4. Draw the logic gate circuitry for the following Boolean expressions using NOR gates only a) F=(A+B)C b) F=A+BC+D c) F=A+B(C+D)T F=AB+AC d)
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A: Below is the answer to above question. I hope this will be helpful for you..
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- 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'Draw 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)’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 endmodule
- Create a non-abbreviated logic diagram for the following Boolean Expressions. You can use all gates. a’b + ab’ ((ab)’(b’c)’ + a’b’c’) (a +b)(a’ + c)(b’ + c’)Draw the non-abbreviated logic diagram for the following Boolean expressions. (You may use XOR gates.) K) (abc’) + (a’ b’ c’)’N) (((a + b)’ + c)’ + d)’Being X and Y of a consecutive circuit with 2 D flip flops, A and B.It has 2 inputs and 1 output as Z. DA= X'Y + XADB = X'B + XAZ=B a) Draw the logic diagram of the circuit.b) Create the state table of the circuit.c) Draw the state diagram of the circuit.
- Draw the logic circuit of the full adder circuit block. the circuit that will perform the operation given below. Implement using a minimum number of full adder circuit blocks and a minimum number of logic gates. Here a1a0, b1b0 and c1c0 are two-bit binary numbers.Write the Boolean equations and draw the logic diagram of the circuit whose outputs are defined by the following truth table: Table P2.27f 1 f 2 a b c1 1 0 0 00 1 0 0 11 0 0 1 01 1 0 1 11 0 1 0 00 1 1 0 11 0 1 1 1A combinational logic circuit has inputs A, B, C, D and E. The output Y is given by: fΠ M(15,31) where A is the Most Significant Bit. Give Y in terms of inputs A, B, C, D and E in a Product of Sum format.
- Truth Tables and Logic Gates /Circuits~ - NOT1. Derive truth tables for the following Boolean expressions:a) ~(A + B)b) ~ A + (~ B)c) (~ A) + Bd) (~ A). (~ B)e) (~ A). (A + (~ B)) 2. Derive truth tables for:a) A.B + (~C)b) A. (BC)c) (AB).Cd) (~ A) .(B + C)e) ~(AB) + C 3. Derive truth tables for the following; inputs are A, B and C.a) W is True if an even amount of inputs is True, and False otherwise.b) W is True if exactly one input is true, and False otherwise.c) W is true if A and C are the same, and False otherwise. 4. Draw the logic circuits for the following Boolean expressions:a) W = (AB) + (NOT C)b) X = (~A). (B + C)c) Y = ~(AB) + CCreate a 1-bit XOR circuit using only AND, OR, and NOT gates. Explicitly show all steps starting from the truth table for XOR, then listing the logical expressions for when XOR is 1, and then translating the expressions into a circuit.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’