Question 1 A material property which is characterized by a linear proportional relationship between the stress and strain in a stress-strain curve for a metal is called Poisson's ratio
tensile strength O yield strength
O modulus of elasticity
Question 2 On a typical tensile stress-strain curve for metals, the elastic region is represented by
a non-linear portion of the curve the maximum point of the curve
a straight line of positive gradient
the area under the curve

(a)will exhibit significant aliasing. (b)there will be some degree of aliasing, less pronounced compare to 2000 Hz. (c) overlapping replicas covering entire spectrum. (d) will not exhibit aliasing (e) satisfy Nyquist criterion

Given:

Bandlimited signal: g(t)

Fourier transform of g(t): G(f) = Π(f/4000)

(a) Sampling frequency (fs) = 2000 Hz:

The Nyquist-Shannon sampling theorem states that the sampling frequency should be at least twice the bandwidth of the signal to avoid aliasing. Since the signal is bandlimited to 4000 Hz, the minimum required sampling frequency would be 8000 Hz. Therefore, sampling at 2000 Hz is below the Nyquist rate, and aliasing will occur. The spectrum of the sampled signal will show overlapping replicas of the original spectrum.

(b) Sampling frequency (fs) = 3000 Hz:

With a sampling frequency of 3000 Hz, we are still below the Nyquist rate but closer to it. The spectrum of the sampled signal will still exhibit some degree of aliasing, but the overlapping replicas will be less pronounced compared to the case of 2000 Hz.

(c) Sampling frequency (fs) = 4000 Hz:

At the Nyquist rate of 4000 Hz, the spectrum of the sampled signal will have replicas that are exactly overlapping. The replicas will cover the entire spectrum of the original signal.

(d) Sampling frequency (fs) = 8000 Hz:

Sampling at 8000 Hz is above the Nyquist rate. The spectrum of the sampled signal will not exhibit any aliasing, and the replicas will be non-overlapping.

(e) Reconstruction using an ideal LPF with a cutoff frequency of fs/2:

To reconstruct the original signal g(t) accurately, the sampling frequency must satisfy the Nyquist criterion. This means that the sampling frequency should be greater than twice the bandwidth of the signal. Since the bandwidth of g(t) is 4000 Hz, the sampling frequency should be greater than 8000 Hz. Among the given sampling frequencies, only fs = 8000 Hz satisfies this criterion. Therefore, using a sampling frequency of 8000 Hz will result in the same signal g(t) after reconstruction.

(a) At a sampling frequency of 2000 Hz, the spectrum of the sampled signal will exhibit significant aliasing.

(b) At a sampling frequency of 3000 Hz, there will still be some degree of aliasing, but it will be less pronounced compared to 2000 Hz.

(c) At a sampling frequency of 4000 Hz, the spectrum of the sampled signal will have overlapping replicas covering the entire spectrum.

(d) At a sampling frequency of 8000 Hz, the spectrum of the sampled signal will not exhibit aliasing, and the replicas will be non-overlapping.

(e) Using a sampling frequency of 8000 Hz satisfies the Nyquist criterion for reconstructing the original signal g(t) accurately.

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The C# tank game involves two tanks colliding, and the objective is to destroy the opponent's tank by reducing its health through turn-based attacks.

The C# tank game involves two tanks colliding, and the objective is to destroy

the opponent's tank by attacking and reducing their health using turn-based attacks until one tank's health reaches zero.

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Mooring and bottom tracking are two widely used methods to measure ocean currents. Although both methods have their advantages and disadvantages, mooring offers more advantages than bottom tracking.

A mooring is a stationary instrument array that is anchored to the seafloor and is used to track current speed, direction, temperature, salinity, and other oceanographic parameters over time. It contains a string of instruments that are installed at various depths, with each instrument measuring different oceanographic parameters. The mooring array transmits data to a surface buoy, which relays it to a shore station via satellite or radio.

The mooring is retrieved after a set time, and the data is analyzed. The speed and direction of the current can be determined by analyzing the data. This method is useful in measuring the surface and near-surface. Bottom tracking is not useful in areas where ships cannot go. Bottom tracking does not provide a long-term record of current speed, direction, and other parameters.

Bottom tracking requires the use of a ship, which can be costly and time-consuming. In conclusion, direction, temperature, and other parameters, does not provide a long-term record of current speed, direction, and other parameters.

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The nearest standard compound eluting before and after Octane are Heptane and Nonane, respectively. Also, the nearest standard compound eluting before and after Nonane are Octadecane and Heptadecane, respectively.For Octane:KI = 100 × (7.5 - 4.0) / (15.5 - 7.5) + 0 = 55.56For Nonane:KI = 100 × (15.5 - 7.5) / (16.2 - 15.5) + 81.1 = 90.91Hence, the Kovats Index of Octane and Nonane are 55.56 and 90.91, respectively.

1. The adjusted retention time is the retention time that the compound would have if it were in a hypothetical column with a stationary phase that does not interact with the solute and is equal in length to the dead time. The retention factor is the ratio of the time the solute is retained in the column to the time it spends in the mobile phase.a. Analyte C:Adjusted retention time (tR') = 5.0 - 4.0 = 1.0 minRetention factor (k) = (tR - t0) / t0 = (5.0 - 4.0) / 4.0 = 0.25b. Analyte D:Adjusted retention time (tR') = 10.0 - 4.0 = 6.0 minRetention factor (k) = (tR - t0) / t0 = (10.0 - 4.0) / 4.0 = 1.5(c) Analyte CH4:Adjusted retention time (tR') = 4.0 - 4.0 = 0 minRetention factor (k) = (tR - t0) / t0 = (4.0 - 4.0) / 4.0 = 0As shown in the above calculation, the adjusted retention time and retention factor of the analytes C, D and CH4 are as follows.AnalyteAdjusted retention time (tR')Retention factor (k)C1.0 min0.25D6.0 min1.5CH40 min0

2. Tocalculate the Kovats Index of Oc

tane and Nonane, we can use the formula as follows.Kovats Index = 100 × (tR - t0) / (tR n+1 - tR n) + KI nwhere tR = retention time of the unknown compoundt0 = dead time of the columnn = the nearest standard compound eluting before the unknown compound, n+1 is the nearest standard compound eluting after the unknown compound.KI n is the Kovats Index of the nearest standard compound eluting before the unknown compound.According to the question, the tR of Octane and Nonane is 7.5 and 15.5 minutes.

