logic function defined by the truth table of Figure below. b. What is the minimum expression for this function? c. Realize F, using AND, OR, and NOT gates.
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a. Buildt he Karnaugh map for the logic function defined by the truth table of Figure below.
b. What is the minimum expression for this function?
c. Realize F, using AND, OR, and NOT gates.
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- Problem: Derive the logic expressions for a circuit that compares two unsigned numbers: X = x2x1xo and Y = = y2y1yo and generates three outputs: XGY, XEY, and XLY. One of these outputs is set to 1 to indicate that X is greater than, equal to, or less than Y, respectively.Using truth table method prove the following equality (P+QR) + PQ = P + Q ii. For the logic diagram shown in Figure Q23 prove it is working as Ex-OR gate. 4x1 MUX I, .F Y I2 I3 S So A BIf a fault occurs in connections C5 (stuck-at-1) for the following logic circuit, the appropriate fault-test by using Boolean difference technique .......... C1 X1 Cs C2 X2 Cs C6 C10 C9 C7 t0 t2 t3 t1 O
- Logic Gates:* 7404LS (NOT)* 7408LS (AND)* 7432LS (OR)* 7400LS (NAND)* 7402LS (NOR)* 7486LS (EX-OR)Or you can use 74HCxx versions. Task 2: 4 INPUT PRIORITY ENCODERa) Write the truth table.b) Find the outputs in terms of min terms using minimal expression.c) By using K map, find the simple/simplest expression of theoutputs.d) Draw the circuit diagram. (Simulation design will be accepted.)e) Simulate the circuit & explain your results. (Please do notdesign separate simulations for each output. You should design ONEsimulation including all inputs and outputs.)An X-input exclusive-OR gate and a Y-input exclusive-OR gate (where X=3, Y=4 have their outputs connected to a 2-input exclusive-NORgate. Do the following:a) Draw the logic diagram and analyze the logic expression of the output (in standard SOPform).b) List out all essential prime implicants.a) For the given logic circuit diagram write the program by using the gate level modeling. b) For the given truth table write the program by using the data flow Modelling. c) Write the test bench of the given logic circuit with all possibilities Y1 Y2 Y3 Y4 Y5 Y6 Y7 A2 A1 A0
- 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 combinational logic circuit that compares between two 2-bit numbers A (A1 A0) and B(B1 B0) is designed. Output F is high when ? > ? and low when ? < ?. 1) Are there any undefined outputs? If there are any undefined outputs what are the inputs?Design a 4-bit arithmetic circuit, with two selection variables S1 and S0, that generates the arithmetic operations in the following table. Draw the logic diagram for a single bit stage. Note that B’ represents “Not B”. Draw the logic diagram for a single bit stag
- 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).a) Create a 4 Variable Karnaugh Map in paper by mapping 1’s for given standard SOP Boolean expression. After mapping , make relevant groups within Karnaugh Map by considering rules for making groups for 4 variable Karnaugh Map. After making relevant grouping , extract the minimum SOP expression by considering rules for extracting minimum SOP using Karnaugh Map. * Standard SOP: *Create Circuit Diagram using logic gates and logic converter in Multisim for given standard SOP and minimum SOP which you have solved. Do make sure that truth table for both expressions should evaluate same result.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'C