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ICS331 Homework 2 1. Design a circuit for a 2-to-4 decoder. Draw the circuit using LogicWorks. X Y O1 O2 O3 O4 this decoder is active high 00 1 0 0 0 01 0 1 0 0 10 0 0 1 0 11 0 0 0 1 XY 00 01 10 11 O1’ 0 1 1 1 O2’ 1 0 1 1 O3’ 1 1 0 1 O4’ this decoder is active low 1 1 1 0 2. Provide the expression for the following truth table. X, Y and Z are inputs, O is the output. What is the purpose of Z? X 1 1 0 0 Y 1 0 1 0 Z 1 0 0 0 O 0 1 1 1 O = xy’z’ + x’yz’+ x’y’z’ = (x’ + y’ + z’). as SoP as PoS We can think of Z in many different ways, depending what we “see” in the table. z is a control input of some kind. We can think that it works so that O = z’; or O = (xy)’ only when z = 1, i.e. O=(xy)’z. Check: O = xy’z’ + x’yz’+ x’y’z’ = z’(xy’+ x’y + x’y’) = z’(y’(x + x’) + x’y) = z’(y’ + x’y) = z’(y’ + x’) = z’(xy)’. For this example, we can also argue that O = z’, because we never encounter the input x=0, y = 0, z =1, or 1 1 0, 011, 101, and 110. 3. Construct a circuit to calculate Y = XZ + X’Z + Z’. Simplify. The solution shows how the K map works: Y = XZ + X’Z + Z’ = Z(X + X’) + Z’ = Z + Z’ = 1 4. Provide the expression for the following circuit: