In the sum of products function f (X, Y, Z) = ∑ (2, 3, 4, 5) , the prime implicants are
X̅Y, XY̅
Prime Implicants of Boolean Functions
Understanding prime implicants is crucial in digital logic design, especially for simplifying Boolean expressions. A prime implicant is a product term obtained by combining the maximum possible number of adjacent squares in a Karnaugh map (K-map) that covers a '1' (minterm) of the function. A simplified sum of products (SOP) expression consists of a sum of some of these prime implicants.
Function Analysis and Minterms
The given sum of products (SOP) function is \(f (X, Y, Z) = \sum (2, 3, 4, 5)\). This means the function outputs '1' for the minterms m2, m3, m4, and m5. We need to find the prime implicants by grouping these minterms.
First, let's list the binary representations of these minterms for three variables (X, Y, Z):
- m2: \((010)_2\) (X=0, Y=1, Z=0 \(\Rightarrow \bar{X}Y\bar{Z}\))
- m3: \((011)_2\) (X=0, Y=1, Z=1 \(\Rightarrow \bar{X}YZ\))
- m4: \((100)_2\) (X=1, Y=0, Z=0 \(\Rightarrow X\bar{Y}\bar{Z}\))
- m5: \((101)_2\) (X=1, Y=0, Z=1 \(\Rightarrow X\bar{Y}Z\))
Karnaugh Map Construction and Filling
We will construct a 3-variable Karnaugh Map (K-map) and place a '1' in the cells corresponding to the given minterms (2, 3, 4, 5) and '0' in the other cells.
| YZ\X | 0 (X̅) | 1 (X) |
|---|---|---|
| 00 (Y̅Z̅) | 0 | 1 (m4) |
| 01 (Y̅Z) | 0 | 1 (m5) |
| 11 (YZ) | 1 (m3) | 0 |
| 10 (YZ̅) | 1 (m2) | 0 |
Identifying Prime Implicants through Grouping
Now, we group adjacent '1's in the K-map to form the largest possible groups that are powers of 2 (1, 2, 4, 8, etc.). These groups represent the prime implicants.
-
Group 1: Minterms m2 (010) and m3 (011) are adjacent.
- Both m2 and m3 have \(X=0\) (i.e., \(\bar{X}\)) and \(Y=1\) (i.e., \(Y\)).
- The variable Z changes from 0 to 1 across these two minterms, so Z is eliminated.
- This group forms the term \(\bar{X}Y\).
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Group 2: Minterms m4 (100) and m5 (101) are adjacent.
- Both m4 and m5 have \(X=1\) (i.e., \(X\)) and \(Y=0\) (i.e., \(\bar{Y}\)).
- The variable Z changes from 0 to 1 across these two minterms, so Z is eliminated.
- This group forms the term \(X\bar{Y}\).
All '1's in the K-map are covered by these two groups. Also, these groups cannot be combined with any other adjacent '1's to form larger groups. Therefore, these are the prime implicants.
Conclusion
Based on the K-map simplification, the prime implicants for the function \(f (X, Y, Z) = \sum (2, 3, 4, 5)\) are \(\bar{X}Y\) and \(X\bar{Y}\).
Comparing this with the given options:
- \(\bar{X}Y, X\bar{Y}\)
- \(\bar{X}Y, X\bar{Y}\bar{Z}, X\bar{Y}Z\) (Not prime implicants as \(X\bar{Y}\) is a larger group)
- \(\bar{X}Y\bar{Z}, \bar{X}YZ, X\bar{Y}\) (Not prime implicants as \(\bar{X}Y\) is a larger group)
- \(\bar{X}Y\bar{Z}, \bar{X}YZ, X\bar{Y}\bar{Z}, X\bar{Y}Z\) (These are the original minterms, not prime implicants)
Thus, the prime implicants are \(\bar{X}Y\) and \(X\bar{Y}\).
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 \({\rm{M}}\left( {\overline {{\rm{M}}\left( {{\rm{a}},{\rm{b}},{\rm{c}}} \right)} ,{\rm{\;M}}\left( {{\rm{a}},{\rm{b}},\overline {{\rm{c\;}}} } \right),{\rm{c}}} \right)?\)
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:
