Implement the following Boolean function F, using the two-level forms of logic (a) NAND- AND, (b) AND-NOR, (c) OR-NAND, and (d) NOR-OR: F(A, B, C, D) = (0, 4, 8, 9, 10, 11, 12, 14)
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- 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.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 theseDesign 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
- 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)(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)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.
- 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.4. For the NOR gate function shown below a) Write the switching expression for the output, F(A,B,C,D) b) Simplify this switching function so that the only gates involved are AND, OR, and NOT gates. c) Draw the logic diagram of this simplified expression using only AND, OR, and NOT gates. am 1, S..pdf DII PrtScn F8 Home F9 End F10 F3 F4 F5 F6 F7 &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.
- Design counter that counts from 00 to 59, using the IC 74LS90 ripple counter and use two 7 segment display to display the result count. You can also use 7447 binary to 7-segment Display Decoder in logicworks.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.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 th