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WorksheetsQuiz: Boolean Algebra, Functions and Implementation
Total questions: 48
Worksheet time: 24mins
A city traffic controller uses a 3-input logic set: P (pedestrian request), V (vehicle sensor), N (night mode). The controller must give WALK signal when a pedestrian has requested and it is safe: either (a) vehicle sensor is OFF, or (b) it is night mode (which uses reduced vehicle flow). The WALK output should not activate if there is an active emergency override E (active = 1) that forces STOP regardless. Boolean expression for WALK (ignore E) that is true when P=1 and (V=0 or N=1) is:
A. P(V̅ + N)
B. PV' + PN'
C. P(V + N)
D. P(V + N)
A city traffic controller uses a 3-input logic set: P (pedestrian request), V (vehicle sensor), N (night mode). The controller must give WALK signal when a pedestrian has requested and it is safe: either (a) vehicle sensor is OFF, or (b) it is night mode (which uses reduced vehicle flow). The WALK output should not activate if there is an active emergency override E (active = 1) that forces STOP regardless. If emergency override E must force WALK = 0 (i.e., inhibit walk when E=1), the combined expression is:
A. P(V̅ + N) + E'
B. P(V̅ + N) . E̅
C. P(V̅ + N) + E
D. P(V̅ + N) ⊕ E
Which Boolean theorem justifies factoring P(V̅ + N) into sum-of-products for K-map minimization?
A. De Morgan’s theorem
B. Distributive law
C. Consensus theorem
D. Absorption law
Convert F=PV̅+PN to a minimal SOP using algebraic simplification — what is minimal?
P(V̅ + N) (already minimal)
P
P(V̅N)
V̅ + N
A city traffic controller uses a 3-input logic set: P (pedestrian request), V (vehicle sensor), N (night mode). The controller must give WALK signal when a pedestrian has requested and it is safe: either (a) vehicle sensor is OFF, or (b) it is night mode (which uses reduced vehicle flow). The WALK output should not activate if there is an active emergency override E (active = 1) that forces STOP regardless. Suppose you must implement F=PV̅+PN using only NAND gates. The minimal number of two-input NAND gates required to realize the function (without using multi-input NANDs) is:
(a)
If P=1, V=1, N=0, E=0 what is WALK?
1
0
Undefined
Oscillating
An elevator has card reader (C), overload sensor (O), and floor selector valid (F). Door unlock U only when C=1 AND F=1 AND O=0.
CF'O
C + F + O'
(C + F)O'
C(F + O')
An elevator has card reader (C), overload sensor (O), and floor selector valid (F). Door unlock U only when C=1 AND F=1 AND O=0. Using De Morgan, express U̅:
C̅ + F̅ + O
C̅ · F̅ · O
C'F'O'
(CF) + O̅
An elevator has card reader (C), overload sensor (O), and floor selector valid (F). Door unlock U only when C=1 AND F=1 AND O=0. 9. The designer wants a fail-safe using NAND only; tying inputs appropriately, a NAND that receives C and F and O' (inverter on O) yields U directly. True or false?
True — single NAND with O inverted works
False — NAND produces complement, extra inversion needed
True only if C and F are tied together first
False unless using three-input NAND
An elevator has card reader (C), overload sensor (O), and floor selector valid (F). Door unlock U only when C=1 AND F=1 AND O=0. If the minimization yields U=CF O̅, the canonical minterm form (three variables) is:
Σ(7) (where minterm numbers assume order C,F,O)
Σ(3, 5)
Σ(4, 5, 6, 7)
Σ(2, 3, 6, 7)
An elevator has card reader (C), overload sensor (O), and floor selector valid (F). Door unlock U only when C=1 AND F=1 AND O=0. A K-map for CFO̅ would show how many minterms?
1
2
4
8
Which Boolean identity is X+XY=XX + XY = XX+XY=X?
Distributive law
Absorption law
Complement law
Associative law
The consensus theorem states: XY+X'Z+YZ=XY+X'Z. The term YZ is called:
Absorbed term
Consensus term
Redundant term
Essential prime implicant
Using Boolean algebra, simplify AB'+A'B+AB to simplest form:
A. A+B
B. A⊕B
C. A+B'
D. AB'+A'B
The dual of expression A+BCA + BCA+BC is:
A. A · (B + C)
B. A · (B · C)
C. A · (B + C)'
D. A(B + C)
Which Boolean theorem is used to convert sum-of-products into product-of-sums?
A. De Morgan’s theorem and distributive law
B. Consensus theorem
C. Absorption law
D. Commutative law
Which of the following functions is self-dual (i.e., f'=f(x̅) )?
A
A⊕1
A⊕B (XOR)
No single-variable function is self-dual
(A+B)(A+B̅) simplifies to:
AAA
A+BB̅
A+B
AB̅
Convert F(A,B)=A+B to canonical POS (product of maxterms) for two variables:
(A+B)(A'+B')
(A+B')(A'+B')
(A+B)(A'+B)
(A+B')(A'+B)
21. F(A,B,C)=Σ(1,3,5,7) is equivalent to which simple expression?
A. A⊕B⊕C
B. C⊕B
C. A+C
D. B
If a boolean function’s minterms are 2,3,6,7 (for 3 variables A,B,C with binary order 0..7), the canonical SOP includes how many product terms?
