11. Use AND gates, OR gates, and inverters as needed to implement the following logic expressions as stated: a. X = AB + BC b. X = A(B+C) c. X = AB+ AB d. X = ABC + B(EF+G) e. X = A[BC(A + B + C + D)] f. X = B(CDE + EFG)(AB+C)
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![11. Use AND gates, OR gates, and inverters as needed to implement the following logic expressions
as stated:
a. X = AB + BC
b. X = A(B+C)
c. X = AB + AB
d. X = ABC + B(EF +G)
e. X = A[BC(A + B + C + D)]
f. X = B(CDE + EFG)(AB+C)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa25512a3-f638-4bdc-aeec-799a9d6f35b6%2Fb2876dc8-0444-496a-9382-63f81cf1172a%2Fvuwnslc_processed.png&w=3840&q=75)
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- Draw the equivalent logic circuit diagram of the following expressions : a. XY = F b. X + Y = F XÝZ = F c. d. XY + XZ = F e. XYZ + XÝZ = FA three input logic functipn will provide a logic high output only when twp (and two only) of the inputs are logic highs. For all other input possibilities, a logic zero is provided on the output. What is the logic expression? A. Y=A'B'C+A'BC'+AB'C' B. Y=AB'C+A'BC+ABC' C. Y=AB'C+ABC+A'B'C' D. Y=ABC'+AB'C+A'B'C'1. a. i. Draw the gates required to build a half adder are ii. When simplified with Boolean Algebra (x + y)(x + z) simplifies to : iii. The output of a logic gate is 1 when all its inputs are at logic 0, the gate is either :
- Using AND gates OR gates and inverters, draw a schematic for the function F = A&B + A&C + B&C. You do not need to minimize the function.(NEED NEAT HANDWRITTEN SOLUTION ONLY OTHERWISE DOWNVOTE )Design the following combinational logic circuit with a four-bit input and a three-bit output. The input represents two unsigned 2-bit numbers: A1 A0 and B1 B0. The output C2 C1.C0 is the result of the integer binary division A1 A0/B1 B0 rounded down to three bits. The 3-bit output has a 2-bit unsigned whole part C2 C1 and a fraction part CO. The weight of the fraction bit CO is 21. Note the quotient should be rounded down, i.e. the division 01/11 should give the outputs 00.0 (1/3 rounded down to 0) not 00.1 (1/3 rounded up to 0.5). A result of infinity should be represented as 11.1. A minimal logic implementation is not required. (Hint: start by producing a truth table of your design).Assuming we are given a Logic Gate network, the output of the circuit is given by the following equation Q=A.B+A.C+ A.B.C By using the Boolean Logic rules for simplification, select which of the following equations is the correct simplified one. Note that, the answer None of the above may sometimes be the correct response. Q=A+(B.C) O Q=A.(B.C) O Q=Ā. (B+C) ⒸQ=A. (B+C) None of the above O Not answered
- Discussion: 1. Simplify the following logical expression and implement them using suitable logic gates a = E2,4,6,10,14 b. F = I12,3,6 2 Determine whether or not the following equalities corect: a. A+B.C+Õ C = BC b. B(AO) +B C+ ACBOc) = AC 3. Convert the following expressions to SOP forms: A (A +B. C) -B b. (A + C)(Ã-B-Č+A.C-D) * 4. Write a Boolean expression for the following statement: Fisa "1" if A, B&C are all 1's or if only two of the variable is a"0". %3D * 5. Fing F2.1 Combinational logic circuits. Tabulates a truth table for the following Boolean expression shown in Equation 1.1. f = A.B.C + A.B.C + A.B.C (1.1) 2.2 Half adder. A half adder is a circuit that adds two binary digits, A and B. It has two outputs, sum (S) and carry (C). The carry signal represents an overflow into the next digit of a multi-digit addition. Figure 1.2 depicted a logic diagram for a half adder. a. derives the Boolean expression for s and c. b. tabulates a truth table for the half adder. Ao Bo Figure 1.2: Half adder os S CSimplify the following Boolean expressions using Karnaugh Map and draw the logic circuits. f = wxyz + wxyz + wxyż + wxỹz + wxyz + wxyz + wxỹz + wãyz
- (B)- Show the Ladder logic program and the equivalent Function Block Diagram for the following Boolean expressions without any simplification: - 1) Y= A.B+C+ (C+ B).D 2) Y= A.B.C+B. D. (A OB)+C 3) Y= A.B (C+ D) +D+ A OBProblem #04] Using AND and OR gates develop the logic circuit for the Boolean equation shown below. Y =AB(C + DEF) + CE(A + B +F) Problem #05] Using AND and OR gates develop the logic circuit for the Boolean equation shown below. X-A(CD+B)Simplify the given Boolean expression and then draw a logic circuit using NOR Gates. X (A, B, C, D) = AB’C’ + AC + A’CD’
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