A combinational circuit is defined by the following three Boolean functions. Design the circuit with a decoder and external gates. F1 = x' y' z' + xz F2 = x y' z' + x'y F3 = x' y' z + xy
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A combinational circuit is defined by the following three Boolean functions. Design the circuit with a decoder and external gates.
F1 = x' y' z' + xz
F2 = x y' z' + x'y
F3 = x' y' z + xy
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- 3. A COMBINATIONAL CIRCUIT IS DEFINED BY THE FOLLOWING THREE BOOLEAN FUNCTIONS. DESIGN THE CIRCUIT WITH A DECODER AND EXTERNAL GATES F1= X'Y'Z' + XZ F2= XY'Z' + X'Y F3= X'Y'Z + XYUsing a decoder and external gates, design the combinational circuit defined by the following three Boolean functions:1. (a) F 1 = x ′ y z ′ + x z F 2 = x y ′ z ′ + x ′ y F 3 = x ′ y ′ z ′ + x y2. (b) F 1 = ( y ′ + x ) z F 2 = y ′ z ′ + x ′ y + y z ′ F 3 = ( x + y ) zImplement the following Boolean expression using only XOR gates: A(B + C) + B(C + A) + C(A + B)
- Write the three outputs of X, Y and Z in terms of the four inputs A, B, C and D for the follow logic gates configuration ---This is my answer: I am unsure if it is right. X = A + (A’B’ * (B’+C’) = A + (A’+B’)*(B’*C’) Y = ((A’+B’)*(B’*C’))*((B’*C’)+CD)Z = (B+C)*(C’+D’)*D’Use Boolean algebra to simplify the following expression, then draw a logic gate circuit for the simplified expression: A’B’C’+A’B’C+AB’C’+AB’CDraw the non-abbreviated logic diagram for the following Boolean expressions. (You may use XOR gates.) A) ((a’)’)’C) a’b + ab’F) ((ab XOR b’) + a’b)’K) (abc’) + (a’ b’ c’)’N) (((a + b)’ + c)’ + d)’
- Design a combinational logic circuit that takes a 3–bit input and has one output P. The P output should be active high only when the inputs corresponds to a prime number Note: the prime numbers: Prime numbers are 2, 3, 5, 7… Select one: a. P= AC+B b. P= A'C+A'B c. P= AC+A'B d. P= AC+A'B'Design a logic circuit with input signal A, control input B, and outputs X and Y to operate as follows: When B = 1, output X will follow input A, and output Y will be 0. When B = 0, output X will be 0, and output Y will follow input A.A 1bit 4 to 1 multiplexer, the 4 inputs of the multiplexer will be the output of another combinational circuit with A and B as an input:00 = 1bit Adder01 = 1bit Subtractor10 = 1bit Comparator (equals)11 = XOR • truth table• Boolean Expression• Logic Circuit Here is the truth table guide:
- Design a combinational circuit with three inputs A, B, & C. The output Z is HIGH when the binary value of the inputs is an EVEN number and is greater than TWO. Draw the circuit using ONLY NOR gates.Given a 4-bit signed integer, design a circuit that outputs its absolute value. You can assume that the input will always have a valid output. (a) Draw a logic diagram of this circuit. You may use 4-bit half adder(s), 2x1 4-bit multiplexer(s), and any logic gate(s) in your design. (b) With the following Verilog code, implement your design above in Verilog. module half_adder (input [3:0] a, input [3:0] b, output [3:0] s); assign s = a + b; endmodule module mux(input [3:0] D0, input [3:0] D1, input S, output reg [3:0] O); always @(*) begin if (S == 0) O = D0; else if (S == 1) O = D1; else O = 4’bx; end endmoduleDesign a combinational circuit with three inputs, x , y , and z , and three outputs, A, B, and C . When the binary input is 0, 1, 2, or 3, the binary output is two greater than the input. When the binary input is 4, 5, 6, or 7, the binary output is two less than the input.