CS 61C (Fall 2007) |
Quiz 22Two questions, submit as "quiz22". Due 2:45pm before lecture 10/19/2007. |
x = (a&b) | (a & ((b|~a) | ~b)) = ab+(a((b+~a)+~b))
(given in Verilog and Boolean equation syntax) to the
input values a
and b
. Your expression should be immediately translatable to a circuit with
at most three gates. The gates of the circuit should include only (one-input) NOT and two-input AND and two-input
OR gates. You may give you answer in either Boolean algebra syntax or Verilog syntax, whichever is more comfortable
for you.