The correct option is (d) a = (A-1) and b =1,A is some constant
Easiest explanation: The transfer function of High frequency emphasis is given as: Hhfe(u, v) = a + b Hhp(u, v) and the transfer function for High-boost filtering is Hhb(u, v) = (A-1) + Hhp(u, v), A being some constant. So, for a = (A-1) and b =1, Hhfe(u, v) = Hhb(u, v).