DSP: Unit V — FIR Filters

Unit V: FIR Filters

FIR Filter Realization

Rajini M
Dept. of Elect. & Comm. Engineering,
PES University;

UE18EC253: DSP
DSP: Unit V — FIR Filters

Overview

1 Introduction
2 Structures for Realization of Discrete-Time Systems
Basic Building Blocks
3 Filter Structures
4 IIR Filter Structures
5 FIR Filter Structures
Direct-form structure
Cascade-form structure
Lattice structure
6 Lattice structure (IIR)
7 Summary
DSP: Unit V — FIR Filters
Introduction

Digital Filters

A Linear time-invariant discrete-time systems are

characterised by the the general linear constant coefficient
difference equation
N−1
X M−1
X
y (n) = − ak y (n − k) + bk x(n − k)
k=1 k=0

By means of z -transform, linear time-invariant discrete-time

systems are characterised by ration transfer function
PM−1 −k
k=0 bk z
H(z) =
1 + N−1 −k
P
k=1 ak z
DSP: Unit V — FIR Filters
Introduction

Digital Filters

Depending on the structure of H(z), we have two types of

digital filter:
DSP: Unit V — FIR Filters
Introduction

Digital Filters

Depending on the structure of H(z), we have two types of

digital filter:
1 FIR filters: All zero systems (also called as moving average
system).
M
X
y (n) = bk x(n − k)
k=1
N
X
H(z) = 1 + ak z −k
k=1
DSP: Unit V — FIR Filters
Introduction

Digital Filters
Depending on the structure of H(z), we have two types of
digital filter:
1 FIR filters: All zero systems (also called as moving average
system).
M
X
y (n) = bk x(n − k)
k=1
N
X
H(z) = 1 + ak z −k
k=1
2 IIR filters: All pole systems (also called as autoregressive
system).
N
X
y (n) = − ak y (n − k) + b0 x(n)
k=1
1
H(z) = PN
1+ −k
k=1 ak z
DSP: Unit V — FIR Filters
Structures for Realization of Discrete-Time Systems

Factors influence the choice of structure

The block diagram representation of a system is referred to as

realization.
The factor that plays an role in deciding the structures:
1. Computational complexity
2. Memory requirements
3. Finite word length effects
DSP: Unit V — FIR Filters
Structures for Realization of Discrete-Time Systems
Basic Building Blocks
DSP: Unit V — FIR Filters
Filter Structures

Filter Structures

The broad category of filter structures:

1. Direct-form structure
2. Cascade form structure
3. Lattice structure
4. Parallel structure
FIR filters: 1, 2 and 3.
IIR filters: 1, 2, 3 and 4.
DSP: Unit V — FIR Filters
IIR Filter Structures

IIR Filters

The transfer function of IIR filter (pole-zero systems)

Y (z) b0 + b1 z −1 + · · · + bM−1 z −(M−1)

H(z) = =
X (z) 1 + a1 z −1 + · · · + aN−1 z −(N−1)

The corresponding difference equation of an IIR filter is given by

difference equation
N−1
X M−1
X
y (n) = − ak y (n − k) + bk x(n − k)
k=1 k=0
DSP: Unit V — FIR Filters
IIR Filter Structures

IIR Filter Structures: Direct-form I

Direct-form I
DSP: Unit V — FIR Filters
IIR Filter Structures

IIR Filter Structure: Direct-form II

Direct-form II
DSP: Unit V — FIR Filters
IIR Filter Structures

IIR Filter Structure: Direct-form II

DSP: Unit V — FIR Filters
IIR Filter Structures

IIR Filter Structure: Cascade form

Cascade form
DSP: Unit V — FIR Filters
IIR Filter Structures

IIR Filter Structure: Parallel form

Parallel form
DSP: Unit V — FIR Filters
FIR Filter Structures

FIR Filters
DSP: Unit V — FIR Filters
FIR Filter Structures

A finite duration impulse response filter has a system function

M−1
X
H(z) = bk z −k = b0 + b1 z −1 + · · · + bM−1 z −(M−1)
k=0
DSP: Unit V — FIR Filters
FIR Filter Structures

A finite duration impulse response filter has a system function

M−1
X
H(z) = bk z −k = b0 + b1 z −1 + · · · + bM−1 z −(M−1)
k=0

The impulse response h(n) is


bn , 0 ≤ n ≤ M − 1;
h(n) =
0, otherwise
DSP: Unit V — FIR Filters
FIR Filter Structures

A finite duration impulse response filter has a system function

M−1
X
H(z) = bk z −k = b0 + b1 z −1 + · · · + bM−1 z −(M−1) (1)
k=0

The impulse response h(n) is


bn , 0 ≤ n ≤ M − 1;
h(n) =
0, otherwise

The difference equation is

y (n) = b0 + b1 x(n − 1) + · · · + bM−1 x(n − M + 1) (2)

Length of the filter (number of taps/ coefficients) = M

Order the filter = M − 1
DSP: Unit V — FIR Filters
FIR Filter Structures
Direct-form structure

Direct-form structure
The direct form structure of FIR filter with difference equation
given by eq.(2) is

This structure requires M − 1 memory locations (to store

M − 1 previous inputs).
Complexity: M multiplications and M − 1 additions per
output point.
This realization is often called as traversal of tapped-delay line
filter.
DSP: Unit V — FIR Filters
FIR Filter Structures
Direct-form structure

Example 1: Direct-form structure

Consider the causal linear time invariant system with system

function H(z)

1 1 2
H(z) = 1 − z −1 + z −2 + z −3 + z −4
3 6 5
DSP: Unit V — FIR Filters
FIR Filter Structures
Direct-form structure

Example 1: Direct-form structure

Consider the causal linear time invariant system with system

function H(z)

1 1 2
H(z) = 1 − z −1 + z −2 + z −3 + z −4
3 6 5
Y (z) 1 −1 1 −2 2
H(z) = = 1 − z + z + z −3 + z −4
X (z) 3 6 5
DSP: Unit V — FIR Filters
FIR Filter Structures
Direct-form structure

