overall system
response

x (n) y(n)
bM
z-1
bM-1 a1 z-1

a2 z-1
b2 z
-1

b1 z
-1

b0
aN-1 z-1

aN


 Cascade form
In this form the system function H(z) is written as a
product of second-order sections with real coefficients
M M1 M2

∑ bk z − k ∏ (1 − pk z −1 )∏ (1 − qk z −1 )(1 − qk z −1 )
∗

H (z) = k =0
N
=A k =1
N1
k =1
N2
1 − ∑ ak z −k ∏ (1 − c k z −1 )∏ (1 − d k z −1 )(1 − d k z −1 )
∗

k =1 k =1 k =1

M1 M2
M = M1 + 2M 2
∏ (1 − p z )∏ (1 + b
k
−1
1k
−1
z + b2 k z ) −2
N = N1 + 2N 2
H (z) = A k =1
N1
k =1
N2

∏ (1 − c k z −1 )∏ (1 − a1k z −1 − a 2 k z − 2 )
k =1 k =1


1 + b1k z −1 + b2 k z −2
H ( z ) = A∏ −1 −2
= A∏ H k ( z )
k 1 − a1 k z − a2k z k

Each second-order section (called biquads) is implemented in
a direct form ,and the entire system function is implemented
as a cascade of biquads.
 If N=M, there are totally  N + 1  biquads.
 2 
 
 If N>M, some of the biquads have numerator coefficients
that are zero, that is, either b2k = 0 or b1k = 0 or both b2k
= b1k = 0 for some k.
 if N>M and N is odd, one of the biquads must have ak2 = 0,
so that this biquad become a first-order section.

biquad
−1 −2
1 + b1k z + b2 k z
a1k z-1 b1k H k (z) =
1 − a1k z −1 − a 2 k z − 2
a2k z-1 b2k

First-order section

a1k z-1 b1k a1k z-1 b1k

a2k z-1

a1k z-1 a1k z-1

a2k z-1

8 − 4 z −1 + 11z −2 − 2 z −3
Example H (z) =
5 3 1
1 − z −1 + z − 2 − z − 3
4 4 8

( 2 − 0.3799 z −1 )(4 − 1.2402 z −1 + 5.2644 z −2 )
H ( z )=
(1 − 0.25 z −1 )(1 − z −1 + 0.5 z − 2 )

x (n) 4 2 y(n)

z-1 -1.2402
0.25
z-1 -0.3799

-0.5 z-1 5.2644


 Parallel form
In this form the system function H(z) is written as a sum
of sections using partial fraction expansion. Each section
is implemented in a direct form. The entire system
function is implemented as a parallel of every section.
Suppose M=N
M

∑ bk z − k N1
Ak N2
b0 k + b1k z −1
H (z) = k =0
= G0 + ∑ −1
+∑ −1
N
k =1 1 − c k z k =1 1 − a1 k z − a2k z −2
1 − ∑ ak z −k
k =1

N = N 1 + 2N 2


 N +1 
 2 
 
b0 k + b1k z −1
H ( z ) = G0 + ∑
k =1 1 − a1k z −1 − a 2 k z −1

 if N is odd, the system has one first-order
N −1
section and 2 second-order sections.
N
 if N is even, the system has 2 second-order
sections.

Example
1 −1 2
10(1 − z )(1 − z −1 )(1 + 2 z −1 )
H (z) = 2 3
(1 − z −1 )(1 − z −1 )1 − ( + j ) z −1  1 − ( − j ) z −1 
3 1 1 1 1 1
4 8 
 2 2 
 2 2 


A1 A2 A3 A4
H (z) = + + +
3 −1 1 −1 1 1 −1 1 1 −1
(1 − z ) (1 − z ) 1 − ( + j ) z 1 − ( − j )z
4 8 2 2 2 2
A1 = 2.93, A2 = −17.68, A3 = 12.25 − j14.57, A4 = 12.25 + j14.57

− 14.75 − 12.90 z −1 24.50 + 26.82 z −1
H (z) = +
7 −1 3 − 2 −1 1 −2
1− z + z 1− z + z
8 32 2

− 14.75 − 12.90 z −1 24.50 + 26.82 z −1
H (z) = +
7 −1 3 − 2 −1 1 −2
1− z + z 1− z + z
8 32 2
-14.75

7/8 z-1 -12.9
x (n) y(n)
-3/32 z-1

24. 5

1 z-1 26.82

-1/2 z-1


 Transposition theorem
If we reverse the directions of all branch transmittances
and interchange the input and output in the flow graph,
the system function remains unchanged.
The resulting structure is called a transposed structure
or a transposed form.


