To solve this problem, let's break it down into the following steps:

1. Analyze the 4:1 Multiplexer Setup: The multiplexer (MUX) has 4 inputs: $I_0$, $I_1$, $I_2$, and $I_3$, and two selection inputs, $A$ and $B$.

   • The output of the MUX depends on the values of $A$ and $B$ according to this selection:
     • $AB = 00$ selects $I_0$
     • $AB = 01$ selects $I_1$
     • $AB = 10$ selects $I_2$
     • $AB = 11$ selects $I_3$

2. Determine Inputs $I_0$, $I_1$, $I_2$, and $I_3$:

   • From the diagram:
     • $I_0 = C$
     • $I_1 = D$
     • $I_2 = C \cdot D$ (AND gate)
     • $I_3 = C + D$ (OR gate)

3. Formulate the Output $F(A, B, C, D)$:

   • The output function $F(A, B, C, D)$ will be a combination of $I_0$, $I_1$, $I_2$, and $I_3$ based on the values of $A$ and $B$.
   • Therefore, we can express $F$ as: \begin{cases} C & \text{if } AB = 00 \\ D & \text{if } AB = 01 \\ C \cdot D & \text{if } AB = 10 \\ C + D & \text{if } AB = 11 \end{cases}$$

4. Construct the Truth Table:

   • We'll construct the truth table by listing all combinations of $A$, $B$, $C$, and $D$, then applying the selection logic to determine $F$.

5. Identify the Minterms:

   • Minterms are the rows in the truth table where the output $F$ is 1.

I'll proceed to create the truth table and determine the minterms.### Truth Table for $F(A, B, C, D)$

Minterms

The rows where $F = 1$ are the minterms:

Would you like further details on how these values were derived or have any other questions?

Related Questions

1. How would you simplify the logic function using Boolean algebra?
2. How does the multiplexer select lines impact the truth table?
3. What would happen if the AND and OR gate outputs were swapped?
4. Can you create a Karnaugh map to further simplify this logic?
5. How do changes in selection inputs affect the resulting minterms?

Tip

When designing logic circuits, breaking down the functionality of each component (like the MUX here) helps in understanding and constructing truth tables accurately.