2
3
4
6
Convert the function F(A,B,C)=ABC+A'BC+AB'C into canonical SOP (list minterm indices):
Σ(6,7)
Σ(3,6,7)
Σ(4,6,7)
Σ(3,5,6,7)
The standard POS form is useful when:
Simplifying via K-map SOP groupings
You want to implement circuit with NANDs only
The function has many 0 outputs and few 1s
The function is monotonic
Express F(A,B)=AB'+A'B in canonical POS:
(A+B)(A'+B')
(A+B')(A'+B)
(A+B')(A'+B')
(A'+B')(A+B)
Which statement about prime implicants is TRUE?
Prime implicants are always essential
Every prime implicant must appear in minimal SOP
Essential prime implicants cover minterms not covered by other implicants
Prime implicants are single literals only
The difference between canonical SOP and minimal SOP is:
Canonical uses all variables in each term; minimal uses simplified implicants
Minimal is less general; canonical is minimal
There is no difference
Canonical contains fewer terms
A 3-variable K-map grouping can combine how many adjacent 1s at maximum for a single group?
(a)
For function F(A,B,C,D) with minterms Σ(0,1,2,3,8,9,10,11), the minimal SOP by K-map will produce which simplified structure?
Two 4-term groups producing two implicants each covering half the map
Four 2-term groups only
Single 8-term implicant
No simplification possible
Using K-map, the function F(A,B,C)=Σ(1,2,5,6) simplifies to:
A. A⊕B
B. C⊕A
C. B⊕C
D. A'B'+AB'
A K-map shows isolated 1s that cannot be grouped with others: those implicants are:
Essential prime implicants always
Don’t-care terms
Single-minterm prime implicants (must be covered individually)
Redundant terms
Identify the correct K-map adjacency property: cell 0 is adjacent to:
cell 1 only
cell 8 only (for 4-variable map)
cells 1,2,4,8 depending on dimension
cells 1 and 8 (wraparound adjacency)
If a K-map has don't-care cells at minterms 3 and 5, how can they be used?
They must be assigned 0 always
They can be treated as 1 or 0 to maximize grouping for simplification
They are illegal minterms
They replace prime implicants
A 4-variable K-map grouping of 4 adjacent ones reduces the product term by how many literals?
Removes 1 literal
Removes 2 literals
Removes 3 literals
Removes 4 literals
A K-map simplification yields two equal minimal solutions; which tie-breaking rule is best?
Choose any — both minimal are OK
Choose the one with fewer literals per implicant (overall literal count minimal)
Choose the lexicographically first implicant set
Choose the one with more implicants
For the function F=Σ(0,2,5,7) using a 3-var K-map, the minimal SOP is:
A. A'C'+AB'C
B. B'C'+A'B'C'
C. C'A'+BC'
D. A'B'+AC
In K-map, a group of 2 covering cells differing by one variable eliminates that variable. Which law explains this?
De Morgan’s theorem
Consensus theorem
Distributive law
Combining law
Which gates are called universal?
AND and OR
NAND and NOR
XOR and XNOR
NOT and BUFFER
Implement Y=A+B using only NAND gates (two-input NANDs). Minimal gate count is:
2
3
4
5
Implement Y=A·B·C̅ using 2-input NAND gates only. The minimal number of NAND gates (counting required input fan-ins realized by cascades) is:
1 (if multi-input NAND available)
2
3
4
A NOR-only implementation of F=AB+C requires how many two-input NOR gates minimally (including necessary inverters realized via NOR)?
2
3
4
5
Using NAND implementations, an inverter can be made by:
A NAND with tied inputs to the same signal
A NAND followed by a NOT gate
A NAND with an input tied to logic 1
A NAND with inputs cross-connected
Implement F=(A+B)·C using NAND gates. Which approach is minimal?
Realize OR via NANDs, then NAND with C, then NAND invert — total 4 NAND gates
Convert to (A+C)(B+C) and implement – fewer gates
Use NAND to create A+B and feed directly with C into a 3-input NAND — 2 gates
Impossible with only NANDs
A design requires both NAND and NOR implementations side-by-side. The primary design reason to prefer NAND over NOR in CMOS is:
NAND has fewer transistors than NOR for same fan-in
NAND offers better pull-down network and speed balancing in CMOS (faster)
Using only NOR gates, implement F=A'B'. This function is equivalent to:
(A'+B')
A'+B'
AB
A+B
Given F(A,B,C,D)=Σ(0,1,2,5,6,7,8,9,14) and don't-care d(3,4,11,13), what is a minimal SOP (expressed with literals)?
A. A'C'+BD'+CD'
B. A'B'+C'D'+AB
C. C'+BD'
D. A'D'+BC'
For G=AB+AB'C+A'BC, which simplification is correct?
AB+BC
AB+C
B(A+A'C)
AB+AC
A boolean network has hazard risk for static-1 (output should stay 1 during single input changes but may glitch). Which practice eliminates static-1 hazard?
Add consensus terms to cover adjacent minterm transitions
Remove redundant terms to reduce logic levels
Insert inverters at outputs
Reduce fan-out
Given H=(A+B)(A+C)(B+C). Which expression is equivalent after expansion and simplification?
AB+AC+BC
A+B+C
(A+B+C)(AB+AC+BC)
ABC