 The characteristics of the FIR filter
 FIR filters have Finite-duration Impulse Responses,
thus they can be realized by means of DFT
 The system function H(z) has the ROC of | z |> 0 ,
thus it is a causal system
 An FIR filter is a nonrecursive system
 FIR filters can be designed to have a linear-phase
response
N −1 It has N-1 order poles at z = 0
H ( z ) = ∑ h( n) z −n
and N-1 zeros in | z |> 0
n=0

The order of such an FIR filter is N-1


 Direct form
In this form the difference equation is implemented
directly as given. N −1
y( n) = ∑ h( m ) x( n − m )
m =0

z-1 z-1 z-1
x (n)
h(0) h(1) h(2) h(N-2) h(N-1)
y(n)

It requires N multiplications


 Cascade form
In this form the system function H(z) is converted into
products of second-order sections with real coefficients
N It requires (3N/2) multiplications
N −1 2
 
H ( z ) = ∑ h( n) z −n
= ∏ (b0 k + b1k z + b2 k z )
−1 −2

n=0 k =1

x (n) b01 b02 b0x
y(n)
z-1 b11 z-1 b12 z-1 b1x

z-1 b21 z-1 b22 z-1 b2x


 Linear-phase form
 Linear phase:
The phase response is a linear function of frequency
 Linear-phase condition
h( n) = h( N − 1 − n) Symmetric impulse response
h( n) = − h( N − 1 − n) Antisymmetric impulse response
 When an FIR filter has a linear phase response, its
impulse response exhibits the above symmetry
conditions. In this form we exploit these symmetry
relations to reduce multiplications by about half.

If N is odd
N −1
H ( z ) = ∑ h( n) z −n

n=0
N −1
−1
N −1 N −1
2
N −1 −
= ∑ h(n)z
n=0
−n
+ h(
2
)z 2
+ ∑ h(n)z
N −1
−n

n= +1
2
N −1
−1
N −1
−1 let n = N − 1 − m
N −1
2
N −1 − 2
= ∑ h( n) z − n + h(
n=0 2
)z 2
+ ∑ h( N − 1 − m ) z −( N −1− m )
m =0
N −1
−1 let n ← m
2
N − 1 − N2−1
= ∑ h( n)[ z − n ± z −( N −1− n ) ] + h( )z
n=0 2
h( n) = ± h( N − 1 − n)

If N is odd
N −1
−1
N −1
2
N −1 −
H (z) = ∑ h(n)[ z
n=0
−n
±z − ( N −1− n )
] + h(
2
)z 2

x (n)
z-1 z-1 z-1

±1 ±1 ±1 ±1
z-1
z-1
z-1

h(0) h(1) h( 2) N −1 N −1
h( − 1) h( )
2 2
y(n)

1 for symmetric
-1 for antisymmetric

If N is even
N
−1
N −1 2 N −1
H ( z ) = ∑ h( n) z − n = ∑ h( n) z − n + ∑ h( n) z − n
n=0 n=0 N
n=
2
N
2
−1
N
2
−1 let n = N − 1 − m
= ∑ h( n) z − n + ∑ h( N − 1 − m ) z −( N −1− m )
n=0 m =0
N
−1 let n ← m
2
= ∑ h( n)[ z − n ± z −( N −1− n ) ]
n=0
h( n) = ± h( N − 1 − n)

N
If N is even 2
−1

H (z) = ∑ h( n)[ z − n ± z −( N −1− n ) ]
n=0

x (n) z-1
z-1 z-1

±1 ±1 ±1 ±1 ±1 z-1
z-1 z-1 z-1

h(0) h(1) h( 2) N N
h( − 2) h( − 1)
2 2
y(n)

1 for symmetric
-1 for antisymmetric

The linear-phase filter structure requires 50% fewer
multiplications than the direct form.


 Frequency sampling form
This structure is based on the DFT of the impulse
response h(n) and leads to a parallel structure. It is also
suitable for a design technique based on the sampling
of frequency response H(z)

N −1 N −1
1 H (k ) 1
H ( z ) = (1 − z − N ) ∑ − k −1
= H c ( z )∑ H k ( z )
′
N k =0 1 − W N z N k =0

Hc (z) = 1 − z − N
− z−N

It has N equally spaced j
2π
k
zeros on the unit circle zk = e N
, k = 0,1,  , N − 1
ωN
jω − jω N −j ωN
H c (e ) = 1 − e = 2 je 2
sin( )
2
ωN
H c (e jω ) = 2 sin( ) All-zero filter
2
H c ( e jω ) or comb filter
2

 
2π 4π
ω
0
N N

N −1 N −1
H (k )
∑ H k′ ( z ) = ∑ 1 − W −k z −1
k =0 k =0
resonant filter
N

It has N equally spaced poles on the unit circle
2π
j k
zk = e N
, k = 0,1,  , N − 1

