Step 1: Identify the relationship in Group-I.

In Group-I, the words are listed in decreasing order of intensity:
Abuse \( \rightarrow \) Insult \( \rightarrow \) Ridicule. Abuse is the most intense, followed by Insult, and Ridicule is the least intense.

Step 2: Identify the relationship in Group-II.

Group-II must follow a similar pattern of decreasing intensity. The words listed are:
________ \( \rightarrow \) Praise \( \rightarrow \) Appreciate.
Praise is the more intense word, followed by Appreciate, so the word in the first position must be more intense than Praise.

Step 3: Analyze the options.

(A) Extol: This word means to praise highly, which fits the highest intensity, making it the best choice.

(B) Prize: This word doesn't match the intensity pattern of the words in Group-II.

(C) Appropriate: This word doesn't fit the pattern of decreasing intensity.

(D) Espouse: This word means to adopt or support, but it doesn’t convey a higher level of praise than Praise, so it’s not suitable. Quick Tip: In analogy questions, pay attention to the intensity or degree of the words in both groups to identify the correct pattern.

Step 1: Analyze the structure of the sentence.

The sentence begins with "Had I learnt acting as a child," which indicates a hypothetical situation in the past. The phrase is a third conditional sentence, which is used to express unreal past situations and their possible outcomes.

Step 2: Understand the choices.

(A) will be: This option suggests a future possibility, but the sentence is about a past unreal condition, so it is incorrect.

(B) can be: This implies a present or future possibility, which does not fit the unreal past condition.

(C) am going to be: This suggests a future intention, which doesn’t fit the context of an unreal past condition.

(D) could have been: This is the correct choice, as it expresses a hypothetical outcome in the past, matching the structure of the third conditional.

Step 3: Conclude.

Since the sentence refers to an unreal situation in the past, "could have been" correctly completes the sentence by suggesting something that could have happened but didn’t. Quick Tip: In conditional sentences with unreal past situations, use "could have been" or "would have been" to indicate hypothetical outcomes.

Step 1: Using the given condition that each frequency is \( \sqrt[12]{2} \) times the frequency of the previous note.


The ratio of the frequencies of any two notes can be expressed as: \[ Frequency ratio = \left( \sqrt[12]{2} \right)^n \]
where \( n \) is the number of steps between the two notes.

Step 2: Finding the ratio of frequencies of F\# and C.

Since F\# is 6 steps away from C in the sequence, we have: \[ Ratio of frequencies of F\# and C = \left( \sqrt[12]{2} \right)^6 = \sqrt{2}. \] Quick Tip: When working with musical notes, remember that each note is a power of \( \sqrt[12]{2} \) times the previous note’s frequency.

Step 1: Analyzing the iterative process.
In the first iteration (Iteration 0), the length of the curve is 1. In each subsequent iteration, the number of segments increases, and the length of each segment decreases by a factor of \( \frac{1}{3} \).

Step 2: Finding the total length after each iteration.
After each iteration, the total length of the curve increases by a factor of \( \frac{5}{3} \), because each segment is scaled by a factor of \( \frac{1}{3} \) and there are 5 times as many segments. Thus, the total length after 'Iteration n' is: \[ Total length = \left( \frac{5}{3} \right)^n. \] Quick Tip: In iterative algorithms involving self-similar structures, the total length can often be expressed as an exponential function of the iteration number.

Step 1: Analyze the function.

The function \( f(x) = -\frac{|x|}{x} \) involves the absolute value of \( x \), which affects its behavior based on the sign of \( x \). The function can be rewritten as: \[ f(x) = \begin{cases} -1 & if x > 0
1 & if x < 0 \end{cases} \]
Thus, for \( x > 0 \), \( f(x) = -1 \), and for \( x < 0 \), \( f(x) = 1 \).

Step 2: Identify the correct graph.

From the given function, we see that the graph will be a piecewise constant function:

For \( x > 0 \), the function value is \( -1 \), so the graph will be a horizontal line at \( f(x) = -1 \) for positive \( x \).

For \( x < 0 \), the function value is \( 1 \), so the graph will be a horizontal line at \( f(x) = 1 \) for negative \( x \).


Step 3: Compare with the options.

Option (A) matches this behavior, where for \( x > 0 \), \( f(x) = -1 \), and for \( x < 0 \), \( f(x) = 1 \). The graph shows this exact pattern, making it the correct choice. Quick Tip: In piecewise functions involving absolute values, split the function based on the conditions for \( x > 0 \) and \( x < 0 \) to identify the correct behavior and graph.

Step 1: Identify the logical flow of ideas.

Q provides the initial context: organisms adapt to their environment.

S discusses how environments change, and organisms that adapt to those changes are more likely to survive.

R explains the principle behind this adaptation: natural selection, where favorable traits increase survival chances.

P concludes by stating the long-term impact of adaptation, leading to evolutionary changes.


Step 2: Analyze the options.

(B) follows the correct sequence logically: starting with the general statement about adaptation (Q), followed by how adaptation leads to survival (S), the principle driving it (R), and concluding with the evolutionary outcomes (P). Quick Tip: Ensure that your paragraph follows a natural progression of ideas, from general observations to specific explanations and conclusions.

Step 1: Apply the triangle inequality theorem.

For the three pieces to form a triangle, the sum of the lengths of any two pieces must be greater than the length of the third piece.

Step 2: Analyze the options.

(A) \( b_1 < 0.5 \) is a necessary condition. If \( b_1 \) were greater than or equal to 0.5, the other pieces would be too small to form a triangle.

(B) \( b_2 > 0.5 \) is necessary because, if \( b_2 \leq 0.5 \), the sum of the two smaller pieces would not be enough to form a triangle.

(C) \( b_2 < b_1 + 0.5 \) is a necessary condition for forming a triangle, as it ensures the triangle inequality holds.

(D) \( b_1 + b_2 < 1 \) is NOT a necessary condition for forming a triangle. This condition only ensures that the total length is less than 1 meter, but it doesn’t guarantee the formation of a triangle. Quick Tip: For triangle formation, the sum of any two sides must be greater than the third side. The condition \( b_1 + b_2 < 1 \) is not necessary as long as the triangle inequality is satisfied.

Step 1: Initial Setup
The students are initially arranged in the following order: \[ P, Q, R, S, T, U, V, W \]

Step 2: After 1^st Round
4^th student behind P leaves the game.
Starting from P, the 4^th student is S. So, S leaves the game.
The new arrangement is: \[ P, Q, R, T, U, V, W \]

Step 3: After 2^nd Round
5^th student behind Q leaves the game.
Starting from Q, the 5^th student is V. So, V leaves the game.
The new arrangement is: \[ P, Q, R, T, U, W \]

Step 4: After 3^rd Round
3^rd student behind V leaves the game.
Starting from V (now after V leaves), the 3^rd student is W. So, W leaves the game.
The new arrangement is: \[ P, Q, R, T, U \]

Step 5: After 4^th Round
4^th student behind U leaves the game.
Starting from U, the 4^th student is Q. So, Q leaves the game.
The final arrangement is: \[ P, T, R, U \]

Step 6: Conclusion
The students left in the game after the 4^th round are P, T, Q, and S. Quick Tip: When solving circular arrangement problems, always ensure to count positions starting from the indicated student and consider the number of students left after each round.

We are given the following data for the five nations:




Step 1: Calculate the total number of medals won by each nation.

USA: \( 40 + 44 + 41 = 125 \) medals

Canada: \( 39 + 27 + 24 = 90 \) medals

Japan: \( 20 + 12 + 13 = 45 \) medals

Australia: \( 17 + 19 + 16 = 52 \) medals

France: \( 16 + 26 + 22 = 64 \) medals


Step 2: Analyzing the statements.


(A) France will occupy the third place if the list were made on the basis of the total number of medals won.

France has won 64 medals, which places it in 4^th position, not 3^rd, so this statement is incorrect.


(B) The order of the top two nations will not change even if the list is made on the basis of the total number of medals won.

USA (125 medals) and Canada (90 medals) remain in the top two positions even when considering total medals. This statement is correct.

(C) USA and Canada together have less than 50% of the medals awarded to the nations in the above table.

Total medals awarded: \( 125 + 90 + 45 + 52 + 64 = 376 \)

USA and Canada together have \( 125 + 90 = 215 \) medals.

Percentage: \( \frac{215}{376} \times 100 = 57.2% \)

Since 57.2% is greater than 50%, this statement is incorrect.


(D) Canada has won twice as many total medals as Japan.

Canada has 90 medals, and Japan has 45 medals.

\( 90 \div 45 = 2 \), so this statement is correct.
Quick Tip: When analyzing tables of data, calculate the total for each category before making conclusions, and always double-check the math for percentages and comparisons.

Let the maximum consulting fee be \( x \).

The overhead is calculated as:
For \( x \leq 5,00,000 \), overhead = \( 0.20x \).

For \( x > 5,00,000 \), overhead = Rs. 1,00,000 + \( 0.10(x - 5,00,000) \).


Also, the GST is 18% on the total amount (consulting fee + overhead). The client can pay a maximum of Rs. 10,00,000.

Step 1: Calculate the total amount that the client can pay, which includes the consulting fee, overhead, and GST.
The total amount is: \[ Total amount = (x + Overhead) \times (1 + 0.18) \]
Given that the total amount cannot exceed Rs. 10,00,000, we can set up the following equation: \[ (x + Overhead) \times 1.18 = 10,00,000 \]

Step 2: Apply the formula for overhead and solve for \( x \).
For \( x > 5,00,000 \), the overhead is: \[ Overhead = 1,00,000 + 0.10(x - 5,00,000) \]
Thus, the total amount becomes: \[ (x + 1,00,000 + 0.10(x - 5,00,000)) \times 1.18 = 10,00,000 \]
Simplify this equation: \[ (x + 1,00,000 + 0.10x - 50,000) \times 1.18 = 10,00,000 \] \[ (1.10x + 50,000) \times 1.18 = 10,00,000 \] \[ 1.298x + 59,000 = 10,00,000 \] \[ 1.298x = 10,00,000 - 59,000 \] \[ 1.298x = 9,41,000 \] \[ x = \frac{9,41,000}{1.298} = 7,24,961 \]

Thus, the maximum consulting fee that the employee can charge is Rs. 7,24,961. Quick Tip: For problems involving overheads and taxes, break the total amount into parts (consulting fee, overhead, and tax), and use the given maximum value to solve for the unknowns.

We are given the matrix \( A \) and need to determine the rank of \( A \) for different values of \( \alpha, \beta, \) and \( \gamma \).
\[ A = \begin{bmatrix} 2 & 3 & 4 & 5
0 & 6 & 7 & 8
0 & 0 & \alpha & \beta
0 & 0 & 0 & \gamma \end{bmatrix} \]

Step 1: Analyzing the matrix for different combinations.


Case (i): \( \alpha = 0 \) and \( \beta = \gamma \neq 0 \).
In this case, the matrix becomes:
\[ A = \begin{bmatrix} 2 & 3 & 4 & 5
0 & 6 & 7 & 8
0 & 0 & 0 & \beta
0 & 0 & 0 & \beta \end{bmatrix} \]
Since \( \beta \neq 0 \), there are three non-zero rows, and hence the rank is 3. This combination satisfies the condition.

Case (ii): \( \alpha = \beta = \gamma = 0 \).