Example 1: Direct-form structure

Consider the causal linear time invariant system with system

function H(z)

1 1 2
H(z) = 1 − z −1 + z −2 + z −3 + z −4
3 6 5
Y (z) 1 −1 1 −2 2
H(z) = = 1 − z + z + z −3 + z −4
X (z) 3 6 5
1 −1 1 −2 2
Y (z) = X (z) − z X (z) + z X (z) + z −3 X (z) + z −4 X (z)
3 6 5
DSP: Unit V — FIR Filters
FIR Filter Structures
Direct-form structure

Example 1: Direct-form structure

Consider the causal linear time invariant system with system

function H(z)

Y (z) 1 1 2
H(z) = = 1 − z −1 + z −2 + z −3 + z −4
X (z) 3 6 5
1 1 2
Y (z) = X (z) − z −1 X (z) + z −2 X (z) + z −3 X (z) + z −4 X (z)
3 6 5
1 1 2
y (n) = x(n) − x(n − 1) + x(n − 2) + x(n − 3) + + x(n − 4)
3 6 5
DSP: Unit V — FIR Filters
FIR Filter Structures
Direct-form structure

Example 1: Direct-form structure

The direct form structure of the FIR filter

with b0 = 1, b1 = − 13 , b2 = 16 , b3 = 1, b4 = 2
5
DSP: Unit V — FIR Filters
FIR Filter Structures
Direct-form structure

Example 1: Direct-form structure

The equivalent structure is

with b0 = 1, b1 = − 13 , b2 = 16 , b3 = 1, b4 = 2
5
DSP: Unit V — FIR Filters
FIR Filter Structures
Cascade-form structure

The cascade realization follows from eq.(1).

Factor the system function H(z) into second order FIR system
so that

b1 −1 bM−1 −(M−1)
H(z) = b0 (1 + z + ··· + z )
b1 b0
K
Y
= b0 Hk (z)
k=1
DSP: Unit V — FIR Filters
FIR Filter Structures
Cascade-form structure

The cascade realization follows from eq.(1).

Factor the system function H(z) into second order FIR system
so that
b1 −1 bM−1 −(M−1)
H(z) = b0 (1 + z + ··· + z )
b1 b0
K
Y
= b0 Hk (z)
k=1

  
 (M−1)/2
Y
(1 + B1k z −1 + B2k z −2 )

b0  M odd






 k=1
H(z) =  
(M−2)/2


 Y
(1 + B1k z −1 + B2k z −2 ) M even

b (1 + b10 z −1 )

 0



k=1
DSP: Unit V — FIR Filters
FIR Filter Structures
Cascade-form structure

Cascade-form structure

Figure: (a) M even (Order = M − 1 odd) (b) M odd (Order = M − 1

even)
DSP: Unit V — FIR Filters
FIR Filter Structures
Cascade-form structure

Example 2: Cascade-form structure

Obtain the cascade-form structure realization of the given FIR

filter system function
6 7 26 1
H(z) = 1 + z −1 + z −2 + z −3 + z −4
5 5 25 5
FIR filter length M = 5 (Order = 4). With factorization we get
  
1 −1 −2 −1 1 −2
H(z) = 1 + z + z 1+z + z
5 5
DSP: Unit V — FIR Filters
FIR Filter Structures
Cascade-form structure

Example 2: Cascade-form structure

The cascade-form realization of H(z)

DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Lattice structures for FIR filter

Lattice structures offers many advantages over other filter

structures, making them popular structure for digital speech
processing and adaptive filters.
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Lattice structures for FIR filter

Lattice structures offers many advantages over other filter

structures, making them popular structure for digital speech
processing and adaptive filters.
Modular in nature: If filter order is changed then we need to
extra stages to the old stages.
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Lattice structures for FIR filter

Lattice structures offers many advantages over other filter

structures, making them popular structure for digital speech
processing and adaptive filters.
Modular in nature: If filter order is changed then we need to
extra stages to the old stages.
Robust: less sensitive to coefficient quantization effects.
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Lattice structures for FIR filter

Lattice structures offers many advantages over other filter

structures, making them popular structure for digital speech
processing and adaptive filters.
Modular in nature: If filter order is changed then we need to
extra stages to the old stages.
Robust: less sensitive to coefficient quantization effects.
Computationally efficient filter structure.
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Lattice structures for FIR filter

Lattice structures offers many advantages over other filter

structures, making them popular structure for digital speech
processing and adaptive filters.
Modular in nature: If filter order is changed then we need to
extra stages to the old stages.
Robust: less sensitive to coefficient quantization effects.
Computationally efficient filter structure.
We begin the development of lattice structure by considering
sequence of FIR filters. We then extend this to IIR filters (All pole
systems only).
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Consider the FIR filter system functions (all zero system)

m
X
Hm (z) = hm (k)z −k = Am (z), m = 0, 1, 2, · · · , M − 1
k=0

Hm (z) is an mth degree polynomial.

DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Consider the FIR filter system functions (all zero system)

m
X
Hm (z) = hm (k)z −k = Am (z), m = 0, 1, 2, · · · , M − 1
k=0

Hm (z) is an mth degree polynomial.

By definition Am (z) is a polynomial
m
X
Am (z) = 1 + αm (k)z −k , m≥1
k=1

and A0 (z) = 1.
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Consider the FIR filter system functions (all zero system)

m
X
Hm (z) = hm (k)z −k = Am (z), m = 0, 1, 2, · · · , M − 1
k=0

Hm (z) is an mth degree polynomial.

By definition Am (z) is a polynomial
m
X
Am (z) = 1 + αm (k)z −k , m≥1
k=1

and A0 (z) = 1.
By comparing the coefficients of Hm (z) and Am (z), we have
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Consider the FIR filter system functions (all zero system)

m
X
Hm (z) = hm (k)z −k = Am (z), m = 0, 1, 2, · · · , M − 1
k=0

Hm (z) is an mth degree polynomial.

By definition Am (z) is a polynomial
m
X
Am (z) = 1 + αm (k)z −k , m≥1
k=1

and A0 (z) = 1.
By comparing the coefficients of Hm (z) and Am (z), we have
hm (0) = 1 = αm (0).
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Consider the FIR filter system functions (all zero system)

m
X
Hm (z) = hm (k)z −k = Am (z), m = 0, 1, 2, · · · , M − 1
k=0

Hm (z) is an mth degree polynomial.

By definition Am (z) is a polynomial
m
X
Am (z) = 1 + αm (k)z −k , m≥1
k=1

and A0 (z) = 1.
By comparing the coefficients of Hm (z) and Am (z), we have
hm (0) = 1 = αm (0).
hm (k) = αm (k), k = 1, 2, · · · , m.
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Lattice structure (FIR) cont.

If {x(n)} is the input sequence to the filter Am (z) and {y (n)}
is the output sequence, we have
m
X
y (n) = x(n) + αm (k)x(n − k)
k=1

where, αm (k) are the direct-form structure coefficients.

DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Lattice structure (FIR) cont.

If {x(n)} is the input sequence to the filter Am (z) and {y (n)}
is the output sequence, we have
m
X
y (n) = x(n) + αm (k)x(n − k)
k=1

where, αm (k) are the direct-form structure coefficients.

DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Now suppose that we have a first-order filter (m = 1). The

output of such a filter is
y (n) = x(n) + α1 (1)x(n − 1) (4)
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Now suppose that we have a first-order filter (m = 1). The

output of such a filter is
y (n) = x(n) + α1 (1)x(n − 1) (4)
This output can also be obtained form a first-order or
single-stage lattice filter as
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Now suppose that we have a first-order filter (m = 1). The

output of such a filter is
y (n) = x(n) + α1 (1)x(n − 1) (4)
This output can also be obtained form a first-order or
single-stage lattice filter as

f1 (n) = x(n) + K1 x(n − 1)

DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Now suppose that we have a first-order filter (m = 1). The

output of such a filter is
y (n) = x(n) + α1 (1)x(n − 1) (4)
This output can also be obtained form a first-order or
single-stage lattice filter as

f1 (n) = x(n) + K1 x(n − 1)

g1 (n) = K1 x(n) + x(n − 1)
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Now suppose that we have a first-order filter (m = 1). The

output of such a filter is
y (n) = x(n) + α1 (1)x(n − 1) (4)
This output can also be obtained form a first-order or
single-stage lattice filter as

f1 (n) = x(n) + K1 x(n − 1)

g1 (n) = K1 x(n) + x(n − 1)
The output from the top branch is exactly eq.(4), if we select
K1 = α1 (1), K1 is called reflection coefficient.
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Next consider an FIR filter with m = 2. The output from

direct-form filter structure is

y (n) = x(n) + α2 (1)x(n − 1) + α2 (2)x(n − 2) (5)

DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Next consider an FIR filter with m = 2. The output from

direct-form filter structure is

y (n) = x(n) + α2 (1)x(n − 1) + α2 (2)x(n − 2) (5)

By cascading two lattice structures, we can obtain eq.(5)

DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

The output of the first stage is

f1 (n) = x(n) + K1 x(n − 1) (6)
g1 (n) = K1 x(n) + x(n − 1) (7)
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

The output of the first stage is

f1 (n) = x(n) + K1 x(n − 1) (6)
g1 (n) = K1 x(n) + x(n − 1) (7)
The output from second stage is
f2 (n) = f1 (n) + K2 g1 (n − 1) (8)
g2 (n) = K2 f1 (n) + g1 (n − 1)
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

The output of the first stage is

f1 (n) = x(n) + K1 x(n − 1) (6)
g1 (n) = K1 x(n) + x(n − 1) (7)
The output from second stage is
f2 (n) = f1 (n) + K2 g1 (n − 1) (8)
g2 (n) = K2 f1 (n) + g1 (n − 1)
Substitute f1 (n) and g1 (n − 1) from eq.(6) and (7) in f2 (n),
we obtain
f2 (n) = x(n) + K1 x(n − 1) + K2 (K1 x(n − 1) + x(n − 2))
= x(n) + K1 (1 + K2 )x(n − 1) + K2 x(n − 2) (9)
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

The output of the first stage is

f1 (n) = x(n) + K1 x(n − 1) (6)
g1 (n) = K1 x(n) + x(n − 1) (7)
The output from second stage is
f2 (n) = f1 (n) + K2 g1 (n − 1) (8)
g2 (n) = K2 f1 (n) + g1 (n − 1)
Substitute f1 (n) and g1 (n − 1) from eq.(6) and (7) in f2 (n),
we obtain
f2 (n) = x(n) + K1 x(n − 1) + K2 (K1 x(n − 1) + x(n − 2))
= x(n) + K1 (1 + K2 )x(n − 1) + K2 x(n − 2) (9)
Now comparing eq.(5) with (9), we get
α2 (2) = K2 , α2 (1) = K1 (1 + K2 )
α2 (1)
or equivalently, K2 = α2 (2), K1 = (1+α2 (2))
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

By continuing this process for an M − 1 stages,

f0 (n) = g0 (n) = 1
fm (n) = fm−1 (n) + Km gm−1 (n − 1), m = 0, 1, 2, · · · , M − 1
gm (n) = Km fm−1 (n) + gm−1 (n − 1), m = 0, 1, 2, · · · , M − 1
Then the output of (M − 1)th -stage
filter corresponds to the
output of (M − 1) order FIR filter, y (n) = fM−1 (n).
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

The top branch gives the filter output, which is the output of
an FIR filter of order m, that is
m
X
y (n) = fm (n) = αm (k)x(n − k), αm (0) = 1
k=0
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

The top branch gives the filter output, which is the output of
an FIR filter of order m, that is
m
X
y (n) = fm (n) = αm (k)x(n − k), αm (0) = 1
k=0

The z-transform relationship is

m
X
Fm (z) = αm (k)z −k X (z) = Am (z)X (z)
k=0
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

The top branch gives the filter output, which is the output of
an FIR filter of order m, that is
m
X
y (n) = fm (n) = αm (k)x(n − k), αm (0) = 1
k=0

The z-transform relationship is

m
X
Fm (z) = αm (k)z −k X (z) = Am (z)X (z)
k=0

Equivalently,
Fm (z) Fm (z)
Am (z) = =
X (z) F0 (z)
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Now, consider the lower branch of the lattice structure

g2 (n) = K2 f1 (n) + g1 (n − 1)

Substituting,
f1 (n) = x(n) + K1 x(n − 1)
and
g1 (n) = K1 x(n) + x(n − 1)
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Now, consider the lower branch of the lattice structure

g2 (n) = g2 (n) = K2 f1 (n) + g1 (n − 1)

= K2 (x(n) + K1 x(n − 1)) + K1 x(n − 1) + x(n − 2)
= K2 x(n) + K1 (1 + K2 )x(n − 1) + x(n − 2)
= α2 (2)x(n) + α2 (1)x(n) + x(n − 2)
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Now, consider the lower branch of the lattice structure

g2 (n) = K2 x(n) + K1 (1 + K2 )x(n − 1) + x(n − 2)

= α2 (2)x(n) + α2 (1)x(n) + x(n − 2)

Consequently, the filter coefficients are {α2 (2), α2 (1), 1}, whereas
the coefficients for the filter that produces output

f2 (n) = x(n) + K1 (1 + K2 )x(n − 1) + K2 x(n − 2)

are {1, α2 (1), α2 (2)}. Hence the two sets of filter coefficients are
in reverse order.
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Now, consider the lower branch of the lattice structure

g2 (n) = K2 x(n) + K1 (1 + K2 )x(n − 1) + x(n − 2) (10)

= α2 (2)x(n) + α2 (1)x(n) + x(n − 2)

Consequently, the filter coefficients are {α2 (2), α2 (1), 1}, whereas
the coefficients for the filter that produces output
f2 (n) = x(n) + K1 (1 + K2 )x(n − 1) + K2 x(n − 2) are
{1, α2 (1), α2 (2)}. Hence the two sets of filter coefficients are in
reverse order.
Define,
m
X m
X
gm (n) = αm (m − k)x(n − k) = βm (k)x(n − k)
k=0 k=0

where, βm (k) = αm (m − k).

DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

In the z-transform domain,

m
X
Gm (z) = βm (k)z −k X (z) = Bm (z)X (z)
k=0
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

In the z-transform domain,

m
X
Gm (z) = βm (k)z −k X (z) = Bm (z)X (z)
k=0

Gm (z) Gm (z)
Bm (z) = =
X (z) G0 (z)
where, Bm (z) represents the system function of the FIR filter
with coefficients {βm },
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

In the z-transform domain,

m
X
Gm (z) = βm (k)z −k X (z) = Bm (z)X (z)
k=0

Gm (z) Gm (z)
Bm (z) = =
X (z) G0 (z)
where, Bm (z) represents the system function of the FIR filter
with coefficients {βm },

m
X
Bm (z) = βm (k)z −k
k=0
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Since βm (k) = αm (m − k),

m
X
Bm (z) = αm (m − k)z −k
k=0
m
X
= αm (l)z l−m
l=0
m
X
−m
= z αm (l)z l
l=0
−m
= z Am (z −1 )

Hence, Bm (z) is called the reciprocal or reverse polynomial of

Am (z).
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

So far, we have established the following relationships

Fm (z) Fm (z)
Am (z) = =
X (z) F0 (z)

Gm (z) Gm (z)
Bm (z) = =
X (z) G0 (z)
Bm (z) = z −m Am (z −1 )
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Taking z-transform on the following equations

f0 (n) = g0 (n) = 1

fm (n) = fm−1 (n) + Km gm−1 (n − 1), m = 0, 1, 2, · · · , M − 1

gm (n) = Km fm−1 (n) + gm−1 (n − 1), m = 0, 1, 2, · · · , M − 1
We get,
F0 (z) = G0 (z) = 1
Fm (z) = Fm−1 (z) + Km z −1 Gm−1 (z), m = 0, 1, 2, · · · , M − 1
Gm (n) = Km Fm−1 (z) + z −1 Gm−1 (z), m = 0, 1, 2, · · · , M − 1
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

If we divide each equation by X (z), we obtain

A0 (z) = B0 (z) = 1

Am (z) = Am−1 (z) + Km z −1 Bm−1 (z), m = 0, 1, 2, · · · , M − 1

−1
Bm (n) = Km Am−1 (z) + z Bm−1 (z), m = 0, 1, 2, · · · , M − 1
Thus the lattice stage is described in the z-domain by the matrix
equation     
Am (z) 1 Km Am−1 (z)
=
Bm (z) Km 1 z −1 Bm−1 (z)
Remember:
Am (z) = m −k and K = α (m)
P
k=0 αm (k)z m m
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Conversion from lattice coefficients to direct-form

coefficients
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Conversion from lattice coefficients to direct-form coefficients

The direct-form FIR filter coefficients {αm (k)} can be obtained

from the lattice coefficients {Ki } by using the following relations:

A0 (z) = B0 (z) = 1

Am (z) = Am−1 (z) + Km z −1 Bm−1 (z), m = 0, 1, 2, · · · , M − 1

Bm (z) = z −m Am (z −1 )

The solution is obtained recursively beginning with m = 1.

