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Taking the inverse ztransform of Eq. (4.90) and using the timeshifting property (4.181, we obtain in VS .NET
Taking the inverse ztransform of Eq. (4.90) and using the timeshifting property (4.181, we obtain Recognize QR Code JIS X 0510 In Visual Studio .NET Using Barcode Control SDK for .NET Control to generate, create, read, scan barcode image in Visual Studio .NET applications. QR Code 2d Barcode Creation In .NET Framework Using Barcode creation for Visual Studio .NET Control to generate, create Denso QR Bar Code image in .NET applications. y [ n ]  2 y [ n  1 1 = x [ n ]+ 3 x [ n  1 1 Decode QRCode In .NET Using Barcode reader for Visual Studio .NET Control to read, scan read, scan image in .NET framework applications. Bar Code Maker In VS .NET Using Barcode generation for .NET framework Control to generate, create barcode image in .NET applications. which is the same as Eq. (2.148). Bar Code Decoder In .NET Framework Using Barcode recognizer for .NET Control to read, scan read, scan image in .NET applications. Denso QR Bar Code Creation In C#.NET Using Barcode creation for VS .NET Control to generate, create QR Code image in .NET framework applications. 4.34. Consider the discretetime system shown in Fig. 48. For what values of k is the system BIB0 stable QR Code JIS X 0510 Creator In .NET Framework Using Barcode creator for ASP.NET Control to generate, create QR Code ISO/IEC18004 image in ASP.NET applications. QR Code ISO/IEC18004 Generation In VB.NET Using Barcode maker for .NET framework Control to generate, create Quick Response Code image in VS .NET applications. THE ZTRANSFORM AND DISCRETETIME LTI SYSTEMS
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Drawing Code39 In Visual Studio .NET Using Barcode generator for .NET framework Control to generate, create ANSI/AIM Code 39 image in Visual Studio .NET applications. Encoding UPC  E1 In VS .NET Using Barcode drawer for .NET framework Control to generate, create UPC  E0 image in Visual Studio .NET applications. From Fig. 48 we see that
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which shows that the system has one zero at z =  k / 3 and one pole at z = k / 2 and that the ROC is J z l > lk/2l. Thus, as shown in Prob. 4.30, the system will be BIB0 stable if the ROC contains the unit circle, lzl= 1. Hence the system is stable only if IkJ< 2. UNILATERAL ZTRANSFORM
4.35. Find the unilateral ztransform of the following x[n]: ( a ) x[n] = a n u [ n ] ( b ) x[n] = a n 'u[n
+ 11 = X(z) (a) Since x[nl = 0 for n < 0, X,(z) and from Example 4.1 we have
( b ) By definition (4.49) we have
1az' Izl > la1
Note that in this case x[n] is not a causal sequence; hence X,(z) Prob. 4.101.
+ X(z) [see Eq. (4.73) in
CHAP. 41
THE zTRANSFORM AND DISCRETETIME LTI SYSTEMS
4.36. Verify Eqs. (4.50) and (4.51), that is, for m 2 0, ( a ) x [ n  m ] z"x,(z) + Z  ~ + ' X [  11 + Z"" 'x[2] +x[m] ( a ) By definition (4.49) with m 2 0 and using the change in variable k = n
 m, we have
( b ) With m 1 0 8,{x[n
+m ] )= C x[n +m]z" = C x [ k ] ~  ( ~  ~ ) n =O
4.37. Using the unilateral ztransform, redo Prob. 2.42.
The system is described by
~ [ n ay[n  11 = x [ n ] ] with y [  11 = y  and x[n 1 = Kbnu[n].Let
~ [ " *l Y I ( z ) Then from Eq. (4.50) 11 ~ [ n  z  ~ Y , ( z ) + y [  I ] = Z  ' & ( Z ) + y  I
From Table 41 we have
x [ n ] X,(Z) =KZ
Izl> lbl
Taking the unilateral ztransform of Eq. (4.931, we obtain
&(z)  a { z  ' ~ ( z + y  J = KZ ) zb
THE ZTRANSFORM AND DISCRETETIME LTI SYSTEMS
[CHAP. 4
Thus, y(z) = ay\ + K
z (za)(zb) Using partialfraction expansion, we obtain Y,(z) =ay, z za
z z  ~ ( ~ z  b za
Taking the inverse ztransform of Y,(z), we get b Y [ n ] = a y  ,anu[n] + Kbnu[n] ba
 Kanu[n] which is the same as Eq. (2.158). 4.38. For each of the following difference equations and associated input and initial conditions, determine the output y[n I: ( a ) y[nl  f y [ n  11 = x [ n l , with x [ n ] = (:In, y[ 11 = 1 ( b ) 3 y [ n ]  4 y [ n  11 + y[n  21 = A n ] , with x[n] = (i)",  11 = 1, y[ 21 = 2 y[ Taking the unilateral ztransform of the given difference equation, we get Y,(z)  +{z'Y,(z) + y [  I]} = X , ( z ) Substituting y[ 11 = 1 and X,(z) into the above expression, we get Thus, Hence, y [ n ) = 7(;)"+'  2 ( f ) n (b) x[n] ++ X,(z) n 2 1 z   7 Taking the unilateral ztransform of the given difference equation, we obtain
2Y, 3Y,(z)  4{z ' 5( z ) + Y [  I ] } + { ~  ( z ) + z  ' y [  11 + ~ [  2 ] }= X I ( Z ) CHAP. 41
THE ZTRANSFORMAND DISCRETETIME LTI SYSTEMS
Substituting y[ 1 = 1, y[2]= 2, and X,(z) into the above expression, we get 1
Thus, Hence, y[n]=$
 (f)"+ +(f)" n r 2 439. Let x [ n ] be a causal sequence and
x b l X(z) Show that
x[O] = lim X(z) Equation (4.94) is called the initial value theorem for the ztransform.
Since x [ n ] = 0 for n < 0, we have
As z
+ oa, 0 for n > 0. Thus, we get
lim X ( z ) = x[O] 2.m
4.40. Let x [ n ] be a causal sequence and Show that if X(z) is a rational function with all its poles strictly inside the unit circle except possibly for a firstorder pole at z = 1, then Ntm
lim x [ N ] = lim (1 2')X(z) (4.95) Equation (4.95) is called the final value theorem for the ztransform.
From the timeshifting property (4.19) we have The lefthand side of Eq.(4.96) can be written as
T H E ZTRANSFORM AND DISCRETETIME LTI SYSTEMS
[CHAP. 4
If we now let z
, 1, then from Eq. (4.96) we have
lim ( I  2  ' ) X ( z ) z+ 1
N+m n = O
( x [ n ] x [ n  I]} lim x [ N ] Supplementary Problems
Find the ztransform of the following x [ n ] : (a) (b) (c) (d) Am.
x [ n l = ( ; , I ,  $1 x [ n ]= 2 S [ n + 21  3S[n  21 x [ n ]= 3 (  f)"u[n]  2(3)"u[n  I ] x [ n l = 3 ( i ) " u [ n l  2(a)"u[n  11 (a) X(z)= f + z'