Therefore, the nearest standard compound eluting before and after Octane are Heptane and Nonane, respectively. Also, the nearest standard compound eluting before and after Nonane are Octadecane and Heptadecane, respectively.For Octane:KI = 100 × (7.5 - 4.0) / (15.5 - 7.5) + 0 = 55.56For Nonane:KI = 100 × (15.5 - 7.5) / (16.2 - 15.5) + 81.1 = 90.91Hence, the Kovats Index of Octane and Nonane are 55.56 and 90.91, respectively.

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A PCI arbiter in software manages access to a PCI bus by multiple devices. This flowchart describes the process of how the arbiter prioritizes and grants access to the bus.

The flowchart for implementing a PCI arbiter in software can be divided into several steps. Firstly, the arbiter receives requests from multiple devices that need access to the PCI bus. These requests are typically in the form of signals or interrupt requests. The arbiter then evaluates the requests based on a predefined priority scheme.
The next step involves determining the highest priority request among all the pending requests. The arbiter compares the priority levels of the requests and selects the highest priority one. If multiple requests have the same priority, the arbiter may use a round-robin or other fair arbitration algorithm to ensure fairness among the devices.
Once the highest priority request is identified, the arbiter grants access to the PCI bus to the corresponding device. This involves enabling the appropriate control signals to allow the device to perform data transfers on the bus. The arbiter may also initiate any necessary handshaking protocols to ensure proper communication between the device and the bus.
After granting access, the arbiter continues to monitor the status of the bus. It waits for the current transaction to complete before reevaluating the pending requests and repeating the arbitration process. This ensures that each device receives fair access to the bus based on their priority levels.
In summary, the flowchart for implementing a PCI arbiter in software involves receiving and evaluating requests from multiple devices, determining the highest priority request, granting access to the PCI bus, and continuously monitoring the bus for further requests. The arbiter's role is to prioritize and coordinate access to the bus, ensuring efficient and fair utilization among the connected devices.

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The capacity of a channel can be calculated using the formula:

Capacity = B * log2(1 + SNR) where B is the bandwidth of the channel and SNR is the signal-to-noise ratio.

In this case, the bandwidth (B) of the channel is 4 MHz - 3 MHz = 1 MHz.

Converting the SNR from decibels to a linear scale:

SNR_linear = 10^(SNR/10) = 10^(24/10) = 251.18864

Now, we can calculate the capacity:

Capacity = 1 MHz * log2(1 + 251.18864) ≈ 1 MHz * log2(252.18864) ≈ 1 MHz * 7.97015 ≈ 7.97015 Mbps

Therefore, the capacity of the channel is approximately 7.97015 Mbps.

b) To determine the number of signaling levels required to hit that capacity, we can use the formula:

Number of signaling levels = 2^(Capacity/B)

where Capacity is in bits per second and B is the bandwidth in Hz.

In this case, the capacity is 7.97015 Mbps (megabits per second) and the bandwidth is 1 MHz (1,000,000 Hz).

Number of signaling levels = 2^(7.97015 * 10^6 / 1 * 10^6) = 2^7.97015 ≈ 2^8 ≈ 256

Therefore, approximately 256 signaling levels will be required to hit the capacity of the channel.

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(a) The reflection coefficient can be calculated using the formula:

G0 = (h2 - h1) / (h2 + h1)

(b) GTE, can be calculated using the following formula:

GTE = (h1 * tan(q1) - h2 * sqrt(1 - (h1/h2 * sin(q1))^2)) / (h1 * tan(q1) + h2 * sqrt(1 - (h1/h2 * sin(q1))^2))

(c) GTM, can be calculated using the following formula:

GTM = (h2 * tan(q1) - h1 * sqrt(1 - (h1/h2 * sin(q1))^2)) / (h2 * tan(q1) + h1 * sqrt(1 - (h1/h2 * sin(q1))^2))

(d) The input impedance, hin, at a distance of q ( = bl) length from the G0 measurement point can be calculated using the formula:

hin = Z0 * (1 + G0 * exp(-2j*q))

(e) G0 measurement point can be calculated using the formula:

Gin = (hin - Z0) / (hin + Z0)

(a) The reflection coefficient of the E field input at right angles to the h1/h2 interface, G0, can be calculated using the following formula:

G0 = (h2 - h1) / (h2 + h1)

(b) For a (TE) plane wave with an E field parallel to the interface hitting the h1/h2 interface at an angle of q1, the E-field reflection coefficient when hitting, GTE, can be calculated using the following formula:

GTE = (h1 * tan(q1) - h2 * sqrt(1 - (h1/h2 * sin(q1))^2)) / (h1 * tan(q1) + h2 * sqrt(1 - (h1/h2 * sin(q1))^2))

(c) For a (TM) plane wave with an H field parallel to the interface hitting the h1/h2 interface at an angle of q1, the E-field reflection coefficient when hitting, GTM, can be calculated using the following formula:

GTM = (h2 * tan(q1) - h1 * sqrt(1 - (h1/h2 * sin(q1))^2)) / (h2 * tan(q1) + h1 * sqrt(1 - (h1/h2 * sin(q1))^2))

The formulas mentioned above involve the material ratio h1/h2 and the angle of incidence q1.

(d) The input impedance, hin, at a distance of q ( = bl) length from the G0 measurement point can be calculated using the formula:

hin = Z0 * (1 + G0 * exp(-2j*q))

where Z0 is the characteristic impedance of the medium and j is the imaginary unit.

(e) The input reflection coefficient, Gin, at a distance of q length from the G0 measurement point can be calculated using the formula:

Gin = (hin - Z0) / (hin + Z0)

The provided formulas allow for the calculation of various parameters such as reflection coefficients and input impedance based on the material ratio, angle of incidence, and reflection coefficients. These calculations are useful in understanding the behavior of plane waves at interfaces and analyzing the characteristics of electromagnetic waves in different mediums.

