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barcode lib ssrs Let A ( z ) be an FIR filter with lattice filter coefficients in Software
Let A ( z ) be an FIR filter with lattice filter coefficients Code 128 Code Set B Decoder In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Encoding Code128 In None Using Barcode generator for Software Control to generate, create ANSI/AIM Code 128 image in Software applications. ( a ) Find the zeros of the system function A ( z ) . Code 128 Code Set C Recognizer In None Using Barcode decoder for Software Control to read, scan read, scan image in Software applications. Generating Code 128 Code Set B In Visual C#.NET Using Barcode encoder for .NET framework Control to generate, create Code 128 Code Set C image in Visual Studio .NET applications. ( h ) Repeat for the case when
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Thus.
A(z) = 1 + + z  ' + t ~+ z3 ~ 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 DISCRETETIME SYSTEMS
(h) If r3= I. follows: [CHAP. 8
the system function may be found by modifying the last step of the stepup recursion in pan (a) as
which are again on the unit circle. (c) If A,,(z) is a pthorder 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 SchurCohn 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 allpole 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 stepdown 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 DISCRETETIME SYSTEMS
This filter structure has three delays and requires five multiplications and five additions to evaluate each value of the output, y(n). A directform 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~" 044 We may easily check the stability of this filter using the SchurCohn 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 stepdown recursion to find A 3 ( z ) , we have
Therefore, 1 r3I = I . I > 1, and it follows that the filter is unstable.
Use the SchurCohn stability test to derive the stability conditions
for a secondorder 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 DISCRETETIME 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 stepdown recursion to find the reflection coefficients corresponding to the denominator polynomial, A3(z)= I  0.82I First, we note that Az(z) =

