Solved HW: Logic Gate Worksheet 1. A logic circuit has two | Chegg.com - Free Printable

Let's solve each problem step by step.

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1. A logic circuit has two inputs being X and Y. Fill in all the possible combinations.

For two binary inputs (X and Y), each can be either 0 or 1. So there are $2^2 = 4$ possible combinations.

| X | Y |
|---|---|
| 0 | 0 |
| 0 | 1 |
| 1 | 0 |
| 1 | 1 |

✔ This is the complete list of input combinations for two binary inputs.

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2. The figure shows a logic circuit and its incomplete truth table. Complete the truth table.

Circuit:
- Inputs A and B go into an OR gate, output is C.
- Output C goes into a NOT gate, producing output Q.

So:
- $ C = A \lor B $
- $ Q = \neg C = \neg (A \lor B) $

This is equivalent to a NOR gate: $ Q = \neg (A \lor B) $

Now fill in the truth table:

| A | B | C = A ∨ B | Q = ¬C |
|---|---|----------|--------|
| 0 | 0 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 |

✔ Final truth table:

| A | B | C | Q |
|---|---|---|---|
| 0 | 0 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 |

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3. The figure shows a logic circuit and its incomplete truth table. Complete the truth table.

Circuit:
- Inputs A and B go into an OR gate, output is C.
- Input B goes directly to D (so D = B).
- Then C and D go into an AND gate, output is Q.

So:
- $ C = A \lor B $
- $ D = B $
- $ Q = C \land D = (A \lor B) \land B $

We can simplify this:
$$
Q = (A \lor B) \land B = B \quad \text{(because } B \land (A \lor B) = B)
$$

But let’s compute it step by step.

| A | B | C = A ∨ B | D = B | Q = C ∧ D |
|---|---|----------|-------|-----------|
| 0 | 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 | 1 |

✔ Final truth table:

| A | B | C | D | Q |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 | 1 |

Note: Q depends only on B, so $ Q = B $. But we follow the circuit as given.

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4. The figure shows a logic circuit and its incomplete truth table. Complete the truth table.

Circuit:
- A and B go into an OR gate, output is C.
- B goes into a NOT gate, output is D.
- C and D go into an OR gate, output is Q.

So:
- $ C = A \lor B $
- $ D = \neg B $
- $ Q = C \lor D = (A \lor B) \lor (\neg B) $

Simplify:
$$
(A \lor B) \lor (\neg B) = A \lor (B \lor \neg B) = A \lor 1 = 1
$$
So Q should always be 1? Let's check with values.

Wait — that would mean Q is always 1 regardless of A and B?

Let’s test:

| A | B | C = A ∨ B | D = ¬B | Q = C ∨ D |
|---|---|----------|--------|-----------|
| 0 | 0 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 1 | 0 | 1 |

Yes! In every case, Q = 1.

So Q is always 1.

But the truth table has only two rows listed? Wait — look at the table:

| A | B | C | D | Q |
|---|---|---|---|---|
| 0 | | | | |
| 0 | | | | |
| 1 | | | | |
| 1 | | | | |

Wait — actually, the table has four rows but only A is filled in some cases.

Looking closely:

| A | B | C | D | Q |
|---|---|---|---|---|
| 0 | | | | |
| 0 | | | | |
| 1 | | | | |
| 1 | | | | |

It seems like B is missing from the table, but based on the pattern, likely the full table is expected.

Wait — perhaps it's meant to have all combinations.

But the table shows only A values: 0, 0, 1, 1 — suggesting B might be missing.

But since it's a two-input circuit, we assume both A and B vary.

So the correct way is to consider all 4 combinations.

Let’s recompute:

| A | B | C = A ∨ B | D = ¬B | Q = C ∨ D |
|---|---|----------|--------|-----------|
| 0 | 0 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 1 | 0 | 1 |

So Q is always 1.

Therefore, no matter what A and B are, Q = 1.

✔ Final truth table:

| A | B | C | D | Q |
|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 1 | 0 | 1 |

Even though the worksheet shows only A values, we must assume it's asking for all combinations.

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✔ Final Answers:

#### 1. All possible combinations of X and Y:

| X | Y |
|---|---|
| 0 | 0 |
| 0 | 1 |
| 1 | 0 |
| 1 | 1 |

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#### 2. Truth table for OR → NOT (NOR):

| A | B | C | Q |
|---|---|---|---|
| 0 | 0 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 |

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#### 3. Truth table for (A ∨ B) ∧ B:

| A | B | C | D | Q |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 | 1 |

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#### 4. Truth table for (A ∨ B) ∨ (¬B):

| A | B | C | D | Q |
|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 1 | 0 | 1 |

> Note: In problem 4, Q is always 1 because $ (A \lor B) \lor (\neg B) = A \lor (B \lor \neg B) = A \lor 1 = 1 $

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Let me know if you'd like diagrams or further explanation!