Boolean Algebra and Logic Gates Questions in English

(C) $1$. The first gate is a $NOR$ gate with inputs $A$ and $1$. The output of this $NOR$ gate is $\overline{A+1}$. Since $A+1 = 1$,the output is $\overline{1} = 0$.
$2$. The second gate is an $OR$ gate with inputs from the first gate (which is $0$) and input $B$. The output of this $OR$ gate is $0+B = B$.
$3$. The third gate is an $OR$ gate with inputs from the second gate (which is $B$) and input $C$. The output $Y$ is $B+C$.

(C) Let us analyze each gate:
$(A)$ $NAND$ gate with inputs $1, 1$: Output $X = \overline{1 \cdot 1} = \overline{1} = 0$.
$(B)$ $NOR$ gate with inputs $0, 1$: Output $X = \overline{0 + 1} = \overline{1} = 0$.
$(C)$ $NAND$ gate with inputs $0, 1$: Output $X = \overline{0 \cdot 1} = \overline{0} = 1$.
$(D)$ $XOR$ gate with inputs $0, 0$: Output $X = 0 \oplus 0 = 0$.
Thus,the gate with an output of $1$ is the $NAND$ gate with inputs $0$ and $1$.

(A) By observing the waveforms at different time intervals,we can construct the truth table:

Time Interval	Input $A$	Input $B$	Output $Y$
$0 - T_1$	$0$	$0$	$0$
$T_1 - T_2$	$0$	$1$	$0$
$T_2 - T_3$	$1$	$0$	$0$
$T_3 - T_4$	$1$	$1$	$1$

The truth table shows that the output $Y$ is $1$ only when both inputs $A$ and $B$ are $1$. This is the characteristic behavior of an $AND$ gate.

(B) An $AND$ gate performs the logical multiplication operation. Its output is $1$ only if all its inputs are $1$.
In terms of switching circuits,an $AND$ gate is equivalent to two switches connected in series.
If both switches are closed (input $1$),current flows (output $1$). If either switch is open (input $0$),the circuit is broken (output $0$).

(C) To identify the logic gate,we analyze the truth table derived from the given voltage waveforms:

$A$	$B$	$Y$
$1$	$1$	$0$
$0$	$0$	$1$
$0$	$1$	$1$
$1$	$0$	$1$

From the truth table,we observe that the output $Y$ is $0$ only when both inputs $A$ and $B$ are $1$. In all other cases,the output is $1$. This behavior corresponds to the $NAND$ gate,which performs the operation $Y = \overline{AB}$.

(D) $NAND$ gate is formed by connecting a $NOT$ gate to the output of an $AND$ gate.
In the given options,option $D$ shows an $AND$ gate followed by a $NOT$ gate,which is the standard representation of a $NAND$ gate.

(B) $NAND$ gate produces an output of $0$ only when both inputs are $1$. Otherwise,it produces an output of $1$.
Based on the input waveforms for $A$ and $B$:
- For $t = 0$ to $2 \ s$: $A=0, B=0 \implies Y=1$.
- For $t = 2$ to $4 \ s$: $A=1, B=0 \implies Y=1$.
- For $t = 4$ to $6 \ s$: $A=0, B=0 \implies Y=1$.
- For $t = 6$ to $8 \ s$: $A=1, B=1 \implies Y=0$.
- For $t > 8 \ s$: $A=0, B=0 \implies Y=1$.
Comparing this sequence $(1, 1, 1, 0, 1)$ with the given options,the waveform in option $B$ matches this output.

(B) The given circuit consists of two $NAND$ gates connected in series.
Let the inputs to the first $NAND$ gate be $A$ and $B$. The output of the first $NAND$ gate is $X = \overline{A \cdot B}$.
This output $X$ acts as the input to the second $NAND$ gate. Since both inputs of the second $NAND$ gate are connected to $X$, its output $Y$ is given by $Y = \overline{X \cdot X} = \overline{X}$.
Substituting the value of $X$, we get $Y = \overline{(\overline{A \cdot B})} = A \cdot B$.
The expression $Y = A \cdot B$ represents the logic function of an $AND$ gate.
Therefore, the given circuit performs the function of an $AND$ gate.

(D) $NOT$ gate is an inverter that requires an active component like a transistor to perform the inversion logic. $A$ diode is a passive,unidirectional device that allows current to flow in only one direction and cannot perform the logical inversion required for a $NOT$ gate.
Furthermore,a diode does not introduce a $180^o$ phase shift between the input and output voltages. Therefore,both the Assertion and the Reason are incorrect.

