$\overline{X}Y$, $X\overline{Y}$
This solution explains how to find the prime implicants for a given sum of products (SOP) function using a Karnaugh map (K-map).
Identify Minterms and Binary Representation
The function is given as $f(X, Y, Z) = \sum(2, 3, 4, 5)$. The minterms are 2, 3, 4, and 5. We convert these decimal values to their 3-bit binary representations (X=MSB, Z=LSB):
- Minterm 2: $2_{10} = 010_2 \implies \overline{X}Y\overline{Z}$
- Minterm 3: $3_{10} = 011_2 \implies \overline{X}YZ$
- Minterm 4: $4_{10} = 100_2 \implies X\overline{Y}\overline{Z}$
- Minterm 5: $5_{10} = 101_2 \implies X\overline{Y}Z$
Construct Karnaugh Map (K-map)
A 3-variable K-map is used to visually represent the minterms. We place a '1' in the cells corresponding to the minterms 2, 3, 4, and 5.
| X | YZ | |||
|---|---|---|---|---|
| 00 | 01 | 11 | 10 | |
| 0 | 0 | 0 | 1 | 1 |
| 1 | 1 | 1 | 0 | 0 |
Find Prime Implicants using Groupings
We group adjacent 1s in the K-map. Groups must contain a power of 2 number of cells (1, 2, 4, ...) and should be as large as possible. Each group represents an implicant.
- Group 1: The 1s corresponding to minterms 4 ($X\overline{Y}\overline{Z}$) and 5 ($X\overline{Y}Z$) can be grouped. These are in the row X=1, columns YZ=00 and YZ=01. This group covers the variable Z. The resulting implicant is $X\overline{Y}$.
- Group 2: The 1s corresponding to minterms 2 ($\overline{X}Y\overline{Z}$) and 3 ($\overline{X}YZ$) can be grouped. These are in the row X=0, columns YZ=10 and YZ=11. This group covers the variable Z. The resulting implicant is $\overline{X}Y$.
These two groups are the largest possible groupings and they cover all the 1s in the K-map. Therefore, they represent the prime implicants.
Determine the Prime Implicants
The prime implicants derived from the maximal groupings are:
- $\overline{X}Y$
- $X\overline{Y}$
These correspond to the terms that cannot be simplified further by combining with other terms or groups.
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
