National Institute of Technology Delhi Entrance Exam Set

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its implications in the context of real-world signal acquisition for advanced engineering applications at institutions like the National Institute of Technology Delhi. The theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be greater than twice the highest frequency component (\(f_{max}\)) present in the signal. This minimum sampling rate is known as the Nyquist rate, \(f_{Nyquist} = 2f_{max}\). In this scenario, a sensor is designed to capture atmospheric pressure fluctuations. The critical information lies in the fact that the most rapid significant pressure change that needs to be reliably detected corresponds to a sinusoidal variation with a frequency of 5 Hz. This 5 Hz represents the highest frequency component (\(f_{max}\)) that the sensor system must accurately represent. According to the Nyquist-Shannon sampling theorem, the sampling frequency (\(f_s\)) must be strictly greater than twice this maximum frequency. Therefore, the minimum required sampling frequency is \(f_s > 2 \times f_{max}\). Substituting the given \(f_{max} = 5\) Hz, we get \(f_s > 2 \times 5\) Hz, which simplifies to \(f_s > 10\) Hz. The question asks for the *minimum* sampling frequency that *guarantees* accurate reconstruction. This means we need the smallest integer value of \(f_s\) that satisfies the inequality \(f_s > 10\) Hz. The smallest integer greater than 10 is 11. Therefore, the minimum sampling frequency required is 11 Hz. This ensures that the sampling process captures enough information to avoid aliasing and allows for the faithful reconstruction of the 5 Hz pressure fluctuation. Understanding this principle is crucial for designing effective data acquisition systems in fields like environmental monitoring and control systems, areas of active research and study at the National Institute of Technology Delhi.

The question probes the understanding of fundamental principles in digital logic design, specifically related to combinational circuits and their implementation using basic logic gates. The scenario describes a circuit that outputs a ‘1’ only when a majority of its three inputs are ‘1’. This is the definition of a majority function. To implement a 3-input majority function, we can derive its truth table. Let the inputs be A, B, and C, and the output be Y. | A | B | C | Y | |—|—|—|—| | 0 | 0 | 0 | 0 | | 0 | 0 | 1 | 0 | | 0 | 1 | 0 | 0 | | 0 | 1 | 1 | 1 | | 1 | 0 | 0 | 0 | | 1 | 0 | 1 | 1 | | 1 | 1 | 0 | 1 | | 1 | 1 | 1 | 1 | From the truth table, the output Y is ‘1’ when at least two of the inputs are ‘1’. This can be expressed in Sum of Products (SOP) form as: \(Y = A \cdot B \cdot \bar{C} + A \cdot \bar{B} \cdot C + \bar{A} \cdot B \cdot C + A \cdot B \cdot C\) We can simplify this expression using Boolean algebra. Notice that the terms \(A \cdot B \cdot C\) can be absorbed. A simpler way to represent the majority function is that the output is ‘1’ if A AND B are ‘1’, OR A AND C are ‘1’, OR B AND C are ‘1’. This is because if any two inputs are ‘1’, the output is ‘1’, and if all three are ‘1’, the output is also ‘1’. So, the simplified SOP form is: \(Y = (A \cdot B) + (A \cdot C) + (B \cdot C)\) This expression directly translates to an implementation using AND gates for each product term and an OR gate to combine these terms. Specifically, we need two-input AND gates for \(A \cdot B\), \(A \cdot C\), and \(B \cdot C\), and a three-input OR gate to combine their outputs. Therefore, the minimum number of gates required is three 2-input AND gates and one 3-input OR gate, totaling 4 gates. This understanding of Boolean algebra and circuit simplification is fundamental for designing efficient digital systems, a core aspect of electrical engineering and computer science programs at institutions like the National Institute of Technology Delhi. The ability to derive and simplify Boolean expressions from functional requirements is crucial for minimizing hardware complexity, reducing power consumption, and improving circuit speed, all of which are key considerations in advanced digital design.

This question probes the understanding of the Nyquist-Shannon sampling theorem, a cornerstone of digital signal processing, which is fundamental to various engineering disciplines taught at the National Institute of Technology Delhi. The theorem dictates the minimum sampling rate required to accurately capture an analog signal without losing information due to aliasing. In this scenario, an analog signal with a maximum frequency component of \(15 \text{ kHz}\) is sampled at \(25 \text{ kHz}\). The critical concept here is the Nyquist frequency, which is half the sampling frequency, representing the highest frequency that can be unambiguously reconstructed from the sampled data. For a sampling frequency \(f_s\), the Nyquist frequency is \(f_s/2\). Any frequency component in the original signal exceeding this Nyquist frequency will be aliased, meaning it will appear as a lower frequency in the sampled signal, leading to distortion and loss of original information. The calculation \(f_s/2 = 25 \text{ kHz} / 2 = 12.5 \text{ kHz}\) reveals the maximum frequency that can be faithfully represented. Since the original signal contains frequencies up to \(15 \text{ kHz}\), which is greater than \(12.5 \text{ kHz}\), aliasing will occur. The frequency \(15 \text{ kHz}\) will alias to \(|15 \text{ kHz} – 1 \cdot 25 \text{ kHz}| = 10 \text{ kHz}\). The highest frequency component that will *not* be aliased is the highest frequency present in the original signal that is less than or equal to the Nyquist frequency. Therefore, the highest frequency that can be unambiguously represented without distortion is \(12.5 \text{ kHz}\). This understanding is crucial for students at NIT Delhi, particularly in courses related to communications, control systems, and embedded systems, where signal integrity and accurate data acquisition are paramount. Misapplication of sampling principles can lead to significant errors in system performance and data interpretation, underscoring the importance of mastering this concept for future research and development.

The question probes the understanding of the fundamental principles governing the operation of a basic semiconductor diode under varying voltage conditions, specifically focusing on the concept of forward bias and its implications for current flow. When a diode is forward-biased, the applied voltage overcomes the built-in potential barrier of the p-n junction. This allows majority charge carriers (electrons in the n-type material and holes in the p-type material) to diffuse across the junction. As the forward voltage increases beyond the threshold (typically around 0.7V for silicon diodes and 0.3V for germanium diodes), the depletion region narrows significantly, and the resistance of the diode drops dramatically. This leads to an exponential increase in the forward current. The question asks about the state of the diode when the applied voltage is less than the threshold voltage but still positive. In this scenario, the applied voltage is insufficient to completely neutralize the built-in potential barrier. While some minimal current might flow due to thermal excitation of minority carriers, the primary characteristic is that the diode is not conducting significant current. It is in a state of “cut-in” or “threshold” voltage, where the resistance is still very high, effectively blocking substantial current flow. Therefore, the diode acts as an open circuit or a very high resistance element.

The question probes the understanding of the fundamental principles governing the operation of a basic semiconductor diode in a forward-biased configuration, specifically focusing on the voltage drop across it. When a diode is forward-biased, current flows through it. However, this flow is not instantaneous with the application of a voltage. There is a characteristic voltage, known as the cut-in voltage or threshold voltage, below which the current is negligible. Once this voltage is exceeded, the diode begins to conduct significantly, and the voltage across it stabilizes at a relatively constant value, which is dependent on the semiconductor material. For silicon diodes, this value is typically around 0.7V, and for germanium diodes, it’s around 0.3V. The question asks about the voltage across the diode when it is conducting. In a forward-biased scenario where current is flowing, the voltage drop across the diode is essentially its characteristic forward voltage drop. This is not the applied voltage, nor is it zero, nor is it directly proportional to the current in a simple linear fashion (though it does increase slightly with current). The most accurate description of the voltage across a conducting forward-biased diode is its forward voltage drop.

The question assesses understanding of the fundamental principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its implications for aliasing. The theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency \(f_s\) must be at least twice the highest frequency component \(f_{max}\) present in the signal, i.e., \(f_s \ge 2f_{max}\). This minimum sampling rate is known as the Nyquist rate. In the given scenario, a continuous-time signal with a maximum frequency component of 15 kHz is being sampled. To avoid aliasing, the sampling frequency must be greater than or equal to twice this maximum frequency. Therefore, the minimum required sampling frequency is \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). If the signal is sampled at a frequency lower than this Nyquist rate, higher frequency components in the original signal will be misrepresented as lower frequencies in the sampled signal, a phenomenon called aliasing. This distortion makes accurate reconstruction impossible. The National Institute of Technology Delhi, with its strong focus on electronics and communication engineering, emphasizes a deep understanding of these foundational concepts for students to effectively design and analyze digital systems. Understanding the Nyquist criterion is crucial for anyone working with analog-to-digital conversion, digital filtering, and the subsequent processing of sampled data, ensuring the integrity and accuracy of information.

The core principle being tested here is the understanding of how a firm’s strategic decisions regarding its supply chain and production location impact its competitive advantage, particularly in the context of the Indian market and the specific focus of institutions like the National Institute of Technology Delhi, which emphasizes technological innovation and national development. A firm aiming to leverage India’s growing domestic market and potentially its skilled workforce, while also mitigating logistical complexities and import duties for certain components, would likely adopt a strategy that balances local manufacturing with strategic sourcing. Consider a scenario where a hypothetical technology firm, “IndoTech Innovations,” is planning to manufacture advanced semiconductor components for the Indian market. The firm has identified two primary strategic options for its production facility: Option 1: Establish a fully integrated manufacturing plant within India, sourcing raw materials and intermediate components both domestically and internationally. Option 2: Assemble the final product in India using components primarily manufactured and imported from Southeast Asia, with a smaller, specialized R&D center in India. To determine the most advantageous strategy for IndoTech Innovations, we must analyze the implications of each option concerning market access, cost structure, technological integration, and long-term sustainability within the Indian economic landscape. Option 1, establishing a fully integrated plant in India, allows for greater control over the entire production process, potentially leading to higher quality assurance and the ability to adapt quickly to local market demands. It also aligns with the “Make in India” initiative, potentially attracting government incentives and fostering domestic technological capabilities. However, it might involve higher initial capital investment and a longer ramp-up period, especially if critical raw materials or specialized machinery are not readily available domestically. The firm would need to invest in developing local supply chains for raw materials, which can be a complex and time-consuming process. Option 2, focusing on assembly in India with imported components, offers a faster market entry and potentially lower initial investment. It leverages existing manufacturing strengths in Southeast Asia. However, it makes the firm more susceptible to global supply chain disruptions, currency fluctuations, and trade policies. Furthermore, it might limit the depth of technological integration and local value creation, which could be a disadvantage in a market that increasingly values indigenous innovation, a key focus for institutions like the National Institute of Technology Delhi. Given the National Institute of Technology Delhi’s emphasis on fostering indigenous innovation and contributing to India’s technological self-reliance, a strategy that prioritizes local value creation and deep technological integration would be more aligned with its ethos and the broader national objectives. Therefore, establishing a more integrated manufacturing process within India, even with its challenges, represents a more robust long-term strategy for a technology firm aiming to build a sustainable competitive advantage and contribute meaningfully to the Indian technological ecosystem. This approach allows for greater control over intellectual property, fosters local skill development, and reduces reliance on external supply chains, thereby enhancing resilience and long-term growth potential. The firm would need to carefully manage its sourcing strategy, identifying reliable domestic suppliers for key inputs while strategically importing specialized materials or equipment where local alternatives are not yet viable. This balanced approach maximizes the benefits of local presence while mitigating immediate supply chain risks. The calculation is conceptual, focusing on strategic alignment and impact: Strategic Alignment Score (Conceptual): Option 1 (Integrated Indian Manufacturing): High alignment with national goals, potential for deep technological integration, greater supply chain control. Option 2 (Assembly with Imported Components): Lower alignment with national goals, shallower technological integration, higher supply chain dependency. Conclusion: Option 1 is strategically superior for a firm aiming for long-term success and alignment with national technological development goals, as emphasized by institutions like the National Institute of Technology Delhi.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its implications for aliasing. The Nyquist-Shannon sampling theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the signal. Mathematically, this is expressed as \(f_s \ge 2f_{max}\). In this scenario, the analog signal has a maximum frequency component of 15 kHz. To avoid aliasing, the sampling frequency must be greater than or equal to twice this maximum frequency. Therefore, the minimum required sampling frequency is \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). If a sampling frequency of 25 kHz is used, which is less than the required minimum of 30 kHz, aliasing will occur. Aliasing is the phenomenon where high-frequency components in the original analog signal are misinterpreted as lower frequencies in the sampled digital signal. This distortion makes it impossible to accurately reconstruct the original analog signal. The specific frequency that appears as a false lower frequency depends on the difference between the sampling frequency and twice the original frequency. For a frequency \(f\) above \(f_s/2\), it will appear as \(|f – k \cdot f_s|\) for some integer \(k\), such that the apparent frequency is within the range \([0, f_s/2]\). In this case, with \(f_s = 25 \text{ kHz}\), a 15 kHz component would be sampled. Since \(15 \text{ kHz} > 25 \text{ kHz} / 2 = 12.5 \text{ kHz}\), aliasing will occur. The aliased frequency would be \(|15 \text{ kHz} – 1 \cdot 25 \text{ kHz}| = |-10 \text{ kHz}| = 10 \text{ kHz}\). This means the original 15 kHz signal would be incorrectly represented as a 10 kHz signal in the digital domain, leading to a loss of fidelity and an inability to recover the original signal’s characteristics. This understanding is crucial for students at the National Institute of Technology Delhi, particularly in programs related to electronics and communication engineering, where signal integrity and accurate data acquisition are paramount.

