Construct digital circuit for Boolean expression Q=A’B+C using only NAND gates.
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Construct digital circuit for Boolean expression Q=A’B+C using only NAND gates.
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- (c) Figure Q3(c)(i) shows a register and Figure Q3(c)(ii) shows the input waveforms (CLOCK and Data in) to the circuit. A1 A9 A10 A2 Function generator A3 A11 A12 AS A13 A6 A14 A7 A15 Data in Bop.7) ip.r 82p.7) Logic analyser U1 U2 U3 U4 UO 6. 1. 6 1 6 INVERTER 3 CLK 3 CLK oCLK CLK 5 K K 5 K K 4027 Clock Function generator Figure Q3(c)(i) (i) Determine the type of register as shown in Figure Q3(c)(i).USE DIGITAL LOGIC AND DESIGN Part 1: In Figure_4; we have 4-bit Comparator using 2-bit Comparators block. You have to satisfy given condition by applying all data on figure 4. At the end, given condition should produce HIGH output and other two should be LOW. A3 A2 A1 A0 = 1101 and B3 B2 B1 B0 = 1110 Figure_4 Part 2: The serial data-input waveform (Data in) and data-select inputs (S0 and S1) are shown in Figure_5. Determine the data-output waveforms from D0 through D3. Figure_5 Part 3: Decoder can be useful when we have to decode some specific numbers from their equivalent code. Figure 6 has a concept of 3 to 8 line decoder from which you have to generate output waveform from D0 to D7 with proper relationship to input. Figure_6 Part 4: The data-input and…Design a 4 input multiplexer using logic diagram and truth table.
- 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.)IH.W: Draw a logic eircuit of the following Boolean expression before and after simplification using karnough map and Boolean expression. Y-AB+ AB A B Y 1 1Find the minimum output expression for circuit of figure below and draw logic circuit. A B
- 2.1 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 CH.W :- 1) A four logic-signal A,B,C,D are being used to represent a 4-bit binary number with A as the LSB and D as the MSB. The binary inputs are fed to a logic circuit that produces a logic 1 (HIGH) output only when the binary number is greater than 01102-610. Design this circuit. 2) repeat problem 1 for the output will be 0 (LOW) when the binary input is less than 01112-710- Saleem LateefI was able to fill out the truth table for part A. I ONLY need help for PART B. B) Implement the logic function (z) using the multiplexer 74HC151 shows in the picture.
- parity generator design, construct and test a circuit that generates an even parity bit ffrom four messages bits . use XOR gates. adding one more XOR gate, expand the circuit so that it generates an odd parity bit also.ehcu.org/pluginfile 100% 10 / 11 locations, count how many times is 0 and how many times 1 is. Questions:- 1- Write a program in assembly language to perform the following logic ci BL CL DL [5100]- 2- How we can perform the NEG and NOT instructions by using different instructions. 3- Write the following program by using different instruction or instructions for each instruction on the program. MOV AL , 00 MOV BX , FFFF XOR CL , FF NEG BYTE PTR [DI] AND CX , LGTask 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