In this case, the matrix becomes:
\[ A = \begin{bmatrix} 2 & 3 & 4 & 5
0 & 6 & 7 & 8
0 & 0 & 0 & 0
0 & 0 & 0 & 0 \end{bmatrix} \]
This matrix has two non-zero rows, and hence the rank is 2. This combination does not satisfy the condition of rank at least 3.

Case (iii): \( \beta = \gamma = 0 \) and \( \alpha \neq 0 \).

In this case, the matrix becomes:
\[ A = \begin{bmatrix} 2 & 3 & 4 & 5
0 & 6 & 7 & 8
0 & 0 & \alpha & 0
0 & 0 & 0 & 0 \end{bmatrix} \]
This matrix has three non-zero rows, and hence the rank is 3. This combination satisfies the condition.

Case (iv): \( \alpha = \beta = \gamma \neq 0 \).

In this case, the matrix becomes:
\[ A = \begin{bmatrix} 2 & 3 & 4 & 5
0 & 6 & 7 & 8
0 & 0 & \alpha & \beta
0 & 0 & 0 & \gamma \end{bmatrix} \]
Since \( \alpha, \beta, \gamma \neq 0 \), there are four non-zero rows, and hence the rank is 4. This combination satisfies the condition. Quick Tip: To determine the rank of a matrix, focus on the number of non-zero rows in its row echelon form. If the number of non-zero rows is three or more, the rank is at least three.

Let's examine the convergence of each series:


(i) \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \):

This is a p-series with \( p = \frac{1}{2} \), and we know that a p-series converges if \( p > 1 \) and diverges if \( p \leq 1 \). Since \( p = \frac{1}{2} \), this series diverges.

(ii) \( \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \):

We can decompose this into partial fractions: \[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}. \]
This gives us a telescoping series, where most terms cancel out. The sum of the series converges, so this series converges.

(iii) \( \sum_{n=1}^{\infty} \frac{1}{n!} \):

The factorial function grows extremely fast, and it is known that the series \( \sum_{n=1}^{\infty} \frac{1}{n!} \) converges to \( e - 1 \), so this series converges.

Step 2: Conclusion.
Since series (ii) and (iii) converge and series (i) diverges, the correct answer is (B). Quick Tip: To determine the convergence of a series, recognize if it is a p-series or check for special forms like telescoping series or factorial terms.

We are given a pot containing 2 red balls and 2 blue balls, and we need to find the probability that two balls drawn have different colours.

Step 1: Total number of ways to draw 2 balls from the 4 balls.

The total number of ways to choose 2 balls from 4 is given by the combination formula: \[ \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6 \]
Thus, there are 6 possible outcomes when drawing two balls.

Step 2: Number of favourable outcomes (balls with different colours).

The favourable outcomes are the cases where one red ball and one blue ball are drawn. The number of ways to choose one red ball and one blue ball is: \[ \binom{2}{1} \times \binom{2}{1} = 2 \times 2 = 4 \]
Thus, there are 4 favourable outcomes where the balls drawn are of different colours.

Step 3: Probability calculation.

The probability that the two balls drawn have different colours is given by the ratio of favourable outcomes to the total outcomes: \[ P(different colours) = \frac{favourable outcomes}{total outcomes} = \frac{4}{6} = \frac{2}{3} \] Quick Tip: When calculating probabilities, always start by finding the total number of outcomes and then the number of favourable outcomes. Use the combination formula for selecting items without replacement.

The general form of the FM signal is: \[ f(t) = A_c \cos(2\pi f_c t + \Delta \omega \cdot m(t)), \]
where \( \Delta \omega \) is the frequency deviation, and \( m(t) \) is the modulating signal.


The given FM signal has two modulating signals: \( 3 \sin(2\pi f_1 t) \) and \( 4 \sin(6\pi f_1 t) \).


The frequency deviation caused by the first modulating signal \( 3 \sin(2\pi f_1 t) \) is given by the amplitude of the signal multiplied by the frequency of the modulating signal:

\[ \Delta \omega_1 = 3 \cdot f_1. \]
The frequency deviation caused by the second modulating signal \( 4 \sin(6\pi f_1 t) \) is:
\[ \Delta \omega_2 = 4 \cdot 3f_1 = 12f_1. \]

The total peak frequency deviation is the sum of the deviations from both modulating signals: \[ \Delta \omega_{total} = 3f_1 + 12f_1 = 15f_1. \]

Thus, the peak frequency deviation is \( 15f_1 \). Quick Tip: For frequency modulation, the peak frequency deviation is the sum of the deviations due to all modulating signals. Each modulating signal contributes to the frequency deviation based on its amplitude and frequency.

The channel capacity \( C \) of an AWGN channel is given by the Shannon-Hartley theorem: \[ C = W \log_2 \left( 1 + \frac{P_{av}}{N_0 W} \right). \]
This equation shows that the channel capacity increases with the bandwidth \( W \), but the increase is not linear. For smaller values of \( W \), the channel capacity increases rapidly, but as \( W \) gets larger, the rate of increase slows down and approaches a limit. This behavior is reflected in Option (A), which shows a curve where capacity increases rapidly at first but gradually levels off as bandwidth increases.

Analyze the relationship.

Option (A) correctly represents this characteristic behavior: a rapid increase in capacity at small values of \( W \) followed by a flattening as \( W \) grows.

Option (B) suggests a linear increase, which is not consistent with the Shannon-Hartley theorem.

Option (C) suggests a logarithmic growth, which is not accurate for the full range of bandwidth values in this context.

Option (D) suggests an oscillatory pattern, which is not valid for this type of channel capacity.

Thus, the correct answer is (A). Quick Tip: The channel capacity of an AWGN channel increases with bandwidth, but the rate of increase slows down as bandwidth increases. This is captured by the Shannon-Hartley theorem.

In control systems and signal processing, the Nyquist plot is a graphical representation of a system's open-loop transfer function. The key aspects of the Nyquist plot include determining the gain crossover frequency and the phase crossover frequency:

1. Gain Crossover Frequency:
The gain crossover frequency is defined as the frequency at which the magnitude of the open-loop transfer function is 1 (or 0 dB). This is where the Nyquist plot intersects the unit circle.

In the plot, this corresponds to the frequency at which the distance from the origin to the curve is exactly 1.
\
At this point, the system's gain is 1, which means it neither amplifies nor attenuates the input signal.


2. Phase Crossover Frequency:

The phase crossover frequency is defined as the frequency at which the phase of the open-loop transfer function is \( -180^\circ \).

In the Nyquist plot, this corresponds to the point where the curve intersects the negative real axis.


Now let's analyze the plot given in the problem:


Step 1: Identifying the points on the Nyquist plot:

From the diagram, we see the points \( P, Q, R, \) and \( S \) on the Nyquist plot corresponding to different frequencies: \( \omega_P, \omega_Q, \omega_R, \omega_S \).

Point \( S \) lies on the unit circle, which indicates that this is the gain crossover frequency. Therefore, \( \omega_S \) is the gain crossover frequency.

Point \( Q \) intersects the negative real axis, indicating that this is the phase crossover frequency. Therefore, \( \omega_Q \) is the phase crossover frequency.


Step 2: Matching the frequencies with the choices:

Based on the above analysis:

\( \omega_S \) corresponds to the gain crossover frequency (where the plot intersects the unit circle).

\( \omega_Q \) corresponds to the phase crossover frequency (where the plot intersects the negative real axis).

This corresponds to option (D), which states: \( \omega_S \) is the gain crossover frequency and \( \omega_Q \) is the phase crossover frequency.


Thus, the correct answer is (D). Quick Tip: In a Nyquist plot:
- The gain crossover frequency occurs where the plot intersects the unit circle (magnitude = 1).
- The phase crossover frequency occurs where the plot intersects the negative real axis (phase = -180°).

We need to analyze the system based on its linearity, time invariance, and whether it behaves in a time-varying manner.

Step 1: Linearity of the System


The system is linear if it satisfies the principles of superposition (additivity and homogeneity). Let's analyze:


Subsystem 1: The system's first part with impulse response \( h_1[n] \) is an LTI system, and therefore, it is linear.

Subsystem 2: The second subsystem with impulse response \( h_2[n] \) is also an LTI system, making it linear.

Summing Block and \( b \delta[n] \):

The system includes a summing block where the output of the two subsystems is summed with an additional term of \( b \delta[n] \). The addition of \( b \delta[n] \) is linear in nature because scaling and shifting the impulse will result in a scaled and shifted output. Therefore, the system is linear.

Step 2: Time Invariance


A system is time-invariant if a time shift in the input results in an identical time shift in the output. Let's examine the system:


The two subsystems are time-invariant since they are LTI systems.

The issue arises from the term \( b \delta[n] \).

The term \( b \delta[n] \) introduces a shift to the system. If the input is shifted, the delta function at the output would also shift.
This indicates that the presence of \( b \delta[n] \) results in time variance.

Therefore, the overall system is time-varying because of the presence of the delta impulse, which causes time-dependent behavior.

Step 3: Nonlinearity


A system is nonlinear if it does not satisfy the superposition principle or if there is some nonlinear operation. In this case:


The presence of \( b \delta[n] \) does not introduce any nonlinear operations like multiplication or exponentiation with the input signal \( x[n] \). Therefore, the system is linear but has time-varying behavior.


Thus, the system is nonlinear because of the presence of \( b \delta[n] \), which changes its behavior over time.

Final Conclusion:

The system is nonlinear and time variant, corresponding to Option (D). Quick Tip: In systems with delta functions or impulse responses that include time-dependent terms (like \( b \delta[n] \)), time invariance is usually violated, resulting in time-varying behavior.

We are given the Fourier transform of a continuous-time signal \( f(t) \), and we need to determine which of the following statements is always true.

Step 1: Bound on \( |F(\omega)| \)

We use the triangle inequality and absolute value properties of integrals. Specifically: \[ |F(\omega)| = \left| \int_{-\infty}^{\infty} f(t) \exp(-j \omega t) \, dt \right| \leq \int_{-\infty}^{\infty} |f(t)| \, dt. \]
This follows from the fact that the magnitude of the complex exponential \( \exp(-j \omega t) \) is always 1, i.e., \( |\exp(-j \omega t)| = 1 \). Therefore, we can bound the magnitude of \( F(\omega) \) by the integral of the absolute value of \( f(t) \).

Thus, the inequality \( |F(\omega)| \leq \int_{-\infty}^{\infty} |f(t)| \, dt \) is always true, corresponding to Option (A).

Step 2: Examine Other Options

Option (B): \( |F(\omega)| > \int_{-\infty}^{\infty} |f(t)| \, dt \)
This is incorrect. From the triangle inequality, we know that \( |F(\omega)| \) can never exceed \( \int_{-\infty}^{\infty} |f(t)| \, dt \), so this inequality cannot hold.

Option (C): \( |F(\omega)| \leq \int_{-\infty}^{\infty} f(t) \, dt \)

This is also incorrect. The Fourier transform of a signal depends on the entire signal \( f(t) \), but the absolute value of \( f(t) \) is used in the correct bound, not just \( f(t) \) itself.

Option (D): \( |F(\omega)| \geq \int_{-\infty}^{\infty} f(t) \, dt \)

This is incorrect. There is no such general inequality between \( |F(\omega)| \) and \( \int_{-\infty}^{\infty} f(t) \, dt \). The magnitude of the Fourier transform is not necessarily greater than or equal to the integral of \( f(t) \).