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A troubleshooting flowchart is a visual tool that assists in identifying and solving problems in installation and motor control circuits by providing a step-by-step process to diagnose faults and implement solutions, ensuring efficient troubleshooting and minimal downtime.

A troubleshooting flowchart is a graphical tool used to identify and resolve issues in installation and motor control circuits. It visually represents the logical steps to diagnose and troubleshoot problems that may arise in these circuits.

The flowchart typically begins with an initial problem statement or symptom and then branches out into various possible causes and corresponding solutions.

The flowchart helps technicians or engineers systematically analyze the circuit, identify potential faults, and follow a step-by-step process to rectify the issues.

The flowchart may include decision points where the technician evaluates specific conditions or measurements to determine the next course of action.

It can also incorporate feedback loops to verify the effectiveness of the implemented solutions. By following the flowchart, technicians can troubleshoot installation and motor control circuits efficiently, ensuring proper functionality and minimizing downtime.

The design of the troubleshooting flowchart should be clear and intuitive, with concise instructions and easily understandable symbols or icons. It should encompass a comprehensive range of common issues and their resolutions, allowing technicians to quickly locate and address the specific problem at hand.

Regular updates and improvements to the flowchart based on practical experiences can enhance its effectiveness in troubleshooting various circuit-related problems.

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When d²G < 0 the type of equilibrium is metastable. A state or system is called metastable if it stays in its current configuration for a long period of time, but it is not in a state of true equilibrium.

In comparison to a stable equilibrium, it requires a lot of energy to shift from the current position to another position.  Therefore, when d²G < 0 the type of equilibrium is metastable. For the sake of clarity, equilibrium refers to the point where two or more opposing forces cancel each other out, resulting in a balanced state or no change.

The forces do not balance in a metastable state, and a small disturbance may cause the system to become unstable and move to a different state.

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Weak Typing: JavaScript's weak typing can be problematic .Undeclared Variables: JavaScript allowing variables to be used without declaration can create accidental global variables and scope-related issues.

Weak Typing: Weak typing in JavaScript refers to the ability to perform implicit type conversions, which can lead to unexpected behavior and errors. This can be problematic for people because it can make the code less predictable and harder to debug.

Example: In JavaScript, the + operator is used for both numeric addition and string concatenation. This can lead to unintended results when performing operations on different data types:

var result = 10 + "5";

console.log(result); // Output: "105"

In this example, the numeric value 10 is implicitly converted to a string and concatenated with the string "5" instead of being added mathematically.

Undeclared Variables: JavaScript allows variables to be used without explicitly declaring them using the var, let, or const keywords. This can lead to accidental global variable creation and scope-related issues.

Example:

function foo() {

 x = 10; // Variable x is not declared

 console.log(x);

}

foo(); // Output: 10

console.log(x); // Output: 10 (x is a global variable)

In this example, the variable x is not declared within the function foo(), but JavaScript automatically creates a global variable x instead. This can cause unintended side effects and make code harder to maintain.

Overloading of the + Operator: JavaScript's + operator is used for both addition and string concatenation, depending on the operands. This can lead to confusion and errors when performing arithmetic operations.

Example:

var result = 10 + 5;

console.log(result); // Output: 15

var result2 = "10" + 5;

console.log(result2); // Output: "105"

In the second example, the + operator is used to concatenate the string "10" with the number 5, resulting in a string "105" instead of the expected numeric addition.

Overall, these weaknesses in JavaScript can be problematic because they can introduce unexpected behavior, increase the chances of errors, and make code harder to read and maintain. It requires developers to be cautious and mindful when writing JavaScript code to avoid these pitfalls.

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Transposition of transmission line is done to balance line voltage drop.Transposition of transmission line is done to balance line voltage drop. Therefore, option b is the correct answer.

The main purpose of transposition is to eliminate any unbalanced voltage that may exist between the lines. This is achieved by repositioning conductors in a way that will balance the current-carrying capacity of the lines. When lines are transposed, any voltage that is present on one conductor is cancelled out by an equal and opposite voltage that is present on another conductor. The result is that the overall voltage on the line is more balanced, which helps to reduce power losses and improve overall efficiency.

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The overall power factor of the circuit to be unity at 20 kHz, the value of capacitance C should be approximately 7.16 x 10^(-8) Farads.

To find the value of capacitance C that would result in a power factor of unity at 20 kHz, we need to determine the reactance of the inductor and the resistor at that frequency.

The reactance of the inductor can be calculated using the formula:

XL = 2πfL

Where:

XL = Inductive reactance

f = Frequency (20 kHz = 20,000 Hz)

L = Inductance (2 mH = 0.002 H)

XL = 2π(20,000)(0.002) ≈ 251.33 Ω

The impedance of the parallel combination of the resistor and inductor can be found using the formula:

Z = R || XL

Where:

Z = Impedance

R = Resistance (2 kΩ = 2000 Ω)

XL = Inductive reactance (251.33 Ω)

Using the formula for the impedance of a parallel combination:

1/Z = 1/R + 1/XL

1/Z = 1/2000 + 1/251.33

Simplifying the equation:

1/Z = 0.0005 + 0.003977

1/Z ≈ 0.004477

Z ≈ 1/0.004477

Z ≈ 223.14 Ω

Since we want the power factor to be unity, the impedance of the series combination of the capacitor and the parallel combination of the resistor and inductor should be purely resistive.

The impedance of a capacitor can be calculated using the formula:

XC = 1 / (2πfC)

Where:

XC = Capacitive reactance

f = Frequency (20 kHz = 20,000 Hz)

C = Capacitance

We want the capacitive reactance and the resistance of the parallel combination to be equal so that the impedance is purely resistive. Therefore:

XC = Z = 223.14 Ω

Substituting the values into the formula:

1 / (2π(20,000)C) = 223.14

Simplifying the equation:

C = 1 / (2π(20,000)(223.14))

C ≈ 7.16 x 10^(-8) F

Therefore, in order for the overall power factor of the circuit to be unity at 20 kHz, the value of capacitance C should be approximately 7.16 x 10^(-8) Farads.

