11. Design a simple circuit from the function by reducing it using appropriate k-map, draw orresponding Logic Diagram for the simplified Expression F (P,Q,R,S,T) = E (0,7,4,12,14,8,16,24,23,31) + d ( 6, 13,15,20,28,22,30) mplement the simplified logical expression, universal gates (NAND) How many NAND gates equired as well specify how many ICs are needed.
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- 5. Simplify the following function using K-Map and draw logic diagram for that. E(A, B,CD)=Em(0,1,2,3,4,5,7,8,10,11,12,13,14,15)An equation in reduced SOP form, is F=AB+B'C+A'C'. I need to draw a logic circuit F using NOT/AND/OR and logoc circuit F using all NAND gates. Thank you for the help. I understood the previous types of gates but I am confused on how to draw these circuits.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.
- Design a 3-bit counter that counts the following sequence: 7,5, 3. 1.0.7, 5. 3, 1, 0, 7. etc. Using the sequential design technique that starts from a state diagram, draw the state table. minimize the logic. and draw the final circuit. The outputs of logic circuit are 2 = Qo Q1. I, = Qo.Qi + Qo.Qi, Io = Qo.Q2, Cont2 = Qj Q2 Cont1 = Qu Q2. Cont0 = Q2 Qo.Q1. h = Qo.Qi + Qo.Q1, Io = Qo Qz Cont2 = Q, Q2 Contl = Qo Q2 Cont0 = Q2 Qo Qı Ij = Qo.Q, + Q».Qı, Io = Qo. Q2. Cont2 = Qj Q2. Contl = Qo.Q2. Cont) = Q2 L = Qo.Qı. I¡ = Q. Qj + Qu Q Io = Qv.Qz Comt2 = Q, Q, Contl = Q Q2 Cont0 = Q2 !! fefsto How much will be per-product cost and thPlease design a 6:1 multiplexer following the below procedures with data inputs of D5, D4, D3,D2, D1, D0 and output of Y.1 How many select signals are needed for this Mux.2) List a truth table for this Mux. Note: for all the unused combinations of select signals, Y=D53) Develop an optimized function for this Mux.4) Sketch the logic diagram of implementing this 6:1 Mux.5) Write a complete VHDL structural model to implement the above 6:1 multiplexer. Assume allthe required sub-component (standard gates) VHDL models are given/known that you can useConstruct a circuit diagram that checks whether the two numbers A and B are in the ratio of 2:3. Also, derive the final Boolean equation for the function. F = 1 if A: B = 2: 3,0 otherwise Here, A and B both are 3 bit binary numbers. NB: You cannot use the IC of comparator, meaning for the comparison part, you need to draw the gate level diagram. You can use block level diagrams for the rest of the parts.
- T: Answer thne f. questions: 1) The hexadecimal number ´Al' has the decimal value equivalent to (A) 80 (B) 161 (C) 100 (D) 101 2) The output of a logic gate is 0 when all its inputs are logic 1. The logic is either (A) a NAND or an EX-OR (B) an OR or an EX-NOR (C) an AND or an EX-OR (D) an NOR or an EX-NOR 3) The Gray code of the Binary number 1100111 is (A) 1011011 (B) 1010100 (C) 1001001 (D) 101101 4) When simplified with Boollean Algebra (a+b)(a+c) simplifies to (A) a (B) a+a(b+c) (C) a(1+bc) (D) a+bc 5) -31 is represented as a sign Binary number ( using Sign-magnitude form ) equal to (A) 00011111 (B) 10101001 (C) 01110010 (D) 00101101 6) The Binary number 110111 is equivalent to decimal number (A) 25 (B) 55 (C) 26 (D) 34 7) With 4 bit, what the range of decimal values if the number is 2's complement signed number. (A) -32 to +31 (B) -2 to +1 (C) -8 to +7 (D) None of theseA shaft position encoder has 4 output. The outputs corresponding to each position are given in thetable below. It is desired to design a circuit thatshaft position between and → the output f=1at the other position → f=0The values not determined by the shaft position will be taken as 0.a) Fill the truth table. b) Reduce the output function as sum of minterms using Karnaugh map. c) Realize this function using logic gates.Minimize the following Equation by using Karnaugh Map, then draw the final Logic Circuit of the minimized equation. Y(A,B,C,D)= E (1,4,5,9,10,11,14,15) * 1 Add file
- 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 stagQ4 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.Minimize the Boolean expression F=AB’C’+C’D+BD’+A’C using K -map and implement the logic circuit using NAND gates only.