 Home
 Products
 Integration
 Tutorial
 Barcode FAQ
 Purchase
 Company
barcode generator for ssrs where B , and B2 are given by in Software
where B , and B2 are given by Code 128 Recognizer In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Code 128 Code Set C Printer In None Using Barcode printer for Software Control to generate, create Code128 image in Software applications. EXAMPLE 4.4.1
Code 128C Decoder In None Using Barcode recognizer for Software Control to read, scan read, scan image in Software applications. Painting Code 128B In C# Using Barcode creator for Visual Studio .NET Control to generate, create Code 128A image in .NET framework applications. Suppose that a sequence x ( n ) has a ztransform
ANSI/AIM Code 128 Creation In .NET Framework Using Barcode printer for ASP.NET Control to generate, create Code 128 Code Set B image in ASP.NET applications. Code 128 Code Set A Encoder In VS .NET Using Barcode encoder for VS .NET Control to generate, create USS Code 128 image in Visual Studio .NET applications. with a region of convergence l z l z the form
Code 128B Creation In Visual Basic .NET Using Barcode encoder for Visual Studio .NET Control to generate, create ANSI/AIM Code 128 image in VS .NET applications. Draw UCC  12 In None Using Barcode creation for Software Control to generate, create UPC Symbol image in Software applications. f . Because p = q
Creating Data Matrix In None Using Barcode drawer for Software Control to generate, create Data Matrix 2d barcode image in Software applications. Draw Code 128 Code Set A In None Using Barcode creator for Software Control to generate, create Code 128 Code Set C image in Software applications. = 2, and the two poles are simple, the partial fraction expansion has
Paint Code 39 In None Using Barcode encoder for Software Control to generate, create USS Code 39 image in Software applications. Barcode Maker In None Using Barcode maker for Software Control to generate, create bar code image in Software applications. The constant C is found by long division: Paint Identcode In None Using Barcode maker for Software Control to generate, create Identcode image in Software applications. Code 3 Of 9 Encoder In Visual Basic .NET Using Barcode encoder for Visual Studio .NET Control to generate, create Code 39 Extended image in VS .NET applications. THE 2TRANSFORM
UPCA Supplement 5 Recognizer In None Using Barcode scanner for Software Control to read, scan read, scan image in Software applications. Print UPCA In None Using Barcode maker for Font Control to generate, create GS1  12 image in Font applications. [CHAP. 4
Printing Code 128B In None Using Barcode creator for Font Control to generate, create Code128 image in Font applications. UCC.EAN  128 Printer In Visual Studio .NET Using Barcode maker for ASP.NET Control to generate, create EAN 128 image in ASP.NET applications. Therefore, C = 2 and we may write X(z) as follows: Encode USS Code 39 In None Using Barcode generator for Online Control to generate, create Code39 image in Online applications. Making UPC A In VS .NET Using Barcode creator for ASP.NET Control to generate, create UCC  12 image in ASP.NET applications. Next, for the coefficients A , and Az we have
and Thus, the complete partial fraction expansion becomes
Finally, because the region of convergence is the exterior of the circle Izl >
i, x(n) is the rightsided sequence
4.4.2 Power Series
The ztransform is a power series expansion, where the sequence values x ( n ) are the coefficients of z" in the expansion. Therefore, if we can find the power the series expansion for X(z), sequence values x ( n ) may be found by simply picking off the coefficients of z". EXAMPLE 4.4.2
Consider the ztransform X(:) = log(l The power series expansion of this function is
+ a:') Izl > la1
Therefore, the sequence x(n) having this ztransform is
CHAP. 41
THE zTRANSFORM
4.4.3 Contour Integration
Another approach that may be used to find the inverse ztransform of X(z) is to use contour integration. This procedure relies on Cauchy's integral theorem, which states that if C is a closed contour that encircles the origin in a counterclockwise direction, With
X(z) = n=w
x(n)zPn
Cauchy's integral theorem may be used to show that the coefficients .x(n) may be found from X(z) as follows: where C is a closed contour within the region of convergenceof X(z) that encircles the origin in acounterclockwise direction. Contour integrals of this form may often by evaluated with the help of Cauchy's residue theorem, x ( z ) z n ldz = [residues of x ( z ) z n ' a t the poles inside C] If X(z) is a rational function of z with a firstorder pole at z = ak, ~es[x(z)z"l at z = a k ] = [(I  cmzl)~(z)zn']z=ak Contour integration is particularly useful if only a few values of x(n) are needed. 4.5 THE ONESIDED ZTRANSFORM
The ztransform defined in Sec. 4.2 is the twosided, or bilateral, ztransform. The onesided, or unilateral, ztransform is defined by The primary use of the onesided ztransform is to solve linear constant coefficient difference equations that have initial conditions. Most of the properties of the onesided ztransform are the same as those for the twosided ztransform. One that is different, however, is the shift property. Specifically, if x(n) has a onesided ztransform X 1 ( ~ the onesided ztransform of x(n  1) is ), It is this property that makes the onesided ztransform useful for solving difference equations with initial conditions. EXAMPLE 4.5.1
Consider the linear constant coefficient difference equation
Let us find the solution to this equation assuming that x ( n ) = S(n  I ) with y (  I ) = y(2) = 1 . We begin by noting that if the onesided ztransform of y(n) is Y l ( z ) , onesided ztransform of y(n the  2 ) is
THE zTRANSFORM
[CHAP. 4
Therefore, taking the ztransform of both sides of the difference equation, we have
YI = 0.25[y(2) (z) + y(1)zI + Z  ~ Y ~ ( +)XI(z) Z ] where X 1 (z) = z' . Substituting for y (  1) and y(2), and solving for Yl we have (z), To find y(n), note that Yl may be expanded as follow^:^ (z) Therefore.
Solved Problems
Computing z'hansforms
The ztransform of a sequence x ( n ) is
If the region of convergence includes the unit circle, find the DTFT of x ( n ) at w = n.
If X(z) is the ztransform of x(n), and the unit circle is within the region of convergence, the DTFT of x(n) may be found by evaluating X(z) around the unit circle: X(eJW) x(z)J;_, = Therefore. the DTFT at o = 7r is and we have
Find the ztransform of each of the following sequences: (a) x ( n ) = 3S(n)+ S(n  2 ) + S(n
(b) x ( n ) = u ( n )  u ( n  1 0 ) (a) Because this sequence is finite in length, the ztransform is a polynomial, and the region of convergence is 0 < Izl < m. Note that because x(n) has nonzero values for n < 0, the region of convergence does not include IzI = co,and because x(n) has nonzero values for n 0, the region of convergence does not include the point z = 0. 2 ~ e the discussion in Sec. 4.4.1 on partial fraction expansions. e
CHAP. 41 ( h ) For this sequence, THE ZTRANSFORM
which converges for all lzl > 0. Note that the roots of the numerator are solutions to the equation
z10 = These roots are
= ej2nkl10
k = 0 , 1 , ..., 9 which are 10 equally spaced points around the unit circle. Thus, the pole at z = 1 in the denominator of X(z) is canceled by the zero at z = 1 in the numerator, and the ztransform may also be expressed in the form Find the ztransform of each of the following sequences: (a) x ( n ) = 2"u(n) 3 ( ; ) " u ( n ) (b) x ( n ) = cos(noo)u(n). (a) Because x(n) is a sum of two sequences of the form anu(n), using the linearity property of the ztransform, and the ztransform pair ffnu(n)