We obtain a sequence of (M − 1) FIR filters, one for each
value of m.
Lets take an example.
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Example 3: Lattice structure (FIR)

Given a three stage lattice filter with the coefficients

K1 = 1/4, K2 = 1/4, K3 = 1/3, determine the FIR filter
coefficients for the direct-form structure.
Solution: Begin with m = 1 and wkt A0 (z) = B0 (z) = 1, we
have
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Example 3: Lattice structure (FIR)

Given a three stage lattice filter with the coefficients

K1 = 1/4, K2 = 1/4, K3 = 1/3, determine the FIR filter
coefficients for the direct-form structure.
Solution: Begin with m = 1 and wkt A0 (z) = B0 (z) = 1, we
have

A1 (z) = A0 (z) + K1 z −1 B0 (z)

= 1 + 1/4z −1
= α1 (0) + α1 (1)z −1
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Example 3: Lattice structure (FIR)

Given a three stage lattice filter with the coefficients

K1 = 1/4, K2 = 1/4, K3 = 1/3, determine the FIR filter
coefficients for the direct-form structure.
Solution: Begin with m = 1 and wkt A0 (z) = B0 (z) = 1, we
have

A1 (z) = A0 (z) + K1 z −1 B0 (z)

= 1 + 1/4z −1
= α1 (0) + α1 (1)z −1
Thus the first-order FIR filter coefficients are α1 (0) = 1 and
α1 (1) = 1/4.
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Example 3: Lattice structure (FIR)

Given a three stage lattice filter with the coefficients

K1 = 1/4, K2 = 1/4, K3 = 1/3, determine the FIR filter
coefficients for the direct-form structure.
Solution: Begin with m = 1 and wkt A0 (z) = B0 (z) = 1, we
have

A1 (z) = A0 (z) + K1 z −1 B0 (z)

= 1 + 1/4z −1
= α1 (0) + α1 (1)z −1
Thus the first-order FIR filter coefficients are α1 (0) = 1 and
α1 (1) = 1/4.
Since B1 (z) is the reversal of A1 (z),
B1 (z) = 1/4 + z −1
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Now, m = 2
A2 (z) = A1 (z) + K2 z −1 B1 (z)
= 1 + 3/8z −1 + 1/2z −2
= α2 (0) + α2 (1)z −1 + α2 (2)z −2
Also,
B2 (z) = 1/2 + 3/8z −1 + z −2
Finally, the addition of third stage
A3 (z) = A2 (z) + K3 z −1 B2 (z)
= 1 + 13/24z −1 + 5/8z −2 + 1/3z −3
= α3 (0) + α3 (1)z −1 + α3 (2)z −2 + α3 (3)z −3
The desired direct-form FIR filter is characterized by the
coefficients,
α3 (0) = 1, α3 (1) = 13/24, α3 (2) = 5/8, α3 (3) = 1/3
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Conversion form direct-form coefficients to lattice

coefficients
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Conversion form direct-form coefficients to lattice coefficients

Suppose that we are given the FIR coefficients for the

direct-form structure or equivalently Am (z), then wkt
Km = αm (m).
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Conversion form direct-form coefficients to lattice coefficients

Suppose that we are given the FIR coefficients for the

direct-form structure or equivalently Am (z), then wkt
Km = αm (m).
Similarly, to obtain Km−1 we need polynomial Am−1 (z).
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Conversion form direct-form coefficients to lattice coefficients

Suppose that we are given the FIR coefficients for the

direct-form structure or equivalently Am (z), then wkt
Km = αm (m).
Similarly, to obtain Km−1 we need polynomial Am−1 (z).
Consequently, we need to find polynomials Am (z) starting
from m = M − 1 and stepping down successively to m = 1.
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Conversion form direct-form coefficients to lattice coefficients

Suppose that we are given the FIR coefficients for the

direct-form structure or equivalently Am (z), then wkt
Km = αm (m).
Similarly, to obtain Km−1 we need polynomial Am−1 (z).
Consequently, we need to find polynomials Am (z) starting
from m = M − 1 and stepping down successively to m = 1.
The desired recursive relation for the polynomial is
Am (z) = Am−1 (z) + Km z −1 Bm−1 (z)
= Am−1 (z) + Km (Bm (z) − Km Am−1 (z))
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Conversion form direct-form coefficients to lattice coefficients

Suppose that we are given the FIR coefficients for the

direct-form structure or equivalently Am (z), then wkt
Km = αm (m).
Similarly, to obtain Km−1 we need polynomial Am−1 (z).
Consequently, we need to find polynomials Am (z) starting
from m = M − 1 and stepping down successively to m = 1.
The desired recursive relation for the polynomial is
Am (z) = Am−1 (z) + Km z −1 Bm−1 (z)
= Am−1 (z) + Km (Bm (z) − Km Am−1 (z))
Solving for Am−1 (z), we obtain
Am (z) − Km Bm (z)
Am−1 (z) = , m = M − 1, M − 2, · · · , 1
1 − Km2
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Conversion form direct-form coefficients to lattice coefficients

Suppose that we are given the FIR coefficients for the

direct-form structure or equivalently Am (z), then wkt
Km = αm (m).
Similarly, to obtain Km−1 we need polynomial Am−1 (z).
Consequently, we need to find polynomials Am (z) starting
from m = M − 1 and stepping down successively to m = 1.
The desired recursive relation for the polynomial is
Am (z) = Am−1 (z) + Km z −1 Bm−1 (z)
= Am−1 (z) + Km (Bm (z) − Km Am−1 (z))
Solving for Am−1 (z), we obtain
Am (z) − Km Bm (z)
Am−1 (z) = , m = M − 1, M − 2, · · · , 1
1 − Km2
Observe that this procedure works as long as |Km | =
6 1.
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Example 4: Lattice structure (FIR)

Determine the lattice coefficients corresponding to the FIR filter

with system function
H(z) = A3 (z) = 1 + 13/24z −1 + +5/8z −2 + 1/3z −3
First we note that K3 = 1/3.
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Example 4: Lattice structure (FIR)

Determine the lattice coefficients corresponding to the FIR filter

with system function
H(z) = A3 (z) = 1 + 13/24z −1 + +5/8z −2 + 1/3z −3
First we note that K3 = 1/3.
Furthermore, B3 (z) = 1/3 + 5/8z −1 + +13/24z −2 + z −3
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Example 4: Lattice structure (FIR)

Determine the lattice coefficients corresponding to the FIR filter

with system function
H(z) = A3 (z) = 1 + 13/24z −1 + +5/8z −2 + 1/3z −3
First we note that K3 = 1/3.
Furthermore, B3 (z) = 1/3 + 5/8z −1 + +13/24z −2 + z −3
The step down relationship with m = 3 yields
A3 (z) − K3 B3 (z)
A2 (z) = 2
= 1 + 3/8z −1 + 1/2z −2
1 − K3
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Example 4: Lattice structure (FIR)

