1- Implement ( without simplification) F= (A+B).(C+A.D) using NAND gates only. 2. Desion a logic eates circuit for P OS F(A BC D)=S(1 247 8 11 13 14)
Q: Logic Gates: * 7404LS (NOT) * 7408LS (AND) * 7432LS (OR) * 7400LS (NAND) * 7402LS (NOR) * 7486LS…
A: Given: The logic gates given are, The truth table for 2-1 multiplexer is asked and circuit diagram…
Q: A B 2
A: The given data is as follows:-
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Q: Q.4 Draw the logic diagram to implement the following expression with minimum number of NAND gates.…
A: The solution is provided in the following section:
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A: According to the question, we need to implement the given function using NAND Gate.
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- (c) Figure Q5(c) shows a logic circuit which has three inputs A, B, C and two outputs F and G. i) Obtain the logic expression for the outputs G and F. ii) Redesign the circuit using only 3-to-8 decoder (with active high outputs) and OR gates. G A B F Figure Q5(c)answere fast please question from DIGITAL LOGIC DESIGN TOPIC : Designing Combinational Logic You are designing a water level circuit using 74ALS151 (8 to 1 Multiplexer IC)* When input is 0000 that means tank is empty.* When input is 1111 that means tank is full.* When input is below 5, that means water level is low.* So, make a circuit using 74ALS151 Multiplexer IC that shows a "low water" indicator light(by setting an output L to 1) when the water level drops below level 5.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.
- 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) 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 A0Convert the following logic gate circuit into a Boolean expression, writing Boolean sub-expressions next to each gate output in the diagram: C DD
- 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.)Problem #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)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 :
- 6. For the follow logic circuit system, the output f is: 5 (A) ab. (B) a + b. (C) a'+b'. (D) a'b'. 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 ODesign a combinational circuit using multiplexer for a car chime based on thefollowing system: A car chime or bell will sound if the output of the logic circuit(X) is set to a logic ‘1’. The chime is to be sounded for either of the followingconditions:• if the headlights are left on when the engine is turned off and• if the engine is off and the key is in the ignition when the door is opened.Use the following input names and nomenclature in the design process:• ‘E’ – Engine. ‘1’ if the engine is ON and ‘0’ if the engine is OFF• ‘L’ – Lights. ‘1’ if the lights are ON and ‘0’ if the lights are OFF• ‘K’ – Key. ‘1’ if the key is in the ignition and ‘0’ if the key is not in the ignition• ‘D’ – Door. ‘1’ the door is open and ‘0’ if the door is closed• ‘X’ – Output to Chime. ‘1’ is chime is ON and ‘0’ if chime is OFF