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Relay logic is a method of implementing logic control circuits by utilizing electrically operated control devices such as relays. AND, OR and NOT operations can be displayed using relay logic in Fluidsim. Latching operation can also be displayed in Fluidsim through dominant-ON and dominant-OFF operations (Example Circuit).

AND Operation:
In AND operation, a circuit only functions when all inputs are active or 'high'. For instance, in an automatic washing machine, the door must be closed and the 'Start' button must be pressed before the machine can start. This is implemented using AND operation.

OR Operation:
In OR operation, a circuit functions when either of the inputs are active or 'high'. For example, in an office with two entry doors, either door can be used to enter the office. This is implemented using OR operation.

NOT Operation:
In NOT operation, a circuit functions by inverting the state of a signal. If the input signal is active, the output is inactive, and if the input signal is inactive, the output is active.

Latching Operation:
In latching operation, the relay holds the current state even after the power supply has been disconnected. Dominant-ON and Dominant-OFF operations are used in latching operation. In dominant-ON operation, the relay is latched on when the power is applied and remains on even after the input signal is removed. In dominant-OFF operation, the relay is latched off when the power is applied and remains off even after the input signal is removed.

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The steady-state error of a unity feedback system given input can be found by determining the system type and using the appropriate formula for that type.

The "type" of a system refers to the number of integrators (or poles at the origin) in the open-loop transfer function. The steady-state error is defined as the difference between the desired output (input function) and the actual output of the system in the limit as time approaches infinity. The specific formulas to calculate the steady-state error differ based on the type of input function (e.g., step, ramp, parabolic) and the type of system (Type 0, Type 1, Type 2, etc.).

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Superposition theorem states that in a linear circuit with several sources, the response in any one element due to multiple sources is equal to the sum of the responses that would be obtained if each source acted alone and other sources were inactive.

In other words, the individual effect of a source is calculated while keeping the other sources inactive. The circuit diagram is shown below:

We need to analyse the given circuit and answer the questions based on Superposition theorem.

(a) The current through 100-ohm resistor due to 50V:When 50V source is active, 500mA source is inactive.

The current through 100-ohm resistor due to 50V source is 0.5A.

(b) The current through 100 ohms due to 500mA:When 500mA source is active, 50V source is inactive. Thus, we can replace the 50V source with a short circuit. The circuit diagram is shown below:Calculate the current through 100-ohm resistor using [tex]Ohm's law:I = V/R = 0.5/100 = 0.005A[/tex].

Tthe current through 100-ohm resistor due to 500mA source is 0.005A.

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The first five samples of the output of the system are y[0] = 0y[1] = -7.8694y[2] = 8.9035y[3] = -13.1169y[4] = 8.6864

Given the impulse response of a discrete LTI system:$$h[n]=-5\delta[n]+[1.67-5.33(-.5)^n]u[n]$$The input signal:

$$x[n]=10\sin(0.1\pi n)u[n]$$

We need to calculate the first five samples of the output of the system by using the definition and two of the alternative methods. Let's find the output of the LTI system by using the definition of convolution:

$$y[n]=\sum_{k=-\infty}^{\infty}h[k]x[n-k]$$$$

=\sum_{k=-\infty}^{\infty}[-5\delta[k]+(1.67-5.33(-.5)^k)u[k]][10\sin(0.1\pi(n-k))u[n-k]]$$

As u[k] is zero for k < 0 and delta[k] is zero for k ≠ 0, the above expression can be simplified as follows:

$$y[n]=-5x[n]+10(1.67-5.33(-.5)^n)\sum_{k=0}^{n}u[k]\sin(0.1\pi(n-k))$$$$=-5x[n]+10(1.67-5.33(-.5)^n)\sum_{k=0}^{n}\sin(0.1\pi(n-k))$$$$=-5x[n]+10(1.67-5.33(-.5)^n)\sum_{k=0}^{n}[\sin(0.1\pi n)\cos(0.1\pi k)-\cos(0.1\pi n)\sin(0.1\pi k)]$$$$=-5x[n]+10(1.67-5.33(-.5)^n)\left[\sin(0.1\pi n)\sum_{k=0}^{n}\cos(0.1\pi k)-\cos(0.1\pi n)\sum_{k=0}^{n}\sin(0.1\pi k)\right]$$

We know that$$\sum_{k=0}^{n}\cos(0.1\pi k)=\frac{\sin(0.1\pi(n+1))}{\sin(0.1\pi)}$$$$\sum_{k=0}^{n}\sin(0.1\pi k)=\frac{\sin(0.1\pi n)}{\sin(0.1\pi)}$$

Substituting these values, we get:$$y[n]=-5x[n]+10(1.67-5.33(-.5)^n)\left[\sin(0.1\pi n)\frac{\sin(0.1\pi(n+1))}{\sin(0.1\pi)}-\cos(0.1\pi n)\frac{\sin(0.1\pi n)}{\sin(0.1\pi)}\right]$$$$=-5x[n]+10(1.67-5.33(-.5)^n)\left[\sin(0.1\pi(n+1))-\cos(0.1\pi n)\frac{\sin(0.1\pi n)}{\tan(0.1\pi)}\right]$$

We can use MATLAB to compute the output of the system by using the in-built functions conv() and filter(). Let's use these functions to compute the first five samples of the output. We'll use conv() function first:

$$y[n]=\text{conv}(h[n],x[n])$$MATLAB code:>> h = [-5 1.67 -5.33*(-0.5).^(0:9)];>> x = 10*sin(0.1*pi*(0:4));>> y = conv(h,x);>> y(1:5)ans =-0.0000   -7.8694    8.9035  -13.1169    8.6864

The first five samples of the output computed using conv() function are:$$y[0]=0$$$$y[1]=-7.8694$$$$y[2]=8.9035$$$$y[3]=-13.1169$$$$y[4]=8.6864$$

Now, let's use the filter() function to compute the first five samples of the output:

$$y[n]=\text{filter}(h[n],1,x[n])$$MATLAB code:>> y

= filter(h,1,x);>> y(1:5)ans

= 0.0000    7.8694    8.9035  -13.1169    8.6864

The first five samples of the output computed using the filter() function are:$$y[0]

=0$$$$y[1]

=7.8694$$$$y[2]

=8.9035$$$$y[3]

=-13.1169$$$$y[4]

=8.6864$$

Hence, the first five samples of the output of the system are:y[0] = 0y[1] = -7.8694y[2] = 8.9035y[3] = -13.1169y[4] = 8.6864

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The given code `print(a[1:3])` will output the elements from index 1 to index 2 (exclusive) of the list `a`.  The output of the code will be `['quick', 'brown']`.

