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Taking the inverse z-transform of Eq. (4.90) and using the time-shifting property (4.181, we obtain in VS .NET
Taking the inverse z-transform of Eq. (4.90) and using the time-shifting 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 QR-Code 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 discrete-time system shown in Fig. 4-8. 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 Z-TRANSFORM AND DISCRETE-TIME LTI SYSTEMS
Draw ECC200 In Visual Studio .NET Using Barcode generator for .NET framework Control to generate, create Data Matrix 2d barcode image in Visual Studio .NET applications. Draw GS1 128 In VS .NET Using Barcode generation for VS .NET Control to generate, create GS1 128 image in Visual Studio .NET applications. [CHAP. 4
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. 4-8 we see that
UPCA Generator In Java Using Barcode drawer for Eclipse BIRT Control to generate, create Universal Product Code version A image in Eclipse BIRT applications. Generate Barcode In .NET Framework Using Barcode printer for Reporting Service Control to generate, create bar code image in Reporting Service applications. Taking the z-transform of the above equations, we obtain
Making Bar Code In C#.NET Using Barcode generator for .NET framework Control to generate, create bar code image in .NET framework applications. Data Matrix ECC200 Generation In None Using Barcode creator for Microsoft Word Control to generate, create ECC200 image in Office Word applications. Y(z) = Q ( z ) Rearranging, we have
EAN-13 Decoder In Visual Studio .NET Using Barcode reader for VS .NET Control to read, scan read, scan image in VS .NET applications. Draw Barcode In Java Using Barcode creator for BIRT Control to generate, create bar code image in BIRT applications. + z-'Q(z) Encode EAN-13 Supplement 5 In C#.NET Using Barcode creation for .NET framework Control to generate, create EAN13 image in .NET framework applications. Painting Code 128 In Visual C# Using Barcode drawer for VS .NET Control to generate, create Code 128B image in .NET applications. from which we obtain
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 Z-TRANSFORM
4.35. Find the unilateral z-transform 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
1-az-' 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 z-TRANSFORM AND DISCRETE-TIME 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 z-transform, 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 4-1 we have
x [ n ] -X,(Z) =KZ
Izl> lbl
Taking the unilateral z-transform of Eq. (4.931, we obtain
&(z) - a { z - ' ~ ( z + y - J = KZ ) z-b
THE Z-TRANSFORM AND DISCRETE-TIME LTI SYSTEMS
[CHAP. 4
Thus, y(z) = ay-\ -+ K
z (z-a)(z-b) Using partial-fraction expansion, we obtain Y,(z) =ay-,- z z-a
z z - ~ ( ~ z - b z-a
Taking the inverse z-transform of Y,(z), we get b Y [ n ] = a y - ,anu[n] + K-bnu[n] b-a
- K-anu[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 z-transform 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 z-transform 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 Z-TRANSFORMAND DISCRETE-TIME 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 z-transform.
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 first-order pole at z = 1, then N-tm
lim x [ N ] = lim (1 -2-')X(z) (4.95) Equation (4.95) is called the final value theorem for the z-transform.
From the time-shifting property (4.19) we have The left-hand side of Eq.(4.96) can be written as
T H E Z-TRANSFORM AND DISCRETE-TIME 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 z-transform 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-'
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