Lattice Structures
Lattices
Consider an mth-order FIR filter with
Hm ( z ) =
Y(z)
X(z)
m
= ∑α m [k ] z −k
k =0
where the α 's are the filter coefficients and α m [ 0] = 1 for any m.
m
Then h m [ n ] = ∑ α m [ k ]δ [ n − k ] and y [ n ] = x [ n ] ∗ α m [ n ]
k =0
For m = 1, y [ n ] = α1 [ 0] x [ n ] + α1 [1] x [ n − 1].
Lattices
The recursion relation y [ n ] = α1 [ 0] x [ n ] + α1 [1] x [ n − 1] can be
realized by this lattice structure if K1 = α1 [1] and y [ n ]. = f1 [ n ].
Lattices
The filter can also be described by
⎡ f1 [ n ] ⎤ ⎡ 1
⎢
⎥=⎢
⎣g1 [ n ]⎦ ⎣ K1
or
K1 ⎤ ⎡ f 0 [ n ] ⎤ ⎡ F1 ( z ) ⎤ ⎡ 1
⎢
⎥⇒⎢
⎥=⎢
⎥
1 ⎦ ⎣g 0 [ n − 1]⎦ ⎣ G1 ( z ) ⎦ ⎣ K1
⎡ y [ n] ⎤ ⎡ 1
⎢
⎥=⎢
⎣g1 [ n ]⎦ ⎣ K1
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 ) + K1 z −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 ] = α 2 [ 0] x [ n ] + α 2 [1] x [ n − 1] + α 2 [ 2] x [ n − 2].
N
=1
Lattices
⎡ F1 ( z ) ⎤ ⎡ 1
⎢
⎥=⎢
⎣G1 ( z ) ⎦ ⎣ K1
K1 ⎤ ⎡ F0 ( z ) ⎤
⎢
⎥
1 ⎥⎦ ⎣ z −1 G 0 ( z ) ⎦
⎡1 0 ⎤ ⎡ F1 ( z ) ⎤ ⎡ F1 ( z ) ⎤ ⎡1 0 ⎤ ⎡ 1
⎢0 z −1 ⎥ ⎢G z ⎥ = ⎢ z −1 G z ⎥ = ⎢ 0 z −1 ⎥ ⎢ K
⎣
⎦ ⎣ 1 ( )⎦ ⎣
⎣
⎦⎣ 1
1 ( )⎦
⎡ F1 ( z ) ⎤ ⎡1 0 ⎤ ⎡ 1
⎢ −1
⎥=⎢
−1 ⎥ ⎢
G
z
z
0
z
(
)
⎦ ⎣ K1
1
⎣
⎦ ⎣
K1 ⎤ ⎡ X ( z ) ⎤
⎢
⎥
1 ⎥⎦ ⎣ z −1 X ( z ) ⎦
K1 ⎤ ⎡ F0 ( z ) ⎤
⎢
⎥
1 ⎥⎦ ⎣ z −1 G 0 ( z ) ⎦
Lattices
⎡ Y ( z ) ⎤ ⎡ F2 ( z ) ⎤ ⎡ 1
⎢
⎥=⎢
⎥=⎢
⎣G 2 ( z )⎦ ⎣G 2 ( z )⎦ ⎣ K 2
⎡1
=⎢
⎣ K2
K 2 ⎤ ⎡1 0 ⎤ ⎡ 1
1 ⎥⎦ ⎢⎣0 z −1 ⎥⎦ ⎢⎣ K1
K 2 ⎤ ⎡ F1 ( z ) ⎤
⎢ −1
⎥
⎥
1 ⎦ ⎣ z G1 ( z ) ⎦
K1 ⎤ ⎡ X ( z ) ⎤
⎢
⎥
1 ⎥⎦ ⎣ z −1 X ( z ) ⎦
Multiplying the matrices,
⎡ Y ( z ) ⎤ ⎡1 + z −1 K1 K 2
⎢
⎥=⎢
−1
G
z
+
K
z
K1
(
)
2
⎣
⎦ ⎣ 2
K1 + z −1 K 2 ⎤ ⎡ X ( z ) ⎤
⎥
⎥ ⎢ −1
K1 K 2 + z −1 ⎦ ⎣ z X ( z ) ⎦
In the time domain
Lattices
y [ n ] = x [ n ] + K1 ( K 2 + 1) x [ n − 1] + K 2 x [ n − 2]
g 2 [ 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) = α 2 [1]
and K 2 = α 2 [ 2] the upper lattice signal is the desired response
and
K1 =
α 2 [1]
α 2 [ 2] + 1
and K 2 = α 2 [ 2].
