If a two-input logic gate produces an output of logic HIGH, only if both inputs are different, then the logic gate IC is IC 7432 O IC 7486 IC 7400 IC 7408
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Q: An IC with two-input logic gate produce a output as HIGH, only if both the inputs are different is
A: An IC with two-input logic gate produce a output as HIGH, only if both the inputs are different.
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Q: Negative OR Gate is О а. IС 7400 Оb. IC 7432 O c. О с. IC 7408 O d. IC 7402
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Q: A synchronous state machine has two inputs (X1 and X2) and one output (Z). The relationship between…
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Q: Design a NAND Logic Circuit that is equivalent to the AOI circuit shown below: A B Z = BC +AC
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A: F1=x'y'z'+xz = x'y'z'+xz(y+y') = x'y'z'+xyz+xy'z = x'y'z'+xy'z+xyz = Σ (0,5,7) F2=xy'z'+x'y =…
If a two-input logic gate produces an output of logic HIGH, only if both inputs are different, then the logic gate IC is
IC 7432
O IC 7486
IC 7400
IC 7408
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Solved in 3 steps
- An IC with two-input logic gate produce a output as HIGH, only if both the inputs are different isA 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:Negative OR Gate is 0 IC 7400 O IC 7432 O IC 7408 OIC 7402
- The combinational logic circuit required for this system is only 4x1Design using data selectors (mux) and logic gatesWhat will be output of the given logic circuit if A = 0, B = 1, C = 1, D = 0.Designs a logic circuit that will allow a signal to pass to the output only when control inputs B and C are both HIGH; otherwise, the output will stay LOW
- A combinational circuit is defined by the following three Boolean functions. Design the circuit with a decoder and external gates. F1 = x' y' z' + xzF2 = x y' z' + x'yF3 = x' y' z + xyThe output of a two - input nand gate is HIGH , when the any of inputs areExclusive-OR (XOR) logic gates can be constructed from what other logic gates? Select one: a. OR gates only b. AND gates and NOT gates c. OR gates and NOT gates d. AND gates, OR gates, and NOT gates
- A) Draw a three-input OR gate, its Boolean expression, and its truth table. B) Repeat part (A) for a three-input NAND gate. C) Repeat part (A) for a three-input NOR gate.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’The following circuit is: a. 2-input Static CMOS nor gateb. 2-input Static CMOS nand gatec. 2-input Dynamic CMOS nand gated. 2-input Dynamic CMOS nor gate