Digital Logic · Unit 3 · 5 hrs
Simplification of Boolean Functions
Exam-focused notes for Simplification of Boolean Functions (Digital Logic, CSC116): what the TU syllabus asks and how it has actually been tested, with 8 solved past questions from this unit.
What this unit covers
- K-map
- Two and Three variable maps
- Four variable maps
- product of sum simplification
- NAND and NOR implementation
- Don't Care conditions
- Determinant and selection of Prime Implicants
K-map
De-Morgan's Law and Boolean Function Simplification
De-Morgan's Law relates the complement of a compound Boolean expression to the complements of its parts. Theorem 1: $\overline{A + B} = \bar{A} \cdot \bar{B}$ The complement of a sum equals the product of the complements. Theorem 2: $\overline{A \cdot B} = ...
Full solved answer →Reduce the following function using k-map. F=B'D+A'BC'+AB'C+ABC' [5]
Function of 4 variables A, B, C, D: $$F = B'D + A'BC' + AB'C + ABC'$$ Order: A B C D. B'D (B=0, D=1; A,C free): - 0001 = m1, 0011 = m3, 1001 = m9, 1011 = m11 A'BC' (A=0, B=1, C=0; D free): - 0100 = m4, 0101 = m5 AB'C (A=1, B=0, C=1; D free): - 1010 = m10, 1...
Full solved answer →Four variable maps
Simplify $F(A,B,C,D)=\sum(1,3,4,6,9,11,12,14)$ and realize the equation using NOR gates only. [5]
- Function: $F(A,B,C,D) = \sum(1,3,4,6,9,11,12,14)$ - 4 variables: A (MSB), B, C, D (LSB) - Requirement: simplify, then realize using NOR gates only Minterms in binary (ABCD): - m1 = 0001, m3 = 0011, m4 = 0100, m6 = 0110 - m9 = 1001, m11 = 1011, m12 = 1100,...
Full solved answer →Minimize the following Boolean function using K-map $F(A, B, C, D) = \Sigma m(0,1,2,5,7,8,9,10,13,15)$. [5]
Minterms with value 1: 0, 1, 2, 5, 7, 8, 9, 10, 13, 15 Rows = AB, Columns = CD (Gray code order): Verify 1s: m0✓ m1✓ m2✓ m5✓ m7✓ m8✓ m9✓ m10✓ m13✓ m15✓ (10 ones) Group 1 (Quad): m{0,1,8,9} - AB = 00 and 10 (A varies), B = 0; CD = 00 and 01 (D varies), C = 0...
Full solved answer →Minimize the following Boolean function using K-map. $F(A,B,C,D)=\sum(0,1,3,5,7,8,9,11,13,15)$ [5]
- Function: $F(A,B,C,D) = \sum(0,1,3,5,7,8,9,11,13,15)$ - Minterms with value 1: {0, 1, 3, 5, 7, 8, 9, 11, 13, 15} - All other cells (2, 4, 6, 10, 12, 14) = 0 Rows = AB (00, 01, 11, 10), Columns = CD (00, 01, 11, 10). Minterm-to-cell mapping (row AB, col CD...
Full solved answer →Don't Care conditions
If $f(P,Q,R,S)=\sum(3,4,7,8,14)$, $d(P,Q,R,S)=\sum(1,6,9,13)$ Simplify it using K-map and design circuit using minimum number of NAND gates. [5]
- $f(P,Q,R,S) = \sum(3,4,7,8,14)$ (required minterms) - Don't cares: $d(P,Q,R,S) = \sum(1,6,9,13)$ - P is MSB, S is LSB $$ \begin{array}{ccccc} & RS=00 & 01 & 11 & 10 \\ \hline PQ=00 & 0(m0) & X(m1) & 1(m3) & 0(m2) \\ 01 & 1(m4) & 0(m5) & 1(m7) & X(m6) \\ 1...
Full solved answer →product of sum simplification
Express the Boolean function F = x+yz as a product of max-terms. [5]
Given data: - Boolean function: $F = x + yz$ - Variables: $x, y, z$ (3 variables, so $2^3 = 8$ combinations) - Required form: Product of max-terms (POS canonical form) All data present. --- Evaluate $F = x + yz$ for all 8 combinations: Row x y z yz F = x + ...
Full solved answer →NAND and NOR implementation
Simplify the following function and implement them with two level NOR gate circuit, $F(w, x, y, z) = wx' + y'z' + w'yz'$. [5]
$$F(w,x,y,z) = wx' + y'z' + w'yz'$$ Term $wx'$ ($w=1, x=0$, y,z any): - $1000=8,\ 1001=9,\ 1010=10,\ 1011=11$ Term $y'z'$ ($y=0, z=0$, w,x any): - $0000=0,\ 0100=4,\ 1000=8,\ 1100=12$ Term $w'yz'$ ($w=0, y=1, z=0$, x any): - $0010=2,\ 0110=6$ Combined: $$F ...
Full solved answer →Make Unit 3 stick
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