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A circuit has three inputs and one output. The output is 1 if at least two of the three input variables are 1, otherwise it is zero. The minimum number of basic gates required to implement the output (Y) are
- a)3
- b)4
- c)7
- d)8
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A circuit has three inputs and one output. The output is 1 if at least...
The truth table for the given condition is shown below:
The K-map for above truth table is shown below.
Thus, Y = AB + BC + CA which can be implemented using 3 AND gates and 1 OR gate (total 4 basic gates).
The K-map for above truth table is shown below.
Thus, Y = AB + BC + CA which can be implemented using 3 AND gates and 1 OR gate (total 4 basic gates).
Most Upvoted Answer
A circuit has three inputs and one output. The output is 1 if at least...
Explanation:
The given circuit has three input variables, and the output is based on the combination of these input variables. The output is 1 only if at least two of the input variables are 1. Otherwise, the output is 0. This operation can be implemented using logic gates.
To implement this function using logic gates, we can use the following steps:
Step 1: Draw the truth table for the given function.
The truth table for the given function is shown below:
| A | B | C | Y |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 |
Step 2: Simplify the boolean expression for the given function.
From the truth table, we can observe that the output is 1 for the input combinations (0,1,1), (1,0,1), and (1,1,0) and (1,1,1). We can simplify the boolean expression using Boolean algebra as follows:
Y = AB'C + A'BC + ABC' + ABC
Y = AB'C + A'BC + C(AB' + A'B + AB)
Y = AB'C + A'BC + ABC + AB'C' + A'B'C
Y = AB'C + A'BC + ABC + AB'C' + A'B'C' + AB'C'
Y = AB'C + A'BC + ABC + AB'C' + A'B'C'
Step 3: Implement the function using basic gates.
We can implement the simplified boolean expression using basic gates such as AND, OR, and NOT gates. The minimum number of gates required to implement the function is 4, which includes 3 AND gates and 1 OR gate.
Therefore, the correct answer is option B, which states that 4 basic gates are required to implement the output.
The given circuit has three input variables, and the output is based on the combination of these input variables. The output is 1 only if at least two of the input variables are 1. Otherwise, the output is 0. This operation can be implemented using logic gates.
To implement this function using logic gates, we can use the following steps:
Step 1: Draw the truth table for the given function.
The truth table for the given function is shown below:
| A | B | C | Y |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 |
Step 2: Simplify the boolean expression for the given function.
From the truth table, we can observe that the output is 1 for the input combinations (0,1,1), (1,0,1), and (1,1,0) and (1,1,1). We can simplify the boolean expression using Boolean algebra as follows:
Y = AB'C + A'BC + ABC' + ABC
Y = AB'C + A'BC + C(AB' + A'B + AB)
Y = AB'C + A'BC + ABC + AB'C' + A'B'C
Y = AB'C + A'BC + ABC + AB'C' + A'B'C' + AB'C'
Y = AB'C + A'BC + ABC + AB'C' + A'B'C'
Step 3: Implement the function using basic gates.
We can implement the simplified boolean expression using basic gates such as AND, OR, and NOT gates. The minimum number of gates required to implement the function is 4, which includes 3 AND gates and 1 OR gate.
Therefore, the correct answer is option B, which states that 4 basic gates are required to implement the output.
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