Design a compound logic circuit with three inputs and one output.

a. Let the output be 1 when the binary value at the input is less than 3, and 0 otherwise.

b. Let the output be 1 when the input is an odd number.

Question

Design a compound logic circuit with three inputs and one output.

a. Let the output be 1 when the binary value at the input is less than 3, and 0 otherwise.

b. Let the output be 1 when the input is an odd number.

Accepted Answer

⚹ To design a compound logic circuit with three inputs (let's call them A, B, and C) and one output (let's call it Y), we need to consider the two conditions given: ⚹ ⚹ Condition a: The output Y should be 1 when the binary value at the input is less than 3. This means that the binary inputs 000, 001, 010, and 011 should result in an output of 1, since their decimal equivalents are 0, 1, 2, and 3, respectively. ⚹ ⚹ Condition b: The output Y should also be 1 when the input is an odd number. This means that the binary inputs 001, 011, 101, and 111 should result in an output of 1, since they represent the odd numbers 1, 3, 5, and 7, respectively. ⚹ ⚹ To derive the minimal Boolean expression, we can use a truth table to list all possible input combinations and their corresponding outputs based on the conditions above. ⚹ ⚹ Here is the truth table for the given conditions: ⚹ | A | B | C | Y (Condition a) | Y (Condition b) | Y (Final Output) | |---|---|---|-----------------|-----------------|------------------| | 0 | 0 | 0 | 1 | 0 | 1 | | 0 | 0 | 1 | 1 | 1 | 1 | | 0 | 1 | 0 | 1 | 0 | 1 | | 0 | 1 | 1 | 0 | 1 | 1 | | 1 | 0 | 0 | 0 | 0 | 0 | | 1 | 0 | 1 | 0 | 1 | 1 | | 1 | 1 | 0 | 0 | 0 | 0 | | 1 | 1 | 1 | 0 | 1 | 1 | ⚹ From the truth table, we can see that the final output Y is 1 for the input combinations 000, 001, 010, 011, 101, and 111. ⚹ ⚹ To simplify the Boolean expression for Y, we can use a Karnaugh map (K-map). ⚹ ⚹ Here is the K-map for the final output Y: ⚹ | AB \ CD | 00 | 01 | 11 | 10 | |---------|----|----|----|----| | 00 | 1 | 1 | 1 | 0 | | 01 | 1 | 0 | 1 | 0 | | 11 | 0 | 1 | 1 | 0 | | 10 | 0 | 0 | 0 | 0 | ⚹ By grouping the ones in the K-map, we can derive the simplified Boolean expression for Y: ⚹ ⚹ $ Y = \overline{A} \cdot \overline{B} + \overline{A} \cdot C + B \cdot C $ ⚹ ⚹ This is the minimal Boolean expression for the given conditions. To build the circuit, you would use NOT gates to obtain $\overline{A}$ and $\overline{B}$, and then use AND gates and an OR gate to combine the terms as per the expression above.⚹

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