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barcode generator for ssrs IzI > a in Software
IzI > a Code128 Reader In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Printing Code 128C In None Using Barcode drawer for Software Control to generate, create Code 128 Code Set A image in Software applications. Taking the limit of X(z) as z + oo, we see that X(z) + 1. Because the limit exists, x(n) is causal, and x(0) = 1. Recognizing Code 128 Code Set A In None Using Barcode reader for Software Control to read, scan read, scan image in Software applications. Painting Code 128 Code Set B In Visual C# Using Barcode generator for Visual Studio .NET Control to generate, create Code 128 image in .NET framework applications. Find the value of x ( 0 ) for the sequence that has a ztransform
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Data Matrix ECC200 Drawer In None Using Barcode maker for Software Control to generate, create DataMatrix image in Software applications. Barcode Creation In None Using Barcode generator for Software Control to generate, create bar code image in Software applications. we see that X(z) + oo as Izl + oo.Therefore, x(n) is not causal. However, because x(n) is rightsided, it may be delayed so that it is causal. Specifically, if we delay x(n) by 1 to form the sequence y(n) = x(n  I), Create Code27 In None Using Barcode generation for Software Control to generate, create ABC Codabar image in Software applications. Drawing UPC A In ObjectiveC Using Barcode creation for iPad Control to generate, create GS1  12 image in iPad applications. Y (z) = Scan USS128 In VB.NET Using Barcode reader for Visual Studio .NET Control to read, scan read, scan image in .NET applications. ECC200 Creation In None Using Barcode creator for Word Control to generate, create Data Matrix ECC200 image in Microsoft Word applications. which approaches 1 as lzl
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X(z)  x(l)z is the ztransform of a causal sequence, and it follows from the initial value theorem that With
we have
x(0) = lim [X(z)  x( I )Z1 =
Izlbm
Generalize the initial value theorem to find the value of a causal sequence x ( n ) at n = 1 , and find x ( 1 ) when If x(n) is causal, X(z) = x(0) + x(1)zI + , ~ ( 2 ) z + ~
Therefore, note that if we subtract x(0) from X(z), Multiplying both sides of this equation by z, we have
If we let z + m, we obtain the value for x(l), x ( l ) = lirn (z[X(z)  x(O)]J
Izlrm
For the given ztransform we see that x(0) = lim X(z) = $
(21+m
Therefore, x(1) = lim (z[X(z)  x(0)I) = Izl+m
THE zTRANSFORM
[CHAP. 4
Let x ( n ) be a leftsided sequence that is equal to zero for n > 0 . If
find x(0). For a leftsided sequence that is zero for n > 0, the ztransform is
Therefore, it follows that x(0) = lim X(z) For the given ztransform, we see that 32' 2zp2 32 2 = lim =2 x(0) = lim X(z) = lim iro 20 3  z' + z2 20 3z2  z 1 If x(n) is real and even with a rational ztransform, show that
and describe what constraints this places on the poles and zeros of X ( z ) . If x(n) is even, x(n) = x(n) Therefore, it follows immediately from the timereversal property that
If X(z) has a zero at z = zO, X(z0) = 0 then
x (z,') which implies that X(z) will also have a zero at z = 1/20, The same holds true for poles. That is, if there is a pole at zO,there must also be a pole at z = I /zo. Use the derivative property to find the ztransform of the following sequences: (a) x ( n ) = n(;)"u(n  2 ) 1 (b) x(n) = ; (  2 )  " 4  n
 1) (a) The derivative property states that if X(z) is the ztransform of x(n), If we let x(n) = nw(n), where w(n) = (;)"u(n from the delay property and the ztransform pair
 2) = f (i)  2) CHAP. 41 it follows that
THE 2TRANSFORM
Therefore, using the derivative property, we have the ztransform of x(n), (b) Evaluating the ztransform of this sequence directly is difficult due to the factor of nI. However, if we define a new sequence, y(n), as follows, y(n) = nx(n) = (2)"4n the ztransform of y(n) is easily determined to be
Y (z) =  l + ;z' II < z
Noting the relationship between x(n) and y(n), we can apply the derivative property to set up a differential equation for X(z), The solution to this differential equation is
X(z)= log (z + f ) and the region of convergence is Iz 1 <
Upsampling is an operation that stretches a sequence in time by inserting zeros between the sequence values. For example, upsampling a sequence x ( n ) by a factor of L results in the sequence y(n> =
otherwise
Express the ztransform of y(n) in terms of the ztransform of x(n). Because y(n) isequal to zero for all n signal is
+ kL, with y(n) equal tox(n/L) forn = kL, the ztransform of the upsampled
If X(z) converges for cu < Izl < /3, Y (z) will converge for cu < lzlL < p, or
a'/'. <
(zI < pIIL
42 .2 Find the ztransform of the sequence
an/10 x(n) = n = 0, 10,20, . . . else
where l 1 = 1. a
THE zTRANSFORM
[CHAP. 4
We recognizex(n) as an exponential sequence that has been upsampled by a factor of 10 (see Prob. 4.21). Therefore, because the ztransform of x(n) is
Inverse z'Ransforms
Find the inverse of each of the following ztransforms: Because X(z) is a finiteorder polynomial, x(n) is a finitelength sequence. Therefore, x(n) is the coefficient that multiplies z" in X(z). Thus, x(0) = 4 and x(2) = x(2) = 3. This ztransform is a sum of two firstorder rational functions of z. Because the region of convergence of X(z) is the exterior of a circle, x(n) is a rightsided sequence. Using the ztransform pair for a rightsided exponential, we may invert X(z) easily as follows: Here we have a rational function of z with a denominator that is a quadratic in z. Before we can find the inverse ztransform, we need to factor the denominator and perform a partial fraction expansion: Because x(n) is rightsided, the inverse ztransform is
One way to invert this ztransform is to perform a partial fraction expansion. With X(z) = I (Iz)(~z~ (I the constants A , B I, and B2 are as follows:

