1. Using Boolean algebra to simplify the following expressions. Then connect the circuits using NAND gates only F=(AOB)O(BỌC)
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- 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 1Using Boolean algebra theorems, simplify the logic expression above as far as possible. Create a Circuit Diagram for the new expression Then create a truthtable for the simplified circuit(m. Simplify the given expression using Boolean algebra and implement the simphifie expression using basic logic gates. Y = A.B.C+ A.B.C + A.B.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 :Simplify the following Boolean expressions using Karnaugh Map and draw the logic circuits. f = wxyz + wxyz + wxyż + wxỹz + wxyz + wxyz + wxỹz + wãyzA logic function is expressed by the equation = 0.1.8.9.4. Plot the the values on the K-map below Note: w is the msb (most significant bit) 3 x 10 0- - W X 1 00 01 Write the simplified Sum of Products (SOP) expression. Use yz WIM or "" for NOT; 11 10 for AND; "+" for OR (without the quotation marks.
- 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.)Consider F(A,B,C) = AB'C + B'C' + A'BC + A'C' 1. Determine how many logic gate inputs would be needed before any simplification. Do not count inputs to NOT gates 2. Use Boolean algebra rules to get the most simplified expression of F(A,B,C). Then determine how many logic gate inputs would be needed after simplification. Again, do not count inputs to NOT gates. 3. Expand the original expression into its canonical SOP representation. 4. Fill out the K-map below using the SOP canonical representation. Group the 1-cells according to the K-map simplification rules. Translate each group into its product term, OR these product terms together, and verify that the expression you get matches the one in Step 2. 5. Draw two circuits in CircuitVerse, one from the original expression for F(A,B,C), the other from the simplified expression in Step 2 or Step 4. Connect the inputs to both circuits, but separate their outputs. Verify through simulation that these two circuits are indeed equivalent. Take…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.
- 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)A2. Using Boolean algebra, reduce the output equation of the function in Table 1. (You can use the “Karnaugh Map of Table 2" to simplify the logic function obtained). CD AB Table 2 00|01|11| 10 00 01 11 A3. Construct the simplified expression obtained in A2 by only NOR gates. Design and draw the NOR gate 10 connection diagram of the simplified function using only one 4001 IC. usingQ4 A-Design Parallel load/Shift right 4 bit register with LD/SR signal. If LD/SR =1 then register parallel load from D3D2DIDO inputs. If LD/SR =0 then it shifts right. Use standard logic gates and D F-F.