Thanks FvM,
The reason I said signed number, is that as long as I know we leverage the 2's complement wrap around in integrator sections to avoid overflow problem. I don't know how can we do it with unsigned number.
Also if we do not convert 0 to -1, there will be no subtraction in integrator sections. because when input is 0, nothing will be added or subtracted.
Finally I dont understand why you say converting 1b'0 to -1 = 16'hFFFF, and 1'b1 to +1 = 16'h0001, is like multiplying the input bit by two.
Please correct me if I am wrong.
Thanks in advance.