Lattices
In the general ( M − 1) th order case,
f0 [ n] = g 0 [ n] = x [ n]
f m [ n ] = f m−1 [ n ] + K m g m−1 [ n − 1] , m = 0,1, 2," , M − 1
g m [ n ] = K m f m−1 [ n ] + g m−1 [ n − 1] , m = 0,1, 2," , M − 1
y [ n ] = f M −1 [ n ]
Lattices
The output signal from the mth stage is
f m [ n ] = x [ n ] ∗ α m [ n ].
Therefore
Fm ( z ) = X ( z ) Α m ( z ) = F0 ( z ) Α m ( z )
m
Z
where α m [ n ] ←⎯
→ Αm ( z ) = ∑α 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
g m [ n] = ∑α m [ m − k ] x [ n − k ]
k =0
Let β m [ k ] = α m [ m − k ]. Then
m
g m [ n] = ∑ β m [ k ] x [ n − k ] = β m [ n] ∗ x [ n]
k =0
Z
and G m ( z ) = X ( z ) Βm ( z ) where β m [ n ] ←⎯
→Βm ( z ) .
Lattices
m
From the definition β m [ k ] = α m [ m − k ] ⇒ Β m ( z ) = ∑ α m [ m − k ] z − k . Let
k =0
0
m
q =m
q =0
q = m − k . Then Β m ( z ) = ∑ α m [ q ] z q−m = z − m ∑ α m [ q ] z q . It was shown
m
above that Α m ( z ) = ∑ α m [ k ] z − k therefore Βm ( z ) = z − m Α m (1/ z ) which
k =0
implies that the zeros of Β m ( z ) are the reciprocals of the zeros of Α 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
⎡ Αm ( z )⎤ ⎡ 1
⎢
⎥=⎢
⎣ Βm ( z ) ⎦ ⎣ K m
K m ⎤ ⎡ Α m−1 ( z ) ⎤
⎢
⎥
1 ⎥⎦ ⎣ Β m−1 ( z ) ⎦
Lattices
The coefficients α of the Direct Form II filter and the reflection
coefficients of the lattice structure are related and the α 's can be
found from the K's recursively by using
Α0 ( z ) = Β0 ( z ) = 1
Α m ( z ) = Α m−1 ( z ) + K m z −1Β m−1 ( z )
Β m ( z ) = z − m Α m (1/ z )
Lattices
Example
Let the reflection coefficients be K1 = 1/ 2 , K 2 = 1/ 5 and K 3 = 3/ 4.
Find the Direct Form II filter coefficients α .
Solution
Α0 ( z ) = Β0 ( z ) = 1
Α1 ( z ) = Α0 ( z ) + K1 z −1Β0 ( z ) = 1 + z −1 K1 = 1 + 0.5 z −1
Β1 ( z ) = z −1Α1 (1/ z ) = z −1 (1 + 0.5 z ) = 0.5 + z −1
Example (cont.)
Solution
Lattices
Α 2 ( z ) = Α1 ( z ) + K 2 z −1Β1 ( z ) = 1 + 0.5 z −1 + 0.2 z −1 ( 0.5 + z −1 )
= 1 + 0.6 z −1 + 0.2 z −2
Β 2 ( z ) = z −2 Α 2 (1/ z ) = z −2 (1 + 0.6 z + 0.2 z 2 ) = 0.2 + 0.6 z −1 + z −2
Α3 ( z ) = Α 2 ( z ) + K 3 z −1Β 2 ( z ) = 1 + 0.6 z −1 + 0.2 z −2
+ 0.75 z −1 ( 0.2 + 0.6 z −1 + z −2 )
= 1 + 0.75 z −1 + 0.65 z −2 + 0.75 z −3
Lattices
Example (cont.)
Solution
Β3 ( z ) = z −3 Α3 (1/ z ) = z −3 (1 + 0.75 z + 0.65 z 2 + 0.75 z 3 )
= 0.75 + 0.65 z −1 + 0.75 z −2 + z −3
So α 3 [ 0] = 1, α 3 [1] = 0.75, α 3 [ 2] = 0.65 and α 3 [3] = 0.75
Lattices
We can also find the reflection coefficients from the Direct Form II
coefficients using
Α m ( z ) = Α m−1 ( z ) + K m z −1Β m−1 ( z )
and
g m [ n ] = K m f m−1 [ n ] + g m−1 [ n − 1] , m = 0,1, 2,", 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
Β m ( z ) = K m Α m−1 ( z ) + z −1Β m−1 ( z ) ⇒ z −1Β m−1 ( z ) = Β m ( z ) − K m Α m−1 ( z )
Combining this with
Α m ( z ) = Α m−1 ( z ) + K m z −1Β m−1 ( z )
we get
Α m ( z ) = Α m−1 ( z ) + K m ⎣⎡Β m ( z ) − K m Α m−1 ( z ) ⎤⎦
Lattices
Finally, solving for Α m−1 ( z ) we get
Α m−1 ( z ) =
Αm ( z ) − K mΒm ( z )
1 − K m2
where K m = α m [ m ]. Iterating on this equation we can find all
the coefficients α .