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: