Digital Logic · Unit 3
Combinational Logic Simplification
Exam-focused notes for Combinational Logic Simplification (Digital Logic, BIT103): what the TU syllabus asks and how it has actually been tested, with 8 solved past questions from this unit.
What this unit covers
- Karnaugh map simplification
- Sum of products and product of sums
- Minterms and maxterms representation
- Don't care conditions in K-maps
- Boolean function minimization techniques
Karnaugh map simplification
Simplify (using K-map): F=(A+B+C+D′)(A+B+C′+D)(A+B′+C′+D′)(A+B′+C′+D)(A′+B′+C′+D)(A′+B+C+D′)(A′+B+C′+D)F = (A + B + C + D')(A + B + C' + D)(A + B' + C' + D')(A + B' + C' + D)(A' + B' + C' + D)(A' + B + C + D')(A' + B + C' + D)F=(A+B+C+D′)(A+B+C′+D)(A+B′+C′+D′)(A+B′+C′+D)(A′+B′+C′+D)(A′+B+C+D′)(A′+B+C′+D)[5]
$$F = (A+B+C+D')(A+B+C'+D)(A+B'+C'+D')(A+B'+C'+D)(A'+B'+C'+D)(A'+B+C+D')(A'+B+C'+D)$$ For a maxterm, uncomplemented variable = 0, complemented variable = 1. Sum Term A B C D Maxterm ------------------ $A+B+C+D'$ 0 0 0 1 $M1$ $A+B+C'+D$ 0 0 1 0 $M2$ $A+B'+C'...
Full solved answer →Simplify: F = X Y Z' + X Z' + X' Y + X'' Y Z' using K-Map in both SOP and POS. [5]
Function of three variables $X, Y, Z$: $$F = XYZ' + XZ' + X'Y + X''YZ'$$ Note $X'' = X$ (double complement), so: $$F = XYZ' + XZ' + X'Y + XYZ'$$ Expand each term over minterms $m(X,Y,Z)$: - $XYZ'$: $X=1,Y=1,Z=0 \Rightarrow m6$ - $XZ'$: $X=1,Z=0 \Rightarrow$...
Full solved answer →Minterms and maxterms representation
Express the given function in sum of minterms. F = y'z + wxy' + wzx' + w'x'z' [5]
Variables (4): $w, x, y, z$ in order (MSB = $w$, LSB = $z$). $$F = y'z + wxy' + wzx' + w'x'z'$$ Total possible minterms: $2^4 = 16$ ($m0$ to $m{15}$). Use $A = A(B + B') = AB + AB'$ for missing variables. $$y'z = (w+w')(x+x')y'z = wxy'z + wx'y'z + w'xy'z + ...
Full solved answer →Express the Boolean Function F=AB+B′CF = AB + B'CF=AB+B′C to sum of max terms with required truth tables.List two uses of sum of max terms.[8+2]
- Boolean function: $F = AB + B'C$ - Variables: $A, B, C$ (3 variables, so $2^3 = 8$ combinations) - Required: express as product (sum) of maxterms, with truth table; list two uses. Row A B C AB B'C F Type -------------------------------- 0 0 0 0 0 0 0 M₀ 1...
Full solved answer →Given is a logic (switching) function F1 in the decimal list sum-of-minterms representation. $F_1(A,B,C,D)=\Sigma(0,2,3,5,7)$, $d(A,B,C,D)=\Sigma(8,10,13,15)$ [5]
- Minterms (=1): 0, 2, 3, 5, 7 - Don't cares (=X): 8, 10, 13, 15 - Zeros: 1, 4, 6, 9, 11, 12, 14 Binary of each cell (A B C D), AB rows / CD columns: AB\CD 00 01 11 10 ----------------------- 00 1 (m0) 0 (m1) 1 (m3) 1 (m2) 01 0 (m4) 1 (m5) 1 (m7) 0 (m6) 11 ...
Full solved answer →Don't care conditions in K-maps
Simplify the Boolean function using don't care conditions d in sum of products form. and product of sums form $F(A, B, C, D) = \pi(1, 3, 7, 8, 12) \text{ and } \pi d(5, 10, 13, 14)$ [5]
The function is given in Product of Sums (POS) form (uses $\pi$): - Maxterms (F = 0): $\pi(1, 3, 7, 8, 12)$ - Don't care terms: $d(5, 10, 13, 14)$ - Variables: $A, B, C, D$ (4 variables, cells 0-15) Minterms (F = 1) = remaining cells = $0, 2, 4, 6, 9, 11, 1...
Full solved answer →Simplify the Boolean Function $F$ in sum of products using the don't-care conditions $d$. $F = B'C'D' + BCD' + ABCD'$, $d = B'CD' + A'BC'D$. [5]
Function (SOP): $$F = B'C'D' + BCD' + ABCD'$$ Don't-care conditions: $$d = B'CD' + A'BC'D$$ Variables: A, B, C, D (4 variables → minterms 0 to 15) --- Order of bits: A B C D. - $B'C'D'$ → B=0, C=0, D=0, A free → A=0: 0000 = m₀, A=1: 1000 = m₈ - $BCD'$ → B=1...
Full solved answer →A logic circuit implements the following Boolean function. F=A'C+AC'D It is found that the required input combination A=C=1 can never occur. Using K-map and proper don't care condition find simpler expression for F and implement it using not gate only. [5]
Function: $F = A'C + AC'D$ Variables: The function contains A, C, D. The term $A'C$ has no B or D dependence, $AC'D$ has no B dependence. So variables are A, C, D only. Actually the presence of D means we should check variable count. The literals used are A...
Full solved answer →Make Unit 3 stick
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