The Boolean function realized by the logic circuit shown is F (A,B,C,D) 4x1 MUX Y I, S, S, A' B Select one: O a. F=Em (0,1 ,3,5,9,10, 14) O b. F= Em (1,2 , 4 ,5,11,14,15) O c.F=Em (2, 3,5,7,8,9,12) O d. F=Em(1,2,4,5,1114, 15)
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- Simplify the following Boolean functions using K-Map and Design the Logic diagram. a) F (A,B,C) = Em(0,1,2,4,7) b) F (A,B,C,D) = m(1,3,9,11,12,13,14,15) c) F (A,B,C,D) = Em(3,7,11,13,14,15) %3D %3DBoolean Function F(A,B,C,D) = { m (1,2,5,8,11,15), don't cares d(A,B,C,D) = { m (3,10) a. Using a K-map simply F in S.O.P. form b. Draw the logic circuit c. Using a K-map simply F in P.O.S. form d. Draw the logic circuit e. Which form has a lower gate input cost?Draw the equivalent logic circuit diagram of the following expressions : a. XY = F b. X + Y = F XÝZ = F c. d. XY + XZ = F e. XYZ + XÝZ = F
- DO NOT COPY ANSERWS IT'S INCORRECT A very detailed solution and if you can use a program to design after the work please do.Problem : Design a circuit that takes a 3-bit number and increments it by two using a minimum number of 4x1 Mux's and a minimum number of logic gates the output is 4 bits. Show your work and label all inputs/outputs appropriately.The Boolean functions f (x2,x1,xO) realized by the digital logic circuit shown in the given figure is Do Decoder D 3-to-8 D. D3 D. Ds De D X2 a. f(x2.x1.x0) = Em(0, 1, 2. 3, 4, 5, 6, 7) b. f(x2,x1.x0)-= Em(1, 2, 4. 5, 7) None of the given options DCDAL NO TCC DDOSimplify the following expression using Karnaugh map and implement. Draw simplified logic diagram as well. Implement on Multisim software. (a) Y=A.B.C'.D+A.B'.C'.D+A'.B'.C'.D+A'.B.C'.D+A'.B'.C'.D'+A'.B.C'.D'+A'.B.C.D'+A'.B.C.D+A'.B'.C.D
- Simplify the following Boolean expressions using Karnaugh Map and draw the logic circuits. f = wxyz + wxyz + wxyż + wxỹz + wxyz + wxyz + wxỹz + wãyz(a) A logic circuit shown in Figure Q.3 has a 4-bit input A and B, three 4-bit wide 2:1 muxes, a 4-bit adder, a 4-bit output F, and a carry flag C. For the given Table Q.3, fill in the value of output F and carry flag C for the given value of A, B, S0, S1 and S2. 51 52 1001 Flag C 0011 Figure Q.3 Table Q.3 A So S1 S2 F Flag C 0001 1000 0010 1001 1 1 0011 1101 0100 1101 1110 0111 1Design a logic circuits w/ 4inputs (A(MSB), B,C,D(L,S,B)) and 1 output x. The 4 inputs represent switches in a vending machine. The switches can be either open (0) or closed (1) The X output becomes high if and only there are two or more switches closed at the same time. Furthermore it is impossible for the first A and the last switch D to be closed at the same time. Reqd: 1.) complete and labeled truth table 2.) canonical sop form expression 3.) grouped kmap 4.) minimum SOP expression
- 1. Given the Boolean expression (b + d)(a’+ b’ + c),a. Convert the expression to the other standard form. What do you call this standard form?b. Derive its canonical form. What do you call this canonical form?c. Derive the other canonical form. What do you call this canonical form?d. Provide the truth table of the expressione. Draw the logic circuit diagrams of the 2 standard formsDesign 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).1) You want to design an arithmetic adder / subtractor logic circuit. a) List the steps you will apply in the design approach. 8-bit BCD full adder Design the circuit. Explain each step. Realize with AND, OR, NOT gates. In the circuit you designed the figures 0,3,5,8 discuss the result by summing up.