The pole locations are identical to the zero locations and that
2π
both occur at ω= k , which are the frequencies at which
N
the designed frequency response is specified. The gains of
the filter at these frequencies are simply the complex-valued
parameters H (k )

N −1 N −1
1 H (k ) 1
H ( z ) = (1 − z −N
)
N ∑ 1 − W −k z −1 = N H c ( z ) ∑ H k′ ( z )
k =0 k =0
N

H ( 0)
0
WN z-1
H (1)
x (n) y(n)
−
WN1 z-1
− z−N

H ( N − 1)

W N ( N −1) z-1
−


Problems
 It requires a complex arithmetic implementation
It is possible to obtain an alternate realization in which only
real arithmetic is used. This realization is derived using the
−
symmetric properties of the DFT and the W N k factor.

 It has poles on the unit circle, which makes this filter
critically unstable
We can avoid this problem by sampling H(z) on a circle |z|
=r where the radius r is very close to one but is less than
one.

jIm[z]
− N N −1
1− z H (k )
∑ 1 − W − k z −1
H (k )
H (z) = unit circle
N k =0 N
r
Re[z]
N −1
1 − r N z−N H r (k )
H (z) =
N ∑ 1 − rW − k z −1
k =0
H r (k )
N

H r (k ) ≈ H (k )

− N N −1
1− r z N
H (k )
H (z) ≈
N ∑ 1 − rW − k z −1
k =0 N

N −1
1 − r N z−N H (k )
H (z) ≈
N ∑ 1 − rW − k z −1
k =0 N

−
By using the symmetric properties of the DFT and the W N k
factor, a pair of single-pole filters can be combined to form
a single two-pole filter with real-valued parameters.
2π
j k
For poles zk = e N
, k = 0,1,  , N − 1
∗
z N −k = zk
2π 2π
−( N − k ) j ( N −k ) j k ∗ −k ∗
rW N = re N
= r (e N
) = r (W N ) = rW k
N

h(n) is a real-valued sequence, so
∗
H ( k ) = H (( N − k )) N RN ( k )

So the kth and (N-k)th resonant filters can be combined to form
a second-order section H k (z ) with real-valued coefficients

H (k ) H(N − k)
H k (z) = − k −1
+
1 − rW N z 1 − rW N ( N − k ) z −1
−

H (k ) ∗
H (k ) b0 k + b1k z −1
= + =
− k −1
1 − rW N z k −1
1 − rW N z −1 2π 2 −2
1 − z 2r cos( k ) + r z
N
k = 1,2,  , N − 1 , N is odd
 2 b0 k = 2 Re[ H ( k )]
 N
 k = 1,2, , − 1, N is even b1k = −2r Re[ H ( k )W N ]
k

 2

N is even jIm[z] N is odd jIm[z]
|z|=r |z|=r

k=N k=0 k=N k=0
2 2
Re[z] Re[z]

If N is even, the filter has a pair of real-valued poles z = ±r
H ( 0) H(N )
H 0 (z) = H N (z) = 2
1 − rz −1 2 1 + rz −1
If N is odd, the filter has a single real-valued pole z=r
H ( 0)
H 0 (z) = −1
1 − rz

second-order section
b0k

2π z-1 b1k
2r cos( k)
N
− r2 z-1

first-order sections
H0 (z) H N (z)
2

H ( 0) H(N )
2
r z-1 r z-1

1 −r z
N −N  N −1
2 
N is even H (z) =  H 0 ( z ) + H N ( z ) + ∑ H k ( z )
N  2 k =1 
 
−N  ( N −1 )

N is odd 1 −r z N 2
H 0 ( z ) + ∑H k ( z )
H (z) =
N  k =1 
 

H0 (z)
1
H1 ( z)
N
x (n) y(n)

H k (z )
− r N z−N 
H N (z)
2


 Fast convolution form
x(n): N1-point sequence
h(n): N2-point sequence
L ≥ N1 + N2 - 1
x (n) X (k )
L-point
DFT
Y (k ) y(n)
L-point
IDFT
h(n)
L-point
DFT
H (k )

h(n)=h(N-1-n) h(n)=-h(N-1-n)
4
6 N = 11 N = 11
2

4
0

2 -2

0 -4
0 1 2 3 4 5 6 7 8 9 10 n 0 1 2 3 4 5 6 7 8 9 10 n

h(n)=h(N-1-n) h(n)=-h(N-1-n)
4
6 N = 10 N = 10
2

4 return
0

2 -2

0 -4
0 1 2 3 4 5 6 7 8 9 n 0 1 2 3 4 5 6 7 8 9 n

Copyright © 2005. Shi Ping CUC

Digital fiiter

More Related Content

What's hot (20)

Viewers also liked (17)

Similar to Digital fiiter (20)

Recently uploaded (20)

Digital fiiter

Editor's Notes