(A) $NAND$ and $NOR$ gates are known as universal gates or digital building blocks.
This is because any logic gate,such as $AND, OR, NOT, XOR,$ or $XNOR$,can be constructed using only $NAND$ gates or only $NOR$ gates.
Since the Assertion states they are building blocks and the Reason correctly explains that they can produce all other basic or complex gates,the Reason is the correct explanation of the Assertion.

(C) In the given circuit,the switches $A$ and $B$ are connected in parallel with the ground. When a switch is at position $0$,it is open,and when it is at position $1$,it is closed (connected to ground).
$1$. If both $A=0$ and $B=0$,both switches are open. The current flows through the $LED$ $(Y)$,so $Y=1$.
$2$. If $A=0$ and $B=1$,switch $B$ is closed (grounded). The current flows through the $LED$ $(Y)$,so $Y=1$.
$3$. If $A=1$ and $B=0$,switch $A$ is closed (grounded). The current flows through the $LED$ $(Y)$,so $Y=1$.
$4$. If $A=1$ and $B=1$,both switches are closed (grounded). The current bypasses the $LED$ through the switches to the ground,so $Y=0$.
The truth table is:

$A$	$B$	$Y$
$0$	$0$	$1$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$0$

This truth table corresponds to the $NAND$ gate operation.

(A) In the given circuit,switches $A$ and $B$ are connected in parallel. The output $Y$ is the state of the $LED$.
When both switches $A$ and $B$ are at position $0$ (connected to ground),the $LED$ is connected across the $+6V$ supply through resistor $R$,so the $LED$ glows $(Y=1)$.
If either switch $A$ or $B$ is moved to position $1$,the circuit is shorted to ground,and the potential difference across the $LED$ becomes zero,so the $LED$ does not glow $(Y=0)$.
The truth table is:

$A$	$B$	$Y$
$0$	$0$	$1$
$0$	$1$	$0$
$1$	$0$	$0$
$1$	$1$	$0$

This truth table corresponds to the $NOR$ gate.

(B) logic gate is considered reversible if the input data can be uniquely recovered from the output data.
In a $NOT$ gate,the output is the complement of the input $(Y = \bar{A})$. If the output is $0$,the input must have been $1$,and if the output is $1$,the input must have been $0$. Thus,the input can be uniquely determined from the output.
Other gates like $OR$,$NOR$,$AND$,and $NAND$ have two inputs,meaning there are $2^2 = 4$ possible input combinations,but only $2$ possible output states ($0$ or $1$). Since multiple input combinations can produce the same output,these gates are not reversible.

(D) Let the inputs be $A=1$ and $B=0$.
$1$. The input $A=1$ passes through a $NOT$ gate,becoming $0$.
$2$. This $0$ is fed into the top $NAND$ gate (both inputs are $0$),so the output of the top $NAND$ gate is $\overline{0 \cdot 0} = 1$.
$3$. The $0$ from the first $NOT$ gate also passes through another $NOT$ gate,becoming $1$.
$4$. This $1$ is fed into the middle $NAND$ gate.
$5$. The bottom $NAND$ gate has inputs $B=0$ and the feedback from $Y$.
$6$. Analyzing the circuit,the output $Y$ is connected to the output of the top and middle $NAND$ gates.
$7$. Given the configuration,the output $Y$ stabilizes at $1$.

(C) The circuit consists of two switches $A$ and $B$ connected to diodes,which act as an $OR$ gate input stage. When either switch $A$ or $B$ is closed (logic $1$),the base of the $NPN$ transistor receives a high voltage,causing it to conduct. When the transistor conducts,the collector voltage drops to near $0 \ V$ (logic $0$). If both switches are open (logic $0$),the base is grounded through the resistor,the transistor is in cutoff $(OFF)$,and the output $Y$ is pulled up to $+5 \ V$ (logic $1$). This behavior corresponds to a $NOR$ gate,where $Y = \overline{A+B}$. According to De Morgan's theorem,$\overline{A+B} = \overline{A} \cdot \overline{B}$.

$A$	$B$	$Y$
$0$	$0$	$1$
$1$	$0$	$0$
$0$	$1$	$0$
$1$	$1$	$0$

(N/A) For an $OR$ gate,the output $Y = A + B$. This means the output is $1$ if either input $A$ or $B$ is $1$,and $0$ only if both inputs are $0$.
$1$. For $t < t_{1}$: $A=0, B=0$,so $Y = 0 + 0 = 0$.
$2$. For $t_{1}$ to $t_{2}$: $A=1, B=0$,so $Y = 1 + 0 = 1$.
$3$. For $t_{2}$ to $t_{3}$: $A=1, B=1$,so $Y = 1 + 1 = 1$.
$4$. For $t_{3}$ to $t_{4}$: $A=0, B=1$,so $Y = 0 + 1 = 1$.
$5$. For $t_{4}$ to $t_{5}$: $A=0, B=0$,so $Y = 0 + 0 = 0$.
$6$. For $t_{5}$ to $t_{6}$: $A=1, B=0$,so $Y = 1 + 0 = 1$.
$7$. For $t > t_{6}$: $A=0, B=1$,so $Y = 0 + 1 = 1$.
Thus,the output waveform $Y$ follows the logic of the $OR$ gate as shown in the figure.

(N/A) For an $AND$ gate,the output $Y$ is $1$ only when both inputs $A$ and $B$ are $1$. Otherwise,the output $Y$ is $0$. Based on the provided input waveforms,we analyze the intervals:

Interval	Inputs $(A, B)$	Output $(Y = A \cdot B)$
$t \leq t_{1}$	$A=0, B=0$	$Y=0$
$t_{1}$ to $t_{2}$	$A=1, B=0$	$Y=0$
$t_{2}$ to $t_{3}$	$A=1, B=1$	$Y=1$
$t_{3}$ to $t_{4}$	$A=0, B=1$	$Y=0$
$t_{4}$ to $t_{5}$	$A=0, B=0$	$Y=0$
$t_{5}$ to $t_{6}$	$A=1, B=0$	$Y=0$
$t > t_{6}$	$A=0, B=0$	$Y=0$

Thus,the output waveform remains at a low state $(0)$ for all intervals except between $t_{2}$ and $t_{3}$,where it goes to a high state $(1)$.

(N/A) The truth table for a $NAND$ gate is:
If $A=0, B=0$,then $Y=1$.
If $A=0, B=1$,then $Y=1$.
If $A=1, B=0$,then $Y=1$.
If $A=1, B=1$,then $Y=0$.
Based on the input waveforms:
For $t < t_{1}$; $A=1, B=1$; hence $Y=0$.
For $t_{1}$ to $t_{2}$; $A=0, B=0$; hence $Y=1$.
For $t_{2}$ to $t_{3}$; $A=0, B=1$; hence $Y=1$.
For $t_{3}$ to $t_{4}$; $A=1, B=0$; hence $Y=1$.
For $t_{4}$ to $t_{5}$; $A=1, B=1$; hence $Y=0$.
For $t_{5}$ to $t_{6}$; $A=0, B=0$; hence $Y=1$.
For $t > t_{6}$; $A=1, B=1$; hence $Y=0$.

(N/A) and $B$ are the inputs and $Y$ is the output of the given circuit. The left part of the circuit is a $NOR$ gate,and the right part is a $NOT$ gate.
The output of the $NOR$ gate is $\overline{A+B}$.
This output serves as the input for the $NOT$ gate. The output of the $NOT$ gate is $\overline{\overline{A+B}} = A+B$.
Therefore,$Y = A+B$,which is the Boolean expression for an $OR$ gate. Hence,this circuit functions as an $OR$ gate.
$(b)$ $A$ and $B$ are the inputs and $Y$ is the output. The inputs $A$ and $B$ first pass through two $NOT$ gates,producing $\overline{A}$ and $\overline{B}$ respectively.
These are then fed as inputs to a $NOR$ gate.
The output of the $NOR$ gate is $Y = \overline{\overline{A} + \overline{B}}$.
Using De Morgan's theorem,$\overline{\overline{A} + \overline{B}} = \overline{\overline{A}} \cdot \overline{\overline{B}} = A \cdot B$.
Therefore,$Y = A \cdot B$,which is the Boolean expression for an $AND$ gate. Hence,this circuit functions as an $AND$ gate.

(N/A) In the given circuit,the input $A$ is connected to both inputs of the $NAND$ gate. Let the inputs be $A_1 = A$ and $A_2 = A$. The output $Y$ of a $NAND$ gate is given by $Y = \overline{A_1 \cdot A_2}$.
Substituting the inputs,we get:
$Y = \overline{A \cdot A}$
Using the Boolean identity $A \cdot A = A$,we have:
$Y = \overline{A}$
The truth table for this operation is:

$A$	$Y = \overline{A}$
$0$	$1$
$1$	$0$

Since the output $Y$ is the complement of the input $A$,this circuit functions as a $NOT$ gate.

(N/A) In both the given circuits,$A$ and $B$ are the inputs and $Y$ is the output.
$(a)$ The output of the first $NAND$ gate is $\overline{A \cdot B}$. This output is fed into a second $NAND$ gate where both inputs are connected together. $A$ $NAND$ gate with both inputs connected acts as a $NOT$ gate. Therefore,the final output is $Y = \overline{(\overline{A \cdot B})} = A \cdot B$. Hence,this circuit functions as an $AND$ gate.
$(b)$ In this circuit,the first two $NAND$ gates have their inputs connected together,so they act as $NOT$ gates. The outputs are $\overline{A}$ and $\overline{B}$ respectively. These are then fed into a third $NAND$ gate. The final output is $Y = \overline{\overline{A} \cdot \overline{B}}$. By De Morgan's theorem,$\overline{\overline{A} \cdot \overline{B}} = \overline{\overline{A}} + \overline{\overline{B}} = A + B$. Hence,this circuit functions as an $OR$ gate.

(N/A) Let $A$ and $B$ be the inputs of the given circuit. The output of the first $NOR$ gate is $\overline{A+B}$.
It can be observed from the figure that the output of the first $NOR$ gate acts as the input for both terminals of the second $NOR$ gate.
Therefore,the final output $Y$ is given by:
$Y = \overline{(\overline{A+B}) + (\overline{A+B})}$
Using the Boolean identity $\overline{X+X} = \overline{X}$,we get:
$Y = \overline{(\overline{A+B})} = A+B$
The truth table for this operation is:

$A$	$B$	$Y (= A + B)$
$0$	$0$	$0$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$1$

This is the truth table of an $OR$ gate. Hence,this circuit functions as an $OR$ gate.

(N/A) For circuit $(a)$:
Input $A$ is connected to both inputs of the $NOR$ gate. The output $Y$ is given by $Y = \overline{A+A} = \bar{A}$.
The truth table is:

$A$	$Y(=\bar{A})$
$0$	$1$
$1$	$0$

This is the truth table of a $NOT$ gate. Hence,circuit $(a)$ functions as a $NOT$ gate.
For circuit $(b)$:
$A$ and $B$ are inputs. The first two $NOR$ gates act as $NOT$ gates,producing outputs $\bar{A}$ and $\bar{B}$.
These are inputs to the final $NOR$ gate. The output $Y$ is:
$Y = \overline{\bar{A}+\bar{B}} = \overline{\overline{A \cdot B}} = A \cdot B$ (using De Morgan's theorem).
The truth table is:

$A$	$B$	$Y(=A \cdot B)$
$0$	$0$	$0$
$0$	$1$	$0$
$1$	$0$	$0$
$1$	$1$	$1$

This is the truth table of an $AND$ gate. Hence,circuit $(b)$ functions as an $AND$ gate.

(C) The given Boolean expression is $Y = \bar{A} \cdot B + A \cdot \bar{B}$.
This is the standard Boolean expression for an $XOR$ (Exclusive $OR$) gate.
To construct this using basic gates:
$1$. Use two $NOT$ gates to obtain $\bar{A}$ and $\bar{B}$ from inputs $A$ and $B$.
$2$. Use two $AND$ gates: the first $AND$ gate takes inputs $\bar{A}$ and $B$ to produce $\bar{A} \cdot B$. The second $AND$ gate takes inputs $A$ and $\bar{B}$ to produce $A \cdot \bar{B}$.
$3$. Use one $OR$ gate to combine the outputs of the two $AND$ gates: $Y = (\bar{A} \cdot B) + (A \cdot \bar{B})$.

(N/A) In electronic circuits such as amplifiers and oscillators,signals exist in the form of continuous changes in current or voltage over time. Such signals are called continuous or analog signals.
$A$ typical analog signal is shown in figures $(a)$ and $(b)$.
If a current or voltage signal has only two discrete values (minimum and maximum),the signal is called a digital signal.
The signal in figure $(c)$ is in the form of a waveform or pulse,which has only two signal levels.
$A$ binary system for displaying such signals is convenient. Binary uses only two digits,$0$ and $1$. The maximum value (e.g.,$5 \ V$) of voltage is expressed as $1$,and the minimum value (e.g.,$0 \ V$) is expressed as $0$.
The input and output voltages indicated by these two values ($0$ and $1$) in a digital circuit are valid.

(N/A) There are two types of systems used in logic circuits:
$(1)$ Positive logic system: In this type of system,the higher voltage level is assigned as logic $'1'$ (high) and the lower voltage level is assigned as logic $'0'$ (low).
$(2)$ Negative logic system: In this type of system,the lower voltage level is assigned as logic $'1'$ (high) and the higher voltage level is assigned as logic $'0'$ (low).

(N/A) logic gate is a digital circuit that implements a specific logical function. It operates based on a defined logical relationship between the input and output voltages.
They are called 'gates' because they control the flow of information based on logical conditions.
The five common logic gates are $NOT$,$AND$,$OR$,$NAND$,and $NOR$. Each logic gate is represented by a unique symbol,and its behavior is defined by its truth table.
$A$ truth table is a table that lists all possible input combinations and their corresponding output values for a logic gate.
The Boolean expression provides the mathematical representation of the operation performed by a specific logic gate.
Logic gates are constructed using semiconductor devices,such as diodes and transistors.

(N/A) $NOT$ Gate:
Symbol: Shown in figure $(a)$. It has one input $A$ and one output $Y$.
Function: If input is $1$,output is $0$. If input is $0$,output is $1$. It acts as an inverter.
Boolean Equation: $Y = \overline{A}$.
Truth Table:
| Input $A$ | Output $Y$ |
| :--- | :--- |
| $0$ | $1$ |
| $1$ | $0$ |
$OR$ Gate:
Symbol: Shown in figure $(a)$. It has two or more inputs $(A, B)$ and one output $Y$.
Function: The output is $1$ if at least one input is $1$.
Boolean Equation: $Y = A + B$.
Truth Table:
| Input $A$ | Input $B$ | Output $Y$ |
| :--- | :--- | :--- |
| $0$ | $0$ | $0$ |
| $0$ | $1$ | $1$ |
| $1$ | $0$ | $1$ |
| $1$ | $1$ | $1$ |
$AND$ Gate:
Symbol: Shown in figure $(a)$. It has two or more inputs $(A, B)$ and one output $Y$.
Function: The output is $1$ only if both inputs $A$ and $B$ are $1$.
Boolean Equation: $Y = A \bullet B$.
Truth Table:
| Input $A$ | Input $B$ | Output $Y$ |
| :--- | :--- | :--- |
| $0$ | $0$ | $0$ |
| $0$ | $1$ | $0$ |
| $1$ | $0$ | $0$ |
| $1$ | $1$ | $1$ |

(N/A) $NAND$ gate is constructed by combining an $AND$ gate and a $NOT$ gate. $\therefore AND + NOT = NAND$.
This is an $AND$ gate followed by a $NOT$ gate. The symbol of this gate is represented by an $AND$ gate with a small circle at the output. It has two inputs $(A, B)$ and one output $(Y)$.
Function of $NAND$ gate: The output is $'0'$ only when all inputs are $'1'$,otherwise the output is $'1'$.
Boolean equation: $Y = \overline{A \cdot B}$.
Truth Table for $NAND$ gate:
| $A$ | $B$ | $Y$ |
|---|---|---|
| $0$ | $0$ | $1$ |
| $0$ | $1$ | $1$ |
| $1$ | $0$ | $1$ |
| $1$ | $1$ | $0$ |
$NAND$ gates are called universal gates because they can be used to realize other basic gates like $OR$,$AND$,and $NOT$.
$NOR$ gate is constructed by combining an $OR$ gate and a $NOT$ gate. $\therefore OR + NOT = NOR$.
This is an $OR$ gate followed by a $NOT$ gate. The symbol is an $OR$ gate with a small circle at the output.
Function of $NOR$ gate: The output is $'1'$ only when all inputs are $'0'$,otherwise the output is $'0'$.
Boolean equation: $Y = \overline{A + B}$.
Truth Table for $NOR$ gate:
| $A$ | $B$ | $Y$ |
|---|---|---|
| $0$ | $0$ | $1$ |
| $0$ | $1$ | $0$ |
| $1$ | $0$ | $0$ |
| $1$ | $1$ | $0$ |
$NOR$ gates are also called universal gates.

(N/A) digital signal is a signal that can take only two discrete values,typically represented as $0$ (low) and $1$ (high). These values correspond to specific voltage levels in electronic circuits.
The mathematical method used to express digital signals is known as $Boolean$ $Algebra$. It is a branch of algebra in which the values of the variables are the truth values $true$ and $false$,usually denoted by $1$ and $0$ respectively.

(B) logic gate is a fundamental building block of a digital circuit. It is a digital circuit that performs a logical operation on one or more binary inputs and produces a single binary output. The relationship between the input and output is based on certain logical rules,which are represented by Boolean algebra. Common examples include $AND$,$OR$,$NOT$,$NAND$,and $NOR$ gates.

(C) logic gate is a digital circuit that performs a logical operation on one or more inputs and produces a single output.
Among the basic logic gates,the $NOT$ gate is the only one that has exactly one input and one output.
The $NOT$ gate performs the operation of inversion,meaning if the input is $0$,the output is $1$,and if the input is $1$,the output is $0$.
Therefore,the correct option is $C$.

(C) An inverter gate is a logic gate that implements logical negation. It takes a single input and produces the opposite output. The $NOT$ gate is known as an inverter because if the input is $1$ (high),the output is $0$ (low),and if the input is $0$ (low),the output is $1$ (high). Therefore,the $NOT$ gate is the inverter gate.

(A) The $AND$ gate is a basic digital logic gate that implements logical conjunction.
It produces an output of $1$ (high) only if all its inputs are $1$ (high).
If any of the inputs are $0$ (low),the output will be $0$ (low).
Mathematically,for two inputs $A$ and $B$,the output $Y$ is given by $Y = A \cdot B$.

(N/A) $NAND$ gate is a universal logic gate that produces a low output $(0)$ only when all its inputs are high $(1)$. Otherwise, it produces a high output $(1)$.
For a two-input $NAND$ gate with inputs $A$ and $B$ and output $Y$, the Boolean expression is $Y = \overline{A \cdot B}$.
The truth table is as follows:
| Input $A$ | Input $B$ | Output $Y$ |
| :--- | :--- | :--- |
| $0$ | $0$ | $1$ |
| $0$ | $1$ | $1$ |
| $1$ | $0$ | $1$ |
| $1$ | $1$ | $0$ |

(A) $NOR$ gate is a universal logic gate that performs the logical $OR$ operation followed by the logical $NOT$ operation.
Therefore, the output of a $NOR$ gate is the complement of the output of an $OR$ gate.
Mathematically, if the inputs are $A$ and $B$, the output $Y$ is given by $Y = \overline{A + B}$.
This is equivalent to an $OR$ gate connected in series with a $NOT$ gate.

(N/A) Basic gates are the fundamental building blocks of digital logic circuits. They perform simple logical operations. The three basic logic gates are:
$1$. $NOT$ gate: It acts as an inverter,producing an output that is the inverse of the input.
$2$. $AND$ gate: It produces a high output $(1)$ only if all its inputs are high $(1)$.
$3$. $OR$ gate: It produces a high output $(1)$ if at least one of its inputs is high $(1)$.
Universal gates are logic gates that can be used to implement any Boolean function or construct any other type of logic gate without the need for other types of gates. The two universal gates are:
$1$. $NAND$ gate: It is a combination of an $AND$ gate followed by a $NOT$ gate.
$2$. $NOR$ gate: It is a combination of an $OR$ gate followed by a $NOT$ gate.

(N/A) In this scenario,the gate needs to open if input $A$ is high $OR$ input $B$ is high. This logic corresponds to an $OR$ gate. $A$ diode-based $OR$ gate circuit is used here,where the cathodes of two diodes are connected together to the output $C$,and the anodes serve as inputs $A$ and $B$. When either input $A$ or $B$ is at a high potential (logic $1$),the corresponding diode becomes forward-biased,allowing current to flow and raising the potential at output $C$ to a high state (logic $1$). If both are high,the output remains high. If both are low (logic $0$),the output is low.
Truth table for $OR$ gate:

