The question requires simplifying the Boolean function $F (W, X, Y, Z) = \Sigma (4, 5, 10, 11, 12, 13, 14, 15)$ to its minimal Sum of Products (SOP) form using the minimum number of terms and literals. A Karnaugh map (K-map) is the standard method for this.
Boolean Function Minterms Identification
The function is defined by the sum of minterms:
- Minterm 4 = $0100$
- Minterm 5 = $0101$
- Minterm 10 = $1010$
- Minterm 11 = $1011$
- Minterm 12 = $1100$
- Minterm 13 = $1101$
- Minterm 14 = $1110$
- Minterm 15 = $1111$
Karnaugh Map (K-map) Grouping
A 4-variable K-map is constructed with WX representing rows and YZ representing columns. The cells corresponding to the minterms listed above are filled with '1'.
The goal is to group adjacent '1's in the largest possible power-of-2 block sizes (1, 2, 4, 8) to cover all the '1's.
Identify the minimal groupings that cover all minterms:
- Group 1: Consider the cells for minterms 4 ($0100$) and 5 ($0101$). These cells represent $X=1$ and $Y=0$. This group simplifies to the term $X\overline{Y}$.
- Group 2: Consider the cells for minterms 10 ($1010$), 11 ($1011$), 14 ($1110$), and 15 ($1111$). These cells represent $Y=1$. Let's analyze this grouping more carefully.
- Cells corresponding to $W=1, Y=1$: Minterms 10 ($1010$), 11 ($1011$), 14 ($1110$), 15 ($1111$). This combines into the term $WY$.
Let's verify if these two terms cover all the required minterms:
- $X\overline{Y}$ covers minterms where X=1 and Y=0: 4 ($0100$), 5 ($0101$), 12 ($1100$), 13 ($1101$).
- $WY$ covers minterms where W=1 and Y=1: 10 ($1010$), 11 ($1011$), 14 ($1110$), 15 ($1111$).
The union of the minterms covered by $X\overline{Y}$ and $WY$ is {4, 5, 12, 13} $\cup$ {10, 11, 14, 15} = {4, 5, 10, 11, 12, 13, 14, 15}. This exactly matches the minterms specified for the function $F$.
The expression $F = X\overline{Y} + WY$ consists of two terms, each with two literals. This represents the minimal form in terms of the number of terms and literals.
Conclusion
The simplified form of the Boolean function $F (W, X, Y, Z) = \Sigma (4, 5, 10, 11, 12, 13, 14, 15)$ is $X\overline{Y} + WY$.
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