Determine the lattice coefficients corresponding to the FIR filter

with system function
H(z) = A3 (z) = 1 + 13/24z −1 + +5/8z −2 + 1/3z −3
First we note that K3 = 1/3.
Furthermore, B3 (z) = 1/3 + 5/8z −1 + +13/24z −2 + z −3
The step down relationship with m = 3 yields
A3 (z) − K3 B3 (z)
A2 (z) = 2
= 1 + 3/8z −1 + 1/2z −2
1 − K3
Hence, K2 = 1/2.
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Example 4: Lattice structure (FIR)

Determine the lattice coefficients corresponding to the FIR filter

with system function
H(z) = A3 (z) = 1 + 13/24z −1 + +5/8z −2 + 1/3z −3
First we note that K3 = 1/3.
Furthermore, B3 (z) = 1/3 + 5/8z −1 + +13/24z −2 + z −3
The step down relationship with m = 3 yields
A3 (z) − K3 B3 (z)
A2 (z) = 2
= 1 + 3/8z −1 + 1/2z −2
1 − K3
Hence, K2 = 1/2.
and B2 (z) = 1/2 + 3/8z −1 + z −2 .
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Example 4: Lattice structure (FIR)

Determine the lattice coefficients corresponding to the FIR filter

with system function
H(z) = A3 (z) = 1 + 13/24z −1 + +5/8z −2 + 1/3z −3
First we note that K3 = 1/3.
Furthermore, B3 (z) = 1/3 + 5/8z −1 + +13/24z −2 + z −3
The step down relationship with m = 3 yields
A3 (z) − K3 B3 (z)
A2 (z) = 2
= 1 + 3/8z −1 + 1/2z −2
1 − K3
Hence, K2 = 1/2.
and B2 (z) = 1/2 + 3/8z −1 + z −2 .

A2 (z) − K2 B2 (z)
A1 (z) = = 1 + 1/4z −1
1 − K22
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Example 4: Lattice structure (FIR)

Determine the lattice coefficients corresponding to the FIR filter

with system function
H(z) = A3 (z) = 1 + 13/24z −1 + +5/8z −2 + 1/3z −3
First we note that K3 = 1/3.
Furthermore, B3 (z) = 1/3 + 5/8z −1 + +13/24z −2 + z −3
The step down relationship with m = 3 yields
A3 (z) − K3 B3 (z)
A2 (z) = 2
= 1 + 3/8z −1 + 1/2z −2
1 − K3
Hence, K2 = 1/2.
and B2 (z) = 1/2 + 3/8z −1 + z −2 .

A2 (z) − K2 B2 (z)
A1 (z) = = 1 + 1/4z −1
1 − K22
Hence, K1 = 1/4.
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Example 5: FIR filter

Realize the following FIR system H(z) = 1 + 3z −1 + 2z −2 in the

following forms (a) cascade form (b) lattice form.
(a) Cascade-form: H(z) = (1 + z −1 )(1 + 2−2 ) = H1 (z)H2 (z)
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Example 5: FIR filter

Realize the following FIR system H(z) = 1 + 3z −1 + 2z −2 in the

following forms (a) cascade form (b) lattice form.
(a) Cascade-form: H(z) = (1 + z −1 )(1 + 2−2 ) = H1 (z)H2 (z)
(b) Lattice form:

H(z) = A2 (z) = 1 + 3z −1 + 2z −2 = 1 + α2 (1)z −1 + α2 (2)z −2

DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Example 5: FIR filter

Realize the following FIR system H(z) = 1 + 3z −1 + 2z −2 in the

following forms (a) cascade form (b) lattice form.
(a) Cascade-form: H(z) = (1 + z −1 )(1 + 2−2 ) = H1 (z)H2 (z)
(b) Lattice form:

H(z) = A2 (z) = 1 + 3z −1 + 2z −2 = 1 + α2 (1)z −1 + α2 (2)z −2

Hence, K2 = α2 (2) = 2.
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Example 5: FIR filter

Realize the following FIR system H(z) = 1 + 3z −1 + 2z −2 in the

following forms (a) cascade form (b) lattice form.
(a) Cascade-form: H(z) = (1 + z −1 )(1 + 2−2 ) = H1 (z)H2 (z)
(b) Lattice form:

H(z) = A2 (z) = 1 + 3z −1 + 2z −2 = 1 + α2 (1)z −1 + α2 (2)z −2

Hence, K2 = α2 (2) = 2.
And B2 (z) = 2 + 3z −1 + z −2
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Example 5: FIR filter

Realize the following FIR system H(z) = 1 + 3z −1 + 2z −2 in the

following forms (a) cascade form (b) lattice form.
(a) Cascade-form: H(z) = (1 + z −1 )(1 + 2−2 ) = H1 (z)H2 (z)
(b) Lattice form:

H(z) = A2 (z) = 1 + 3z −1 + 2z −2 = 1 + α2 (1)z −1 + α2 (2)z −2

Hence, K2 = α2 (2) = 2.
And B2 (z) = 2 + 3z −1 + z −2
For m = 1,
A2 (z) − K2 B2 (z)
A1 (z) = = 1 + z −1 = α1 (1) + α1 (1)z −1
1 − K22
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Example 5: FIR filter

Realize the following FIR system H(z) = 1 + 3z −1 + 2z −2 in the

following forms (a) cascade form (b) lattice form.
(a) Cascade-form: H(z) = (1 + z −1 )(1 + 2−2 ) = H1 (z)H2 (z)
(b) Lattice form:

H(z) = A2 (z) = 1 + 3z −1 + 2z −2 = 1 + α2 (1)z −1 + α2 (2)z −2

Hence, K2 = α2 (2) = 2.
And B2 (z) = 2 + 3z −1 + z −2
For m = 1,
A2 (z) − K2 B2 (z)
A1 (z) = = 1 + z −1 = α1 (1) + α1 (1)z −1
1 − K22

