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r3= -0.5
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12. Then, we find Az(z) as follows:
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[A3(z) - T ~ Z - ~ A ~ ( Z - ' ) ] I - r:
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r2= 0.3123. Finally, we have
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Thus. the second reflection coefficient is
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I A,(z) = [Az(z)- r 2 ~ - 2 ~ ~ ( ~ - ' ) 1
1 - (0.3 123)2
[I - 0.6401~-' 0.3123~-' - 0 . 3 1 2 3 ~ - ~ ( 1 0.64012 -
+ 0.3123z2)]
= 1 - 0.4878~-'
and, therefore,
r = -0.4878. 1
Thus, a lattice implementation of this allpass filter is as shown in the figure below.
CHAP. 81
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS
Find the system function for the lattice filter given in the figure below.
x(n)
This structure implements a third-order IIR filter with three poles and three zeros:
To find the system function, first we use the step-up recursion to find A3(z)from the reflection coefficients and r3. With rI = 0.2, we have
r,,r2,
Next, for a2(k) have we
Finally, for a3(k)we have
Therefore, the denominator polynomial is
To find the numerator B3(2), use Eq. (8.8), we
Thus, we have
Therefore, the system function is
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS Sketch a lattice filter structure for each of the following system functions:
[CHAP. 8
(a) To implement the filter
using a lattice filter structure, we must first find the reflection coefficients corresponding to the denominator polynomial. Using the step-down recursion, we find
Next, we find the coefficients c l(k) that produce the numerator polynomial
Using the recursion
we have and
cl(l)= bl(l)= - I c,(0) = bl(0) - c l ( l ) a 1 ( l ) = 2.4698
(note that a l ( l ) = r , ) . Therefore. the lattice filter structure for this system is as shown in the figure below.
(b) To find the lattice filter structure for
we first use the step-down recursion to find the reflection coefficients for the denominator, which are
Next, we use the recursion
CHAP. 81
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS to find the coefficients c3(k).which are
Thus, the lattice filter structure for this system is as shown in the figure below.
xfn)
Note that the system function
is an allpass filter. Because the denominator is the same as the system function in part (b), the reflection coefficients are rl = 0.5 and T2 = 0.75, and the lattice filter is as shown in the figure below.
Finite Word-Length Effects
8.33 Express the fractions using 6 bits.
With B
in sign-magnitude, one's complement, and two's complement notation
+ 1 = 6 bits, I bit will be a sign bit, and 5 will be the fractional bits. Because
the 6-bit representation for x =
$ in all three binary forms is
For x = -
& we have, in sign-magnitude form,
and in one's complement form.
x = 0.001 11 = 1.1 1000
and in two's complement form
x = 0.001 1 1
+ 0.00001 = 1.1 1001
IMPLEMENTATION O F DISCRETE-TIME SYSTEMS Consider the following implementation of a second-order filter:
[CHAP. 8
r cos wo
r cos wo
sinwo
(a) Find the system function corresponding to this network.
(b) If the coefficients ( r cos wo) and
( I . sinwo) are quantized to 4 bits, draw the set of allowable pole locations in the z-plane. For what types of filters would this filter structure be preferred over a direct form structure'
(a) This filter structure is called the coupled form realization. The system function for this filter may be found as follows. The difference equations relating .r(n), v ( n ) .and v(n) are u(n) = x(n) - r sin(q)y(n - 1 )
+ r cos(wIJ)v(ri- I )
y (n) = r sin(wo)v(n - I ) + r cos(w,,)v(n- I ) Taking the z-transform of the first equation. we have
Solving for V(z) yields V(z) = X(z) - r sin(wo)z-' Y (z) I - r cos(wo)z-~
Substituting this into the :-transform of the second difference equation gives
Solving this equation for Y (z). we have
Therefore, the system function is
(b) This filter has poles at
-.. -
rc*llull
- r cos q , j r sin q,
Thus, the coefficients in this structure are the real and imaginary parts of the pole locations. Therefore, if the I bits. the poles will lie at the intersections of 2'+ evenly spaced horizontal coefficients are quantized to B and vertical lines in the i-plane. These positions are illustraled in the figure below for the first quadrant when B+1=4.
CHAP. 81
IMPLEMENTATION O F DISCRETE-TIME SYSTEMS
Note that, compared with the direct form structure, the allowable pole locations are uniformly distributed within the unil circle. The cost for this uniform spacing is four multiplies and three additions per output value, compared to only two multiplications and two additions for a direct form implementation. This extra cost may be worthwhile, particularly for low-pass filters that have poles close to the unit circle in the vicinity of z = 1, where the density of the allowable pole locations is sparse in the direcl form implementation compared to the coupled form implementation.
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