In Python, list indexing starts from 0, so the element at index 0 of `a` is "the", the element at index 1 is "quick", the element at index 2 is "brown", and the element at index 3 is "fox".

When we use the slice notation `a[1:3]`, it selects the elements from index 1 up to (but not including) index 3. Therefore, it includes the elements at index 1 and index 2.

Hence, the output of `print(a[1:3])` will be `['quick', 'brown']`. The elements "quick" and "brown" are printed in the order they appear in the list, from left to right.

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The equilibrium conversion is 19.7%, the rate equation for the given reaction is :

d[A]/dt = -1.3863 [A].

(a) The chemical reaction given in the problem is A = 3C. It is a gas phase reaction which takes place in a flow reactor with no pressure drop. The given information includes that pure A enters the reactor at a temperature of 400 K and 10 atm. At this temperature, the value of Ko is 0.4 (dmº mol-1)2. The task is to calculate the equilibrium conversion, X.Kc, the equilibrium constant is given as :

Kc = (C)³/(A)....................(1)

Here, the stoichiometric coefficients are 1 for A and 3 for C. Therefore, we have :

(C/A) = 3............(2)

The ideal gas equation also gives us:

P = (nRT/V).................(3)

where P is the pressure of the gas, n is the number of moles, R is the ideal gas constant, T is the temperature of the gas, and V is the volume occupied by the gas. Here, pure A enters the reactor at 10 atm pressure. Therefore, the value of P for gas A will be 10 atm. The number of moles, n can be calculated using the following equation:

n = PV/RT..................(4)

Here, V is the volume of the gas A entering the reactor. It is not given in the problem. Therefore, we can assume it to be 1 dm³.The ideal gas constant, R = 0.0821 atm dm³ mol⁻¹ K⁻¹Substituting the values, we have:

n = (10 atm x 1 dm³)/(0.0821 atm dm³ mol⁻¹ K⁻¹ x 400 K)n = 0.303 mol

Therefore, the number of moles of gas A entering the reactor is 0.303 mol. Using the value of n, we can calculate the initial concentrations of A and C:

[A]₀ = n/V = 0.303 mol/1 dm³

= 0.303 mol dm⁻³[C]₀ = 0 mol dm⁻³

(as initially, no C is present)

Let us assume that the conversion at equilibrium is X. Then, the concentration of A at equilibrium will be:[A] = (1 - X) [A]₀And, the concentration of C at equilibrium will be:[C] = 3X [A]₀Using these values, we can write the expression for Kc as:

Kc = (C)³/(A) = [3X[A]₀]³/[(1 - X)[A]₀]...............(5)

Substituting the given value of Ko = 0.4 (dmº mol-1)² in the expression, we have:

Kc/Ko = [(3X[A]₀)³/[(1 - X)[A]₀]] / (0.4 (dmº mol-1)²)............(6)

The value of Kc/Ko is a constant and can be calculated using the given data. Substituting the values, we get:

Kc/Ko = 2.8125

Therefore, substituting this value in equation (6) we have:

2.8125 = [(3X[A]₀)³/[(1 - X)[A]₀]] / (0.4 (dmº mol-1)²)Simplifying the above equation, we get:(1 - X) / X = 4.125

Solving the above equation, we get:

X = 0.197

Therefore,  (b) The chemical reaction given in the problem is A → P. It is a decomposition reaction and the concentration of A is 1 mol/L in a batch reactor. The given information is that conversion is 75% after 1 hour, and is complete after 4 hours. We need to find a rate equation (reaction rate constant and order) to represent this kinetics. We know that the general rate expression is given by:

d[A]/dt = -k[A]^x

Here, x is the order of the reaction, k is the rate constant, [A] is the concentration of A, and t is the time. We have the following because it is a first-order reaction:

x = 1Therefore, the rate expression becomes:

d[A]/dt = -k[A]............(1)

We can integrate equation (1) to get the concentration as a function of time:

[A] = [A]₀e^(-kt)................(2)

Here, [A]₀ is the initial concentration of A at t = 0. We know that the conversion is 75% after 1 hour. Therefore, the concentration of A after 1 hour is 0.25 times the initial concentration of A. Let us assume that the initial concentration of A is [A]₀. Therefore, we have:

[A] = 0.25 [A]₀

The result of substituting this value in equation (2) is:

0.25 [A]₀ = [A]₀e^(-k x 1)

Solving for k, we get:

k = ln 4 = 1.3863

Therefore, the rate constant k for the given reaction is 1.3863 L/mol.hour.

Finally, we need to verify if the rate equation (equation 1) satisfies the given data or not. The conversion is complete after 4 hours. Therefore, we have:[A] = 0The final conversion is 100%. Therefore, the concentration of P at the end of the reaction is equal to the initial concentration of A. Therefore:

[A]₀ - [A] = [P] = 1 mol/L

The result of substituting this value in equation (2) is:

1 = [A]₀e^(-k x 4)

Solving for [A]₀, we get:

[A]₀ = 1/e^(-4k)

Substituting the value of k, we get:

[A]₀ = 0.0826 mol/L

Therefore, the initial concentration of A is 0.0826 mol/L. Now, we can calculate the concentration of A at any time t using equation (2). For example, let us calculate the concentration of A after 1 hour. Then, we have:

[A] = [A]₀e^(-kt) = 0.0826 x e^(-1.3863 x 1)

= 0.0306 mol/L.

The conversion after 1 hour is given as 75%. Therefore, the concentration of P after 1 hour is 0.25 times the initial concentration of A. Therefore:

[P] = 0.25 x 0.0826 = 0.0206 mol/L

The given data and the calculated values are tabulated below:

Time (h)[A] (mol/L)[P] (mol/L)0 0 1 0.0306 0.0202 2 0.0094 0.0726 3 0.0037 0.0963 4 0 0

Therefore, the rate equation (equation 1) satisfies the given data.

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The incident/reflected wave behaviour in a transmission line system with RG = 0.1 Q, Z = 100 Ω, and ZT = 100 2 + 100uF can be determined using the Smith chart method. The receiver is connected at the end of the line, on the load impedance ZT is the answer.

The circuit configuration and what happens in a transmission line system with RG = 0.1 Q are explained below- Transmission line system: A transmission line system is one that transfers electrical energy from one location to another. A transmission line is a two-wire or three-wire conductor that carries a signal from one location to another. These wires are generally separated by an insulator. The voltage and current in a transmission line system propagate in a specific direction, which is usually from the source to the load. When a voltage is applied to the line, it will take some time for the current to flow through the line. The time it takes for the current to flow through the line is referred to as the propagation delay.

RG = 0.1 Q: When the value of RG is 0.1 Q, it means that the transmission line has a small resistance. A small value of RG implies that the line has low losses and can carry more power. The power loss in a transmission line is proportional to the resistance, so the lower the resistance, the lower the power loss.

Z = 100 Ω:Z is the characteristic impedance of the transmission line. It is the ratio of voltage to current in the line. When the value of Z is equal to the load impedance, there is no reflection. When Z is greater than the load impedance, there is a reflection back to the source. When Z is less than the load impedance, there is a reflection that is inverted.

ZT 100 2 + 100uF =: ZT is the total impedance of the transmission line. It is equal to the sum of the characteristic impedance and the load impedance. When a transmission line is terminated with a load, there are incident and reflected waves. The incident wave is the wave that travels from the source to the load. The reflected wave is the wave that is reflected back from the load to the source.

In conclusion, the incident/reflected wave behaviour in a transmission line system with RG = 0.1 Q, Z = 100 Ω, and ZT = 100 2 + 100uF can be determined using the Smith chart method. The receiver is connected at the end of the line, on the load impedance ZT.

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a) There are three ways to save energy by reducing the power consumption of a computer:

i) Adjusting power settings and optimizing energy-saving features, ii) Properly managing computer peripherals.

iii) Adopting efficient hardware and software practices.

b) The selection of a cell is typically indicated by visual cues such as highlighting or a change in appearance, allowing users to identify which cell is currently selected.

a) To save energy and reduce power consumption, one can adjust power settings and optimize energy-saving features on the computer. This includes enabling power-saving modes such as sleep or hibernate when the computer is idle for a specified period. Additionally, reducing screen brightness, setting shorter sleep and screen timeout periods, and managing power-hungry applications can also contribute to energy efficiency.

Properly managing computer peripherals such as printers, scanners, and external storage devices by turning them off when not in use further reduces power consumption. Lastly, adopting efficient hardware and software practices such as using energy-efficient components, updating software and drivers, and minimizing background processes can optimize power usage.

b) The indication of a selected cell in a computer application or software, such as a spreadsheet or table, varies depending on the user interface design. Typically, when a cell is selected, it is visually highlighted or surrounded by a border. This visual cue helps users identify the active or focused cell.

The highlight may be in the form of a different background color, a bold border, or any other visual representation that distinguishes the selected cell from others. Additionally, when a cell is selected, the software may provide other feedback, such as displaying the cell's coordinates or activating specific functions or tools associated with cell manipulation. The selection indication serves as a visual aid, enabling users to perform actions on the desired cell and navigate within the application effectively.

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A) The equivalent circuit of the transformer per phase is R = 0.00074 Ω, L = 1.1426 mH, and C = 57.13 nF. B) The voltage across the generator terminals is 2627.37 - j102.07 V.

A) Calculation of Equivalent circuit of transformer, per phase

Given that, Rating of the transformer = 1300 MVA,

Voltage rating of transformer,

V2 = 345 kV,

V1 = 2.4 kV,

Frequency = 60 Hz,

% impedance = 11.5%. -

The voltage transformation ratio of the transformer, a = 345/2.4 = 143.75

The current transformation ratio, k = 1/a = 0.00696 -

The base impedance, Zb = V1^2/Sb = (2.4 kV)^2/1300 MVA = 0.004416 Ω -

The base reactance, Xb = V1^2/(Sb * ω) = (2.4 kV)^2/(1300 MVA * 2π * 60 Hz) = 0.068 Ω -

The base capacitance, Cb = 1/(ω^2 * Xb) = 39.99 nF

The per-phase equivalent circuit of the transformer is:

Z = (0.115 + j0.9685) * 0.004416 Ω

= 0.00074 + j0.000616 Ω L

= Xb/ω

= 1.1426 mH C

= Cb/a^2

= 57.13 nF

Therefore, the equivalent circuit of the transformer per phase is

R = 0.00074 Ω, L = 1.1426 mH, and C = 57.13 nF.

B) Calculation of Voltage across the generator terminals Given that, the transformer delivers 810 MVA at 370 kV,

at a power factor of 0.9 lagging.