Hence, K1 = α1 (1) = 1.
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Example 6: FIR filter

Realize the following FIR system

H(z) = 1 − 1.3343z −1 + 0.9025z −2
in the following forms (a) direct form (b) lattice form.
Solution:
Y (z)
(a)Direct-form: H(z) = 1 − 1.3343z −1 + 0.9025z −2 = X (z)

y (n) = x(n) − 1.3343x(n − 1) + 0.9025x(n − 2)

(b)Lattice form:
H(z) = A2 (z) = 1−1.3343z −1 +0.9025z −2 = 1+α2 (1)z −1 +α2 (2)z −2
Hence, K2 = α2 (2) = 0.9025. And Bm (z) = z −m Am (z −1 )

B2 (z) = 0.9025 − 1.3343z −1 + z −2

DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

For m = 2,
A2 (z) − K2 B2 (z)
A1 (z) = 2
= 1 − 0.7013z −1 = α1 (1) + α1 (1)z −1
1 − K2

Hence, K1 = α1 (1) = −0.7013

DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

Example 7: FIR filter

An FIR filter is described by the difference equation

y (n) = 2x(n) + 13/12x(n − 1) + 5/4x(n − 2) + 2/3x(n − 3)

Determine its lattice form.

Solution:

Y (z) = 2X (z) + 13/12z −1 X (z) + 5/4z −2 X (z) + 2/3z −3 X (z)

Y (z)
= H(z) = 2 + 13/12z −1 + 5/4z −2 + 2/3z −3
X (z)
H(z) = K0 A3 (z) = 2(1 + 13/24z −1 + 5/8z −2 + 1/3z −3 )
A3 (z) = 1 + 13/24z −1 + 5/8z −2 + 1/3z −3
Hence,
K3 = α3 (3) = 1/3
DSP: Unit V — FIR Filters
FIR Filter Structures
Lattice structure

B3 (z) = 1/3 + 5/8z −1 + 13/24z −2 + z −3

For m = 2,
A2 (z) = 1 − 3/8z −1 + 1/2z −2
B2 (z) = 1/2 − 3/8z −1 + z −2
Hence, K2 = 1/2 For m = 1,
A1 (z) = 1 + 1/4z −1
Hence, K1 = 1/4.
DSP: Unit V — FIR Filters
Lattice structure (IIR)

Lattice structure (IIR)

DSP: Unit V — FIR Filters
Lattice structure (IIR)

IIR Filter (All pole systems)

Let an N th order all-pole system function H(z) is given by
1 1
H(z) = PN =
1 + k=1 aN (k)z −k A N (z)

The difference equation and direct-form realization of this system

are
N
X
y (n) = − aN (k)y (n − k) + x(n) (11)
k=1
DSP: Unit V — FIR Filters
Lattice structure (IIR)

If we interchange the roles of input and output , we obtain

N
X
x(n) = − aN (k)x(n − k) + y (n)
k=1

or, equivalently
N
X
y (n) = x(n) + aN (k)x(n − k) (12)
k=1

We observe that
Eq.(12) describes an FIR system having the system function
H(z) = AN (z).
Eq.(11) represents an IIR system with the system function
H(z) = AN1(z) .
One system can be obtained from the other by interchanging
the roles of input and output.
DSP: Unit V — FIR Filters
Lattice structure (IIR)

FIR IIR
x(n) = f0 (n) = g0 (n) x(n) = f1 (n)
y (n) = f1 (n) y (n) = f0 (n) = g0 (n)
Therefore, in IIR filter given the input fN (n) = x(n), we must
successively find fN−1 (n), fN−1 (n), · · · , f0 (n). Rearranging FIR
filter fm (n) equation, we get,
fm−1 (n) = fm (n) − Km gm−1 (n − 1)
The equation for gm (n) remains unchanged.
DSP: Unit V — FIR Filters
Lattice structure (IIR)

The resulting set of equations

fN (n) = x(n)

fm−1 (n) = fm (n) − Km gm−1 (m − 1), m = N, N − 1, · · · , 1

gm (n) = Km fm−1 (n) + gm−1 (n − 1), m = N, N − 1, · · · , 1
y (n) = f0 (n) = g0 (n)
The corresponding structure is
DSP: Unit V — FIR Filters
Lattice structure (IIR)

Consider single stage lattice structure for IIR filter

x(n) = f1 (n)
f0 (n) = f1 (n) − K1 g0 (n − 1)
g1 (n) = K1 f0 (n) + g0 (n − 1)
y (n) = f0 (n) = x(n) − K1 y (n − 1)
DSP: Unit V — FIR Filters
Lattice structure (IIR)

Similarly for two stage lattice structure for IIR filter

x(n) = f2 (n)

First stage outputs Second stage outputs

f1 (n) = f2 (n) − K2 g1 (n − 1) f0 (n) = f1 (n) − K1 g0 (n − 1)
g2 (n) = K1 f1 (n) + g1 (n − 1) g1 (n) = K1 f0 (n) + g0 (n − 1)
DSP: Unit V — FIR Filters
Lattice structure (IIR)

y (n) = f0 (n) = g0 (n) = −K1 (1 + K2 )y (n − 1) − K2 y (n − 2) + x(n)

g2 (n) = K2 y (n) + K1 (1 + K2 )y (n − 1) + y (n − 2)
Clearly, the above difference equation represents a two pole IIR
system.
DSP: Unit V — FIR Filters
Lattice structure (IIR)

Taking z-transform,

Y (z) 1 1
= H(z) = =
X (z) A2 (z) 1 + K1 (1 + K2 )z −1 − K2 z −2

Gm (z)
= B2 (z) = K2 + K1 (1 + K2 )z −1 + z −2
Y (z)
It must be noted that coefficients of A2 (z) and B2 (z) are in
reverse order and B2 (z) = z −2 A2 (z −1 )
DSP: Unit V — FIR Filters
Lattice structure (IIR)

Therefore, in general the system function for all-pole IIR system is

Y (z) F0 (z) 1
= =
X (z) FN (z) AN (z)

