HW: Logic Gate Worksheet | Exercises Logic | Docsity - Free Printable

Let's solve each problem step by step.

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Problem 1: Fill in all possible combinations for inputs X and Y

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

We fill in the truth table:

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

✔ Answer: These are all possible input combinations for two binary variables.

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Problem 2: Complete the truth table for the logic circuit

Circuit Description:
- Inputs: A and B
- First gate: OR gate → output is C = A OR B
- Second gate: NOT gate → output Q = NOT(C) = NOT(A OR B)

So, this is a NOR gate (NOT of OR).

Let’s compute each row:

| A | B | C = A OR B | Q = NOT(C) |
|---|---|------------|------------|
| 0 | 0 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 |

✔ Completed 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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Problem 3: Complete the truth table for the logic circuit

Circuit Description:
- Inputs: A and B
- First gate: OR gate → C = A OR B
- D = B (directly connected)
- Then AND gate: Q = C AND D = (A OR B) AND B

Let’s compute each row:

| A | B | C = A OR B | D = B | Q = C AND D |
|---|---|------------|-------|-------------|
| 0 | 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 | 1 |

✔ Completed 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: This circuit simplifies to Q = (A OR B) AND B = B, since:
- If B = 0 → Q = 0
- If B = 1 → Q = (A OR 1) AND 1 = 1
So, Q = B. But we'll keep it as per the logic flow.

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Problem 4: Complete the truth table for the logic circuit

Circuit Description:
- Inputs: A and B
- First gate: OR gate → C = A OR B
- Second gate: NOT gate → D = NOT(B)
- Third gate: OR gate → Q = C OR D = (A OR B) OR (NOT B)

Let’s evaluate each case.

We need to consider all combinations of A and B.

| A | B | C = A OR B | D = NOT(B) | Q = C OR D |
|---|---|------------|------------|------------|
| 0 | 0 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 1 | 0 | 1 |

Wait — let's check carefully:

- When A=0, B=0:
- C = 0 OR 0 = 0
- D = NOT(0) = 1
- Q = 0 OR 1 = 1

- A=0, B=1:
- C = 0 OR 1 = 1
- D = NOT(1) = 0
- Q = 1 OR 0 = 1

- A=1, B=0:
- C = 1 OR 0 = 1
- D = NOT(0) = 1
- Q = 1 OR 1 = 1

- A=1, B=1:
- C = 1 OR 1 = 1
- D = NOT(1) = 0
- Q = 1 OR 0 = 1

So in all cases, Q = 1.

That means this circuit outputs always 1, regardless of inputs.

But let’s double-check logic:

Q = (A OR B) OR (NOT B)

Note:
- (A OR B) OR (NOT B) = A OR B OR NOT B
- Since B OR NOT B = 1 (tautology)
- So A OR 1 = 1

Therefore, Q = 1 always

✔ Completed 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 |

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✔ Final Answers Summary:

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

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

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#### 2. NOR Gate Circuit:

| 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. (A OR B) AND B Circuit:

| 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. (A OR B) OR (NOT B) Circuit:

| 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: This circuit acts like a constant 1 output (always true), due to logical tautology.

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