Digital input signals A, B, C with A as the MSB and C as the LSB are used to realize the Boolean function F = m0 + m2 + m3 + m5 + m7, where mi denotes the ith minterm. In addition, F has a don’t care for m1. The simplified expression for F is given by:
Boolean Function Definition and Minterm Representation
The question asks us to simplify a given Boolean function \(F\) using three digital input signals: A, B, and C. A is specified as the Most Significant Bit (MSB), and C as the Least Significant Bit (LSB).
- The Boolean function is defined as a sum of specific minterms: \(F = m0 + m2 + m3 + m5 + m7\).
- Additionally, there is a 'don’t care' condition for minterm \(m1\).
To begin, let's understand what each minterm represents in terms of the binary values of A, B, and C. A minterm is a product term where each variable appears exactly once, either in its true (uncomplemented) or complemented form. For 3 variables, there are \(2^3 = 8\) possible minterms, from \(m0\) to \(m7\).
Here’s a table showing the binary representation for each minterm:
| Minterm | A B C (Binary) | Decimal Equivalent |
|---|---|---|
| \(m0\) | 0 0 0 | 0 |
| \(m1\) | 0 0 1 | 1 |
| \(m2\) | 0 1 0 | 2 |
| \(m3\) | 0 1 1 | 3 |
| \(m4\) | 1 0 0 | 4 |
| \(m5\) | 1 0 1 | 5 |
| \(m6\) | 1 1 0 | 6 |
| \(m7\) | 1 1 1 | 7 |
K-Map Construction for Simplification
A 3-variable Karnaugh Map (K-Map) is an excellent tool for simplifying Boolean expressions. It arranges the minterms in such a way that adjacent cells differ by only one bit, making it easy to spot opportunities for simplification.
We will construct an 8-cell K-Map. We mark a '1' in the cells corresponding to the minterms present in the function F, and an 'X' in the cell corresponding to the don't care condition. Minterms not specified as '1' or 'X' are assumed to be '0'.
- Minterms for F (marked '1'): \(m0, m2, m3, m5, m7\).
- Don’t care (marked 'X'): \(m1\).
The general structure of a 3-variable K-Map is as follows, with A representing the rows and BC representing the columns (in Gray code order: 00, 01, 11, 10):
| BC | |||||
|---|---|---|---|---|---|
| 00 (\(\bar B \bar C\)) | 01 (\(\bar B C\)) | 11 (\(B C\)) | 10 (\(B \bar C\)) | ||
| A | 0 (\(\bar A\)) | \(m0\) | \(m1\) | \(m3\) | \(m2\) |
| 1 (\(A\)) | \(m4\) | \(m5\) | \(m7\) | \(m6\) | |
Now, let's populate the K-Map with the given values:
| BC | |||||
|---|---|---|---|---|---|
| 00 | 01 | 11 | 10 | ||
| A | 0 | 1 (\(m0\)) | X (\(m1\)) | 1 (\(m3\)) | 1 (\(m2\)) |
| 1 | 0 (\(m4\)) | 1 (\(m5\)) | 1 (\(m7\)) | 0 (\(m6\)) | |
K-Map Grouping for Simplified Expression
The goal of K-Map simplification is to form the largest possible groups of adjacent '1's (and 'X's if they help enlarge the group). Groups must be powers of 2 (e.g., 2, 4, 8 cells) and can wrap around the edges of the map.
Let's identify the groups:
- First Group (Quad of 4 cells):
- Observe the entire first row where A is 0. This row contains the values: 1 (for \(m0\)), X (for \(m1\)), 1 (for \(m3\)), and 1 (for \(m2\)).
- All four cells in this row can be grouped together. This is a group of 4.
- In this group, the variable A is consistently 0 (represented as \(\bar A\)). The variables B and C change across the group (e.g., \(\bar B \bar C\) to \(\bar B C\) to \(B C\) to \(B \bar C\)), meaning they cancel out in the simplification.
- Therefore, this group simplifies to \(\bar A\).
- Second Group (Quad of 4 cells):
- After forming the first group, we still need to cover \(m5\) and \(m7\), which are '1's in the second row.
- Consider the two columns where C is 1 (i.e., BC = 01 and BC = 11). These columns contain cells \(m1 (X), m3 (1)\) from the first row and \(m5 (1), m7 (1)\) from the second row.
- These four cells form another valid group of 4.
- In this group, the variable C is consistently 1 (represented as \(C\)). The variables A and B change across the group (A goes from 0 to 1, B goes from 0 to 1), so they cancel out.
- Therefore, this group simplifies to \(C\).
Both required minterms (m0, m2, m3, m5, m7) are covered by these two groups. The 'don't care' minterm \(m1\) was used to make the groups larger, which leads to a more simplified expression.
The simplified Boolean expression for F is the OR combination of the terms from these two groups:
\(F = \bar A + C\)
This simplified expression matches option 2.
Important Questions from Karnaugh Maps
The minimized sum of products expression for f(a,b,c,d) = Ʃm(0,1,5,6,7,8,9) with don’t care Ʃm(10,11,12,13,14,15) is ___________.
A problem detector system produces an alarm in the factory when one of the three conditions occurs. The system is designed as such tha only one condition can occur at a time. If the three conditions are defined as q, r, and s respectively, the output logic for the system is given as
The Boolean expression \({\rm{F}}\left( {{\rm{x}},{\rm{y}},{\rm{z}}} \right) = {\rm{\;\bar xy\;\bar z}} + {\rm{\;x\;\bar y\bar z}} + {\rm{\;x\;y\;\bar z}} + {\rm{\;x\;y\;z}}\) is converted into the canonical product of sum (POS) form is
A 3 - input majority gate is defined by the logic function \({\rm{M}}\left( {{\rm{a}},{\rm{b}},{\rm{c}}} \right) = {\rm{\;ab\;}} + {\rm{\;bc\;}} + {\rm{\;ac}}\) . Which one of the following gate is represented by the function \({\rm{M}}\left( {\overline {{\rm{M}}\left( {{\rm{a}},{\rm{b}},{\rm{c}}} \right)} ,{\rm{\;M}}\left( {{\rm{a}},{\rm{b}},\overline {{\rm{c\;}}} } \right),{\rm{c}}} \right)?\)
In the sum of products function f (X, Y, Z) = ∑ (2, 3, 4, 5) , the prime implicants are