Inputs $(A, B)$	Output $(C)$
$0, 0$	$0$
$0, 1$	$1$
$1, 0$	$1$
$1, 1$	$1$

(N/A) To obtain a $NOT$ gate using an $NPN$ transistor, the transistor is used in the common-emitter configuration. The input $A$ is applied to the base through a resistor $R_B$, and the output $C$ is taken from the collector terminal.
For case $I$:
When $A=0$ (low input), the base current $I_B = 0$. Consequently, the collector current $I_C = \beta I_B = 0$. Using Kirchhoff's voltage law in the output loop, $V_{CC} = I_C R_C + V_{CE}$. Since $I_C = 0$, $V_{CE} = V_{CC} = 5 \text{ V}$. Thus, the output $C = 1$ (high).
For case $II$:
When $A=1$ (high input), the transistor is driven into saturation, making $I_B$ and $I_C$ maximum. In this state, the voltage drop across $R_C$ is nearly equal to $V_{CC}$, so $V_{CE} \approx 0$. Thus, the output $C = 0$ (low).
This confirms the operation $C = \bar{A}$, which is the characteristic of a $NOT$ gate.

Input $A$	Output $C$
$0$	$1$
$1$	$0$

(N/A) The given circuit is an $OR$ gate.
When input $A$ or input $B$ is at a low potential $(0)$,the corresponding diode ($D_1$ or $D_2$) becomes forward-biased and conducts,pulling the output $V_0$ to a low potential $(0)$.
When both inputs $A$ and $B$ are at a high potential $(1)$,both diodes are reverse-biased and do not conduct. The output $V_0$ is then pulled to the high potential $(+5 \text{ V})$ through the resistor.
Thus,the circuit acts as an $OR$ gate.
The truth table is as follows:
| $A$ | $B$ | $V_0 = A + B$ |
|---|---|---|
| $0$ | $0$ | $0$ |
| $0$ | $1$ | $1$ |
| $1$ | $0$ | $1$ |
| $1$ | $1$ | $1$ |

(N/A) For the circuit in figure $(3)$:
$Y_1 = \overline{A}$,$Y_2 = \overline{B}$.
The output of the $NAND$ gate is $Y = \overline{Y_1 \cdot Y_2} = \overline{\overline{A} \cdot \overline{B}} = A + B$.
The final $NOT$ gate gives $C_1 = \overline{Y} = \overline{A + B}$.
This is the Boolean expression for a $NOR$ gate.
Based on the input signals $A$ and $B$,the output $C_1$ is high $(1)$ only when both $A$ and $B$ are low $(0)$,which occurs in the interval $t = 4 \text{ s}$ to $t = 5 \text{ s}$.
For the circuit in figure $(4)$:
$Y_1 = \overline{A}$,$Y_2 = \overline{B}$.
The output of the $NOR$ gate is $C_2 = \overline{Y_1 + Y_2} = \overline{\overline{A} + \overline{B}} = \overline{\overline{A \cdot B}} = A \cdot B$.
This is the Boolean expression for an $AND$ gate.
Based on the input signals $A$ and $B$,the output $C_2$ is high $(1)$ only when both $A$ and $B$ are high $(1)$,which occurs in the interval $t = 1 \text{ s}$ to $t = 2 \text{ s}$.

(C) The provided truth table shows that the output $Y$ is $1$ only when the inputs $A$ and $B$ are different (i.e., $0,1$ or $1,0$).
This is the characteristic truth table of an $XOR$ (Exclusive-$OR$) gate.
The Boolean expression for this gate is $Y = A \oplus B = \overline{A} \cdot B + A \cdot \overline{B}$.
As shown in the circuit diagram, the output $Y$ is obtained by combining the outputs of two $AND$ gates ($Y_1 = \overline{A} \cdot B$ and $Y_2 = A \cdot \overline{B}$) through an $OR$ gate, which results in $Y = Y_1 + Y_2 = \overline{A} \cdot B + A \cdot \overline{B}$.

(N/A) The expression $Y = \bar{A} \cdot B + A \cdot \bar{B}$ represents the $XOR$ (Exclusive $OR$) logic operation.
To construct this using basic gates:
$1$. Use two $NOT$ gates to obtain the complements $\bar{A}$ and $\bar{B}$ from inputs $A$ and $B$.
$2$. Use two $AND$ gates: the first $AND$ gate takes inputs $\bar{A}$ and $B$ to produce $\bar{A} \cdot B$. The second $AND$ gate takes inputs $A$ and $\bar{B}$ to produce $A \cdot \bar{B}$.
$3$. Use one $OR$ gate to combine the outputs of the two $AND$ gates,resulting in $Y = \bar{A} \cdot B + A \cdot \bar{B}$.