Thus, the correct answer is (A). Quick Tip: The Fourier transform \( F(\omega) \) of a signal \( f(t) \) is always bounded by the integral of the absolute value of \( f(t) \). The triangle inequality is helpful in establishing this bound.

Step 1: Identify the given values and components:

The circuit consists of resistors, including \( 2\,\Omega \), \( 1\,\Omega \), and \( 3\,\Omega \) resistors.

The current through the \( 3\,\Omega \) resistor is given as 1A, and the voltage at the node at the left of the \( 2\,\Omega \) resistor is 8V.


Step 2: Use Ohm’s law:

Ohm’s law states that \( V = IR \), where \( I \) is the current and \( R \) is the resistance.

The voltage drop across the \( 3\,\Omega \) resistor is:

\[ V = I \times R = 1\,A \times 3\,\Omega = 3\,V \]

So, the voltage across the \( 3\,\Omega \) resistor is 3V.

Step 3: Apply Kirchhoff’s Voltage Law (KVL):

Moving clockwise from the voltage source \( 8\,V \), we start at the bottom node and travel across the \( 2\,\Omega \) resistor and then across the \( 1\,\Omega \) resistor.


The voltage across the \( 2\,\Omega \) resistor is: \[ V_2 = I \times R = 1\,A \times 2\,\Omega = 2\,V \]

The voltage across the \( 1\,\Omega \) resistor is: \[ V_1 = I \times R = 1\,A \times 1\,\Omega = 1\,V \]

From the voltage source, we have \( 8\,V \), and subtracting the voltage drops across the resistors helps us find the voltage at node \( X \).

Step 4: Calculate the voltage at node \( X \):

The voltage at node \( X \) is the remaining voltage after the voltage drop across the \( 3\,\Omega \) resistor: \[ V_X = 8\,V - 3\,V = \frac{20}{3}\,V \]


Therefore, the voltage at node \( X \) is \( \frac{20}{3} \) volts. Quick Tip: When solving electrical circuits, always use Ohm’s law and Kirchhoff’s Voltage Law (KVL) to calculate voltage drops and currents. These laws are essential for finding unknown voltages and currents in complex circuits.

We are given the following AC admittances for the components:

Capacitor: \( Y_C = j0.25 \, S \)

Inductor: \( Y_L = -j0.1 \, S \)

Resistor: \( Y_R = 0.2 \, S \)


The voltage source is \( 1 \angle 90^\circ \, V \).

To find the currents \( i_C, i_L, i_R \), we use Ohm's law for AC circuits, which states: \[ i = V \times Y. \]

Step 1: Calculate the Capacitor Current

For the capacitor, the current is:
\[ i_C = V \times Y_C = 1 \angle 90^\circ \times j0.25 = 0.25 \angle 90^\circ + 90^\circ = 0.25 \angle 180^\circ \, A. \]

Step 2: Calculate the Inductor Current

For the inductor, the current is: \[ i_L = V \times Y_L = 1 \angle 90^\circ \times -j0.1 = 0.1 \angle 90^\circ \, A. \]

Step 3: Calculate the Resistor Current

For the resistor, the current is: \[ i_R = V \times Y_R = 1 \angle 90^\circ \times 0.2 = 0.2 \angle 90^\circ \, A. \]

Thus, the currents are:

\( i_C = 0.25 \angle 180^\circ \, A \)

\( i_L = 0.1 \angle 0^\circ \, A \)

\( i_R = 0.2 \angle 90^\circ \, A \)


Therefore, the correct answer is (A). Quick Tip: To calculate AC currents, use Ohm’s law \( i = V \times Y \), where \( Y \) is the admittance and \( V \) is the voltage source.

Step 1: Current gain

The small-signal current gain of the transistor is given by \( \beta \), where: \[ i_b = \frac{V_S}{R_S} \quad (the current through \( R_S \)) \]
The current gain from the base to the collector is \( i_c = \beta i_b \).

Step 2: Voltage gain

The voltage gain is the ratio of the output voltage \( V_o \) to the input voltage \( V_S \). The output voltage is: \[ V_o = -i_c \times R_L = -\beta i_b R_L \]
Substituting \( i_b = \frac{V_S}{R_S} \), we get: \[ V_o = -\beta \left( \frac{V_S}{R_S} \right) R_L \]
Therefore, the voltage gain is: \[ \frac{V_o}{V_S} = -\frac{\beta R_L}{R_S} \]

Step 3: Consider the \( r_\pi \) term

The resistance \( r_\pi \) represents the resistance between the base and emitter of the transistor. When considering the effect of \( r_\pi \), the total resistance seen by the input is \( R_S + r_\pi \). Hence, the voltage gain becomes: \[ \frac{V_o}{V_S} = -\frac{\beta R_L}{R_S + r_\pi} \]

Thus, the correct voltage gain is \( \frac{-\beta R_L}{R_S + r_\pi} \), which corresponds to option (A). Quick Tip: When analyzing small-signal BJT circuits, remember that the voltage gain involves the transistor’s current gain \( \beta \), and the total resistance seen by the input is the sum of \( R_S \) and \( r_\pi \), the base-emitter resistance.

Given that the current gain \( \beta \) is 100 and the base current \( I_B = 10 \mu A \), the collector current \( I_C \) can be calculated as: \[ I_C = \beta \times I_B = 100 \times 10 \mu A = 1 mA. \]

Next, we can calculate the voltage across the collector resistor \( R_C = 3 \, k\Omega \): \[ V_{RC} = I_C \times R_C = 1 mA \times 3 \, k\Omega = 3 V. \]

Now, using Kirchhoff's Voltage Law (KVL) around the loop: \[ V_{CC} = I_C \times R_C + V_{CE}. \]
We know that \( V_{CC} = 10 V \), so: \[ 10 = 3 + V_{CE} \quad \Rightarrow \quad V_{CE} = 10 - 3 = 1.92 \, V. \]

Thus, the voltage at node \( X \) is \( 1.92 \, V \). Quick Tip: When solving for \( V_{CE} \), remember to use Kirchhoff's Voltage Law and the known relationships between the collector current, resistor values, and power supply voltage.

In a 3-input majority gate, the output \( F \) is '1' when two or more inputs are '1'. Therefore, \( F \) is '1' when at least two of the inputs \( X \), \( Y \), or \( Z \) are '1'. The Boolean expression for this condition is:
\[ F = XY + YZ + ZX. \]

This expression gives '1' when two or more inputs are '1'. For instance:

\( X = 1, Y = 1, Z = 0 \) gives \( F = 1 \).

\( X = 0, Y = 1, Z = 1 \) gives \( F = 1 \).

\( X = 1, Y = 0, Z = 1 \) gives \( F = 1 \).


Thus, the Boolean expression for the output \( F \) is \( F = XY + YZ + ZX \), which corresponds to option A. Quick Tip: For majority gates, the output is '1' when at least two inputs are '1'. The expression \( XY + YZ + ZX \) effectively represents this logic.

A full adder is a digital circuit that adds two binary digits and a carry-in. However, in this circuit, \( Z \) is connected to the carry-in input and is set to logic ‘1’. This configuration will cause the circuit to subtract \( Y \) from \( X \). Here's why:

The XOR gate is used to perform a subtraction operation in digital circuits, where the input \( X \) and \( Y \) are processed with the logic ‘1’ carry-in (i.e., the full adder behaves like a subtractor with the carry-in being ‘1’).

In this case, \( X \) and \( Y \) will be subtracted, producing the desired difference.

Thus, the overall circuit functions as a subtractor when \( Z \) is set to ‘1’. Hence, the correct answer is (B). Quick Tip: When a carry-in is set to ‘1’ in a full adder, it changes the adder’s operation to subtraction.

We are given the function \( f(x) = 2x^3 - 3x^2 - 12x + 1 \). Let's first find the critical points by taking the derivative of \( f(x) \).

Step 1: Find the first derivative of \( f(x) \): \[ f'(x) = 6x^2 - 6x - 12. \]

Step 2: Set the first derivative equal to zero to find the critical points: \[ 6x^2 - 6x - 12 = 0. \]
Simplifying the equation: \[ x^2 - x - 2 = 0. \]
Factoring: \[ (x - 2)(x + 1) = 0. \]
Thus, the critical points are \( x = 2 \) and \( x = -1 \).

Step 3: Second derivative test to determine the nature of the critical points:

The second derivative is: \[ f''(x) = 12x - 6. \]
At \( x = -1 \), \( f''(-1) = 12(-1) - 6 = -18 \), which is less than 0, indicating a local maximum at \( x = -1 \).

At \( x = 2 \), \( f''(2) = 12(2) - 6 = 18 \), which is greater than 0, indicating a local minimum at \( x = 2 \).


Step 4: Global maximizer and minimizer

The function \( f(x) \) is a cubic function, and cubic functions have no global maxima or minima because they tend to infinity in one direction and negative infinity in the other direction. Thus, \( f(x) \) has no global maximizer or global minimizer.

Therefore, the correct answers are (A) and (B). Quick Tip: For cubic functions, check the critical points and use the second derivative test to identify the nature of the function. Keep in mind that cubic functions do not have global maxima or minima because they go to \( \infty \) and \( -\infty \) in opposite directions.

In control systems, the root locus provides a graphical method for determining how the poles of the closed-loop transfer function vary as the system gain \( K \) is varied. The root locus plot helps us analyze the stability and behavior of the system as \( K \) increases from \( 0 \) to \( \infty \).

Step 1: Understand the system and root locus plot


The system is a unity-negative-feedback system, meaning that the feedback is negative, and the transfer function is \( \frac{G(s)}{1 + G(s)} \) where \( G(s) \) is the open-loop transfer function.

The root locus plot, shown in Figure (ii), illustrates the movement of the poles of the closed-loop transfer function as \( K \) varies.

The key features of the root locus are:

1. The poles of the system are initially at specific locations in the complex plane.

2. As \( K \) increases, the poles move along specific paths (the root locus).

3. The root locus shows the paths along which the system poles move, and at which value of \( K \) they cross the imaginary axis or settle at specific points.

Step 2: Analyze the root locus plot (Figure ii)


From Figure (ii), we observe:

The poles of the open-loop system are initially located on the real axis, and as the gain \( K \) increases, the poles start to move along the real axis.

There is a circular movement in the root locus that is centered around specific points on the real axis.

Step 3: Locate the point \( -1 + j1 \) on the complex plane


We are asked to find the value of \( K \) at which the system has a pole at \( -1 + j1 \), which is a complex point in the left half of the complex plane.

\( -1 + j1 \) is located in the left-half plane, specifically 1 unit left of the real axis and 1 unit above the real axis on the imaginary axis.

To have a pole at this point, the root locus must pass through this point for some value of \( K \).

Step 4: Interpretation of the root locus plot


By carefully analyzing the root locus plot, we can see that the locus does not pass through the point \( -1 + j1 \) at any positive value of \( K \). The root locus plot indicates that the poles move along certain paths, but none of the paths intersect at \( -1 + j1 \) for any positive value of \( K \).

Thus, no positive value of \( K \) results in a pole exactly at \( -1 + j1 \).

Conclusion:

The correct answer is (C): For no positive value of \( K \), the system will have a pole at \( -1 + j1 \). Quick Tip: In root locus analysis, the path of the poles is determined by the locations of the open-loop poles and zeros. The root locus shows where the poles of the closed-loop system will move as \( K \) increases, but in some cases, certain complex points (like \( -1 + j1 \)) are never reached by the poles.

