Final answer:
To draw the logic diagram for the Boolean expression (WZ + X)Y + Z', we use multiple OR and AND gates. To simplify the expression, we use a Karnaugh Map to find the simplified expression Z' + X.
Step-by-step explanation:
To draw the logic diagram for the Boolean expression (WZ + X)Y + Z', we can break it down into smaller components. Let's start with the innermost part, (WZ + X). This can be represented by a OR gate with inputs WZ and X. Then, we take the output of this OR gate and connect it to another OR gate along with the input Y. Finally, we take the output of this second OR gate and connect it to an AND gate along with the input Z'. The complete logic diagram will have three gates: two OR gates and one AND gate.
To simplify the expression using a Karnaugh Map, we need to create a truth table and find the groups of 1s. The truth table for the expression is as follows:
| W | X | Y | Z' | Output |
|---|---|---|----|--------|
| 0 | 0 | 0 | 1 | 1 |
| 0 | 0 | 1 | 1 | 1 |
| 0 | 1 | 0 | 1 | 1 |
| 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 | 1 |
| 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 0 | 1 | 0 |
| 1 | 1 | 1 | 1 | 1 |
From the truth table, we can see that the output is 1 for all combinations except when W=1, X=1, and Y=0. To simplify the expression, we can use the Karnaugh Map:
| 00 | 01 | 11 | 10 |
______|____|____|____|____|
0 0 | 1 | 1 | 0 | 1 |
1 1 | 1 | 1 | 1 | 1 |
From the Karnaugh Map, we can see that the simplified expression is Z' + X. We can represent this as an OR gate with inputs Z' and X. Therefore, the corresponding two-level logic diagram as a sum of products implementation will have one OR gate and one AND gate.