Taking the inverse z-transform of Eq. (4.90) and using the time-shifting property (4.181, we obtain in VS .NET

Encoding QR Code in VS .NET Taking the inverse z-transform of Eq. (4.90) and using the time-shifting property (4.181, we obtain

y [ n ] - 2 y [ n - 1 1 = x [ n ]+ 3 x [ n - 1 1

which is the same as Eq. (2.148).

4.34. Consider the discrete-time system shown in Fig. 4-8. For what values of k is the system BIB0 stable

+ z-'Q(z)

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.

4.35. Find the unilateral z-transform of the following x[n]:

( a ) x[n] = a n u [ n ]

+ 11

= X(z)

(a) Since x[nl = 0 for n < 0, X,(z)

1-az-'

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]

+m ] )= C x[n +m]z-" = C x [ k ] ~ - ( ~ - ~ )

~ [ n- ay[n - 11 = x [ n ] ]

~ [ " *l Y I ( z )

Then from Eq. (4.50)

&(z)

Thus, y(z)

= ay-\

z (z-a)(z-b)

Using partial-fraction expansion, we obtain Y,(z) =ay-,-

- 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

2Y, 3Y,(z) - 4{z- ' 5( z ) + Y [ - I ] } + { ~ - ( z ) + z - ' y [ - 11 + ~ [ - 2 ] }= X I ( Z )

Thus,

Hence,

- (f)"+ +(f)"

n r -2

x b l -X(z)

x[O] = lim X(z)

+ oa,

lim X ( z ) = x[O]

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

lim x [ N ] = lim (1 -2-')X(z)

(4.95)

---, 1,

lim ( I - 2 - ' ) X ( z )

( x [ n ]- x [ n - I]}

lim x [ N ]

Find the z-transform of the following x [ n ] :

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-'