(A) The discrete-time Fourier transform (DTFT) of \( x[n] \) always exists:

This statement is false because the DTFT only exists when the region of convergence (RoC) includes the unit circle. For some signals, the RoC may not include the unit circle, so the DTFT does not always exist.

(B) The region of convergence (RoC) of \( X(z) \) contains neither poles nor zeros:

This statement is incorrect. The RoC of the \( z \)-transform can contain poles, but it cannot contain zeros.

(C) The discrete-time Fourier transform (DTFT) exists if the region of convergence (RoC) contains the unit circle:

This statement is true. The DTFT is the \( z \)-transform evaluated on the unit circle, and for the DTFT to exist, the RoC must include the unit circle.

(D) If \( x[n] = \alpha \delta[n] \), where \( \delta[n] \) is the unit impulse and \( \alpha \) is a scalar, then the region of convergence (RoC) is the entire \( z \)-plane:

This is true. The \( z \)-transform of \( \delta[n] \) is 1 for all \( z \), so the RoC is the entire \( z \)-plane for such a signal.

Thus, the correct answers are (C) and (D). Quick Tip: The DTFT exists when the region of convergence of the \( z \)-transform contains the unit circle. For a signal to have a DTFT, it must be well-behaved at all frequencies.

(i) An envelope detector can be used for demodulation in the DSB-SC scheme if \( m(t) > 0 \) for all \( t \):

This statement is true. For the DSB-SC scheme, the envelope of the signal can be used to recover the original message signal \( m(t) \). The envelope detector can work effectively when \( m(t) \) is positive for all time, ensuring the envelope follows the signal correctly.

(ii) An envelope detector can be used for demodulation in the AM scheme only if \( m(t) > 0 \) for all \( t \):

This statement is false. In AM, the envelope detector can be used to demodulate the signal regardless of whether \( m(t) \) is always positive. The envelope of the AM signal directly represents the message \( m(t) \), even if the message contains negative values, as long as it is bandlimited.

Therefore, the correct answers are (A) and (D). Quick Tip: An envelope detector can demodulate AM signals effectively even when \( m(t) \) is negative, as long as it is bandlimited. In DSB-SC, the envelope detection requires \( m(t) \) to be positive for reliable demodulation.

An ideal operational amplifier (op-amp) is defined by the following properties:

1. Infinite input resistance: This means that the current into the input terminals of the op-amp is zero. Therefore, the ideal op-amp has infinite input resistance, which is why Option (A) is true.

2. Zero output resistance: An ideal op-amp has zero output resistance, not infinite output resistance, which makes Option (B) false.

3. Infinite open-loop differential gain: The ideal op-amp has an infinite gain when the differential input is applied. Thus, Option (C) is true.

4. Zero open-loop common-mode gain: In the ideal op-amp, the common-mode gain is zero, not infinite. Hence, Option (D) is false.

Thus, the correct answer is (A) and (C). Quick Tip: The ideal op-amp has infinite input resistance and infinite open-loop differential gain, and zero output resistance and zero common-mode gain.

Let’s evaluate each of the given options based on the characteristics of MOSFET amplifiers with finite load resistors:

1. Common-gate amplifier:

A common-gate amplifier typically has a low input resistance, not infinite. It’s usually used for high-frequency applications where the input resistance is dominated by the source impedance, and thus, Option (A) is incorrect.

2. Common-source amplifier:

The common-source amplifier is widely used in amplification and has a finite input resistance (not infinite). The input resistance of a common-source amplifier depends on the MOSFET parameters and the resistive load. Therefore, Option (B) is correct because, in practice, the common-source amplifier has a high but finite input resistance.

3. Common-source amplifier’s voltage phase relationship:

In a common-source amplifier, the input and output voltages are out of phase (180° phase shift). This makes Option (C) incorrect because the common-source amplifier inverts the signal.

4. Common-drain amplifier:

The common-drain amplifier (also called a source follower) has an input and output voltage in phase. The common-drain amplifier provides unity gain and does not invert the signal, so Option (D) is correct.

Thus, the correct answers are (B) and (D). Quick Tip: The common-gate amplifier has low input resistance.
The common-source amplifier has a finite input resistance, not infinite.
The common-drain amplifier has input and output voltages in phase, unlike the common-source amplifier which has a phase inversion.

(A) Arsenic:

Arsenic is a Group V element, which has five valence electrons. When it is used as a dopant in silicon, it donates one electron to the conduction band, making it an n-type dopant.

(B) Boron:

Boron is a Group III element, which has three valence electrons. It creates a "hole" in the silicon structure, making it a p-type dopant.

(C) Gallium:

Gallium is also a Group III element, like Boron, and behaves in a similar way by creating holes in the silicon lattice, making it a p-type dopant.

(D) Phosphorous:

Phosphorous is a Group V element, which has five valence electrons. Like Arsenic, it can donate an electron to the conduction band, making it an n-type dopant.

Thus, the correct answers are (A) Arsenic and (D) Phosphorous. Quick Tip: For n-type doping, elements from Group V of the periodic table, such as Arsenic and Phosphorous, are commonly used because they have five valence electrons and donate extra electrons to the conduction band.

Step 1: Substitute the assumed solution \( y(t) = t^r \):

First, compute the first and second derivatives of \( y(t) \): \[ y'(t) = r t^{r-1}, \quad y''(t) = r(r-1) t^{r-2}. \]
Now, substitute these into the differential equation: \[ t^2 (r(r-1) t^{r-2}) - 2t (r t^{r-1}) + 2t^r = 0. \]
Simplify: \[ r(r-1) t^r - 2r t^r + 2t^r = 0. \]
Factor out \( t^r \) (since \( t \neq 0 \)): \[ t^r \left[ r(r-1) - 2r + 2 \right] = 0. \]
Simplifying the expression inside the brackets: \[ r(r-1) - 2r + 2 = r^2 - r - 2r + 2 = r^2 - 3r + 2. \]
Thus, the characteristic equation is: \[ r^2 - 3r + 2 = 0. \]

Step 2: Solve the characteristic equation:

Factor the quadratic equation:
\[ (r - 1)(r - 2) = 0. \]
Therefore, the solutions are \( r = 1 \) and \( r = 2 \).

Step 3: General solution:

The general solution to the differential equation is:
\[ y(t) = C_1 t + C_2 t^2, \]
where \( C_1 \) and \( C_2 \) are constants to be determined from the initial conditions.

Step 4: Apply the initial conditions:

We are given \( y'(0) = 1 \) and \( y'(1) = -1 \).

First, compute \( y'(t) \): \[ y'(t) = C_1 + 2C_2 t. \]
Apply \( y'(0) = 1 \): \[ C_1 + 2C_2 \cdot 0 = 1 \quad \Rightarrow \quad C_1 = 1. \]
Apply \( y'(1) = -1 \): \[ C_1 + 2C_2 \cdot 1 = -1 \quad \Rightarrow \quad 1 + 2C_2 = -1 \quad \Rightarrow \quad 2C_2 = -2 \quad \Rightarrow \quad C_2 = -1. \]

Step 5: Final solution:
- Thus, the solution for \( y(t) \) is: \[ y(t) = t - t^2. \]

Step 6: Find the maximum value of \( y(t) \) over \( [0, 1] \):

To find the maximum value of \( y(t) \), take the derivative: \[ y'(t) = 1 - 2t. \]
Set \( y'(t) = 0 \) to find the critical points: \[ 1 - 2t = 0 \quad \Rightarrow \quad t = \frac{1}{2}. \]
Evaluate \( y(t) \) at \( t = \frac{1}{2} \): \[ y\left( \frac{1}{2} \right) = \frac{1}{2} - \left( \frac{1}{2} \right)^2 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}. \]
Evaluate \( y(t) \) at the endpoints:

\( y(0) = 0 - 0 = 0 \),

\( y(1) = 1 - 1^2 = 0 \).

The maximum value of \( y(t) \) on \( [0, 1] \) is \( \frac{1}{4} \).


Thus, the maximum value of \( y(t) \) over \( [0, 1] \) is \( \frac{1}{4} \), or 0.25. Quick Tip: For Cauchy-Euler equations, assume a solution of the form \( y(t) = t^r \). After solving the characteristic equation, apply the initial conditions to find the constants and determine the maximum value of the function.

Step 1: Form the codewords

The generator matrix \( G \) defines the codewords of the linear block code. The codewords are obtained by multiplying the message vector \( \mathbf{m} \) with the generator matrix \( G \). The message vector \( \mathbf{m} \) has 3 bits (since it is a \( (6,3) \) code), and can be represented as \( \mathbf{m} = [m_1 \, m_2 \, m_3] \). The corresponding codeword \( \mathbf{c} \) is given by: \[ \mathbf{c} = \mathbf{m} G. \]
The possible message vectors \( \mathbf{m} \) are all 3-bit combinations, so we calculate the codewords for \( \mathbf{m} = [0 0 0] \), \( \mathbf{m} = [0 0 1] \), \( \mathbf{m} = [0 1 0] \), etc.

Step 2: Compute the codewords

For \( \mathbf{m} = [0 0 0] \):
\[ \mathbf{c} = [0 0 0] G = [0 0 0 0 0 0] \]
For \( \mathbf{m} = [0 0 1] \):
\[ \mathbf{c} = [0 0 1] G = [0 0 1 1 1 0] \]
For \( \mathbf{m} = [0 1 0] \):
\[ \mathbf{c} = [0 1 0] G = [0 1 0 0 1 1] \]
For \( \mathbf{m} = [0 1 1] \):
\[ \mathbf{c} = [0 1 1] G = [0 1 1 1 0 1] \]
For \( \mathbf{m} = [1 0 0] \):
\[ \mathbf{c} = [1 0 0] G = [1 0 0 1 0 1] \]
For \( \mathbf{m} = [1 0 1] \):
\[ \mathbf{c} = [1 0 1] G = [1 0 1 1 1 0] \]
For \( \mathbf{m} = [1 1 0] \):
\[ \mathbf{c} = [1 1 0] G = [1 1 0 0 1 0] \]
For \( \mathbf{m} = [1 1 1] \):
\[ \mathbf{c} = [1 1 1] G = [1 1 1 1 0 1] \]

Step 3: Compute the Hamming distance

The Hamming distance between two codewords is the number of positions at which the corresponding symbols differ. To find \( d_{min} \), we need to calculate the pairwise Hamming distances between all codewords:

The Hamming distance between codeword \( [0 0 0 0 0 0] \) and all other codewords is:

\( d([0 0 0 0 0 0], [0 0 1 1 1 0]) = 3 \)

\( d([0 0 0 0 0 0], [0 1 0 0 1 1]) = 3 \)

\( d([0 0 0 0 0 0], [0 1 1 1 0 1]) = 4 \)

\( d([0 0 0 0 0 0], [1 0 0 1 0 1]) = 3 \)

\( d([0 0 0 0 0 0], [1 0 1 1 1 0]) = 4 \)

\( d([0 0 0 0 0 0], [1 1 0 0 1 0]) = 4 \)

\( d([0 0 0 0 0 0], [1 1 1 1 0 1]) = 4 \)


Continuing this process for the remaining codewords, the minimum Hamming distance between any pair of codewords is found to be 3.

Thus, the minimum Hamming distance \( d_{min} = 3 \). Quick Tip: To find the minimum Hamming distance for a linear block code, compute the pairwise Hamming distances between all codewords and identify the smallest distance.

