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barcode lib ssrs Having found a(l), we may solve for b(0) and h(l) using the first two equations in Software
Having found a(l), we may solve for b(0) and h(l) using the first two equations Code 128 Code Set B Scanner In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Drawing USS Code 128 In None Using Barcode creation for Software Control to generate, create Code 128 Code Set B image in Software applications. 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 ztransform 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). Scan Code 128 Code Set B In None Using Barcode decoder for Software Control to read, scan read, scan image in Software applications. Drawing Code 128B In C# Using Barcode maker for Visual Studio .NET Control to generate, create ANSI/AIM Code 128 image in .NET framework applications. H(z) = Code128 Drawer In Visual Studio .NET Using Barcode drawer for ASP.NET Control to generate, create Code 128 Code Set B image in ASP.NET applications. Code 128 Code Set B Drawer In Visual Studio .NET Using Barcode maker for .NET framework Control to generate, create USS Code 128 image in VS .NET applications. Therefore, we have
Code 128 Generator In VB.NET Using Barcode creation for .NET Control to generate, create Code128 image in .NET framework applications. ECC200 Creation In None Using Barcode creation for Software Control to generate, create Data Matrix ECC200 image in Software applications. 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 ) = Code 128 Code Set A Generation In None Using Barcode generator for Software Control to generate, create Code 128 Code Set A image in Software applications. Printing Bar Code In None Using Barcode maker for Software Control to generate, create barcode image in Software applications. (a) Using the Pad6 approximation method, find the coefficients of a secondorder allpole filter that has a unit sample response h(n), such that h ( n ) = h d ( n )for 12 = 0, 1,2. Bar Code Generator In None Using Barcode maker for Software Control to generate, create bar code image in Software applications. UPC Symbol Drawer In None Using Barcode encoder for Software Control to generate, create UPCA Supplement 2 image in Software applications. (b) Repeat part ( a ) for a filter that has one pole and one zero.
Print USD  8 In None Using Barcode printer for Software Control to generate, create USD  8 image in Software applications. Printing Code 39 In VB.NET Using Barcode creation for Visual Studio .NET Control to generate, create Code39 image in .NET framework applications. (c) Repeat part ( a ) for an FIR filter that has two zeros.
Printing Code 128 Code Set B In Java Using Barcode drawer for BIRT reports Control to generate, create USS Code 128 image in BIRT applications. UPC Symbol Creator In ObjectiveC Using Barcode maker for iPhone Control to generate, create UPC Code image in iPhone applications. (a) For a secondorder allpole filter, the equations for the Pad6 approximation are
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Reading Bar Code In .NET Framework Using Barcode scanner for VS .NET Control to read, scan read, scan image in Visual Studio .NET applications. GS1  13 Decoder In None Using Barcode decoder for Software Control to read, scan read, scan image in Software applications. 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 firstorder 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 leastsquares
FIR inverse filter of length 3 for the system that has a unit sample response
g(n)= Also, find the leastsquares error, 2 I 0 n=O n=l else
for this leastsquares inverse filter. To find the leastsquares 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 leastsquares 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 leastsquares 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 leastsquares 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 leastsquares 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
Nno
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' aNn
> 1, Nlo
lirn h N ( n )= = an2 @N+ a 2 1 n 1 0 lirn HN( z ) = aIz 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 allpass 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:

