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3. Find the Boolean expression and convert to NAND gates for the following two circuits:

3. Find the Boolean expression and convert to NAND gates for the following two circuits:

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
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**Boolean Expression and Conversion to NAND Gates** **Objective:** Find the Boolean expression and convert it to NAND gates for the following two circuits. ### Circuit 1: - **Input Variables:** A, B, C. - **Configuration:** 1. Inputs B and C are fed into an AND gate. 2. Inputs A and B are also fed into another AND gate. 3. The outputs of both AND gates are fed into a third AND gate. 4. The output of this third AND gate, along with the output from the initial AND gate (B and C), are fed into an OR gate. - **Boolean Expression:** \[ ((B \cdot C) \cdot (A \cdot B)) + (B \cdot C) \] ### Circuit 2: - **Input Variables:** A, B, C, D. - **Configuration:** 1. Inputs B, C, and D are fed into an AND gate. 2. The output of this AND gate is combined with input A in an OR gate. - **Boolean Expression:** \[ A + (B \cdot C \cdot D) \] ### Conversion to NAND Gates: 1. **For Circuit 1:** - Use De Morgan's laws and double negation to design the circuit using only NAND gates. This involves expressing all the AND and OR operations in terms of NAND. 2. **For Circuit 2:** - Similarly, apply transformations using NAND gates equivalents for the AND and OR operations. **Note:** Conversion involves utilizing the properties: - \(A \cdot B\) can be rendered using NAND as \((A \text{ NAND } B) \text{ NAND } (A \text{ NAND } B)\). - \(A + B\) can be achieved using NAND by Demorgans Theorem as \((A \text{ NAND } A) \text{ NAND } (B \text{ NAND } B)\). For more detailed diagrams and step-by-step transformations, please refer to educational resources on digital logic design.
Transcribed Image Text:**Boolean Expression and Conversion to NAND Gates** **Objective:** Find the Boolean expression and convert it to NAND gates for the following two circuits. ### Circuit 1: - **Input Variables:** A, B, C. - **Configuration:** 1. Inputs B and C are fed into an AND gate. 2. Inputs A and B are also fed into another AND gate. 3. The outputs of both AND gates are fed into a third AND gate. 4. The output of this third AND gate, along with the output from the initial AND gate (B and C), are fed into an OR gate. - **Boolean Expression:** \[ ((B \cdot C) \cdot (A \cdot B)) + (B \cdot C) \] ### Circuit 2: - **Input Variables:** A, B, C, D. - **Configuration:** 1. Inputs B, C, and D are fed into an AND gate. 2. The output of this AND gate is combined with input A in an OR gate. - **Boolean Expression:** \[ A + (B \cdot C \cdot D) \] ### Conversion to NAND Gates: 1. **For Circuit 1:** - Use De Morgan's laws and double negation to design the circuit using only NAND gates. This involves expressing all the AND and OR operations in terms of NAND. 2. **For Circuit 2:** - Similarly, apply transformations using NAND gates equivalents for the AND and OR operations. **Note:** Conversion involves utilizing the properties: - \(A \cdot B\) can be rendered using NAND as \((A \text{ NAND } B) \text{ NAND } (A \text{ NAND } B)\). - \(A + B\) can be achieved using NAND by Demorgans Theorem as \((A \text{ NAND } A) \text{ NAND } (B \text{ NAND } B)\). For more detailed diagrams and step-by-step transformations, please refer to educational resources on digital logic design.
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