1What is the main purpose of using a truth table while validating a logic gate?
Implementation and validation of logic gates
Easy
A.To calculate the power supply frequency
B.To identify the color of connecting wires
C.To compare expected and observed outputs
D.To measure the physical size of the gate
Correct Answer: To compare expected and observed outputs
Explanation:
A truth table lists the expected output for each input combination, allowing the implemented gate to be checked.
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2How many possible input combinations must be tested for a two-input logic gate?
Implementation and validation of logic gates
Easy
A.Three combinations
B.Four combinations
C.Eight combinations
D.Two combinations
Correct Answer: Four combinations
Explanation:
Two binary inputs have possible combinations: , , , and .
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3For a two-input AND gate, which input combination produces a HIGH output?
Implementation and validation of logic gates
Easy
A.,
B.,
C.,
D.,
Correct Answer: ,
Explanation:
An AND gate produces HIGH only when all its inputs are HIGH.
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4For a two-input OR gate, which input combination produces a LOW output?
Implementation and validation of logic gates
Easy
A.,
B.,
C.,
D.,
Correct Answer: ,
Explanation:
An OR gate produces LOW only when all its inputs are LOW.
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5If the input of a NOT gate is HIGH, what should its output be during validation?
Implementation and validation of logic gates
Easy
A.HIGH
B.Floating
C.LOW
D.Unchanged
Correct Answer: LOW
Explanation:
A NOT gate complements its input, so an input of produces an output of .
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6When does a two-input NAND gate produce a LOW output?
Implementation and validation of logic gates
Easy
A.When the inputs are different
B.When both inputs are LOW
C.When both inputs are HIGH
D.When either input is LOW
Correct Answer: When both inputs are HIGH
Explanation:
A NAND gate is the complement of an AND gate, so it is LOW only when both inputs are HIGH.
Incorrect! Try again.
7When does a two-input NOR gate produce a HIGH output?
Implementation and validation of logic gates
Easy
A.When the inputs are different
B.When both inputs are LOW
C.When both inputs are HIGH
D.When either input is HIGH
Correct Answer: When both inputs are LOW
Explanation:
A NOR gate is the complement of an OR gate, so it is HIGH only when both inputs are LOW.
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8What is the output of a two-input XOR gate when its inputs are different?
Implementation and validation of logic gates
Easy
A.Always HIGH
B.Always floating
C.Always LOW
D.Always unchanged
Correct Answer: Always HIGH
Explanation:
An XOR gate produces HIGH when its two inputs have different logic levels.
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9Which instrument can be used to check whether a digital signal is HIGH or LOW?
Implementation and validation of logic gates
Easy
A.Solder sucker
B.Screwdriver
C.Wire stripper
D.Logic probe
Correct Answer: Logic probe
Explanation:
A logic probe indicates the logic level present at a point in a digital circuit.
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10Why should unused logic-gate inputs not be left floating during validation?
Implementation and validation of logic gates
Easy
A.They may reduce the input count
B.They may cause unpredictable outputs
C.They may increase the wire length
D.They may change the gate symbol
Correct Answer: They may cause unpredictable outputs
Explanation:
A floating input can assume an uncertain logic level and cause unstable or incorrect outputs.
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11Which pair consists entirely of universal logic gates?
Design of logic gates using universal gates
Easy
A.NAND and NOR
B.OR and XOR
C.AND and OR
D.NOT and XOR
Correct Answer: NAND and NOR
Explanation:
NAND and NOR are universal because either type can be used to construct all basic logic gates.
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12How can a two-input NAND gate be used as a NOT gate for input ?
Design of logic gates using universal gates
Easy
A.Connect to both inputs
B.Connect the output to ground
C.Connect one input to ground
D.Leave both inputs unconnected
Correct Answer: Connect to both inputs
Explanation:
With both inputs connected to , the output is .
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13Which arrangement implements an AND gate using only NAND gates?
Design of logic gates using universal gates
Easy
A.One NAND gate followed by a NOR inverter
B.One NAND gate followed by a NAND inverter
C.One NOR gate followed by a NAND inverter
D.One OR gate followed by a NOR inverter
Correct Answer: One NAND gate followed by a NAND inverter
Explanation:
The first NAND produces , and the NAND inverter complements it to obtain .
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14To implement an OR gate using NAND gates, what is done before applying the inputs to the final NAND gate?
Design of logic gates using universal gates
Easy
A.Both inputs are grounded
B.Both inputs are added
C.Both inputs are inverted
D.Both inputs are disconnected
Correct Answer: Both inputs are inverted
Explanation:
By De Morgan's law, , so the inputs are inverted before the final NAND operation.
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15How can a two-input NOR gate be used as a NOT gate for input ?
Design of logic gates using universal gates
Easy
A.Connect to both inputs
B.Connect one input to HIGH
C.Connect the output to HIGH
D.Leave both inputs floating
Correct Answer: Connect to both inputs
Explanation:
With both inputs connected to , the output is .
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16Which arrangement implements an OR gate using only NOR gates?
Design of logic gates using universal gates
Easy
A.One NOR gate followed by a NAND inverter
B.One NAND gate followed by a NOR inverter
C.One AND gate followed by a NAND inverter
D.One NOR gate followed by a NOR inverter
Correct Answer: One NOR gate followed by a NOR inverter
Explanation:
The first NOR produces , and the NOR inverter complements it to obtain .
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17To implement an AND gate using NOR gates, what is done to inputs and before the final NOR operation?
Design of logic gates using universal gates
Easy
A.Both inputs are removed
B.Both inputs are shorted
C.Both inputs are inverted
D.Both inputs are grounded
Correct Answer: Both inputs are inverted
Explanation:
By De Morgan's law, , so both inputs are inverted first.
Incorrect! Try again.
18Which expression represents an OR gate implemented with NAND operations?
Design of logic gates using universal gates
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
De Morgan's law gives .
Incorrect! Try again.
19What does the term "universal gate" mean?
Design of logic gates using universal gates
Easy
A.A gate that operates without a power supply
B.A gate that can build any Boolean function
C.A gate that accepts only one input signal
D.A gate that always produces a HIGH output
Correct Answer: A gate that can build any Boolean function
Explanation:
A universal gate can be combined with gates of the same type to implement any Boolean function.
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20What is the minimum number of two-input NAND gates needed to implement a NOT gate?
Design of logic gates using universal gates
Easy
A.Four gates
B.One gate
C.Two gates
D.Three gates
Correct Answer: One gate
Explanation:
One NAND gate acts as a NOT gate when its two inputs are connected together.
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21During the validation of a two-input XOR gate, which output sequence should be observed for inputs ?
Implementation and validation of logic gates
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
An XOR gate produces HIGH only when its inputs are different, giving the sequence .
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22A two-input NAND gate produces HIGH for every tested input combination, including . Which fault best explains this result?
Implementation and validation of logic gates
Medium
A.Both inputs are shorted together
B.The output is stuck LOW
C.The output is stuck HIGH
D.The gate behaves as an XOR gate
Correct Answer: The output is stuck HIGH
Explanation:
A working NAND gate must produce LOW for . A continuously HIGH result indicates a stuck-HIGH output.
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23An AND gate is tested with inputs and , but its output is measured as HIGH. What should the expected output be?
Implementation and validation of logic gates
Medium
A.HIGH, because one input is HIGH
B.LOW, because both inputs are not HIGH
C.HIGH, because the inputs are different
D.LOW, because both inputs are not LOW
Correct Answer: LOW, because both inputs are not HIGH
Explanation:
An AND gate produces HIGH only when every input is HIGH. Therefore, .
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24A logic circuit gives the output sequence for inputs . Which gate has been implemented?
Implementation and validation of logic gates
Medium
A.XOR gate
B.XNOR gate
C.NOR gate
D.NAND gate
Correct Answer: XNOR gate
Explanation:
An XNOR gate produces HIGH when its inputs are equal, which occurs for and .
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25While validating a NOT gate, the input is changed from LOW to HIGH. What output transition confirms correct operation?
Implementation and validation of logic gates
Medium
A.LOW to HIGH
B.LOW to LOW
C.HIGH to HIGH
D.HIGH to LOW
Correct Answer: HIGH to LOW
Explanation:
A NOT gate complements its input, so an input transition from LOW to HIGH must produce an output transition from HIGH to LOW.
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26Two switches provide the inputs to an OR gate. An LED connected to the output should turn on under which condition?
Implementation and validation of logic gates
Medium
A.At least one switch is ON
B.Both switches have equal states
C.Exactly one switch is OFF
D.Both switches are OFF
Correct Answer: At least one switch is ON
Explanation:
An OR gate produces HIGH when one or both inputs are HIGH, so the LED turns on when at least one switch is ON.
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27A two-input NOR gate is operating correctly. For which input combination should its output LED turn on?
Implementation and validation of logic gates
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
A NOR gate produces HIGH only when all its inputs are LOW.
Incorrect! Try again.
28A circuit intended to implement produces LOW only for . What does this observation indicate?
Implementation and validation of logic gates
Medium
A.The XOR gate is functioning correctly
B.The NAND gate is functioning correctly
C.The AND gate is functioning correctly
D.The OR gate is functioning correctly
Correct Answer: The OR gate is functioning correctly
Explanation:
The truth table of an OR gate has a LOW output only for the input combination .
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29When checking a logic-gate truth table experimentally, why should every possible input combination be applied?
Implementation and validation of logic gates
Medium
A.To reduce the circuit propagation delay
B.To determine the package pin count
C.To verify the complete logical behavior
D.To increase the output voltage level
Correct Answer: To verify the complete logical behavior
Explanation:
Testing every input combination confirms that the circuit matches the expected truth table under all logical conditions.
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30An XOR gate has inputs and . If is permanently connected to logic HIGH, how does the output depend on ?
Implementation and validation of logic gates
Medium
A.The output is always HIGH
B.The output is always LOW
C.The output is
D.The output is equal to
Correct Answer: The output is
Explanation:
For an XOR gate, , so the circuit behaves as an inverter.
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31How can a two-input NAND gate be connected to function as a NOT gate?
Design of logic gates using universal gates
Medium
A.Connect one input permanently to LOW
B.Connect both inputs to the same signal
C.Connect the output back to one input
D.Connect the two inputs to complementary signals
Correct Answer: Connect both inputs to the same signal
Explanation:
Joining both NAND inputs gives .
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32What is the minimum number of two-input NAND gates required to implement a two-input AND gate?
Design of logic gates using universal gates
Medium
A.One NAND gate
B.Four NAND gates
C.Three NAND gates
D.Two NAND gates
Correct Answer: Two NAND gates
Explanation:
The first NAND produces , and a second NAND used as an inverter produces .
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33Which NAND-only arrangement implements the OR function ?
Design of logic gates using universal gates
Medium
A.NAND both inputs through a single gate
B.Invert both inputs, then NAND them
C.NAND both inputs, then invert the result
D.Invert one input, then NAND both inputs
Correct Answer: Invert both inputs, then NAND them
Explanation:
By De Morgan's law, . Two NAND inverters followed by a NAND gate implement this expression.
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34How can a NOR gate be used as an inverter?
Design of logic gates using universal gates
Medium
A.Apply complementary signals to the inputs
B.Connect the output to both inputs
C.Connect both inputs to the same signal
D.Connect one input permanently to HIGH
Correct Answer: Connect both inputs to the same signal
Explanation:
With both inputs connected to , the output is .
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35What is the minimum number of two-input NOR gates required to implement a two-input OR gate?
Design of logic gates using universal gates
Medium
A.Two NOR gates
B.Three NOR gates
C.Four NOR gates
D.One NOR gate
Correct Answer: Two NOR gates
Explanation:
The first NOR produces , and the second NOR acts as an inverter to produce .
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36The outputs of two NOR gates are and . If and are applied to another NOR gate, what is the final expression?
Design of logic gates using universal gates
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The final output is .
Incorrect! Try again.
37A NAND-only circuit first generates and then applies and to a NAND gate. Which function is produced?
