Minimize the Boolean expression f(A, B, C) = A + BC' + ABC' into min terms. Find the expression for the given Boolean expression using K-map with d(8, 9, 10, 11, 12, 13, 14, 15). F... Minimize the Boolean expression f(A, B, C) = A + BC' + ABC' into min terms. Find the expression for the given Boolean expression using K-map with d(8, 9, 10, 11, 12, 13, 14, 15). Find three inputs A, B, and C such that the output is 1 when...
Understand the Problem
The question asks to minimize a Boolean expression using three variables (A, B, C) and convert it into min terms, as well as to use a Karnaugh map (K-map) for the given Boolean expression.
Answer
The minimized expression is $f(A, B, C) = A + B$ with min terms $m(0, 1, 2, 3, 4, 5, 6, 7)$.
Answer for screen readers
The minimized expression is: $$ f(A, B, C) = A + B $$
The min terms are: $$ m(0, 1, 2, 3, 4, 5, 6, 7) $$
Three inputs (A, B, C) that yield an output of 1 include (1, 0, 0), (0, 1, 0), etc.
Steps to Solve
- Identify the given Boolean expression
We have: $$ f(A, B, C) = A + BC' + ABC' $$
- Apply Boolean algebra to minimize the expression
Using the absorption law: $$ A + ABC' = A(1 + BC') = A + BC' $$ Thus, the expression remains: $$ f(A, B, C) = A + BC' $$
- Convert the expression into min terms
Now we convert $f(A, B, C) = A + BC'$ into min terms.
- The term $A$ corresponds to min terms (1, 3, 5, 7).
- The term $BC'$ corresponds to (0, 2, 6, 4).
Combining these gives:
- Min terms: 1, 3, 5, 7, 0, 2, 4, 6, thus: $$ m(0, 1, 2, 3, 4, 5, 6, 7) $$
- Set up the K-map
To set up the K-map with don’t care conditions $d(8, 9, 10, 11, 12, 13, 14, 15)$, we place the min terms in their respective cells in the K-map.
- Fill in the K-map
The K-map is labeled as follows:
| BC
00 | 01 | 11 | 10
-----------------------
0 | 1 | 1 | 1 | 0 (for A=0)
1 | 1 | 1 | 1 | 1 (for A=1)
The cells filled based on the min terms and don’t cares.
- Group the K-map
Look for groups of 1’s in the K-map:
- One group of four (covering the number 1, 3, 5, 7).
- One group of four for don’t cares.
This gives us the simplified expression.
- Final expression from K-map
From grouping, we find: $$ f(A, B, C) = A + B $$
- Find inputs for output 1
We now identify combinations of $A$, $B$, and $C$ such that the output is 1:
- From $A = 1$, we get (1, 0, 0), (1, 0, 1), (1, 1, 0), (1, 1, 1).
- From $B = 1$, we can have combinations with A as 0 or 1:
- (0, 1, 0), (0, 1, 1)
Thus, any of these combinations will yield an output of 1.
The minimized expression is: $$ f(A, B, C) = A + B $$
The min terms are: $$ m(0, 1, 2, 3, 4, 5, 6, 7) $$
Three inputs (A, B, C) that yield an output of 1 include (1, 0, 0), (0, 1, 0), etc.
More Information
The minimization process simplifies the logic circuit, reducing complexity and power consumption. The K-map is a visual tool that quickly helps in minimizing Boolean expressions effectively.
Tips
- Misidentifying min terms during conversion.
- Overlooking don’t care conditions while filling the K-map.
- Forgetting to group all possible ones in the K-map correctly.
AI-generated content may contain errors. Please verify critical information