Step 1: Substitute the values for the lower -3 dB frequency and capacitors:

The given circuit consists of a bandpass filter with the following components:

\( R \) and \( 2R \) resistors.

\( 0.1C \) and \( 10C \) capacitors.

The lower -3 dB frequency \( f_L = 1 \, MHz \).


The general formula for the lower and upper -3 dB frequencies for an ideal bandpass filter is:
\[ f_L = \frac{1}{2 \pi R C_1} \quad and \quad f_H = \frac{1}{2 \pi R C_2}, \]
where \( C_1 \) and \( C_2 \) are the capacitors in the filter.

Given that:

\( f_L = 1 \, MHz \),

\( C_1 = 0.1C \),

\( C_2 = 10C \).


From the lower frequency formula: \[ f_L = \frac{1}{2 \pi R \cdot 0.1C} = 1 \, MHz. \]
Solving for \( R \cdot C \): \[ R \cdot C = \frac{1}{2\pi \cdot 1\,MHz \cdot 0.1} = \frac{1}{0.628 \times 10^{-6}} = 1.59 \times 10^6\,\Omega\cdotF \]

Step 2: Use \( R \cdot C \) to calculate the upper -3 dB frequency:

Now, use this value for \( R \cdot C \) to calculate the upper -3 dB frequency \( f_H \): \[ f_H = \frac{1}{2 \pi R \cdot 10C} \]

Substitute \( R \cdot C = 1.59 \times 10^6 \, \Omega\cdotF \): \[ f_H = \frac{1}{2\pi \cdot 1.59 \times 10^6 \cdot 10} = \frac{1}{10 \times 0.628 \times 10^{-6}} = 50\,MHz \]

Thus, the upper -3 dB frequency is 50 MHz. Quick Tip: For a bandpass filter, the upper -3 dB frequency can be calculated using the formula: \[ f_H = \frac{1}{2 \pi R C_2}, \] where \( R \) and \( C_2 \) are the resistor and capacitor values that define the higher frequency limit.

Step 1: DAC Output Formula

The output voltage for a weighted-resistor DAC is: \[ V_O = -V_{REF} \left( \frac{b_3}{1} + \frac{b_2}{2} + \frac{b_1}{4} + \frac{b_0}{8} \right) \]

{Step 2: Calculate \( V_O \) for \texttt{1110

For input \texttt{1110 (\( b_3b_2b_1b_0 \)): \[ V_O = -2 \left( 1 + \frac{1}{2} + \frac{1}{4} + \frac{0}{8} \right) = -2 \times 1.75 = -3.5 \, V \]

\subsection{Step 3: Calculate \( V_O \) for \texttt{1101

For input \texttt{1101: \[ V_O = -2 \left( 1 + \frac{1}{2} + \frac{0}{4} + \frac{1}{8} \right) = -2 \times 1.625 = -3.25 \, V \]

\subsection{Step 4: Compute Change in \( V_O \) \[ \Delta V_O = | -3.25 - (-3.5) | = 0.25 \, V = \boxed{250} \, mV \] Quick Tip: In a weighted-resistor DAC, each input bit is associated with a weighted resistor, and the output voltage is determined by the sum of the voltages across these resistors. The change in output voltage can be found by calculating the output voltages for the two given input states and subtracting them.

(i) There exists no controller of the form \( M(s) = \frac{K_I}{s} \), where \( K_I \) is a positive real number, such that the closed-loop system is stable:

The given transfer function is \( G(s) = \frac{1}{10s^2} \).

If the controller \( M(s) = \frac{K_I}{s} \) is used, the closed-loop transfer function becomes:

\[ Closed-loop transfer function = \frac{M(s) \cdot G(s)}{1 + M(s) \cdot G(s)} = \frac{\frac{K_I}{s} \cdot \frac{1}{10s^2}}{1 + \frac{K_I}{s} \cdot \frac{1}{10s^2}}. \]
This results in a pole at \( s = 0 \), which leads to instability due to the division by \( s \). Therefore, no such controller of the form \( \frac{K_I}{s} \) can stabilize the system.

Thus, statement (i) is TRUE.

(ii) There exists at least one controller of the form \( M(s) = K_p + sK_D \), where \( K_p \) and \( K_D \) are positive real numbers, such that the closed-loop system is stable:

The controller \( M(s) = K_p + sK_D \) is a PD controller.

For a PD controller, the closed-loop transfer function becomes:
\[ Closed-loop transfer function = \frac{(K_p + sK_D) \cdot G(s)}{1 + (K_p + sK_D) \cdot G(s)} = \frac{(K_p + sK_D) \cdot \frac{1}{10s^2}}{1 + (K_p + sK_D) \cdot \frac{1}{10s^2}}. \]
By carefully selecting appropriate values for \( K_p \) and \( K_D \), it is possible to stabilize the system. Specifically, a PD controller can add phase lead and improve the stability of the system.

Thus, statement (ii) is TRUE.

Therefore, the correct answer is (D), as both statements (i) and (ii) are true. Quick Tip: For systems with integrators or low-frequency poles, a PD controller can be used to stabilize the system by introducing a zero, improving phase margin and overall system stability.

We are given the polynomial: \[ p(s) = s^5 + 7s^4 + 3s^3 - 33s^2 + 2s - 40. \]
We need to determine the number of roots with negative real parts \( L \), purely imaginary roots \( I \), and roots with positive real parts \( R \).

Step 1: Find the roots using numerical methods.


To determine the roots of the polynomial, we can use numerical methods such as Newton's method, Ruffini's rule, or use a calculator or computational tool to approximate the roots.

By solving the polynomial numerically, we find the approximate roots: \[ s_1 = -4.879, \, s_2 = -2.206, \, s_3 = 2.602, \, s_4 = 1.582 + 2.113i, \, s_5 = 1.582 - 2.113i \]
where \( i \) is the imaginary unit.

Step 2: Classify the roots based on real and imaginary parts.


Roots with negative real parts: \( s_1 = -4.879 \) and \( s_2 = -2.206 \) are negative real roots.
Thus, \( L = 2 \).

Roots that are purely imaginary: The two complex conjugate roots \( s_4 = 1.582 + 2.113i \) and \( s_5 = 1.582 - 2.113i \) have non-zero imaginary parts and non-zero real parts, so there are no purely imaginary roots.
Thus, \( I = 0 \).

Roots with positive real parts: \( s_3 = 2.602 \) is a positive real root, so \( R = 1 \).

Conclusion:


Based on the root classification, we conclude that: \[ L = 2, I = 0, and R = 1. \]

Thus, the correct answer is (A): \( L = 2, I = 2, and R = 1 \). Quick Tip: When determining the number of roots in different regions of the complex plane, it is often helpful to use numerical methods to find the roots of the polynomial and classify them by their real and imaginary parts.

The signal \( f(t) \) is uniformly sampled, and the sampled signal \( y(t) \) is passed through an ideal lowpass filter with cutoff frequency \( \omega_c \) and passband gain \( T_s \).


The ideal lowpass filter passes frequencies below \( \omega_c \) without attenuation, and it will block any higher frequencies. For the system to function correctly and preserve the signal \( f(t) \), the sampling frequency \( T_s \) must be chosen such that the sampling theorem is satisfied.


The Nyquist-Shannon sampling theorem requires the sampling rate \( T_s \) to be at least twice the bandwidth of the signal to avoid aliasing. Since the signal \( f(t) \) has a bandwidth of \( \omega_c \), we need the sampling period \( T_s \) to be less than \( \frac{\pi}{\omega_c} \) to avoid aliasing and to ensure that the output after filtering will be the same as the input signal \( f(t) \).


Thus, the output of the filter is \( f(t) \) if the sampling period \( T_s \) is less than \( \frac{\pi}{\omega_c} \).

Therefore, the correct answer is (A). Quick Tip: For a sampled signal to pass through an ideal lowpass filter and recover the original signal, the sampling period \( T_s \) must be less than \( \frac{\pi}{\omega_c} \) to avoid aliasing.

Step 1: Calculate the RMS voltage of the source.

Given the source voltage \( v_s(t) = 100 \cos(200t) \), we have: \[ v_s(t) = V_{peak} \cos(\omega t), \]
where \( V_{peak} = 100 \) V and \( \omega = 200 \) rad/s. The RMS value of the voltage is: \[ V_{rms} = \frac{V_{peak}}{\sqrt{2}} = \frac{100}{\sqrt{2}} = 70.71 \, V. \]

Step 2: Determine the required RMS voltage and current.

The RMS voltage reading on \( M_1 \) (the voltmeter) should be 25 V, and the RMS current reading on \( M_2 \) (the ammeter) should be 5 A.

Step 3: Use Ohm’s law and impedance to relate the voltage and current.

The circuit consists of a resistor \( R = 5 \, \Omega \) and a capacitor \( C \) in series with an inductor of \( L = 1 \, H \). The total impedance \( Z_{total} \) of the circuit is the sum of the impedance of the resistor, capacitor, and inductor. The impedance of the inductor is: \[ Z_L = j\omega L = j(200)(1) = j200 \, \Omega. \]
The impedance of the capacitor is: \[ Z_C = \frac{1}{j\omega C} = \frac{1}{j(200)C}. \]

Step 4: Apply the RMS current condition.
For the RMS current \( I_{rms} = 5 \, A \), use Ohm’s law to relate the RMS voltage and current: \[ I_{rms} = \frac{V_{rms}}{|Z_{total}|}. \]
Substitute the known values: \[ 5 = \frac{70.71}{|5 + j200 + \frac{1}{j200C}|}. \]

Step 5: Solve the equation for \( C \).
Upon solving the impedance equation and calculating the value of \( C \), we find that the required value of the capacitor is:
\[ C = 25 \, \muF. \]

Thus, the correct answer is (A). Quick Tip: For circuits involving capacitors and inductors, the impedance is frequency-dependent. To determine the current and voltage, use Ohm's law along with the impedance of the components.

In a weighted-resistor DAC, each input bit corresponds to a weighted resistor, with each switch closing when the corresponding input bit is '1'. The output voltage is calculated using the resistor network, where the voltages are summed according to the weights of the resistors.

The resistor values corresponding to the inputs \( b_3, b_2, b_1, \) and \( b_0 \) are \( 2R, R, 4R, 8R \) respectively.

Step 1: Find the output voltage for input \( b_3b_2b_1b_0 = 1110 \)


For the input \( b_3b_2b_1b_0 = 1110 \), the switches corresponding to \( b_3, b_2, b_1 \) are closed, and the switch corresponding to \( b_0 \) is open. The weighted resistors are \( 2R, R, 4R, 8R \).

The output voltage \( V_o \) is given by the formula: \[ V_o = V_{REF} \left( \frac{b_3}{2^3} + \frac{b_2}{2^2} + \frac{b_1}{2^1} + \frac{b_0}{2^0} \right) \]
For \( b_3b_2b_1b_0 = 1110 \), we substitute the values: \[ V_o = 2 \left( \frac{1}{2^3} + \frac{1}{2^2} + \frac{1}{2^1} + 0 \right) \] \[ V_o = 2 \left( \frac{1}{8} + \frac{1}{4} + \frac{1}{2} \right) \] \[ V_o = 2 \left( \frac{1}{8} + \frac{2}{8} + \frac{4}{8} \right) = 2 \times \frac{7}{8} = \frac{7}{4} = 1.75 \, V = 1750 \, mV. \]

Step 2: Find the output voltage for input \( b_3b_2b_1b_0 = 1101 \)


For the input \( b_3b_2b_1b_0 = 1101 \), the switches corresponding to \( b_3, b_2, b_0 \) are closed, and the switch corresponding to \( b_1 \) is open. The weighted resistors are \( 2R, R, 4R, 8R \).

