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barcode generator for ssrs CHAP. 41 in Software
CHAP. 41 Recognizing Code 128C In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Code 128C Encoder In None Using Barcode drawer for Software Control to generate, create ANSI/AIM Code 128 image in Software applications. THE zTRANSFORM Inverse transforming each term, we have x(n) = a[(1)" Recognizing USS Code 128 In None Using Barcode scanner for Software Control to read, scan read, scan image in Software applications. Code 128 Code Set A Generation In Visual C# Using Barcode creation for Visual Studio .NET Control to generate, create Code 128 Code Set C image in VS .NET applications. + 1 + 2(n .+ I)]u(n) Code 128 Code Set A Generator In .NET Framework Using Barcode printer for ASP.NET Control to generate, create Code 128A image in ASP.NET applications. Create Code 128 Code Set B In .NET Using Barcode generator for .NET Control to generate, create Code 128 Code Set B image in VS .NET applications. Another way to invert this ztransform is to note that x(n) is the convolution of the two sequences. x(n) = x ~ ( n* xz(n) ) where xl (n) = u(n) and x2(n) is a step function that is upsampled by a factor of 2. Because xl(n) * x2(n) = { I , 1 , 2 , 2 , 3 , 3 , 4 , 4 , . . .) we have the same result as before. Make Code128 In VB.NET Using Barcode generator for .NET framework Control to generate, create ANSI/AIM Code 128 image in .NET applications. Barcode Generator In None Using Barcode encoder for Software Control to generate, create bar code image in Software applications. Find the inverse ztransform of the secondorder system
GS1128 Drawer In None Using Barcode maker for Software Control to generate, create GS1128 image in Software applications. EAN 13 Creator In None Using Barcode drawer for Software Control to generate, create EAN13 image in Software applications. Here we have a secondorder pole at z = f . The partial fraction expansion for X(z) is
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Print Identcode In None Using Barcode drawer for Software Control to generate, create Identcode image in Software applications. UPCA Supplement 5 Drawer In VS .NET Using Barcode generator for Reporting Service Control to generate, create UPC Symbol image in Reporting Service applications. and the constant A2 is
EAN 128 Scanner In VB.NET Using Barcode reader for .NET Control to read, scan read, scan image in Visual Studio .NET applications. EAN 13 Decoder In VS .NET Using Barcode decoder for Visual Studio .NET Control to read, scan read, scan image in Visual Studio .NET applications. Therefore, Making Bar Code In ObjectiveC Using Barcode encoder for iPhone Control to generate, create barcode image in iPhone applications. Recognize GS1  12 In Java Using Barcode scanner for Java Control to read, scan read, scan image in Java applications. Find the inverse of each of the following ztransforms: Making Barcode In Java Using Barcode printer for Android Control to generate, create bar code image in Android applications. ECC200 Reader In VB.NET Using Barcode decoder for .NET framework Control to read, scan read, scan image in .NET applications. (a) X z = log (1  iz') () Izl >
(b) X(z) = ellZ, with x ( n ) a rightsided sequence
(a) There are several ways to solve this problem. One is to look up or compute the power series expansion of the log function. Another way is to differentiate X(z). Specifically, because if we multiply both sides of this equation by (z), we have
THE zTRANSFORM Note that the region of convergence for X(z) is Izl > as it is for X(z), the inverse ztransform of Y (z) is [CHAP. 4
i. Because the region of convergence for Y (z) is the same
y(n) = ($)"u(n  I ) Now, from the derivative property, y(n) = nx(n), and i t follows that x(n) =  i ( i ) " u ( n  I) (6) For this ztransform, we could determine the inverse by tinding the power series expansion of X(z). However, another approach is to do what we did in part (a) and take the derivative. Differentiating X(z), we find d X(z) dz Multiplying both sides by (z), we have = z~x(z) and taking the inverse ztransform gives nx(n) = x(n  1) which is a recursion for x(n). To solve this recursion, we need an initial condition. Because x(n) is a rightsided sequence, we may use the initial value theorem to find x(0). Specifically, x(0) = lim l:+m
X (z) = I
Thus. the recursion that we want to solve is
with x(0) = 1. The solution for n > 0 is
and we have
42 .6 Find the inverse ztransform of X(z) = sin z.
To find the inverse ztransform of X(z) = sin z, we expand X(z) in a Taylor series about z = 0 as follows: Because
~ ( z= ) ) x(n)z" : )I=N
we may associate the coefficients in the Taylor series expansion with the sequence values x(n). Thus, we have x(n) = ( 1)" I (2lnl+ I)! n = 1,3, 5,... CHAP. 41
THE ZTRANSFORM
Evaluate the following integral: where the contour of integration C is the unit circle. Recall that for a sequence x(n) that has a ztransform X(z), the sequence may be recovered using contour integration as follows: Therefore, the integral that is to be evaluated corresponds to the value of the sequence x(n) at n = 4 that has a ztransform x (z) = + 2zI
 22  +z1)(1  +I) Thus, we may find x(n) using a partial fraction expansion of X(z) and then evaluate the sequence at n = 4. With this approach, however, we are finding the values of x(n) for all n. Alternatively, we could perform long division would then be the value of and divide the numerator of X(z) by the denominator. The coefficient multiplying z  ~ x(n) at n = 4, and the value of the integral. However, because we are only interested in the value of the sequence at n = 4, the easiest approach is to evaluate the integral directly using the Cauchy integral theorem. The value of the integral is equal to the sum of the residues of the poles of x(z)z3 inside the unit circle. Because has poles at z =
and z =
Therefore, we have
Find the inverse ztransform of
Note that the denominator of X(z) is a tenthorder polynomial. Although the roots may be found easily, performing a partial fraction expansion would be time consuming. For this problem, it is much better to exploit the properties of the ztransform. Note, for example, that X(z) = Y (zJO) where I Y(z) = 1  culozl