Similarly,

GN (z) GN (z)
= = BN (z) = z −N AN (z −1 )
Y (z) G0 (z)

Hence, we have the lattice parameters or reflection

coefficients Km , for both the all-pole and all-zero lattice
structure are the same.
The two lattice structures differ only in the interconnections
of their signal flow graphs.
DSP: Unit V — FIR Filters
Lattice structure (IIR)
DSP: Unit V — FIR Filters
Lattice structure (IIR)

Stability of All-pole System (Schur-Cohn Stability

test)

An all-pole system is stable if all the roots of the denominator

polynomial AN (z)are inside the unit circle.
N
X
AN (z) = 1 + = 1 + αN (1)z −1 + · · · + αN (N)z −N
k=1

The Schur-Cohn Stability test states that the polynomial

AN (z) has all its roots inside the unit circle iff

|Km | < 1 m = 1, 2, · · · , N
DSP: Unit V — FIR Filters
Lattice structure (IIR)

Example 8: IIR filter lattice structure

Consider an all-pole IIR filter given by

1
H(z) =
1+ 13/24z −1 + 5/8z −2 + 1/3z −3
Determine the lattice structure. Comment on its stability.
Solution:
We have m = N = 3 and
1 1
H(z) = =
A3 (z) 1 + 13/24z + 5/8z −2 + 1/3z −3
−1

A3 (z) = 1 + 13/24z −1 + 5/8z −2 + 1/3z −3

Hence, K3 = 1/3.

B3 (z) = 1/3 + 5/8z −1 + 13/24z −2 + z −3

DSP: Unit V — FIR Filters
Lattice structure (IIR)

With m = 2,
A3 (z) − K3 B3 (z)
A2 (z) = = 1 + 3/8z −1 + 1/2z −2
1 − K32
Hence, K2 = 1/2
B2 (z) = 1/2 + 3/8z −1 + z −2
With m = 1,
A2 (z) − K2 B2 (z)
A1 (z) = = 1 + 1/4z −1
1 − K22
Hence, K1 = 1/4. Since all the reflection coefficients satisfy
|Km | < 1, the given system is stable.
DSP: Unit V — FIR Filters
Lattice structure (IIR)

Example 9: IIR filter

Determine the system function H(z) of an IIR filter with reflection

coefficients K1 = 0.6, K2 = 0.3, K3 = 0.5, K4 = 0.9.
Solution:
The system transfer H(z) = A41(z) . We find the polynomial A4 (z)
recursively.
Recall:
A0 (z) = B0 (z) = 1.
Am (z) = Am−1 (z) + Km z −1 Bm−1 (z).
Bm (z) = z −m A(z −1 )
A1 (z) = A0 (z) + K1 z −1 B0 (z) = 1 + 0.6z −1
B1 (z) = 0.6 + z −1
DSP: Unit V — FIR Filters
Lattice structure (IIR)

Similarly,

A2 (z) = A1 (z) + K2 z −1 B1 (z) = 1 + 0.78z −1 + 0.3z −2

B2 (z) = 0.3 + 0.78z −1 + z −2

and

A3 (z) = A2 (z) + K3 z −1 B2 (z) = 1 + 0.93z −1 + 0.69z −2 + 0.5z −3

B3 (z) = 0.5 + 0.69z −1 + 0.93z −2 + z −3

Finally,

A4 (z) = A3 (z)+K4 z −1 B3 (z) = 1+1.38z −1 +1.311z −2 +1.337z −3 +0.9z −4

Thus the system function H(z) is

1 1
H(z) = =
A4 (z) 1 + 1.38z −1 + 1.311z −2 + 1.337z −3 + 0.9z −4
DSP: Unit V — FIR Filters
Lattice structure (IIR)

Example 10: IIR filter

Find the lattice structure for the resonator

1
H(z) =
1 − 2r cos ω0 z −1 + r 2 z −2
What happens if r = 1?
Solution:
(a) K2 = r 2 , K1 = − 2r1+r
cos ω0
2 .
(b) When r = 1, the system becomes an oscillator
DSP: Unit V — FIR Filters
Lattice structure (IIR)

Example 11: FIR filter

Determine the impulse response of an FIR filter with reflection

coefficients K1 = 0.6, K2 = 0.3, K3 = 0.5, K4 = 0.9.
Solution:
To find the impulse response h(n), we need the the system
function H(z). But H(z) = A4 (z). We find the polynomial A4 (z)
recursively.
Recall:
A0 (z) = B0 (z) = 1.
Am (z) = Am−1 (z) + Km z −1 Bm−1 (z).
Bm (z) = z −m A(z −1 )
A1 (z) = A0 (z) + K1 z −1 B0 (z) = 1 + 0.6z −1
B1 (z) = 0.6 + z −1
DSP: Unit V — FIR Filters
Lattice structure (IIR)

Similarly,

A2 (z) = A1 (z) + K2 z −1 B1 (z) = 1 + 0.78z −1 + 0.3z −2

B2 (z) = 0.3 + 0.78z −1 + z −2

and

A3 (z) = A2 (z) + K3 z −1 B2 (z) = 1 + 0.93z −1 + 0.69z −2 + 0.5z −3

B3 (z) = 0.5 + 0.69z −1 + 0.93z −2 + z −3

Finally,

H(z) = A4 (z) = A3 (z)+K4 z −1 B3 (z) = 1+1.38z −1 +1.311z −2 +1.337z −3 +

Thus the impulse response is

h(n) = {1, 0.38, 1.311, 1.337, 0.9}

DSP: Unit V — FIR Filters
Summary

To find reflection coefficients Km from system function H(z)

Extract Am (z) from system function H(z).
Am−1 (z) = Am (z)−K m Bm (z)
1−Km2
Km = αm (m).
To find system function H(z) from reflection coefficients Km
A0 (z) = B0 (z) = 1.
Am (z) = Am−1 (z) + Km z −1 Bm−1 (z).
Bm (z) = z −m A(z −1 )
H(z) = Am (z)
DSP: Unit V — FIR Filters

Thank you