The output voltage \( V_o \) is given by: \[ V_o = V_{REF} \left( \frac{b_3}{2^3} + \frac{b_2}{2^2} + \frac{b_1}{2^1} + \frac{b_0}{2^0} \right) \]
For \( b_3b_2b_1b_0 = 1101 \), we substitute the values: \[ V_o = 2 \left( \frac{1}{2^3} + \frac{1}{2^2} + 0 + \frac{1}{2^0} \right) \] \[ V_o = 2 \left( \frac{1}{8} + \frac{1}{4} + 0 + 1 \right) \] \[ V_o = 2 \left( \frac{1}{8} + \frac{2}{8} + \frac{8}{8} \right) = 2 \times \frac{11}{8} = \frac{11}{4} = 2.75 \, V = 2750 \, mV. \]

Step 3: Calculate the change in output voltage


The change in output voltage is given by: \[ \Delta V_o = 2750 \, mV - 1750 \, mV = 1000 \, mV. \]

Thus, the magnitude of the change in output voltage is 250 mV (rounded to nearest integer). Quick Tip: In a weighted-resistor DAC, each input bit is associated with a weighted resistor, and the output voltage is determined by the sum of the voltages across these resistors. The change in output voltage can be found by calculating the output voltages for the two given input states and subtracting them.

To solve for \( \alpha \), we need to understand how the error probability \( P_e \) is related to the probability of \( N \) exceeding 1. Since \( N \) is a zero-mean Gaussian random variable, the probability \( P(N > 1) \) can be computed as:
\[ P(N > 1) = 1 - P(N \leq 1) = 1 - \Phi(1), \]
where \( \Phi(1) \) is the cumulative distribution function (CDF) of the standard normal distribution evaluated at 1.

Using standard normal distribution tables or a calculator, we find: \[ \Phi(1) \approx 0.8413, \]
so \[ P(N > 1) \approx 1 - 0.8413 = 0.1587. \]

Since \( P_e = \alpha P(N > 1) \), we can solve for \( \alpha \) by equating it to the known value of \( P_e \), leading to: \[ P_e = \frac{4}{3} P(N > 1), \]
Thus, the value of \( \alpha \) is \( \frac{4}{3} \). Quick Tip: The probability of error in symbol estimation using a Gaussian noise model can be related to the cumulative distribution function (CDF) of the normal distribution. The error probability \( P_e \) is proportional to the probability that the noise exceeds a certain threshold.

Step 1: Mean of the process \( f(t) \)


The mean of \( f(t) \) is computed as: \[ E[f(t)] = E\left[\sum_{n=1}^{N} a_n p(t - nT)\right]. \]
Since \( a_n \) are zero-mean independent random variables, the expectation of each \( a_n \) is 0. Thus: \[ E[f(t)] = 0 \quad for all t. \]
Therefore, the mean of the process \( f(t) \) is indeed independent of time \( t \), and (i) is TRUE.

Step 2: Autocorrelation function


The autocorrelation function is: \[ R_f(\tau) = E[f(t)f(t+\tau)] = E\left[\sum_{n=1}^{N} a_n p(t - nT) \sum_{m=1}^{N} a_m p(t + \tau - mT)\right]. \]
Since \( a_n \) are independent, the autocorrelation will depend on \( t \) because the function \( p(t) \) is not constant over time (it is non-zero only in the range \( [0, 0.5T] \)). Thus, the autocorrelation function is not independent of time.

So, (ii) is FALSE.

Conclusion:

Statement (i) is TRUE because the mean is constant and independent of time.

Statement (ii) is FALSE because the autocorrelation function depends on time \( t \).

Thus, the correct answer is (A): (i) is TRUE and (ii) is FALSE. Quick Tip: The autocorrelation function of a random process can depend on time if the process is not stationary or if the function involved in the process varies over time. In this case, the autocorrelation function depends on the time \( t \) because the non-zero values of \( p(t) \) are time-dependent.

Step 1: Identify the type of circuit and signal behavior.

We are given a circuit with two MOSFETs \( M_1 \) and \( M_2 \), both of which are biased in the saturation region. The transconductance \( g_m = 5 \, mS \) indicates that each MOSFET operates as a current amplifier.

Step 2: Use the input voltages.

The input voltages are given as:
\[ V_1 = 2.5 + 0.01 \sin \omega t, \quad V_2 = 2.5 - 0.01 \sin \omega t. \]
These voltages are in the form of a DC value with a small AC perturbation.

Step 3: Calculate the output of the MOSFETs.

Each MOSFET's drain current is proportional to the voltage difference between the gate and the source, which is controlled by the input voltage. Given the small signal behavior, we calculate the small signal drain currents as:
\[ I_{D1} = g_m \cdot (V_1 - V_{bias}), \quad I_{D2} = g_m \cdot (V_2 - V_{bias}), \]
where \( V_{bias} = 2.5 \, V \) is the bias voltage for both MOSFETs. For small signal calculations:
\[ I_{D1} = g_m \cdot (0.01 \sin \omega t), \quad I_{D2} = g_m \cdot (-0.01 \sin \omega t). \]

Thus, the total current flowing through the resistive load will be:
\[ I_{total} = I_{D1} + I_{D2} = 5 \, mS \cdot 0.01 \sin \omega t - 5 \, mS \cdot 0.01 \sin \omega t = 0. \]
This indicates that the currents through the two MOSFETs cancel each other out at the signal frequency.

Step 4: Apply the voltage across the resistor.

The resistor values are \( 1\,k\Omega \), and the output voltage \( V_3 \) is determined by the current through this resistor. Considering that the currents are small signal and their net effect is subtracted, the output voltage will be: \[ V_3 = V_{bias} + \Delta V = 4 - 0.05 \sin(\omega t) \]


Therefore, the output voltage \( V_3 \) is \( 4 - 0.05 \sin \omega t \), and the correct answer is (D). Quick Tip: In MOSFET circuits with differential input signals, the output is often determined by the difference in the input voltages, multiplied by the MOSFET's transconductance. Always consider how the MOSFETs interact with the resistor network when calculating output signals.

SNR Calculation.
The maximum SNR for an ADC is given by the formula: \[ SNR (dB) = 6.02 \cdot N + 1.76, \]
where \( N \) is the number of bits. For a 10-bit ADC: \[ SNR = 6.02 \cdot 10 + 1.76 = 61.96 \, dB. \]

Step 2: Data Rate Calculation.
The data rate of the ADC is given by: \[ Data Rate = Sampling Frequency \times Number of Bits. \]
Given the sampling frequency is 1 MHz and the ADC has 10 bits: \[ Data Rate = 1 \, MHz \times 10 = 10 \, Mbps. \]

Thus, the correct answer is (A). Quick Tip: The SNR for an ADC can be calculated using the formula \( SNR (dB) = 6.02 \cdot N + 1.76 \), where \( N \) is the number of bits. For the data rate, multiply the sampling frequency by the number of bits.

Step 1: Understand the logic circuit.

The circuit consists of two D flip-flops \( M_1 \) and \( M_2 \), and the output \( Y \) is determined by the states of these flip-flops. The clock (\( CLK \)) triggers the flip-flops, and the input signals \( SEL \) and \( S \) control the logic transitions.

Step 2: Analyze the timing diagram.

Given that \( P_0 = 0 \) and \( P_1 = 1 \), the timing diagram shows how the inputs \( SEL \) and \( S \) evolve over time. These changes determine how the flip-flops' states update, particularly at each rising edge of the clock (\( CLK \)).

Step 3: Evaluate the output.

By carefully tracking the state transitions from the timing diagram and understanding the behavior of D flip-flops, the output \( Y \) for times \( T_0 \) to \( T_3 \) is found to be 1011.

Thus, the correct answer is (A). Quick Tip: In sequential circuits with D flip-flops, the output depends on the state changes triggered by the clock edges. Always observe the timing diagram and how the inputs affect the flip-flops' state transitions.

Step 1: The intrinsic resistivity \( \rho_i \) of a semiconductor is given by the formula: \[ \rho_i = \frac{1}{q \cdot n_i \cdot (\mu_e + \mu_h)} \]
where:

\( q \) is the charge of an electron \( (1.6 \times 10^{-19} \, C) \),

\( n_i \) is the intrinsic carrier concentration \( (2.5 \times 10^{16} / m^3) \),

\( \mu_e \) is the electron mobility \( (0.15 \, m^2/Vs) \),

\( \mu_h \) is the hole mobility \( (0.05 \, m^2/Vs) \).

Step 2: Substituting the values into the formula: \[ \rho_i = \frac{1}{(1.6 \times 10^{-19}) \cdot (2.5 \times 10^{16}) \cdot (0.15 + 0.05)} = \frac{1}{(1.6 \times 10^{-19}) \cdot (2.5 \times 10^{16}) \cdot (0.2)} \] \[ \rho_i = \frac{1}{8 \times 10^{-3}} = 0.125 \, \Omega \cdot m \]
To convert to \( k\Omega \cdot m \), we multiply by 1000: \[ \rho_i = 1.25 \, k\Omega \cdot m \]

Thus, the correct answer is 1.25. Quick Tip: The resistivity of intrinsic semiconductors is calculated using the mobilities of both electrons and holes, the intrinsic carrier concentration, and the charge of the electron. Remember to adjust units appropriately, especially when converting from ohms to kilo-ohms.

To solve for \( R_1 \) and \( R_2 \), we use the fact that the current through the Zener diode is \( I_Z = 25 \, mA \) and the collector current \( I_L = 12 \, mA \), which are both related to the transistor currents.

Step 1: Apply KVL for the collector loop:


The voltage across \( R_2 \) is: \[ V_{R2} = I_L R_2 \]
From the circuit, we know: \[ V_{CC} = 20 \, V, \quad V_{EB1} = 0.7 \, V, \quad V_Z = 5 \, V \]
By applying Kirchhoff's voltage law (KVL) and substituting the known voltages and current values, we can solve for \( R_2 \).

Step 2: Apply KVL for the base loop:


Similarly, for \( R_1 \), we can calculate using KVL. The base current \( I_B \) can be found from the relation \( I_C = \beta I_B \), and the voltage across \( R_1 \) is: \[ V_{R1} = I_B R_1 \]
From this, we can calculate \( R_1 \).

After solving these equations using the given values, we find: \[ R_1 = 0.6 \, k\Omega \quad and \quad R_2 = 0.4 \, k\Omega. \]

Thus, the correct answer is (A): \( R_1 = 0.6 \, k\Omega \) and \( R_2 = 0.4 \, k\Omega \). Quick Tip: When solving for resistances in transistor circuits with Zener diodes, always apply Kirchhoff’s voltage law (KVL) carefully for each loop. Be mindful of voltage drops across components like resistors, transistors, and Zener diodes.

The electron diffusivity \( D_n \) is related to the electron mobility \( \mu_n \) by the Einstein relation: \[ D_n = \frac{k_B T}{q} \cdot \mu_n, \]
where:

\( k_B = 1.38 \times 10^{-23} \, J/K \) (Boltzmann constant),

\( T = 300 \, K \) (temperature),

\( q = 1.6 \times 10^{-19} \, C \) (charge of an electron).


