7. Determine the Boolean description for the circuit shown below. 7. This circui

Question

7. Determine the Boolean description for the circuit shown below.

7. This circuit is drawn in the Multisim application and shows three inputs, labeled A, B and C going into a circuit that consists entirely of two-input NAND gates, in two stages, with an output labeled Y.

Inputs A, B and C are shown on the left of the diagram denoted using white 1/16 inch squares. Output Y is shown on the right of the diagram denoted using a white 1/16 inch square.

Stage 1:

Inputs A and B are the inputs to the first two-input NAND gate, with the gate marked with the denotation 74LS00N/U1A.

Input C is used as both inputs for the Second two-input NAND gate, with the gate marked with the denotation 74LS00N/U1C.

Stage 2: The outputs for the A/B input two-input NAND gate and the C input two-input NAND gate are used as the inputs to the final two-input NAND gate.

The output of this final Stage 2 two-input NAND gate is the output of the circuit and is marked Y.

8. 8. The diagram below is an attempt to build a 3-bit majority detector (Y = AB + AC + BC). Is the circuit correct? Prove your answer.


This circuit is drawn in the Multisim application and shows three inputs, labeled A, B and C going into a circuit that consists of 5 two-input NAND gates, in three stages, with an output labeled Y.

Inputs A, B and C are shown on the left of the diagram denoted using white 1/16 inch squares. Output Y is shown on the right of the diagram denoted using a white 1/16 inch square.

Stage 1: A and C are the inputs to the first two-input NAND gate, with the gate marked with the denotation 74LS00N/U1A. A and B are the inputs to the second two-input NAND gate, with the gate marked with the denotation 74LS00N/U1B. B and C are the inputs to the third two-input NAND gate with the gate marked with the denotation 74LS00N/U1C.

Stage 2: The output of the first two two-input NAND gates, with A/B and A/C inputs lead to another two-input NAND gate marked with the denotation 74LS00N/U1D. The output for this two-input NAND gate, is the first input into the stage 3 two-input NAND gate.

Stage 3: The output of the stage 2 two-input NAND gate, is the first input into the stage 3 two-input NAND gate, and is marked with the denotation 74LS00N/U2A The output of the stage 1 two-input NAND gate with B and C inputs is the second input to the final two-input NAND gate. The output of this final stage 3 two-input NAND gate is the output of the circuit and is marked Y.

7. Determine the Boolean description for the circuit shown below. This circuit is drawn in the Multisim application and shows three inputs, labeled A, B and C going into a circuit that consists entirely of two-input NAND gates, in two stages, with an output labeled ½Y½. Inputs A, B and C are shown on the left of the diagram denoted using white 1/16 inch squares. Output Y is shown on the right of the diagram denoted using a white 1/16 inch square. Stage 1: Inputs A and B are the inputs to the first two-input NAND gate, with the gate marked with the denotation ½74LS00N/U1A½. Input C is used as both inputs for the Second two-input NAND gate, with the gate marked with the denotation ½74LS00N/U1C½. Stage 2: The outputs for the A/B input two-input NAND gate and the C input two-input NAND gate are used as the inputs to the final two-input NAND gate. The output of this final Stage 2 two-input NAND gate is the output of the circuit and is marked ½Y½. The diagram below is an attempt to build a 3-bit majority detector (Y = AB + AC + BC). Is the circuit correct? Prove your answer. This circuit is drawn in the Multisim application and shows three inputs, labeled A, B and C going into a circuit that consists of 5 two-input NAND gates, in three stages, with an output labeled ½Y½. Inputs A, B and C are shown on the left of the diagram denoted using white 1/16 inch squares. Output Y is shown on the right of the diagram denoted using a white 1/16 inch square. Stage 1: A and C are the inputs to the first two-input NAND gate, with the gate marked with the denotation ½74LS00N/U1A½. A and B are the inputs to the second two-input NAND gate, with the gate marked with the denotation ½74LS00N/U1B½. B and C are the inputs to the third two-input NAND gate with the gate marked with the denotation ½74LS00N/U1C½. Stage 2: The output of the first two two-input NAND gates, with A/B and A/C inputs lead to another two-input NAND gate marked with the denotation ½74LS00N/U1D½. The output for this two-input NAND gate, is the first input into the stage 3 two-input NAND gate. Stage 3: The output of the stage 2 two-input NAND gate, is the first input into the stage 3 two-input NAND gate, and is marked with the denotation ½74LS00N/U2A½ The output of the stage 1 two-input NAND gate with B and C inputs is the second input to the final two-input NAND gate. The output of this final stage 3 two-input NAND gate is the output of the circuit and is marked ½Y½.

Explanation / Answer

Y = ((AB)' C')' = (A'B'C')' = (AB + BC + CA)

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