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 OB
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- Task 6: Simplifying Boolean functions in EWB using the logic converter Simplify the following Boolean expression in EWB using the logic converter F (A, B, C) = AB'C'+ A'B'C'+ A'BC'+ A'B'CA 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'Write the Boolean expression for the logic circuit in Figure 5. Simplify the Boolean expression using DeMorgan’s Theorem and Boolean algebra rules and laws. Show the steps of your simplification below. simplify curcuit.
- 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 OBA three input logic function will provide a logic high output only when all of the inputs are logic highs or when all of the inputs are logic lows. For all other input possibilities, a logic zero is provided on the output. What is the logic expression? A. A'B'C'+AB'C' B. Y=A'B'C'+ABC C. ABC'+AB'C D. A'B'C'+AB'C1. 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 :
- A B Prepare the truth table of the logic circuit (logic circuit) on the right. Then, using the truth table, write the function you have obtained according to the SOP (minterm) without any simplification and draw the logic circuit for the new function you have obtained. Make sure that the circuit you have drawn is understandable, the function you have written and the truth table are readable.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).Draw the logic circuits for the following Boolean expressions and simplify the logic expressions using theorems of Boolean algebra. (Ā+B)(A+ Ē + D)D
- Basic Combinational Logic Circuits 2. Write the output expression for logic circuit: A B B XShow 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. (AOP)+C 3) Y= A.B (C + D) + D+ A BWrite the Boolean expression for the logic circuit in Figure 5. Simplify the Boolean expression using DeMorgan’s Theorem and Boolean algebra rules and laws. Show the steps of your simplification below.