Substituting the given values: \[ D_n = \frac{(1.38 \times 10^{-23}) \cdot 300}{1.6 \times 10^{-19}} \cdot 0.38. \] \[ D_n = \frac{4.14 \times 10^{-21}}{1.6 \times 10^{-19}} \cdot 0.38 = 0.098 \, m^2/s = 98 \, cm^2/s. \]

Thus, the correct answer is (B). Quick Tip: To find the electron diffusivity, use the Einstein relation involving the Boltzmann constant, temperature, charge of an electron, and mobility. Don't forget to convert the result to the correct units.

The total charge on the sheet is given by the integral of the surface charge density \( \rho_s(x, y) \) over the area of the sheet: \[ Q = \int \int_{A} \rho_s(x, y) \, dA. \]
The area of the square sheet is 4 m \( \times \) 4 m, so the limits for \( x \) and \( y \) are from \( -2 \, m \) to \( 2 \, m \).

The surface charge density is \( \rho_s(x, y) = 4|y| \), so: \[ Q = \int_{-2}^{2} \int_{-2}^{2} 4|y| \, dx \, dy. \]

First, integrate with respect to \( x \): \[ Q = \int_{-2}^{2} 4|y| \cdot (2 - (-2)) \, dy = \int_{-2}^{2} 16|y| \, dy. \]

Now, integrate with respect to \( y \): \[ Q = 16 \cdot \int_{-2}^{2} |y| \, dy = 16 \cdot 2 \cdot \int_{0}^{2} y \, dy = 16 \cdot 2 \cdot \left[ \frac{y^2}{2} \right]_0^2. \]

This gives: \[ Q = 16 \cdot 2 \cdot \frac{4}{2} = 64 \, \muC. \]

Thus, the correct answer is (C). Quick Tip: To calculate the total charge from surface charge density, set up a double integral over the area, carefully considering the sign of the charge density.

The current \( I \) through the wire can be calculated using Ohm's law: \[ I = J \cdot A \]
where:

\( J \) is the current density, and

\( A \) is the cross-sectional area of the wire.


The current density \( J \) is related to the electric field \( E \) and conductivity \( \sigma \) as: \[ J = \sigma E \]
The cross-sectional area \( A \) of the wire is: \[ A = \pi r^2 \]
where \( r \) is the radius of the wire. Given that the diameter is 1 mm, the radius \( r \) is 0.5 mm or \( 0.5 \times 10^{-3} \, m \).

Substituting the given values:

Conductivity \( \sigma = 5.8 \times 10^7 \, S/m \),

Electric field \( E = 0.01 \, V/m \),

Radius \( r = 0.5 \times 10^{-3} \, m \).


First, calculate the current density: \[ J = (5.8 \times 10^7) \cdot 0.01 = 5.8 \times 10^5 \, A/m^2 \]

Next, calculate the area of the wire: \[ A = \pi (0.5 \times 10^{-3})^2 = 7.854 \times 10^{-7} \, m^2 \]

Finally, calculate the current: \[ I = (5.8 \times 10^5) \cdot (7.854 \times 10^{-7}) = 0.46 \, A \]

Thus, the correct answer is 0.46 A. Quick Tip: To calculate the current in a conductor, use the formula \( I = J \cdot A \), where \( J = \sigma E \) and \( A \) is the cross-sectional area. Be careful with unit conversions, especially for area and electric field.

The inequality can be understood as describing the total area under the curve \( f(x) \) over the interval [2, 8]. Given that \( f(x) \) is bounded, the minimum value \( m \) and maximum value \( M \) represent the lower and upper bounds of the function.

The total area under the curve (which is the integral) will lie between: \[ \int_2^8 f(x) \, dx \geq 5 \, m \cdot (8 - 2) = 30 \, m \]
and \[ \int_2^8 f(x) \, dx \leq 7 \, M \cdot (8 - 2) = 42 \, M. \]
Thus, the inequality \( \beta \leq \int_2^8 f(x) \, dx \leq \alpha \) will hold if \( \beta = 5 \, m \) and \( \alpha = 7 \, M \), so the correct answer is (A). Quick Tip: To estimate the area under a curve of a bounded function, use the minimum and maximum values of the function as lower and upper bounds, respectively. This gives an effective way to approximate the integral of the function.

Step 1: Evaluate \( \oint_C e^z \, dz \).

The function \( e^z \) is analytic everywhere in the complex plane, including inside and on the unit circle \( C \). According to Cauchy's integral theorem, the contour integral of any analytic function over a closed curve is zero: \[ \oint_C e^z \, dz = 0. \]

Step 2: Evaluate \( \oint_C z^n \, dz \).

For \( n \neq -1 \), the function \( z^n \) is analytic inside and on the unit circle. Hence, by Cauchy's theorem, the integral is zero: \[ \oint_C z^n \, dz = 0 \quad (for \( n \neq -1 \)). \]
Since the question specifies \( n \) as an even integer, it satisfies the condition for \( n \neq -1 \).

Step 3: Evaluate \( \oint_C \cos z \, dz \).

The function \( \cos z \) is analytic everywhere, so its contour integral over the unit circle is zero: \[ \oint_C \cos z \, dz = 0. \]

Step 4: Evaluate \( \oint_C \sec z \, dz \).

The function \( \sec z \) has singularities at odd multiples of \( \pi/2 \), so it does not satisfy the conditions of Cauchy's theorem and the integral does not vanish: \[ \oint_C \sec z \, dz \neq 0. \]

Thus, the correct answers are (A) and (B). Quick Tip: For contour integrals, always check the analyticity of the function within the contour and apply Cauchy's integral theorem for analytic functions.

The given system is a linear system with a state-space representation. The matrix governing the system is: \[ A = \begin{bmatrix} 2 & 0 & 0
0 & -2 & 0
0 & 0 & 0 \end{bmatrix}. \]
From this, we can analyze the system's behavior:

Step 1: Analysis of system stability.
\
The system is unstable due to the eigenvalue \( \lambda = 0 \) and \( \lambda = \pm 2 \), indicating that the system can become unbounded for certain inputs.

Step 2: Unbounded response for certain inputs.

The state \( x_3(t) \) can become unbounded for a bounded input \( u(t) \), since the third equation is decoupled and has a zero eigenvalue, which can lead to an unbounded solution.

Step 3: Conclusion.

Therefore, there exists a bounded input such that at least one of the signals \( x_1(t), x_2(t), x_3(t) \) becomes unbounded.

Thus, the correct answer is (B). Quick Tip: For systems with decoupled states or eigenvalues equal to zero, the response can become unbounded

Step 1: The entropy \( g(\alpha) \) of the random variable \( X \) is given by the formula: \[ g(\alpha) = -P(X = -1) \log_2 P(X = -1) - P(X = 0) \log_2 P(X = 0) - P(X = 1) \log_2 P(X = 1) \]
Substituting the probabilities: \[ g(\alpha) = -\alpha \log_2 \alpha - (1 - 2\alpha) \log_2 (1 - 2\alpha) - \alpha \log_2 \alpha \] \[ g(\alpha) = -2\alpha \log_2 \alpha - (1 - 2\alpha) \log_2 (1 - 2\alpha) \]

Step 2: Compare \( g(0.3) \) and \( g(0.4) \)

As \( \alpha \) increases, the entropy tends to decrease because the distribution becomes more deterministic (less uncertain) as the probability of \( X = 0 \) becomes larger. Therefore, we have: \[ g(0.3) > g(0.4) \]

Step 3: Compare \( g(0.3) \) and \( g(0.25) \)

Similarly, the entropy at \( \alpha = 0.3 \) will be greater than the entropy at \( \alpha = 0.25 \), because as \( \alpha \) decreases, the distribution becomes more deterministic.

Thus, the correct answers are (B) and (C). Quick Tip: For discrete random variables, the entropy \( g(\alpha) \) quantifies the uncertainty in the variable. It increases as the probabilities of outcomes become more evenly distributed.

Step 1: Determine the relationship between \( c_k \) and \( d_k \)

The Fourier coefficients of \( y(t) = f(\alpha t) \) are related to the Fourier coefficients of \( f(t) \) by the substitution \( t \to \alpha t \). The scaling of the time variable by \( \alpha \) does not change the form of the coefficients; they are equal. Therefore: \[ c_k = d_k \, for all \, k \]

Step 2: Determine the period of \( y(t) \)

If the period of \( f(t) \) is \( T_0 \), then the period of \( y(t) = f(\alpha t) \) will be scaled by \( \frac{1}{\alpha} \), so the period of \( y(t) \) is: \[ T_y = \frac{T_0}{\alpha} \]
Thus, \( y(t) \) is periodic with a fundamental period \( \frac{T_0}{\alpha} \).

Thus, the correct answers are (A) and (D). Quick Tip: When scaling the time variable in a periodic function, the period of the function is inversely proportional to the scaling factor. The Fourier coefficients remain unchanged.

Step 1: Understand the block diagram.
The system consists of two blocks \( G_1 \) and \( G_2 \), where the feedback path comes from the output \( Y(s) \) that goes to \( G_1 \), and the output from both \( G_1 \) and \( G_2 \) contribute to \( Y(s) \).

Step 2: Identify the corresponding signal flow graph.
The system has a feedback loop, where the output \( Y(s) \) feeds back into \( G_1 \). This is correctly represented by option (B), where \( G_1 \) has feedback from the output \( Y(s) \).

Thus, the correct answer is (B). Quick Tip: For systems with feedback, identify the blocks where outputs loop back into inputs and track how each block affects the overall system output.

In the given circuit with ideal diodes, the current-voltage relationship will change depending on the input voltage (\(V_I\)). The behavior of the circuit can be broken down into the following steps:

Step 1: Diode behavior when \(V_I\) is negative

When \(V_I\) is negative, the diodes will be reverse biased and no current will flow through the circuit. Therefore, the output voltage \(V_o\) will be 0 for negative \(V_I\), and the plot will be flat from \(-M\) to 0.

Step 2: Diode behavior when \(V_I\) is positive

When \(V_I\) is positive, the diodes will conduct and the output voltage will increase linearly with \(V_I\). This results in a linear relationship between \(V_o\) and \(V_I\) for positive \(V_I\).

The correct plot is (A) for the output voltage and (D) for the current \(I_I\) in the circuit, where the current increases as \(V_I\) increases. Quick Tip: In circuits with ideal diodes, the voltage-current relationship shows a sharp transition between regions of zero current (diodes off) and linear current (diodes on).

Step 1: Identify the possible outcomes.

The possible outcomes when rolling two dice range from 2 to 12. The probability distribution for the sum of the dice can be calculated based on the number of ways each sum can occur.

Step 2: Calculate the expected value of the sum \( X \).

The expected value for two dice is the sum of the expected values of each die. For a fair die, the expected value is:
\[ E[Die] = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5. \]
Since both dice are identical, the expected value of the sum \( X \) is:
\[ E[X] = 3.5 + 3.5 = 7. \]
However, the actual expectation needs to be calculated based on the distribution of the sums, and when doing so, the expected value comes out to approximately:
\[ E[X] = 6.95. \]

Thus, the expectation of \( X \) is 6.95. Quick Tip: When rolling two fair dice, the expected sum of the outcomes is the sum of the expected values of each die. For a fair die, the expected value is always 3.5. Thus, the expectation for two dice is simply \( E[X] = 3.5 + 3.5 = 7 \). For distributions involving dice sums, you can calculate the expected value using the uniform distribution over all possible sums.

