Wiley signals and systems e book TLFe BO 541

526

Appendix A . Solutions to the Exercises

The diffcrrncc is a shift, and a constant factor

Solution 13.12
a) .r;[k] = ~ [ k-] 2 ~ [ -k r‘ - 11

+~

[ -k2,r. - 21

with the shift theorem:

X ( 2 ) = 2 { 5 [ k ] }= A (1- z2-’’-1

p 7 - 2

+

2-1

b) double zeros at z7’+l= 1
‘rhc single pole a.t z

j

zrL= e

2 I,

7/

7.1-1

. n = 0, 1,2, . . . ,r

1 i s riullificd by a. zero at z = 1.

:

pole witallnmltiplicity of 2r.

+ 1al

y?

=0