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barcode generator for ssrs 1  azI in Software
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ANSI/AIM Code 128 Printer In .NET Framework Using Barcode printer for ASP.NET Control to generate, create ANSI/AIM Code 128 image in ASP.NET applications. Creating Code 128 Code Set A In VS .NET Using Barcode creator for VS .NET Control to generate, create Code 128 Code Set A image in .NET applications. (b) For this sequence we write x(n) = cos(nwo)u(n) = [eJnq+ ejnq M n ) Therefore, the ztransform is Making Code 128 In Visual Basic .NET Using Barcode drawer for VS .NET Control to generate, create Code 128 Code Set A image in VS .NET applications. Code128 Generator In None Using Barcode creation for Software Control to generate, create USS Code 128 image in Software applications. with a region of convergence lzl
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THE 2TRANSFORM where the sum converges for 1321 < I
121 i [CHAP. 4
Alternatively, note that the timereversed sequence y(n) = x(n) = (;)"u(n) has a ztransform given by with a region of convergence given by Izl > 3. Therefore, using the timereversal property, Y(z) = X(zI), we obtain the same result. (b) Because x(n) is the sum of two sequences, we will find the ztransform of x(n) by finding the ztransforms of each of these sequences and adding them together. The ztransform of the first sequence may be found easily using the shift property. Specifically, note that because the ztransform of (;)"u(n
+ 2) is 4.2' times the ztransform of (a)"u(n), that is.
which has a region of convergence lzl > The second term is a leftsided exponential and has a ztransform that we have seen before, that is, with a region of convergence Izl < 3. Finally, for the ztransform of .u(n). we have
with a region of convergence
< Izl < 3.
(c) As we saw in Problem 4.3(b), the ztransform of cos(nwo)u(n)is
cos(nwo)u(n) 6 Therefore, using the exponentiation property, I  (cos w")zI I ~(COS O&)z' z2 ' ~zl 1 >
we have with a region of convergence lz 1 > f .
( d ) Writing x(n) as
s ( n ) = a n u ( n ) cr"u(n) we may use the linearity and timereversal properties to write
1 1 X(z) = 1  ffz' I  ffz
S(n) ;<1z1<2
which may be simplified to
CHAP. 41
THE ZTRANSFORM
Without explicitly solving for X(z), find the region of convergence of the ztransform of each of the following sequences: (a, x ( n ) = [(i)n( i ) n ] u ( n 10, +
1 (b) x(n) = 0 10(n( otherwise
( a ) Because the first sequence is rightsided, the region of convergence is the exterior of a circle. With a pole at z = coming from the term and a pole at z = coming from the term (:)" it follows that the region of convergence must be Izl > (i)", i.
(b) This sequence is finite in length. Therefore, the region of convergence is at least 0 < Izl < oo. Because x(n) has nonzero values for n < 0 and for n r 0, z = 0 and z = oo are not included within the ROC. (c) Because this sequence is leftsided, the region of convergence is the interior of a circle. With a pole at z = 2, it follows that the region of convergence is Iz I < 2. Find the ztransform of the sequence y ( n ) = xi=, ( k ) in terms of the ztransform of x ( n ) . x
There are two ways to approach this problem. The first is to note that x(n)may be written in terms of y(n)as follows: Therefore, if we transform both sides of this equation, and use the shift property of the ztransform, we find X ( z ) = Y ( z ) z  ' y ( z ) Solving for Y ( z ) ,we find
Thus. which is referred to as the summation property. The second approach is to note that y(n) is the convolution of x ( n )with a unit step, Therefore, using the convolution theorem, we have
and, with U ( z ) = I / ( I  z  I ) , we obtain the same result as before. For the region of convergence, note that because the ROC of l / ( z )is Iz( > 1, the ROC of Y ( z ) will be at least R, = R, n ( I Z I > I ) where R, is the ROC of X ( z ) . Find the ztransform of the sequence y ( n ) where
and y(n) = 0 for n < 0. Assume that la1 < 1.
THE ZTRANSFORM
[CHAP. 4
For this sequence, we may use a variation of the summation property derived in Prob. 4.6. Specifically, recall that if then Now consider the twosided summation, Y(z)= X(z) I  z' which may be written as
Therefore, if we let
Therefore, we have
Y ( z )= X I( z ) + X2W  x(0) I  z' Finally, with x ( n ) = a " 1 , follows that x l ( n )= x2(n) = a n u ( n ) ,and x ( 0 ) = I. Thus, it