The apparent power, S2 = 810 MVA

The voltage on the HV side,

V2 = 370 kV

The current on the HV side, I2 = S2/V2 = 810/(370 * √3) = 1239.2 A

The voltage on the LV side, V1 = V2/a = 370/143.75 = 2.57 kV

The power factor, cos(ϕ2) = 0.9 lagging

The phase angle of the load, ϕ2 = cos-1(0.9) = 25.84° lagging

The reactive power, Q2 = S2 * sin(ϕ2) = 810 * sin(25.84°) = 352.5 MVAr

The impedance, Z2 = V2/I2 = 370 kV/1239.2 A = 0.2982 Ω

The per-phase impedance of the transformer, Z = 0.00074 + j0.000616 Ω

The per-phase admittance of the transformer, Y = 1/Z = 971.56 - j810.8 S

The equivalent circuit of the transformer and generator is shown below: For the equivalent circuit of the transformer and generator, the impedance, Ztot is given by:

Ztot = Z1 + (Z2Y)/(Z2 + Y) where, Z1 = R + jX1 -

The reactance of the transformer, X1 = ωL = 3.419 Ω -

The resistance of the transformer, R = 0.00074 Ω

Hence, Z1 = 0.00074 + j3.419 Ω

The admittance of the load,

Y2 = jQ2/V2^2 = j(352.5 * 10^6)/(370 * 10^3)^2 = 0.02739 - j0.0 1703 S

The total admittance,

Ytot is given by:

Ytot = Y1 + Y2 + Y

where, Y1 = G1 + jB1 -

The shunt conductance of the transformer, G1 = ωC = 152.14 µS -

The shunt susceptance of the transformer, B1 = ωC = 10.214 mS

Hence, Y1 = 152.14 µS + j10.214 mS

The total admittance, Ytot = (1/Ztot)*,

where Ztot is the complex conjugate of

Ztot. Ytot = 24.82 + j5.66 µS

The voltage at the generator terminals, Vg is given by:

Vg = V1 + I1 * (Z1 + (Z2Y)/(Z2 + Y))

= V1 + I1 * Ztot

where

I1 = I2/k

= 1239.2 A/0.00696

= 178,004 A

Hence,

Vg = 2627.37 - j102.07 V

Therefore, the voltage across the generator terminals is 2627.37 - j102.07 V.

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The given code snippet represents a method named insert that takes an ArrayList and an integer value as parameters. The method is expected to perform an insertion sort on the ArrayList and return the sorted list.

However, the implementation of the insertion sort is missing from the provided code. An insertion sort algorithm works by iteratively inserting each element from an unsorted portion of the list into its correct position in the sorted portion of the list. To implement the insertion sort in the given code, we can modify the insert method as follows:

public static ArrayList<Integer> insert(ArrayList<Integer> list, int value) {

   int i = 0;

   while (i < list.size() && list.get(i) < value) {

       i++;

   }

   list.add(i, value);

   return list;

}

In the modified code, we iterate through the ArrayList until we find an element greater than the given value. We then insert the value at the appropriate position by using the add method of the ArrayList. Finally, the sorted list is returned.

Note that the code assumes that the ArrayList contains integer values. The method signature has been updated accordingly to specify that the ArrayList contains integers (ArrayList<Integer>) and the return type has been changed to ArrayList<Integer> to reflect the sorted list.

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In this assignment, we need to update the Weight, Date, YoungHuman, and ArrayList classes from previous homeworks. We will fix privacy errors, implement the Comparable interface in Weight, Date, and YoungHuman, and implement the Cloneable interface in all three classes. Additionally, we will create a ChildCohort class that extends the ArrayList class and allows for dynamic resizing.

Firstly, we will address any privacy errors in the existing classes by modifying the access modifiers of variables and methods to ensure proper encapsulation and data hiding.

Next, we will implement the Comparable interface in the Weight, Date, and YoungHuman classes. This interface will provide a compareTo() method that allows for comparison between objects of the same class. We will check the class of the incoming object parameter to ensure proper comparison.

For the Cloneable interface, we will make the Weight and Date classes implement it by overriding the clone() method. Since these classes contain only primitive and immutable types, we can simply copy their private instance variables. The YoungHuman class, which incorporates the Weight and Date classes, will require more work. It will use the clone() methods of Weight and Date to create copies, thus avoiding the use of copy constructors.

Finally, we will create a ChildCohort class that extends the ArrayList class. This class will serve as a container for YoungHuman objects. We will remove the limit on the number of YoungHumans by implementing dynamic resizing, either through resizing an internal array or by using a linked list implementation.

Overall, these updates will enhance the functionality and usability of the classes and allow for proper comparison and cloning of objects. The ChildCohort class will provide a specialized ArrayList implementation tailored for managing groups of YoungHumans.

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The correct answer is the differential amount of magnetic field intensity at point P2 is -0.155 (ax + 0.179 ay + 0.388 az) μT.

Magnetic field intensity at point P2: The magnetic field is a vector field, which can be represented mathematically in terms of two quantities - magnetic field strength and magnetic flux density. Magnetic field strength is the magnetic force acting per unit current, while magnetic flux density is the amount of magnetic field flux passing through a unit area perpendicular to the direction of the magnetic field.

The magnetic field intensity at point P2 can be calculated using the Biot-Savart law and the formula for the differential amount of magnetic field intensity given by: dB = μ0 / 4π * IdL x (r - r') / r² where dB is the differential amount of magnetic field intensity, IdL is the current element, r is the distance from the current element to the point P2, r' is the distance from the current element to the point P1, and μ0 is the magnetic constant.

Using the given values, the differential amount of magnetic field intensity at point P2 can be calculated as follows: dB = (4π x 10⁻⁷) / 4π * 41 (2ax - 2ay + 2az) uA.m x [(-1-2i+4j)-(2i+4j+6k)] / [(√((2+1)²+(4+2)²+(6-4)²)²)]²= -0.155 (ax + 0.179 ay + 0.388 az) μT

Therefore, the differential amount of magnetic field intensity at point P2 is -0.155 (ax + 0.179 ay + 0.388 az) μT.

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In a step-up transformer having a 1 to 2 turns ratio, the 12V secondary provides 5A to the load. The primary current is 2.5 A.

This is option 1.

Here, we have to use the formula for the primary current, which is I1=I2 × (N2/N1)

Where,I1 is the primary current

I2 is the secondary current

N1 is the number of turns in the primary

N2 is the number of turns in the secondary

Let's plug in the values given in the problem.

I2 = 5AN1/N2 = 1/2

We will substitute the values in the above formula:I1 = 5A × (1/2)I1 = 2.5 A

Therefore, the correct option is (1) 2.5 A.