Design of logic gates using universal gates
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The output is , which is also the implication function .
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38A standard two-input XOR circuit is constructed using only two-input NAND gates. How many NAND gates are required?
Design of logic gates using universal gates
Medium
A.Two NAND gates
B.Three NAND gates
C.Five NAND gates
D.Four NAND gates
Correct Answer: Four NAND gates
Explanation:
The common NAND-only XOR implementation uses four two-input NAND gates and requires no separate inverter.
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39A designer must implement using only NAND gates. Which simplification is most useful?
Design of logic gates using universal gates
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Using De Morgan's law, .
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40Why are NAND and NOR gates described as universal gates?
Design of logic gates using universal gates
Medium
A.Both types produce identical truth tables
B.Both types always use only two inputs
C.Either type can implement any Boolean function
D.Either type eliminates propagation delay
Correct Answer: Either type can implement any Boolean function
Explanation:
NOT, AND, and OR operations can each be built using only NAND gates or only NOR gates, allowing any Boolean function to be implemented.
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41A NAND-gate implementation is tested with inputs and . The measured output remains at logic 1 for approximately before changing to logic 0. Which conclusion is most appropriate?
Implementation and validation of logic gates
Hard
A.The circuit has a propagation delay of
B.The circuit is permanently stuck at logic 1
C.The circuit has an invalid truth table
D.The circuit implements an XOR gate
Correct Answer: The circuit has a propagation delay of
Explanation:
The temporary persistence of the previous output after an input transition indicates propagation delay. The final output of a NAND gate for is logic 0.
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42Using only 2-input NAND gates, which expression correctly implements the function ?
Design of logic gates using universal gates
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
A NAND inverter gives and . NANDing them produces .
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43A NOR-based circuit is intended to implement . During validation, the output is measured as logic 1 for and logic 0 for all other input combinations. What is the best diagnosis?
Implementation and validation of logic gates
Hard
A.The circuit is behaving as an AND gate
B.The input terminals are electrically shorted
C.The output should be logic 1 for only
D.The measured behavior matches a NOR gate
Correct Answer: The measured behavior matches a NOR gate
Explanation:
A NOR gate produces logic 1 only when every input is 0, exactly matching the observed validation results.
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44What is the minimum number of 2-input NAND gates required to implement without using any other gate type?
Design of logic gates using universal gates
Hard
A.Six
B.Four
C.Three
D.Five
Correct Answer: Four
Explanation:
A standard NAND-only XOR uses , , , and .
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45A 3-input NAND gate is validated by connecting its unused input permanently to logic 1 while applying signals to the other two inputs. Which 2-input function is observed at the output?
Implementation and validation of logic gates
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
With the third input fixed at 1, , so the 3-input NAND behaves as a 2-input NAND.
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46A NOR-only design must implement . Which expression is correct?
Design of logic gates using universal gates
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
NOR used as an inverter gives and . The final NOR gives .
Incorrect! Try again.
47Two cascaded inverters are used to validate logic-level restoration. The first inverter has and ; the second has and . What is the maximum rise-or-fall propagation delay through the chain?
Implementation and validation of logic gates
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
A transition passes through both gates, so the worst-case delay is the largest relevant sum: .
Incorrect! Try again.
48Using NAND gates, which implementation realizes a 2:1 multiplexer with select input , data inputs , and output ?
Design of logic gates using universal gates
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The two first-level NANDs create complemented product terms, and the final NAND performs their OR. The select inverter provides .
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49A NAND gate output is connected to an LED through a resistor to ground. The LED turns on when the output is high. Which input sequence tests all logically distinct output transitions most efficiently?
Implementation and validation of logic gates
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The NAND output changes only when entering or leaving . This sequence exercises both a high-to-low and a low-to-high transition while visiting all input states.
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50A NAND-only implementation of is required using only 2-input NAND gates. What is the minimum number of gates?
Design of logic gates using universal gates
Hard
A.Seven
B.Six
C.Four
D.Five
Correct Answer: Six
Explanation:
Implement as a NAND structure using three input inversions, two NAND stages, and one output inversion, requiring gates.
