Lattice Structures
Lattices
Consider an mth-order FIR filter with
Y z  m
   m k z  k
H m z  
X z  k0
where the  's are the filter coefficients and  m 0   1 for any m.
m
Then h m n     m k  n  k  and y n   x n   m n 
k0
For m  1, y n   1 0 x n  1 1x n  1.
{
1
Lattices
The recursion relation y n   1 0 x n  1 1x n  1 can be
{
1
realized by this lattice structure if K1  1 1 and y n   f1 n .
Lattices
The filter can also be described by
 f1 n   1
g n    K
 1    1
or
K1   f0 n    F1 z    1






1   g0 n  1  G1 z   K1
 y n    1
g n    K
 1    1
K1   x n    Y z    1






1   x n  1  G1 z   K1
K1   F0 z  


1   z 1 G 0 z 
K1   X z  


1   z 1 X z 
Multiplying out the matrices,
Y z   X z   K1z 1 X z   y n   x n  K1 x n  1
G1 z   K1 X z   z 1 X z   g1 n   K1 x n  x n  1
Notice that in these two recursion relations the coefficients occur in
reverse order.
Lattices
Next consider a second-order FIR filter
y n    2 0 x n   2 1x n  1  2 2x n  2 .
{
1
Lattices
 F1 z    1 K1   F0 z  
 1

G z    K

1
  z G 0 z 
 1    1
 1 0   F1 z    F1 z    1 0   1
  1



1  
1  
G
z
0
z
0
z


z
G
z


1

 1  
  K1
 
 F1 z    1 0   1 K1   X z  
 1


  1
1  
K
1
0
z
z
G
z
z
X
z
 

1  

 1

K1   F0 z  


1   z 1 G 0 z 
Lattices
 Y z    F2 z    1 K 2   F1 z  
 1

G z   G z    K

1
  z G1 z 
 2    2    2
 1 K 2   1 0   1 K1   X z  


  0 z 1   K
  1
K
1
1
 2

  z X z 
 1
Multiplying the matrices,
1
 Y z   1  z K1K 2
G z   
1


K

z
K1
2


  2
K1  z 1K 2   X z  

1   1
K1K 2  z   z X z 
In the time domain
Lattices
y n   x n  K1 K 2  1x n  1 K 2 x n  2 
g2 n   K 2 x n  K1 K 2  1x n  1 x n  2 
Again the coefficients occur in reverse order in the two recursion
relations. This occurs for any value of m. If K1 K 2  1   2 1
and K 2   2 2  the upper lattice signal is the desired response
and
K1 
 2 1
 2 2  1
and K 2   2 2 .
Lattices
In the general M  1th order case,
f0 n   g0 n   x n 
fm n   fm1 n  K m gm1 n  1 , m  0,1,2,L , M  1
gm n   K m fm1 n  gm1 n  1 , m  0,1,2,L , M  1
y n   fM 1 n 
Lattices
The output signal from the mth stage is
fm n   x n   m n .
Therefore
Fm z   X z m z   F0 z m z 
m
Z
where  m n 
 m z     m k z  k .
k 0
Lattices
Given that the coefficients for g are always the reverse of
the coefficients for f, the other output signal of the mth stage is
m
gm n     m m  k x n  k 
k0
Let  m k    m m  k . Then
m
gm n    m k x n  k    m n  x n 
k0
Z
and G m z   X z m z  where m n 
 m z .
Lattices
m
From the definition  m k    m m  k  m z     m m  k z  k . Let
k0
0
q  m  k. Then m z     m q z
qm
qm
z
m
m
q

q
z
 m   . It was shown
q0
m
above that m z     m k z  k therefore m z   z  m m 1 / z  which
k0
implies that the zeros of m z  are the reciprocals of the zeros of m z .
Lattices
Now the transfer function of the mth stage can be expressed
in terms of the transfer function of the m  1th stage as
 m z   1
  z   K
 m    m
K m   m1 z  


1   z 1m1 z 
Lattices
The coefficients  of the Direct Form II filter and the reflection
coefficients of the lattice structure are related and the  's can be
found from the K's recursively by using
0 z   0 z   1
m z   m1 z   K m z 1m1 z 
m z   z  m m 1 / z 
Example
Lattices
Let the reflection coefficients be K1  1 / 2 , K 2  1 / 5 and K 3  3 / 4.
Find the Direct Form II filter coefficients  .
Solution
0 z   0 z   1
1 z   0 z   K1z 10 z   1  z 1K1  1  0.5z 1
1 z   z 11 1 / z   z 1 1  0.5z   0.5  z 1
Example (cont.)
Solution
Lattices

2 z   1 z   K 2 z 11 z   1  0.5z 1  0.2z 1 0.5  z 1
 1  0.6z 1  0.2z 2



2 z   z 2 2 1 / z   z 2 1  0.6z  0.2z 2  0.2  0.6z 1  z 2
3 z   2 z   K 3 z 12 z   1  0.6z 1  0.2z 2

 0.75z 1 0.2  0.6z 1  z 2
 1  0.75z 1  0.65z 2  0.75z 3

Lattices
Example (cont.)
Solution

3 z   z 3 3 1 / z   z 3 1  0.75z  0.65z 2  0.75z 3

 0.75  0.65z 1  0.75z 2  z 3
So  3 0   1,  3 1  0.75,  3 2   0.65 and  3 3  0.75
Lattices
We can also find the reflection coefficients from the Direct Form II
coefficients using
m z   m1 z   K m z 1m1 z 
and
gm n   K m fm1 n  gm1 n  1 , m  0,1,2,L , M  1
z transforming we get
G m z   K m Fm1 z   z 1G m1 z 
Lattices
Dividing through by X z  we get
m z   K m m1 z   z 1m1 z   z 1m1 z   m z   K m m1 z 
Combining this with
m z   m1 z   K m z 1m1 z 
we get
m z   m1 z   K m m z   K m m1 z 
Lattices
Finally, solving for m1 z  we get
m z   K m m z 
m1 z  
1  K m2
where K m   m m . Iterating on this equation we can find all
the coefficients  .