X = ABC + BCD is in the form of minterms expression.
Question 1 options:
True 

False 
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Wich of the following sum of products represents the given truth table?
Question 2 options:







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The minterm expression for the function g(A,B,C) given:
g(A,B,C)=A'B + AB' + AC
Question 3 options:







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The maxterm expression for the function g(A,B,C) given:
g(A,B,C)=A'B + AB' + AC
Question 4 options:







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For the function:
f(A,B,C,D) = A'BCD+ A'B + ACD'+ BC
What would be the canonical sum of products expansion?
Question 5 options:







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The waveforms are correct for the logic circuit shown above.
Question 6 options:
True 

False 
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For circuit shown, which option gives the equivalent, minimal circuit?
Note, the bubble shown in the input B to the lower AND gate in the option a is equivalent to an inverter.
Question 7 options:










Derive the expression of the circuit as a sum of products, map into a Karnaugh Map, minimize and apply De Morgans 

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Using Boolean algebra to simplify the expression Z = AB + A(B + C) + B(B + C), the completed first step would result in the expression:
Question 8 options:
Z = AB + ABAC + BB + BC 

Z = AB + AB + C + BB + C 

Z = AA + AB + AB + AC + BB + BC 

Z = AB + AB + AC + BB + BC 
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Using Boolean algebra, the complete simplification of Z = AB + A(B + C) + B(B + C) gives us:
Question 9 options:
Z = AB + AC + B 

Z = B + AC 

Z = AB = AC = BC 

Z = AB + AC + B + BC 
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Using Boolean algebra, the expression given for Y = above simplifies to:
Question 10 options:
Y = BC + A'B'C + AB'C 

Y = BC + B'C 

Y = C 

Y = BC +BC' + A' 
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The implementation of simplified sumofproducts expressions may be easily implemented into actual logic circuits using all ________ with little or no increase in circuit complexity.
Question 11 options:
OR gates 

AND gates 

NAND gates 

multipleinput inverters 
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Table 41
The truth table in Table 41 indicates that:
Question 12 options:
The output (Z) is HIGH only when a single input is HIGH. 

The output (Z) is HIGH only when the majority of the inputs are HIGH. 

The output (Z) is HIGH only when the binary input count is an even number greater than zero. 

The output (Z) is HIGH only when the binary input count is an odd number. 
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Table 41
The circuit implementation of the sumofproducts expression for Table 41 would require (without minimizing):
Question 13 options:
Three 3input OR gates, one 2input AND gate, and five inverters 

Three 3input AND gates, two 3input OR gates, and five inverters 

One 3input OR gate, two 3input AND gates, and five inverters 

Three 3input AND gates, one 3input OR gate, and three inverters 
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Table 41
Circuit implementation of the simplified expression for Table 41 will require (as a minimum):
Question 14 options:
Two 2input AND gates, two 2input OR gates, and two inverters. 

Two 2input AND gates, one 2input OR gate, and one inverter. 

One 2input AND gate, two 2input OR gates, and two inverters. 

Three 2input AND gates, two 2input OR gates, and two inverters. 
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A logic circuit allows a signal (A) to pass to the output without inversion when either (but not both) of the control signals (B_{1} and B_{2}) are HIGH. Which of the following option is the output expression for this circuit?
Question 15 options:
Option a) 

Option b) 

Option c) 

Option d) 
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The logic gates required to implement the logic circuit in the preceding question would be:
Question 16 options:
an Exclusive NOR gate with inputs B_{1} and B_{2} whose output is fed, along with input A, to an AND gate. 

an Exclusive NOR gate with inputs B_{1} and B_{2} whose output is fed, along with input A, to an OR gate. 

an Exclusive OR gate with inputs B_{1} and B_{2} whose output, is fed along with input A, to an AND gate. 

an Exclusive OR gate with inputs B_{1} and B_{2} whose output is fed, along with input A, to an OR gate. 
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Which of the following is the simplest form of the expression Y = ABC[AB + C(BC + AC)]?
Question 17 options:
Y = ABC + BC 

Y = BC 

Y = AC + BC 

V = ABC 
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Simplifying logic circuits results in:
Question 18 options:
fewer potential faults. 

fewer connections. 

fewer gates. 

all of the above 
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Given F_{1} = Σ m(0, 4, 5, 6) and F_{2} = Σ m(0, 3, 4, 6, 7) the minterm expression for F_{1} + F_{2} is:
Question 19 options:
F_{1} + F_{2} = Σ m(0, 3, 4, 5, 6, 7) 

F_{1} + F_{2} = Σ m(0, 6) 

F_{1} + F_{2} = Σ m(1, 2) 

None of the above 
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Given F_{1} = Π M(0, 4, 5, 6) and F_{2} = Π M(0, 4, 7) the maxterm expression for F_{1} + F_{2} is:
Question 20 options:
F_{1} + F_{2} = Π M(0, 4, 5, 6, 7) 

F_{1} + F_{2} = Π M(1, 2, 3) 

F_{1} + F_{2} = Π M(0, 4) 

None of the above 
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A switching circuit has four inputs as shown. A and B represent the first and second bit of a binary number N_{1} and C andD represent the first and second bit of binary number N_{2}. The output is to be 1 only if the product N_{1}xN_{2} is less than or equal to 2. The minterm expansion of F is:
Question 21 options:
F=∑m(0,1,2) 

F=∑m(0,1,2,3,4) 

F=∑m(0,1,2,3,4,5,6,8,9,12) 

None of the above 
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A bank vault has three locks with a different key for each lock. Each key is owned by a different person. To open the door, at least two people must insert their keys into the assignated lock. The signals A, B and C are 1 if there is a key inserted into lock 1, 2 or 3 respectively. The equation for the variable Z which is 1 iff the door should open is:
Question 22 options:
Z=AB + AC + BC 

Z=ABC 

Z=A + B + C 

None of the above 
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A logic circuit realizing the function f has four inputs A, B, C and D. The three inputs A, B, C are the binary representation of the digits 0 through 7 with A being the mostsignificant bit. The input D is an oddparity bit, ie the value of D is such that A, B, C and D allways contain an odd number of 1's. (For example, the digit 1 is represented by ABC = 001 and D = 0 and the digit 3 is represented by ABCD = 0111.) The function f has value 1 if the input digit is a prime number. (A number is prime if it divisible only by itself and 1; 2 is considered to be prime but 0 and 1 are not)
A list of the minterms and don't care minterms of f is :
Question 23 options:
f = ∑m(4,7,11,14) + d(0,3,5,6,9,10,12,15) 

f = ∑m(2,3,5,7) + d(0,1,4,6,8,9,10,11,12,14,15) 

f = ∑m(2,3,5,7,11,13) + d(0,1,2,4,6,8,9,10,12,14,15) 

None of the above 



Build the truth table for f(A,B,C,D) following the specifications given above. Observe that some of the rows of the truth table cannot hold realistic parameters and will never occurr and therefore should be assigned as don't care 

Option cf(A,B,C,D)= A'BC'D'+A&...