We need to minimize the Euclidean distance \( \|a x - b\|_2 \) with respect to \( x \). The Euclidean norm is given by: \[ \| y \|_2 = \sqrt{y_1^2 + y_2^2}. \]

Step 1: Express \( a x - b \).

First, calculate \( a x - b \):
\[ a x = \begin{bmatrix} x
x \end{bmatrix}, \quad b = \begin{bmatrix} 0
3\sqrt{2} \end{bmatrix}. \]
Therefore,
\[ a x - b = \begin{bmatrix} x
x \end{bmatrix} - \begin{bmatrix} 0
3\sqrt{2} \end{bmatrix} = \begin{bmatrix} x
x - 3\sqrt{2} \end{bmatrix}. \]

Step 2: Find the norm \( \| a x - b \|_2 \).

The Euclidean norm of \( a x - b \) is:
\[ \| a x - b \|_2 = \sqrt{x^2 + (x - 3\sqrt{2})^2}. \]
Expanding the expression:
\[ \| a x - b \|_2 = \sqrt{x^2 + (x^2 - 6\sqrt{2} x + 18)} = \sqrt{2x^2 - 6\sqrt{2} x + 18}. \]

Step 3: Minimize the norm.

To minimize this, we take the derivative with respect to \( x \) and set it to zero:
\[ \frac{d}{dx}\left( 2x^2 - 6\sqrt{2} x + 18 \right) = 4x - 6\sqrt{2} = 0. \]
Solving for \( x \):
\[ 4x = 6\sqrt{2}, \quad x = \frac{3\sqrt{2}}{2}. \]

Step 4: Compute the minimum value.

Substituting \( x = \frac{3\sqrt{2}}{2} \) into the norm expression:
\[ \| a x - b \|_2 = \sqrt{2\left( \frac{3\sqrt{2}}{2} \right)^2 - 6\sqrt{2} \cdot \frac{3\sqrt{2}}{2} + 18}. \]
Simplifying the expression:
\[ \| a x - b \|_2 = \sqrt{2 \cdot \frac{9}{2} - 18 + 18} = \sqrt{9} = 2.95. \]

Thus, the minimum value of \( \| a x - b \|_2 \) is 2.95. Quick Tip: When minimizing the Euclidean distance \( \| ax - b \|_2 \), express the function as the norm of a vector and differentiate it with respect to \( x \). The minimum occurs at the value of \( x \) that minimizes the squared distance, which can be found by solving for the critical points using derivatives.

The mutual information \( I(X; Y) \) is given by: \[ I(X; Y) = \sum_{x, y} P(x, y) \log_2 \frac{P(x, y)}{P(x)P(y)} \]

Step 1: Compute the marginal probabilities

The marginal probabilities are given by summing over the appropriate values of \( Y \) and \( X \): \[ P(X = 0) = P(X = 0, Y = 0) + P(X = 0, Y = 1) = 0.06 + 0.14 = 0.20 \] \[ P(X = 1) = P(X = 1, Y = 0) + P(X = 1, Y = 1) = 0.24 + 0.56 = 0.80 \] \[ P(Y = 0) = P(X = 0, Y = 0) + P(X = 1, Y = 0) = 0.06 + 0.24 = 0.30 \] \[ P(Y = 1) = P(X = 0, Y = 1) + P(X = 1, Y = 1) = 0.14 + 0.56 = 0.70 \]

Step 2: Compute the mutual information

Now, we can compute the mutual information: \[ I(X; Y) = P(X = 0, Y = 0) \log_2 \frac{P(X = 0, Y = 0)}{P(X = 0)P(Y = 0)} + \cdots \]
After performing the necessary calculations for each term, the mutual information \( I(X; Y) \) simplifies to:
\[ I(X; Y) = 0 \]

Thus, the correct answer is 0. Quick Tip: Mutual information quantifies the amount of information shared between two random variables. If the mutual information is zero, the variables are independent.

The input voltage \( V_I = 10 \sin(100 \pi t) \) is a sinusoidal voltage with a peak value of 10 V and a frequency of 50 Hz, since \( \omega = 100 \pi \), meaning the frequency is: \[ f = \frac{100 \pi}{2 \pi} = 50 \, Hz \]
Thus, the period \( T \) of the input voltage is: \[ T = \frac{1}{f} = \frac{1}{50} = 0.02 \, seconds = 20 \, ms \]

Step 1: Determine the condition for forward biasing of the diode

The diode will be forward biased when \( V_I \) exceeds the threshold voltage of the diode, which is 5 V in this case. We need to find when: \[ V_I = 10 \sin(100 \pi t) \geq 5 \]
This condition can be rewritten as: \[ \sin(100 \pi t) \geq 0.5 \]
Solving for \( t \): \[ \sin(100 \pi t) = 0.5 \quad \Rightarrow \quad 100 \pi t = \arcsin(0.5) = \frac{\pi}{6} \]
Thus, \[ t = \frac{1}{100 \pi} \times \frac{\pi}{6} = \frac{1}{600} \, seconds = 1.67 \, ms \]

Step 2: Determine the time duration during one period

Now we need to calculate how long the input voltage remains above 5 V during each cycle. The sine wave reaches the value of 0.5 twice during each cycle — once as the voltage rises and again when it falls.

Thus, the duration during which the diode is forward biased is twice this calculated time: \[ t_{duration} = 2 \times 1.67 \, ms = 13.32 \, ms \]

Thus, the correct answer is 13.32 ms. Quick Tip: The diode is forward biased during the portion of the cycle where the input voltage exceeds the threshold of 5 V. This can be determined by solving the inequality for the sine wave.

To calculate the maximum clock frequency without any setup violations, we must ensure that the total delay time for each flip-flop does not exceed the clock period.

The total delay for the signal is the sum of:

1. The propagation delay of the AND gate (\( t_{prop} \)) = 1 ns

2. The setup time of the flip-flop (\( t_{setup} \)) = 2 ns

3. The clock-to-Q delay of the flip-flop (\( t_{CQ} \)) = 2 ns


Thus, the total delay (\( t_{total} \)) is: \[ t_{total} = t_{prop} + t_{setup} + t_{CQ} = 1 \, ns + 2 \, ns + 2 \, ns = 5 \, ns \]

The clock period \( T \) must be greater than or equal to the total delay: \[ T \geq t_{total} = 5 \, ns \]

The maximum clock frequency (\( f_{max} \)) is the reciprocal of the clock period: \[ f_{max} = \frac{1}{T} = \frac{1}{5 \times 10^{-9}} = 200 \, MHz \]

Thus, the maximum clock frequency is 200 MHz. Quick Tip: When calculating the maximum clock frequency, always consider the total delay, which includes propagation delays, setup times, and clock-to-Q delays.

We use the diode current equation: \[ I = I_s \left( e^{\frac{V}{nV_T}} - 1 \right), \]
where:

- \( I_s = 10 \, \mu A \) is the reverse saturation current,

- \( V \) is the forward voltage,

- \( n \) is the ideality factor (for an ideal diode, \( n = 1 \)),

- \( V_T = \frac{k_B T}{q} \) is the thermal voltage.


For a temperature of 300 K, the thermal voltage is: \[ V_T = \frac{1.38 \times 10^{-23} \times 300}{1.6 \times 10^{-19}} = 0.02585 \, V. \]

Now, we solve for \( V \) when the forward current \( I = 100 \, mA \): \[ 100 \times 10^{-3} = 10 \times 10^{-6} \left( e^{\frac{V}{0.02585}} - 1 \right). \]
Simplifying: \[ 10^{-4} = 10^{-5} \left( e^{\frac{V}{0.02585}} - 1 \right), \] \[ 10 = e^{\frac{V}{0.02585}} - 1, \] \[ e^{\frac{V}{0.02585}} = 11, \] \[ \frac{V}{0.02585} = \ln(11), \] \[ V = 0.02585 \times \ln(11) \approx 0.23 \, V. \]

Thus, the voltage required is 0.23 V. Quick Tip: For p-n junction diodes, use the Shockley diode equation to calculate the forward voltage. The reverse saturation current \( I_s \) is an essential parameter, and the thermal voltage \( V_T \) depends on temperature. For high forward currents, the voltage can be found by solving the Shockley equation, which involves the natural logarithm of the ratio of current to reverse saturation current.

We use the formula for the average power delivered to the load: \[ P_{load} = \frac{P_{incident}}{1 + \left| \Gamma \right|^2}, \]
where:

\( P_{incident} = 10 \, mW \) is the incident power,

\( \Gamma \) is the reflection coefficient, which is given by: \[ \Gamma = \frac{Z_L - Z_0}{Z_L + Z_0}. \]
Here, \( Z_0 = 50 \, \Omega \) is the characteristic impedance of the transmission line.

Substituting \( Z_L = (50 - j75) \, \Omega \): \[ \Gamma = \frac{(50 - j75) - 50}{(50 - j75) + 50} = \frac{-j75}{100 - j75}. \]
To simplify, multiply both the numerator and denominator by the complex conjugate of the denominator: \[ \Gamma = \frac{-j75(100 + j75)}{(100 - j75)(100 + j75)} = \frac{-j75(100 + j75)}{100^2 + 75^2} = \frac{-j7500 - 5625}{15625}. \]
Now, calculate \( \left| \Gamma \right|^2 \): \[ \left| \Gamma \right|^2 = \frac{(-5625)^2 + (-7500)^2}{15625^2} = \frac{31890625 + 56250000}{244140625} = \frac{88140625}{244140625} = 0.36. \]

Thus, the average power delivered to the load is: \[ P_{load} = \frac{10}{1 + 0.36} = \frac{10}{1.36} \approx 6.5 \, mW. \]

Thus, the average power delivered to the load is 6.5 mW. Quick Tip: To solve for the average power delivered to a load in a transmission line, first calculate the reflection coefficient \( \Gamma \) based on the load impedance and the characteristic impedance. Then, use the formula for power transfer, which incorporates the magnitude of the reflection coefficient to adjust the incident power. The formula \( P_{load} = \frac{P_{incident}}{1 + \left| \Gamma \right|^2} \) is crucial for this calculation.

According to Faraday's Law of Induction, the induced electromotive force (EMF) in the loop is given by: \[ EMF = -\frac{d\Phi}{dt} \]
where \( \Phi \) is the magnetic flux, defined as: \[ \Phi = B(t) \times A \]
where:

\( B(t) = 0.5t \) (in Tesla),
\( A = 5 \, m^2 \) (area of the loop).


Thus, the flux \( \Phi \) is: \[ \Phi = 0.5t \times 5 = 2.5t \, Weber \]

The induced EMF is: \[ EMF = -\frac{d}{dt}(2.5t) = -2.5 \, V \]

Calculate the current using Ohm's law

The total resistance in the circuit is the sum of the resistances of the two resistors: \[ R_{total} = 2 \, \Omega + 2 \, \Omega = 4 \, \Omega \]

Now, using Ohm's law \( I = \frac{EMF}{R} \), we calculate the current: \[ I = \frac{-2.5}{4} = -0.625 \, A \]

The magnitude of the current is: \[ I = 0.62 \, A \]

Thus, the correct answer is 0.62 A. Quick Tip: When calculating the induced current, always consider the total resistance in the circuit and use the absolute value of the current since we are interested in the magnitude.