Draw the equivalent logic circuit diagram of the following expressions : a. XY = F b. X + Y = F c. XÝZ = F
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- Simplify the following Boolean expressions using Karnaugh Map and draw the logic circuits. f = wxyz + wxyz + wxyż + wxỹz + wxyz + wxyz + wxỹz + wãyzDesign 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. 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 forms
- 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 CQ6: For the following logic function: F(x,y,z) = xyz + xyz + xyz +xyz a. Draw the logic Circuit b. Construct the Truth Table c. Simplify the Logic CircuitWrite a VHDL code for the following simple logic circuit. D- X1 X2 f X3
- Design a logic circuit with four inputs and one output that will produce "l" in the output only if the input patterns have odd number of zeros. a) Write the Boolean equation for the circuit in the simplest SOP form. b) Draw the logic circuit for the above equation in its simplest form. c) Re Design the logic circuit using NANI) gates only?Consider the multiplexer based logic circuit shown in the figure MUX MUX 1 Select one: a. W S1' S2' O b. W + S2 + S1 c. WS1 + WS2 + S1 S2 O d. WeS1es2Simplify 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
- DIGITAL LOGIC DESIGN Are the following addition results Overflow or underflow and why?Palagiaph 1. Find logic finctions for the circuits shown below. FProblem #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)