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barcode generator for ssrs with a region of convergence Izl > I . in Software
with a region of convergence Izl > I . Code 128A Decoder In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Creating Code 128 Code Set C In None Using Barcode maker for Software Control to generate, create Code 128 Code Set C image in Software applications. Let x ( n ) be a finitelength sequence that is nonzero only for 0 5 n 5 N  1, and consider the onesided periodic sequence, y(n), that is formed by periodically extending x(n) as follows: Decode Code 128B In None Using Barcode reader for Software Control to read, scan read, scan image in Software applications. Code 128C Creation In C#.NET Using Barcode drawer for .NET Control to generate, create Code 128 image in Visual Studio .NET applications. Express the ztransform of y(n) in terms of X(z) and find the region of convergence of Y(z). Encoding ANSI/AIM Code 128 In VS .NET Using Barcode creation for ASP.NET Control to generate, create Code128 image in ASP.NET applications. Code 128C Encoder In VS .NET Using Barcode encoder for Visual Studio .NET Control to generate, create Code 128 image in Visual Studio .NET applications. The onesided periodic sequence y(n) may be written as the convolution of x ( n ) with the pulse train Painting Code 128 In Visual Basic .NET Using Barcode creator for .NET Control to generate, create Code128 image in Visual Studio .NET applications. Barcode Encoder In None Using Barcode maker for Software Control to generate, create barcode image in Software applications. CHAP. 41 In other words, Create EAN / UCC  13 In None Using Barcode generation for Software Control to generate, create UCC.EAN  128 image in Software applications. Encoding GS1  12 In None Using Barcode maker for Software Control to generate, create GTIN  12 image in Software applications. THE zTRANSFORM
Data Matrix Generation In None Using Barcode encoder for Software Control to generate, create DataMatrix image in Software applications. Bar Code Creation In None Using Barcode generator for Software Control to generate, create bar code image in Software applications. Therefore, the ztransform of y(n) is the product of the ztransforms of x(n) and pN(n). Because pN(n)is a sum of shifted unit samples, and because the ztransform of S(n  kN) is equal to z  ' ~ , the ztransform of pN(n)is Printing Standard 2 Of 5 In None Using Barcode encoder for Software Control to generate, create Code 2/5 image in Software applications. UPCA Generator In None Using Barcode maker for Office Excel Control to generate, create GTIN  12 image in Excel applications. Thus, the ztransform of the onesided periodic sequence y(n) is
UPCA Printer In ObjectiveC Using Barcode generation for iPad Control to generate, create Universal Product Code version A image in iPad applications. Draw USS Code 39 In Java Using Barcode maker for Android Control to generate, create Code39 image in Android applications. Because x(n) is finite in length and zero for n c 0, the region of convergence for X(z) is J z J> 0. Therefore, the region of convergence of Y (z) is I z 1 > 1. ANSI/AIM Code 128 Creator In Java Using Barcode printer for BIRT Control to generate, create Code 128 image in BIRT reports applications. Make Code 3 Of 9 In ObjectiveC Using Barcode encoder for iPad Control to generate, create Code39 image in iPad applications. Consider the sequence shown in the figure below.
Code 128 Code Set A Printer In .NET Using Barcode maker for ASP.NET Control to generate, create Code 128 Code Set B image in ASP.NET applications. GS1 128 Generator In Java Using Barcode maker for Java Control to generate, create EAN 128 image in Java applications. The sequence repeats periodically with a period N = 4 for n 2 0 and is zero for n < 0. Find the ztransform of this sequence along with its region of convergence. This is a problem that may be solved easily using the property derived in Prob. 4.8. Because
where N = 4 and w(n) = S(n then and we have
 I) + 2S(n  2) + S(n  3) W(z) = zl[l
+ 2z' + z  ~ ] Because x(nj is rightsided and X(z) has four poles at lzl = I, the region of convergence is I z l > 1. Properties
Use the ztransform to perform the convolution of the following two sequences: THE zTRANSFORM The convolution theorem for ztransforms states that if y(n) = h(n) H(z)X (z). With
[CHAP. 4
* x(n), the ztransform of y(n) is Y(z) it follows that Y(z) = H(z)X(z) = ( I Multiplying these two polynomials, we have
+ iz' + : Z  ~ ) ( I+ zI + 4z') By inspection, we then have for the sequence y(n), Evaluate the convolution of the two sequences
h ( n ) = (OS)"u(n) and x ( n ) = 3"u(n) To evaluate this convolution, we will use the convolution property of the ztransform. The ztransform of h(n) is and the ztransform of x(n) may be found from the timereversal and shift properties, or directly as follows: Therefore, the ztransform of the convolution, y(n) = x(n) * h(n), is
The region of convergence is the intersection of the regions JzJ> and lzl < 3, which is inverse ztransform, we perform a partial fraction expansion of Y (z), < lzl < 3. To find the
Y (z) =  + 1  I,I
B 1  321 and Therefore, it follows that
B = [(I  3z')Y (z)],,~ =  6 5 Y(.) = ($)(;)"u(n) + (;)3nu(n
CHAP. 41
THE ZTRANSFORM
Let x(n) be an absolutely summable sequence, with a rational ztransform. If X(z) has a pole at z = and limlzl+m X(z) = 1, what can be said about the extent of x(n) (i.e., finiteinlength, rightsided, etc.) Because x(n) is an absolutely summable sequence, the ROC of X ( z ) includes the unit circle, Izl = I. With a pole at z = f , the region of convergence will either be an annulus of the form r < lzl < r,, or it will be the exterior of a circle, r  < Izl. However, because X (z) converges as Izl + m, the region of convergence will be the exterior of a circle, and it follows that x(n) is rightsided (infinite in length) with x(n) = 0 for n < 0. Find the ztransform of x(n) = lnl(;)lnl.
Using the derivative property and the ztransform pair
it follows that the ztransform of w(n) = n(;)"u(n) is
Because x(n) may be written as
x(n) = \nl(f)"' = n(f)"u(n)  n ( f ) zdn) using linearity and the timereversal property, we have
which has a region of convergence
< Izl < 2.
Let y(n) be a sequence that is generated from a sequence x(n) as follows: (a) Show that y(n) satisfies the timevarying difference equation
and show that z2 dX(z) Y (z) =  Z  I dz where X(z) and Y (z) are the ztransforms of x(n) and y(n), respectively. (b) Use this property to find the ztransform of THE ZTRANSFORM
[CHAP. 4
(a) From the definition of y(n), we see that
and it follows immediately that y(n) y(n
 1) = nx(n) From this difference equation, we may take the ztransform of both sides. Because
then
Y(z)  z'Y (z) = z
dX(z) dz
(b) To find the ztransform of the given sequence, note that
where Because the ztransform of x(n) is
x(n) = (f)"u(n) 1 X(z) = 1  {zI
then
z2 dX(z) z2 :z 1 2 Y(z) = 2 z  1 dz 2  1 (1  iZl) 3 (1  ; z  ~ ) 2 ( l z  ~ ) Because x(n) is rightsided, then the region of convergence is the exterior of a circle. Having poles at z = 1 and z = it follows that the region of convergence is Izl > I. Find the value of x(0) for the sequence that has a ztransform X(z) = 1 1  az'

