I'm working through Joseph Cavanagh's Digital Design and Verilog HDL Fundamentals. In chapter 3, he discusses using (+) and (-) in front of bits to indicate active high and active low signals, i.e.
I understand the concept of active low, where a logical 0 is a TRUE. And I think I understand what he's saying for figure (a), the equation is -z1 = (x1x2), and for (b), in order to get -z1, the active low signal, the equation is (x1x2'). What I want to make sure I understand are the semantics, for lack of a better term, of figure (b). x1 is active high, x2, because of the (-) is active low, so in order to drive z1 low (i.e., -z1), x2 has to be negated so that -x2 becomes x2 (i.e, x2').
I was interpreting figure (b) as NAND(1, 0) = z1, and what we want is NAND(1, 0') = -z1, but the discussion of active high and active low was confusing me. I think what I would like is confirmation my interpretation of figures (a) and (b) are correct. Or that I am over analysing this.