Test: Logic Gates & Boolean Algebra - 1

The given function is,
y = A + B' = (A'B)'
f(A,B) = A+B'
As 0 and 1 are available.
f(0,B) = A + B' = B'= NOT Gate.
f(A,B') = A + (B')' = A + B = OR Gate.
f(o,f(A,B')) = 0 +(A + B)' = NOR Gate.
f(A',B) = A' + B' = (A.B)' = NAND Gate.
f(0,f(A',B)) = 0+((A.B)')' = A.B = AND Gate.
With the combination of OR and NOT, NOR gate can be implemented. Since NOR gate is universal logic gate, so all the functions can be implemented. 
Hence, we can implement NOT Gate we can also implement OR, AND, NOR and NAND Gate.
Hence the correct answer is option 1 and option 3.

output F = (X + Y)'
waveforms are given as

As we can see for the time 35msec to 40msec output F is logic 1 (5 V)

X = [(A + B̅) (B + C)] B
= (AB + AC + 0 +  B̅C)B
= AB + ABC
= AB(1 + C)
= AB

Concept:
Prime implicant: Each square (or) rectangle made up of the group of adjacent min terms is called a sub cube. Each of these sub cubes is called a prime implicant.
Essential prime implicant: The prime implicant which contains at least one ‘1’ which can not be covered by any other prime implicant is called essential prime implicant.
Redundant prime implicant: The prime implicant whose each ‘1’ is covered least by one essential prime implicant is called a redundant prime implicant.

Application:
Given Boolean function is: f(A, B, C, D) = Σ m(0, 1, 4, 6, 7, 8, 10, 14, 15)
The K-map is as shown below

From the above K-map,
The prime implicants: BC, A̅ B̅ C̅, ACD̅, AB̅ D̅, B̅ C̅ D̅, A̅ BD̅, A̅ C̅ D̅
The essential prime implicants: BC, A̅ B̅ C̅
 

After simplification, we get

∵ De Morgan’s law:

∵ De Morgan’s law:

Concept:
Consensus Law is one of the most powerful theorems used in digital electronics for the minimization of Boolean function or equation either in the successive reduction method or in the K-Map method.
Statement:

• The consensus theorem states that the consensus term of a disjunction is defined when the terms in function are reciprocals to each other (such as A and A̅).
• The consensus theorem is defined in two statements (normal form and it's dual). They are
• AB + ĀC+BC = AB+ĀC
• (A+B)(Ā+C)(B+C) = (A+B)( Ā+C)

Calculation:
Y = P̅Q + QR + PR
Y = P̅Q + PR + QR (P̅ + P)
Y = P̅Q + PR + QRP̅ + QRP
Y = P̅Q(1 + R) + PR(1 + Q)
Y = P̅Q + PR where (1 + A = 1) according to Boolean algebra. 

Output expression Q is equivalent to NAND gate.

Analysis:
Y = AB + A(B + C) + B(B + C)
= AB + AB + AC + B + BC
Since AB + AB = AB, we get:
Y = AB + AC + B (1 + C)
Since 1 + X (any variable) = X, we get:
Y = AB + AC + B     
Y = B(1 + A) + AC  
Y = B + AC

Concept:
3 variable K-maps:

• For a 3-variable Boolean function, there is a possibility of 8 output minterms.
• The general representation of all the minterms using 3-variables is shown below.

Calculation:
Given Boolean expression is,
F = AB + AC̅ + BC
= A B C̅ + A B C + A B̅ C̅ + A B C̅ + A B C + A̅ B C


F = BC + AC̅

F(P, Q, R) = PQ + QR' + PR'
= PQ (R + R') + (P + P')QR' + P(Q + Q')R'
= PQR + PQR' + PQR' + P'QR' + PQR' + PQ'R'
= PQR + PQR' + P'QR' + PQ'R'
= m₇ + m₆ + m₂ + m₄
= m₂ + m₄ + m₆ + m₇