Let's break down the problem step by step: --- ## **Given:** **Boolean Function:** \( F(A, B, C, D) = \sum(, 1, 2, 3, 4, 5, 7, 8, 10, 12, 14) \) We need to implement it using: a) 8x1 MUX (using \(A, B, C\) as select lines) b) 4x1 MUX (using \(A, D\) as select lines) --- ## **Step 1: Write the Truth Table** Let's write the minterms (where output = 1): | A | B | C | D | F | |---|---|---|---|---| | | | | | 1 | | | | | 1 | 1 | | | | 1 | | 1 | | | | 1 | 1 | 1 | | | 1 | | | 1 | | | 1 | | 1 | 1 | | | 1 | 1 | 1 | 1 | | 1 | | | | 1 | | 1 | | 1 | | 1 | | 1 | 1 | | | 1 | | 1 | 1 | 1 | | 1 | | Others | | | --- ## **a) Using 8x1 MUX (Select lines: A, B, C)** - Select lines: A, B, C → 3 bits (so 8 combinations) - The inputs to the MUX will be functions of D (since D is left). For each combination of (A,B,C), fill in the input for the corresponding data line as a function of D. Let's list all possible combinations of A,B,C (from to 7) and see for which D the output is 1. | S2 (A) | S1 (B) | S (C) | Combinations (ABC) | Data Input (I) | For which D is F=1? | Expression for Data Input | --------|--------|--------|--------------------|----------------|----------------------|-------------------------| | | | | | I | D=,1 | 1 (since both and 1) | | | | 1 | 1 | I1 | D=,1 | 1 | | | 1 | | 2 | I2 | D=,1 | 1 | | | 1 | 1 | 3 | I3 | D=1 | D | | 1 | | | 4 | I4 | D= | D' | | 1 | | 1 | 5 | I5 | D= | D' | | 1 | 1 | | 6 | I6 | D= | D' | | 1 | 1 | 1 | 7 | I7 | D= | D' | But let’s double-check for each (A,B,C): #### For each (A,B,C): - (,,): minterms (D=), 1 (D=1) → I = 1 - (,,1): minterms 2 (D=), 3 (D=1) → I1 = 1 - (,1,): minterms 4 (D=), 5 (D=1) → I2 = 1 - (,1,1): minterm 7 (D=1) only → I3 = D - (1,,): minterm 8 (D=) only → I4 = D' - (1,,1): minterm 10 (D=) only → I5 = D' - (1,1,): minterm 12 (D=) only → I6 = D' - (1,1,1): minterm 14 (D=) only → I7 = D' **Final inputs to MUX:** | I | I1 | I2 | I3 | I4 | I5 | I6 | I7 | |----|----|----|----|----|----|----|----| | 1 | 1 | 1 | D | D' | D' | D' | D' | **Connect A, B, C to S2, S1, S respectively.** --- ## **b) Using 4x1 MUX (Select lines: A, D)** - Select lines: A (S1), D (S) → 2 bits - The remaining variables: B, C Let’s group the minterms according to the select lines (A, D): We need to write the function in terms of B and C for each (A,D) combination. Let’s write the groups: ### **(A=, D=):** Possible minterms: - (A,B,C,D) = (,,,) → minterm → 1 - (,,1,) → minterm 2 → 1 - (,1,,) → minterm 4 → 1 So for (A=, D=), F = 1 for all combinations except (,1,1,) (minterm 6, not included, so F=). Check (,1,1,): minterm 6 is not in the list, so F=. So, for (A=, D=): - (B=, C=): F=1 - (B=, C=1): F=1 - (B=1, C=): F=1 - (B=1, C=1): F= So, the function is F = B'C' + B'C + BC' = B' + C' ### **(A=, D=1):** Possible minterms: - (,,,1): minterm 1 → 1 - (,,1,1): minterm 3 → 1 - (,1,,1): minterm 5 → 1 - (,1,1,1): minterm 7 → 1 So for all (B,C) values, F=1. So, output = 1 ### **(A=1, D=):** Possible minterms: - (1,,,): minterm 8 → 1 - (1,,1,): minterm 10 → 1 - (1,1,,): minterm 12 → 1 - (1,1,1,): minterm 14 → 1 So, let's write the values for (B,C): - (,): minterm 8 → 1 - (,1): minterm 10 → 1 - (1,): minterm 12 → 1 - (1,1): minterm 14 → 1 So, for all (B,C), F=1 ### **(A=1, D=1):** Possible minterms: - (1,,,1): minterm 9 → not present - (1,,1,1): minterm 11 → not present - (1,1,,1): minterm 13 → not present - (1,1,1,1): minterm 15 → not present So, for all (B,C), F= --- **Summary Table for 4x1 MUX:** | S1 (A) | S (D) | Input (I) | |--------|--------|------------| | | | B' + C' | | | 1 | 1 | | 1 | | 1 | | 1 | 1 | | So, connect A to S1, D to S, and inputs as above. --- ## **Final Answers** ### **a) Using 8x1 MUX** - **Select Lines:** A → S2, B → S1, C → S - **Data Inputs:** - I = 1 - I1 = 1 - I2 = 1 - I3 = D - I4 = D' - I5 = D' - I6 = D' - I7 = D' --- ### **b) Using 4x1 MUX** - **Select Lines:** A → S1, D → S - **Data Inputs:** - I = B' + C' - I1 = 1 - I2 = 1 - I3 = --- **If you need the circuit diagrams as well, let me know!**