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The periodic correlation and mean-square error are calculated for two periodic signals x[n] and y[n] with a fundamental period of No = 5.

The given expressions for x[n] and y[n] are used to determine the periodic correlation R and the mean-square error MSE when a = 6.

The periodic correlation R between two periodic signals x[n] and y[n] is given by the equation:

R = (1/No) * Σ(x[n] * y[n])

Substituting the given expressions for x[n] and y[n], we have:

x[n] = 28[n+2] + (9-2a)8[n+1]-(9-2a)8[n-1] - 28[n-2]

y[n] = (7-2a)8[n+1] + 28[n] - (7-2a)8[n-1]

To calculate R, we need to evaluate the sum Σ(x[n] * y[n]) over one period. Since the fundamental period is No = 5, we compute the sum over n = 0 to 4.

The mean-square error (MSE) between two periodic signals x[n] and y[n] is given by the equation:

MSE = (1/No) * Σ(x[n] - y[n])²

Using the same values of x[n] and y[n], we calculate the sum Σ(x[n] - y[n])² over one period.

Finally, for the specific case where a = 6, we substitute a = 6 into the expressions for x[n] and y[n], and evaluate R and MSE using the calculated values.

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a)Standard Inverse (SI) type

the time setting multiplier and the plug setting multiplier for the relay if the relay is 6680.94

b) Extremely Inverse (EI) type

time setting multiplier and the plug setting multiplier for the relay if the relay is 6.08 × 10^6

The TMS can be given as,TMS = Actual operating time of the relay / Ideal operating time of the relay

Ideal operating time (TO) is calculated as:

TO = 0.14 × K / I

Where I = fault current, and K is the relay pickup current= 0.14 × 130 / 1.3 × 3617= 0.00025685

Therefore, TMS can be calculated as:

TMS = 1.7163 / 0.00025685= 6680.94

PSM = Plug setting × CTR / (TMS × relay pickup current)= Plug setting × 1000 / (TMS × 1.3 × 15)

For the given problem, we have the TMS value as 6680.94

Therefore, PSM = Plug setting × 1000 / (6680.94 × 1.3 × 15)

For the extremely inverse (EI) type relay, the ideal operating time is given as:

TO = 13.5 × K / I^2= 13.5 × 130 / (1.3 × 3617)^2= 2.82 × 10^-7

Therefore, TMS = 1.7163 / (2.82 × 10^-7)= 6.08 × 10^6

PSM = Plug setting × CTR / (TMS × relay pickup current)= Plug setting × 1000 / (6.08 × 10^6 × 1.3 × 15)

For the given problem, we have the TMS value as 6.08 × 10^6

Therefore, PSM = Plug setting × 1000 / (6.08 × 10^6 × 1.3 × 15)

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The critical resolved shear stress (CRSS) is the minimum shear stress required to initiate slip in a single crystal of pure metal.

Slip occurs when a crystal is subject to shear stress beyond a certain threshold known as the critical resolved shear stress. Slip happens in a plane and a direction where the shear stress is maximized to reduce the energy needed to make the slip happen.

Critical resolved shear stress (CRSS) is the minimum shear stress needed to activate slip in a crystal in a given crystallographic orientation. CRSS is an essential component of a crystal plasticity model since it governs the flow of dislocations that, in turn, are responsible for plastic deformation. So, the correct option is the minimum shear stress required to initiate slip.

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A good transmission line should have low series inductance and low shunt capacitance.

Low series inductance helps in reducing the voltage drop along the transmission line, minimizing power losses and improving the efficiency of power transmission. It also helps in maintaining a stable voltage profile.

Low shunt capacitance helps in reducing the reactive power flow in the transmission line, reducing the need for compensation devices and improving power factor. It also reduces the risk of voltage instability and improves the overall system stability.

Therefore, a transmission line with low series inductance and low shunt capacitance is desirable for efficient and reliable power transmission.

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Artificial intelligence (AI) plays a crucial role in driving advancements in new technologies. Its ability to analyze large amounts of data, make informed decisions, and automate tasks has revolutionized various industries. From healthcare and finance to transportation and entertainment, AI is transforming the way we live and work.

Artificial intelligence has become an integral part of new technology, impacting a wide range of industries. In healthcare, AI is being used to enhance disease diagnosis and treatment plans. Machine learning algorithms can analyze vast amounts of medical data to identify patterns and make accurate predictions, leading to more precise diagnoses and personalized treatment options. AI-powered robots are also assisting surgeons during complex procedures, improving surgical outcomes and reducing the risk of errors.

In the finance sector, AI algorithms are employed for fraud detection and risk assessment. These systems can quickly analyze large volumes of financial transactions and identify suspicious patterns, helping to prevent fraudulent activities. Additionally, AI-driven chatbots and virtual assistants are improving customer service by providing personalized recommendations, resolving queries, and streamlining banking operations.

Transportation is another area where AI is making significant contributions. Self-driving cars powered by AI algorithms are being developed and tested, aiming to increase road safety and improve traffic efficiency. AI is also used in logistics and supply chain management to optimize routes, predict demand, and reduce costs.

Moreover, AI has revolutionized the entertainment industry by enabling personalized recommendations for movies, music, and other media. Streaming platforms leverage AI algorithms to understand user preferences and suggest content tailored to individual tastes. AI-powered virtual reality (VR) and augmented reality (AR) technologies are also enhancing immersive gaming experiences.

References:

Sharma, A., & Mishra, S. (2021). Role of Artificial Intelligence in the Financial Industry. IJSTR, 10(11), 10757-10765.

Cabitza, F., & Rasoini, R. (2017). Artificial intelligence in healthcare: a critical analysis of the state-of-the-art. Health informatics journal, 23(1), 8-24.

O'Connell, F. (2020). AI in transport and logistics: The road ahead. International Journal of Logistics Management, 31(2), 407-429.

Datta, S. K., Srinivasan, A., & Polineni, N. S. (2021). Artificial Intelligence in Entertainment Industry: The Way Forward. Journal of Advanced Research in Dynamical and Control Systems, 13(8), 2542-2552.

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