The correct option is D The system H(z) can act as all pass filter for β=1α∗.
For LPF:
limz→1z+11−βz−1=21−β(high)
limz→1z+11−βz−1=01+β=0(low)Correct
(b) For HPF
limz→1z−11−z−1=1 (high)
So system is not high pass filter
(c) H(z)=z2−zαz−β
Since, order of numerator is greater than order of denominator. Therefore system can not be casual.
(d) For all pass filter,
Pole=1Zero∗