# barcode lib ssrs Having found a(l), we may solve for b(0) and h(l) using the first two equations in Software Printing Code 128B in Software Having found a(l), we may solve for b(0) and h(l) using the first two equations

Having found a(l), we may solve for b(0) and h(l) using the first two equations
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3 0.52-' Notice that the unit sample response corresponding to this system exactly matches the given unit sample response. In general, however, this will not be true. A perfect match depends on hd(n)being the inverse z-transform of a rational function of z, and it depends upon an appropriate choice for the order of the Pad6 approximation (the number of poles and zeros).
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H(z) =
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Therefore, we have
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Let the first three values of the unit sample response of a desired causal filter be hd(0) = 3 , h d ( l ) = and h d ( 2 ) =
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(a) Using the Pad6 approximation method, find the coefficients of a second-order all-pole filter that has a unit sample response h(n), such that h ( n ) = h d ( n )for 12 = 0, 1,2.
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(b) Repeat part ( a ) for a filter that has one pole and one zero.
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(c) Repeat part ( a ) for an FIR filter that has two zeros.
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(a) For a second-order all-pole filter, the equations for the Pad6 approximation are
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which, with the given values for hn(n) become
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From the last two equations, we have
Solving for a(1) and a(2), we find a(])= Then, using the first equation, we have
a(2) := -1 72
h(0) = 3 Thus, the system function of the filter is
FILTER DESIGN (b) Using a first-order system to match the given values of hd(n),
[CHAP. 9
the equations that we must solve are as follows,
or, using the given values for h&),
We may solve for a( I) using the last equation.
+ \$ a ( l )= 0
or a(l)=
Next. we solve for h(0) and b(l) using the first two equations.
which gives
I:;[
,II;'[ I;:[
Thus, the system function is
(c) For an FIR filter, the solution is trivial:
Find the least-squares
FIR inverse filter of length 3 for the system that has a unit sample response
g(n)=
Also, find the least-squares error,
2 I 0
n=O n=l else
for this least-squares inverse filter. To find the least-squares inverse, we need to solve the linear equations
CHAP. 91
FILTER DESIGN
where is the deterministic autocorrelation of g(n). With N = 3, these equation:, may be written in matrix form as follows,
For the given sequence g(n),we compute the autocorrelation sequence as follows,
Therefore, the linear equations become
and the solution is
h ( n )=
0.494 -0.235 0.094 0
n .=0 n =1
n =2
else
Performing the convolution of h ( n ) with g(n),we have
From this sequence, we may evaluate the squared error,
Find the FIR least-squares inverse filter of length N for the system having a unit sample response
where cr is an arbitrary real number.
Before we begin. note that if la[ > 1, G ( z )has a zero that is outside the unit circle. In this case, G ( z )is not minimum a phase, and the inverse filter 1 / G ( z )cannot be both causal and stable. However, if l1 < I ,
and the inverse filter is
FILTER DESIGN We begin by finding the least-squares inverse of length N = 2. The autocorrelation sequence r.,*(k) is I+w2 k=O k=fl else
[CHAP. 9
Therefore, the linear equations that we must solve are
The solution for h(O) and h ( l ) is easily seen to be
The system function of this least-squares inverse filter is
which has a zero at
Note that because
the zero of H(z) is inside the unit circle, and H(z) is minimum phase. regardless of whether the zero of G(z) is inside or outside the unit circle. Let us now look at the least-squares inverse, hN(n), length N . In this case, the linear equations have the form of'
Solving these equations for arbitrary a and N may be accomplished as follows. For n = 1.2, . . . , N - 2 these equations may be represented by the homogeneous difference equation,
The general solution to this equation is of the form
where c l and c.2 are constants that are determined by the boundary conditions at 11 = O and n = N last equations in Eq. (9.22)l:
I [the ti rst and
Substituting Eq. (9.23) into Eq. (9.24), we have
CHAP. 91
FILTER DESIGN
which, after canceling common terms, may be simplified to
The solution for cl and c2 is
Therefore, h N ( n )is
Let us now look at what happens asyrnptoticaIly as N
m. If Ial < 1,
a"-N
N-no
lirn h N ( n )= -- a"
-I\'
which is the inverse filter, that is,
lim h N ( n )= a n u ( n )= g - l ( n )
and However, if
la 1
lirn H N ( z )= I - az-'
aN-n
> 1,
N-lo
lirn h N ( n )= -= a-n-2
@N+ a -2 1
n 1 0
lirn HN( z ) =
-a-Iz- 1
which is not the inverse filter. Note that although &n) = h N ( n )* g ( n ) does not converge to S(n) as N + m, taking ) the limit o ~ B N ( zas N + co,we have
which is an all-pass filter, that is,
The first five samples of the unit sample response of a causal filter are h(0) = 3 h(1) = - 1 h(2) = 1
h(3) = 2
h(4) = 0
If it is known that the system function has two zeros and two poles, determine whether or not the filter is stable. The system function of this filter has the form
FILTER DESIGN To determine whether or not this system is stable, it is necessary to find the denominator polynomial,
[CHAP, 9
and check to see whether or not the roots of A(z) lie inside the unit circle. Given that H(z) has two poles and two zeros, we may use the Pad6 approximation method to find the denominator coefficients: