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 ___________.
Minimizing f(a,b,c,d) Sum of Products using a K-map
We are given a boolean function f(a,b,c,d) with its minterms and don't care conditions and asked to find the minimized Sum of Products (SOP) expression. The minterms are Ʃm(0,1,5,6,7,8,9) and the don't cares are Ʃm(10,11,12,13,14,15). We can use a 4-variable Karnaugh map (K-map) to simplify this expression.
First, construct a 4-variable K-map and fill in the cells corresponding to the given minterms with '1' and the don't care terms with 'X'.
| ab\cd | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 00 | 1 (0) | 1 (1) | 0 (3) | 0 (2) |
| 01 | 0 (4) | 1 (5) | 1 (7) | 1 (6) |
| 11 | X (12) | X (13) | X (15) | X (14) |
| 10 | 1 (8) | 1 (9) | X (11) | X (10) |
Next, group adjacent '1's and 'X's in powers of two (2, 4, 8, 16) to cover all the '1's. Use don't cares ('X's) to make the largest possible groups.
- Group 1: A group of 4 covering cells 0, 1, 8, and 9. The binary representations are 0000, 0001, 1000, 1001. In these terms, 'a' changes (0 to 1), 'b' is always 0, 'c' is always 0, and 'd' changes (0 to 1). The variables that remain constant are \(\overline{b}\) and \(\overline{c}\). This group represents the term \(\overline{b}\overline{c}\). This group covers '1's at 0, 1, 8, 9.
- Group 2: A group of 4 covering cells 5, 7, 13(X), and 15(X). The binary representations are 0101, 0111, 1101, 1111. In these terms, 'a' changes (0 to 1), 'b' is always 1, 'c' changes (0 to 1), and 'd' is always 1. The variables that remain constant are b and d. This group represents the term bd. This group covers '1's at 5, 7.
- Group 3: A group of 4 covering cells 6, 7, 14(X), and 15(X). The binary representations are 0110, 0111, 1110, 1111. In these terms, 'a' changes (0 to 1), 'b' is always 1, 'c' is always 1, and 'd' changes (0 to 1). The variables that remain constant are b and c. This group represents the term bc. This group covers '1's at 6, 7.
All '1's at 0, 1, 5, 6, 7, 8, and 9 are now covered by these three groups. These groups are also essential prime implicants:
- Cell 0 is only covered by the group \(\overline{b}\overline{c}\).
- Cell 8 is only covered by the group \(\overline{b}\overline{c}\).
- Cell 5 is only covered by the group bd.
- Cell 6 is only covered by the group bc.
Since the essential prime implicants cover all the '1's, these three terms form the minimized SOP expression.
The minimized sum of products expression is the sum of the terms from the essential prime implicants:
f(a,b,c,d) = \(\overline{b}\overline{c}\) + bd + bc
Comparing this expression with the given options, we find that Option 3 matches our result.
Important Questions from Karnaugh Maps
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
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:
