15.2 Boolean Algebra and Logic Circuits (3)
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1.
A logic circuit has two inputs, P and Q, and a single output R. The Boolean expression for R is given by R = (P + Q) * (~P * Q). Using a Karnaugh map, simplify this expression. Show your working.
Answer:
The Boolean expression is R = (P + Q) * (~P * Q). We can use a 2-variable Karnaugh map to simplify this.
The expression (P + Q) * (~P * Q) can be rewritten as P + Q * (~P * Q). This can be simplified by noticing that Q * (~P * Q) = Q * (~P) * Q = ~P * Q2 = ~P * Q. Therefore, the expression becomes P + ~P * Q.
Now, we can group the '1's in the Karnaugh map:
- The '1's in the first row (00 and 01) can be grouped.
- The '1's in the third column (11 and 10) can be grouped.
This gives us the simplified expression: R = P + Q
2.
A digital circuit is designed to perform a Boolean function with three inputs, A, B, and C. The truth table for this function is given below:
| A | B | C | Output |
|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 |
Using a Karnaugh map, simplify the Boolean expression for the output of this circuit. Show your working.
Answer:
The Boolean expression for the function can be represented using a 3-variable Karnaugh map. The map will look like this:
We can group the '1's as follows:
- A 000 and A 001 can be grouped.
- A 010 and A 011 can be grouped.
- A 101 and A 110 can be grouped.
- A 111 remains as a single '1'.
Grouping these gives us the simplified expression: F = (A & ~B & ~C) | (~A & ~B & C) | (A & B & ~C) | (A & B & C)
Alternatively, this can be expressed as: F = A'C' + A'BC + AB'C + ABC
3.
Question 1: Consider the following Boolean expression: (A → B) ∧ (¬B → ¬A). Simplify this expression using Boolean algebra laws. Show each step of your simplification.
Answer:
- The expression is (A → B) ∧ (¬B → ¬A).
- We can use the implication equivalence: A → B ≡ ¬A ∨ B. Therefore, A → B becomes ¬A ∨ B. Similarly, ¬B → ¬A becomes B ∨ ¬A.
- Substituting these into the original expression, we get: (¬A ∨ B) ∧ (B ∨ ¬A).
- Using the distributive law: (¬A ∨ B) ∧ (B ∨ ¬A) ≡ (¬A ∧ B ∨ ¬A ∧ ¬A) ∨ (B ∧ B ∨ B ∧ ¬A).
- Simplifying the conjunctions: (¬A ∧ B ∨ False) ∨ (B ∨ B ∧ ¬A).
- This simplifies to: (¬A ∧ B) ∨ (True) ∨ (B ∧ ¬A).
- Since True ≡ True, the expression becomes: (¬A ∧ B) ∨ True ∨ (B ∧ ¬A).
- The OR operation with True is idempotent, so the expression simplifies to: True.
- Therefore, the simplified expression is True.