Y=(A,B,C,D) minterms(0,3,4,5,6,7,11,15) minimise the equation by Quine McCluskey 

To minimize the Boolean function \( Y(A, B, C, D) \) using the **Quine-McCluskey method**, we'll proceed with the following steps:

### Problem Setup:

We are given the minterms for the function:
\[
Y(A, B, C, D) = \sum m(0, 3, 4, 5, 6, 7, 11, 15)
\]
where \( m \) stands for the minterms. This represents the Boolean function in terms of minterms.

### Step 1: Binary Representation of Minterms

Each minterm will be converted to its 4-bit binary equivalent:

- \( m(0) \)  → 0000
- \( m(3) \)  → 0011
- \( m(4) \)  → 0100
- \( m(5) \)  → 0101
- \( m(6) \)  → 0110
- \( m(7) \)  → 0111
- \( m(11) \) → 1011
- \( m(15) \) → 1111

### Step 2: Group Minterms by the Number of 1’s

Now, we group these minterms based on the number of 1’s in their binary representation:

- **0 ones:** \( 0000 \) (minterm 0)
- **2 ones:** \( 0100, 0011 \) (minterms 4, 3)
- **3 ones:** \( 0101, 0110, 0111, 1011 \) (minterms 5, 6, 7, 11)
- **4 ones:** \( 1111 \) (minterm 15)

### Step 3: Prime Implicant Table (Pairwise Combination)

We now begin the process of combining the minterms where they differ by only one bit. 

1. Combine 0000 (m0) with 0100 (m4):
   - Result: 0–00 (combined m0 and m4)
   
2. Combine 0011 (m3) with 0111 (m7):
   - Result: 0–11 (combined m3 and m7)
   
3. Combine 0101 (m5) with 0111 (m7):
   - Result: 01–1 (combined m5 and m7)
   
4. Combine 0110 (m6) with 0111 (m7):
   - Result: 011– (combined m6 and m7)

No further combinations are possible.

### Step 4: Prime Implicant Chart

From the results of step 3, the prime implicants are:

- \( 0–00 \) (covers \( m0, m4 \))
- \( 0–11 \) (covers \( m3, m7 \))
- \( 01–1 \) (covers \( m5, m7 \))
- \( 011– \) (covers \( m6, m7 \))
- \( 1011 \) (covers \( m11 \))
- \( 1111 \) (covers \( m15 \))

### Step 5: Essential Prime Implicants

Using a prime implicant chart, we determine which minterms are covered by each implicant:

- \( 0–00 \) → \( m0, m4 \)
- \( 0–11 \) → \( m3, m7 \)
- \( 01–1 \) → \( m5, m7 \)
- \( 011– \) → \( m6, m7 \)
- \( 1011 \) → \( m11 \)
- \( 1111 \) → \( m15 \)

#### Coverage:
- Minterms \( m0 \) and \( m4 \) are covered by \( 0–00 \).
- Minterms \( m3 \) and \( m7 \) are covered by \( 0–11 \) and \( 01–1 \).
- Minterm \( m11 \) is covered only by \( 1011 \).
- Minterm \( m15 \) is covered only by \( 1111 \).

The essential prime implicants are:

1. \( 0–00 \) (covers \( m0, m4 \))
2. \( 1011 \) (covers \( m11 \))
3. \( 1111 \) (covers \( m15 \))

### Step 6: Final Simplified Expression

Now, we convert the prime implicants back to their Boolean forms:

- \( 0–00 \) → \( A'C' \) (since \( A = 0 \) and \( C = 0 \), while \( B \) and \( D \) are irrelevant)
- \( 0–11 \) → \( A'BD \) (since \( A = 0 \), \( B = 1 \), and \( D = 1 \))
- \( 01–1 \) → \( A'B'D' \) (since \( A = 0 \), \( B = 1 \), and \( D = 0 \))
- \( 1011 \) → \( ABCD' \)
- \( 1111 \) → \( ABCD \)

Thus, the simplified Boolean expression is:

\[
Y(A, B, C, D) = A'C' + A'BD + ABCD' + ABCD
\]

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Would you like any further clarification on this method or more details?

### 5 Related Questions:
1. How does the Quine-McCluskey method differ from Karnaugh Maps?
2. Can we use the Quine-McCluskey method for functions with more variables?
3. What are essential and non-essential prime implicants?
4. How can we optimize larger Boolean expressions efficiently?
5. How do don't-care conditions affect minimization in Quine-McCluskey?

### Tip:
When minimizing Boolean functions, always ensure to cover all the minterms while avoiding redundant implicants.

Y=(A,B,C,D) minterms(0,3,4,5,6,7,11,15) minimise the equation by Quine McCluskey

Learn how to minimize the Boolean function Y(A, B, C, D) = Σ m(0, 3, 4, 5, 6, 7, 11, 15) using the Quine-McCluskey method. This detailed guide covers minterms, binary grouping, prime implicants, and the final simplified expression. Ideal for students in Grades 10-12 studying Boolean algebra.

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