526 Appendix A Solutions to the Exercises The diffcrrncc is a shift, and a constant factor Solution 13.12 a) r;[k] = ~ [ k-] ~ [ -k r‘ - 11 +~ [ -k2,r - 21 with the shift theorem: X ( ) = { [ k ] }= A (1- z2-’’-1 p - + 2-1 b) double zeros at z7’+l= ‘rhc single pole a.t z j zrL= e I, 7/ 7.1-1 n = 0, 1,2, ,r i s riullificd by a zero at z = : pole witallnmltiplicity of 2r + 1al y? =0