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51A CMOS NAND input is accidentally left floating during validation. The observed output changes when a nearby jumper wire is moved. Which explanation is most accurate?
Implementation and validation of logic gates
Hard
A.The output is guaranteed to remain high because CMOS inputs are ideal
B.The floating input picks up noise and has an undefined logic level
C.The NAND gate automatically converts the floating input to logic 0
D.The nearby wire changes the Boolean expression of the NAND gate
Correct Answer: The floating input picks up noise and has an undefined logic level
Explanation:
A floating CMOS input can be affected by capacitive coupling and leakage. It must be tied to a defined logic level during validation.
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52A NOR-only circuit implements . Which structural approach is valid using De Morgan transformations?
Design of logic gates using universal gates
Hard
A.NOR all four inputs, then invert only input
B.NOR each pair, then OR the two pair outputs directly
C.NOR each pair, then NOR the two pair outputs
D.Invert each input, then NOR all four inverted inputs
Correct Answer: NOR each pair, then NOR the two pair outputs
Explanation:
Let and . Then .
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53A NAND gate is powered from . Its input thresholds are and . A measured input is . How should this input be classified?
Implementation and validation of logic gates
Hard
A.Undefined logic level
B.Valid logic 0
C.Valid logic 1
D.Short-circuit current state
Correct Answer: Undefined logic level
Explanation:
The voltage lies between and , the undefined transition region. It cannot be reliably interpreted as either logic state.
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54A function is given by . Which simplified expression can be implemented efficiently with NAND gates?
Design of logic gates using universal gates
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The listed minterms correspond to odd parity between and , independent of . Thus .
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55A combinational circuit output briefly glitches when inputs change from to , even though both stable states should produce the same intended output. Which validation method best identifies the cause?
Implementation and validation of logic gates
Hard
A.Replace every gate with an inverter
B.Measure propagation delays along reconvergent paths
C.Test only the all-zero input combination
D.Increase the supply voltage until the glitch disappears
Correct Answer: Measure propagation delays along reconvergent paths
Explanation:
Different path delays can produce static or dynamic hazards when signals reconverge. Timing measurements reveal whether unequal delays cause the transient.
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56A NAND-only implementation of is required. Which expression uses the correct NAND arrangement?
Design of logic gates using universal gates
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The expression . NANDing with each input and NANDing the results implements the OR of and .
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57A 2-input NOR gate is connected with both inputs tied together to signal . What function should be verified at the output, and what test distinguishes it from an inverter wiring error?
Implementation and validation of logic gates
Hard
A.; test both inputs independently
B.; test only
C.; test only
D.; test both and
Correct Answer: ; test both and
Explanation:
Tying both NOR inputs to gives . Both input states are needed to validate inversion.
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58Which NAND-only expression implements the 3-input majority function, where the output is 1 when at least two inputs are 1?
Design of logic gates using universal gates
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The majority function is . The first-level NANDs generate complemented product terms, and the final multi-input NAND realizes their sum.
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59When validating a gate, an oscilloscope shows the output high level at with no load but only after connecting four gate inputs. Which parameter is most directly implicated?
Implementation and validation of logic gates
Hard
A.Boolean absorption law
B.Input truth-table ordering
C.Output drive capability and fan-out
D.Unused input inversion
Correct Answer: Output drive capability and fan-out
Explanation:
Connecting multiple inputs increases the load. The resulting high-level drop indicates that the driver may be exceeding its specified fan-out or output-current capability.
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60A NOR gate is to be built using NAND gates only. Which sequence correctly preserves the required polarity?
Design of logic gates using universal gates
Hard
A.NAND and , then invert the NAND output twice
B.NAND and , then connect the result directly to the output
C.Invert the NAND output once, then invert both inputs
D.Invert and with NANDs, then NAND the inverted signals
Correct Answer: Invert and with NANDs, then NAND the inverted signals
Explanation:
The construction is , which is OR rather than NOR; therefore the stated sequence does not preserve NOR polarity. A final NAND inversion is also required, so none of the listed options is correct.
Incorrect! Try again.
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