2 i egr 2400- hw 1 num. sys./logic gates name 2. for the following circuits, complete the truth table. write the product-of-sums equation for output z of the truth table

General
1

2 i egr 2400- hw 1 num. sys./logic gates name 2. for the following circuits, complete the truth table. write the product-of-sums equation for output z of the truth table in problem 2.a. and a sum-of-products equation for the z output in the table in problem 2.b 3.

2

LectureNotes said 2 i egr 2400- hw 1 num. sys./logic gates name 2. for the following circuits, complete the truth table. write the product-of-sums equation for output z of the truth table in problem 2.a. and a sum-of-products equation for the z output in the table in problem 2.b 3.

To tackle this problem, we need to follow these steps:

  1. Complete the Truth Table for the Given Circuit:

    • Identify the inputs and outputs of the circuit.
    • List all possible combinations of input values.
    • Determine the corresponding output for each combination based on the logic gates used in the circuit.

  2. Write the Product-of-Sums (POS) Equation for Output Z:

    • Identify the rows in the truth table where the output Z is 0.
    • Write the corresponding sum terms (OR terms) for these rows.
    • Combine these sum terms using AND operations to form the POS equation.

  3. Write the Sum-of-Products (SOP) Equation for Output Z:

    • Identify the rows in the truth table where the output Z is 1.
    • Write the corresponding product terms (AND terms) for these rows.
    • Combine these product terms using OR operations to form the SOP equation.

Example Walkthrough

Problem 2.a: Product-of-Sums (POS) Equation

Assume we have a circuit with inputs A, B, and C, and output Z. Let’s complete the truth table for this circuit:

A B C Z
0 0 0 1
0 0 1 0
0 1 0 1
0 1 1 0
1 0 0 1
1 0 1 0
1 1 0 1
1 1 1 0

From the truth table, we see that Z is 0 for the following input combinations:

  • A = 0, B = 0, C = 1
  • A = 0, B = 1, C = 1
  • A = 1, B = 0, C = 1
  • A = 1, B = 1, C = 1

Each of these conditions can be expressed as a sum term (OR term):

  • For A = 0, B = 0, C = 1: ( \overline{A} + \overline{B} + C )
  • For A = 0, B = 1, C = 1: ( \overline{A} + B + C )
  • For A = 1, B = 0, C = 1: ( A + \overline{B} + C )
  • For A = 1, B = 1, C = 1: ( A + B + C )

The POS equation is the AND of these sum terms:

Z = (\overline{A} + \overline{B} + C) \cdot (\overline{A} + B + C) \cdot (A + \overline{B} + C) \cdot (A + B + C)

Problem 2.b: Sum-of-Products (SOP) Equation

From the same truth table, we see that Z is 1 for the following input combinations:

  • A = 0, B = 0, C = 0
  • A = 0, B = 1, C = 0
  • A = 1, B = 0, C = 0
  • A = 1, B = 1, C = 0

Each of these conditions can be expressed as a product term (AND term):

  • For A = 0, B = 0, C = 0: ( \overline{A} \cdot \overline{B} \cdot \overline{C} )
  • For A = 0, B = 1, C = 0: ( \overline{A} \cdot B \cdot \overline{C} )
  • For A = 1, B = 0, C = 0: ( A \cdot \overline{B} \cdot \overline{C} )
  • For A = 1, B = 1, C = 0: ( A \cdot B \cdot \overline{C} )

The SOP equation is the OR of these product terms:

Z = (\overline{A} \cdot \overline{B} \cdot \overline{C}) + (\overline{A} \cdot B \cdot \overline{C}) + (A \cdot \overline{B} \cdot \overline{C}) + (A \cdot B \cdot \overline{C})

By following these steps, you can derive the POS and SOP equations for any given truth table.