barcode lib ssrs Let A ( z ) be an FIR filter with lattice filter coefficients in Software

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Let A ( z ) be an FIR filter with lattice filter coefficients
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( a ) Find the zeros of the system function A ( z ) .
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( h ) Repeat for the case when
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r3= - 1 .
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ITjl < I ITPI = 1
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(c) Can a general result be proved for lattice filters that have reflection coefficients Ti with
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for j = 1,2, ... . p - 1
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( a ) To find the system function A ( z ) for a given set of reflection coefficients, we use the step-up recursion. For a l ( k )we have
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Then, with
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r2 = 4 we have
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Finally, with T3 = I we have
Thus.
A(z) = 1
+ + z - ' + t ~+ z-3 ~ fi3 = C-~().JM6"
The zeros of the system function may be found by factoring A ( z ) . The roots are found to be
pI = - 1
which are on the unit circle.
p2 = e~1).4646n
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS
(h) If r3= -I. follows:
[CHAP. 8
the system function may be found by modifying the last step of the step-up recursion in pan (a) as
which are again on the unit circle. (c) If A,,(z) is a pth-order FIR filter with reflection coefficients r, where (r,I < I for j = 1.. . . . p lr,*l 1, = A,,(z) = A p - I ( ~ ) z - " A ~ - ~ ( ~ - ' ) f and it follows that the polynomial A,(z) is symmetric or antisymmetric, that is.
- I,
Therefore, A,&) has (generalized) linear phase, which implies that all of the zeros of A,@) lie on the unit circle or in conjugate reciprocal p a n However, if Ir, 1 < I for j = 1.2, . . . p - I, the zeros must lie on the unit circle. The reason for this is as follows. The Schur-Cohn stability criterion states that none of the roots of A,(z) may lie outside the unit circle if ITj\ 5 1 for j = 1.2, . . . , p. Therefore, if A&) has generalized linear phase with no zeros outside the unit circle. then all of the zeros must be on the unit circle.
Draw a lattice filter implementation for the all-pole filter
and determine the number of multiplications, additions, and delays required to implement the filter. Compare this structure to a direct form realization of H ( z ) in terms of the number of multiplies, adds, and delays.
To implement this filter usinga lattice filter, we must first derive the reflectioncoefficientsrI. 2 ,and r,3 r corresponding to the denominator polynomial. With
it follows that
r3 0.6. =
Next, using the step-down recursion to find A2(z),we have
I A2(z) = [Adz) - r 3 ~ - ' ~ d - ' ) l I - r: 1 1 l - 0 . 2 ~ ' 0.4zT2 t 0.61-" - 0 . 6 ~ - ~ -10 . 2 ~ 0.4z2 + 0.6z3)] ( + 1 - (0.6)2 = I - 0 . 6 8 7 5 ~ ~ '0.8 1 2 5 8
Thus. for r2we have
r2 = 0.8 125. Finally, for A'(,-),
we have
and, therefore, r I = -0.3793. Thus, the structure is as follows:
CHAP. 81
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS
This filter structure has three delays and requires five multiplications and five additions to evaluate each value of the output, y(n). A direct-form structure also requires three delays but only three multiplications and three additions.
Without factoring any polynomials, determine whether o r not the following causal filter is stable:
H ( z )=
+ 21-' + 2 r 2 + 2 - 3 1 + 1 . 5 8 ~ - '+ 1.638~-2+ 1.556~-"
04-4
We may easily check the stability of this filter using the Schur-Cohn stability test, which involves checking to see whether or not the magnitudes of the reflection coefficients corresponding to the denominator polynomial are bounded by 1 in magnitude. With
it follows that
lr41= 0.4 < 1.
Using the step-down recursion to find A 3 ( z ) , we have
Therefore,
1 r3I = I . I > 1, and it follows that the filter is unstable.
Use the Schur-Cohn stability test to derive the stability conditions
for a second-order filter
H (z)=
+a(1 ) ~ - I + a ( 2 ) ~ - ~
b(0)
In order for H ( z )to be stable, it is necessary and sufficient for the reflection coefficients TI and Tzto have a magnitude that is less than 1. In t e n s of the first two reflection coefficients, the denominator of H(z) is
Therefore, because a(2) = Tz, the constraint that
)r2( 1 gives us the first condition, <
Next, with
a(1) = rl + rlrz = II
+ a(2)lrl
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS it follows that
[CHAP. 8
Because we require that 1 rl1 < I,
These two equations are equivalent to
as was to be shown.
Implement the allpass filter
using a lattice filter structure.
To find the lattice filter structure for this allpass filter, we use the step-down recursion to find the reflection coefficients corresponding to the denominator polynomial, A3(z)= I - 0.82-I First, we note that Az(z) =
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