(C) Let the output of the $NAND$ gate be $P = \overline{AB}$.
Let the output of the $OR$ gate be $Q = A+B$.
The $AND$ gate takes $P$ and $Q$ as inputs,so its output is $R = P \cdot Q = (\overline{AB}) \cdot (A+B)$.
Using Boolean algebra: $R = (\bar{A} + \bar{B}) \cdot (A+B) = \bar{A}A + \bar{A}B + \bar{B}A + \bar{B}B = 0 + \bar{A}B + A\bar{B} + 0 = A \oplus B$ ($XOR$ operation).
The final $NOR$ gate takes $P$ and $R$ as inputs,so $Z = \overline{P+R} = \overline{(\overline{AB}) + (A \oplus B)}$.
Let's evaluate for each input $(A, B)$:
$1$. $(1, 0): P = \overline{1 \cdot 0} = 1, Q = 1+0 = 1, R = 1 \cdot 1 = 1. Z = \overline{1+1} = 0$.
$2$. $(0, 0): P = \overline{0 \cdot 0} = 1, Q = 0+0 = 0, R = 1 \cdot 0 = 0. Z = \overline{1+0} = 0$.
$3$. $(1, 1): P = \overline{1 \cdot 1} = 0, Q = 1+1 = 1, R = 0 \cdot 1 = 0. Z = \overline{0+0} = 1$.
$4$. $(0, 1): P = \overline{0 \cdot 1} = 1, Q = 0+1 = 1, R = 1 \cdot 1 = 1. Z = \overline{1+1} = 0$.
The outputs are $0, 0, 1, 0$.

(A) The circuit consists of three $NOT$ gates (implemented using $NOR$ gates with shorted inputs) followed by a $NOR$ gate.
Let the inputs be $A, B, C$.
The outputs of the three $NOT$ gates are $A' = \bar{A}$,$B' = \bar{B}$,and $C' = \bar{C}$.
These are fed into a $NOR$ gate. The final output $Y$ is given by:
$Y = \overline{A' + B' + C'} = \overline{\bar{A} + \bar{B} + \bar{C}}$.
Using De Morgan's theorem,$\overline{\bar{A} + \bar{B} + \bar{C}} = \overline{\bar{A}} \cdot \overline{\bar{B}} \cdot \overline{\bar{C}} = A \cdot B \cdot C$.
Thus,the circuit performs the $AND$ operation.

(B) The given circuit consists of two $NOT$ gates followed by a $NAND$ gate.
The inputs to the $NAND$ gate are $\bar{A}$ and $\bar{B}$.
The output $Y$ of the $NAND$ gate is given by the Boolean expression:
$Y = \overline{\bar{A} \cdot \bar{B}}$
Using De Morgan's theorem,$\overline{X \cdot Y} = \bar{X} + \bar{Y}$.
Therefore,$Y = \overline{\bar{A}} + \overline{\bar{B}} = A + B$.
This represents an $OR$ gate operation.
By analyzing the input waveforms for $A$ and $B$ at different time intervals,we can determine the output $Y = A + B$:
- For $t < 5$,$A=0, B=1 \implies Y=1$.
- For $5 < t < 7$,$A=1, B=1 \implies Y=1$.
- For $7 < t < 10$,$A=0, B=0 \implies Y=0$.
- For $10 < t < 15$,$A=1, B=1 \implies Y=1$.
- For $15 < t < 17$,$A=0, B=0 \implies Y=0$.
- For $17 < t < 19$,$A=1, B=0 \implies Y=1$.
- For $t > 19$,$A=0, B=1 \implies Y=1$.
Comparing this with the given options,the waveform corresponding to the $OR$ operation is represented by option $B$.

(B) The given circuit consists of two $NOT$ gates followed by a $NOR$ gate.
Let the inputs be $A$ and $B$.
The outputs of the two $NOT$ gates are $\bar{A}$ and $\bar{B}$.
These are the inputs to the $NOR$ gate.
The output $Y$ of the $NOR$ gate is given by the Boolean expression:
$Y = \overline{\bar{A} + \bar{B}}$
Using De Morgan's theorem,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.
Thus,the circuit acts as an $AND$ gate.
The truth table for an $AND$ gate is:

$A$	$B$	$Y$
$0$	$0$	$0$
$0$	$1$	$0$
$1$	$0$	$0$
$1$	$1$	$1$

Therefore,the correct option is $B$.