The question tests the understanding of the fundamental principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its implications for aliasing. The scenario describes a continuous-time signal \(x(t) = \cos(200\pi t) + \sin(500\pi t)\). The highest frequency component in this signal is \(f_{max}\). To determine \(f_{max}\), we analyze the arguments of the cosine and sine functions. For \(\cos(200\pi t)\), the angular frequency is \(\omega_1 = 200\pi\) rad/s. Since \(\omega = 2\pi f\), we have \(f_1 = \frac{200\pi}{2\pi} = 100\) Hz. For \(\sin(500\pi t)\), the angular frequency is \(\omega_2 = 500\pi\) rad/s, which corresponds to a frequency \(f_2 = \frac{500\pi}{2\pi} = 250\) Hz. Therefore, the maximum frequency present in the signal is \(f_{max} = \max(f_1, f_2) = 250\) Hz. The Nyquist-Shannon sampling theorem states that to perfectly reconstruct a continuous-time signal from its samples, the sampling frequency \(f_s\) must be greater than twice the maximum frequency component of the signal, i.e., \(f_s > 2f_{max}\). This minimum required sampling frequency is known as the Nyquist rate. In this case, the Nyquist rate is \(2 \times 250 \text{ Hz} = 500\) Hz. The question asks about the consequences of sampling this signal at \(f_s = 400\) Hz. Since \(400 \text{ Hz} < 500 \text{ Hz}\), the sampling rate is below the Nyquist rate, and aliasing will occur. Aliasing is the phenomenon where high-frequency components in the original signal are incorrectly represented as lower frequencies in the sampled signal. Specifically, a frequency \(f\) sampled at \(f_s\) will appear as \(|f - k f_s|\) for some integer \(k\), where \(|f - k f_s| \le f_s/2\). Let's examine the two components: 1. The 100 Hz component: When sampled at 400 Hz, the sampled frequency is \(100 \text{ Hz}\). Since \(100 \text{ Hz} \le 400/2 = 200 \text{ Hz}\), this component is below the folding frequency and will be represented correctly. 2. The 250 Hz component: When sampled at 400 Hz, the sampling frequency is \(f_s = 400\) Hz. The folding frequency is \(f_s/2 = 200\) Hz. Since \(250 \text{ Hz} > 200 \text{ Hz}\), this component will alias. The aliased frequency \(f_{alias}\) can be found by \(f_{alias} = |f – k f_s|\) such that \(f_{alias} \le f_s/2\). For \(f = 250\) Hz and \(f_s = 400\) Hz, we can choose \(k=1\): \(f_{alias} = |250 – 1 \times 400| = |-150| = 150\) Hz. Since \(150 \text{ Hz} \le 200 \text{ Hz}\), this is the observed frequency. Therefore, the sampled signal will appear to contain a 100 Hz component and a 150 Hz component. The original 250 Hz component is aliased to 150 Hz. This means the sampled signal will be a sum of a cosine wave at 100 Hz and a sine wave at 150 Hz. The question asks what the sampled signal will appear to be. It will appear as a signal composed of these two frequencies. The correct option describes this outcome: the 100 Hz component remains unchanged, and the 250 Hz component aliases to 150 Hz. This is a fundamental concept in digital signal processing taught in undergraduate electrical engineering and computer science programs, emphasizing the importance of choosing an appropriate sampling rate to avoid distortion. Understanding aliasing is crucial for designing systems that accurately capture and process real-world signals, a core skill for graduates of institutions like the National Institute of Technology Delhi.

The question assesses understanding of the fundamental principles of electromagnetic induction and Faraday’s Law, specifically in the context of a changing magnetic flux through a coil. The scenario describes a circular coil of radius \(r\) and \(N\) turns placed in a uniform magnetic field \(B\). The magnetic field is then varied linearly with time, such that \(B(t) = kt\), where \(k\) is a constant. The induced electromotive force (EMF) in a coil is given by Faraday’s Law: \(\mathcal{E} = -\frac{d\Phi_B}{dt}\), where \(\Phi_B\) is the magnetic flux. The magnetic flux through a single turn of the coil is \(\Phi_{B, \text{single}} = B \cdot A \cdot \cos(\theta)\), where \(A\) is the area of the coil and \(\theta\) is the angle between the magnetic field and the normal to the coil’s area. In this case, the magnetic field is uniform and perpendicular to the plane of the coil, so \(\theta = 0\) and \(\cos(\theta) = 1\). The area of the circular coil is \(A = \pi r^2\). Therefore, the flux through a single turn is \(\Phi_{B, \text{single}} = (kt)(\pi r^2)\). For a coil with \(N\) turns, the total magnetic flux is \(\Phi_B = N \Phi_{B, \text{single}} = Nkt\pi r^2\). To find the induced EMF, we differentiate the total flux with respect to time: \[ \mathcal{E} = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}(Nkt\pi r^2) \] Since \(N\), \(k\), \(\pi\), and \(r^2\) are constants, the derivative is: \[ \mathcal{E} = -Nk\pi r^2 \frac{d}{dt}(t) = -Nk\pi r^2 (1) = -Nk\pi r^2 \] The magnitude of the induced EMF is \(| \mathcal{E} | = Nk\pi r^2\). This result indicates that the induced EMF is constant over time because the magnetic field changes linearly with time. This is a core concept in understanding how changing magnetic fields generate electric currents, a principle fundamental to many technologies studied at institutions like the National Institute of Technology Delhi. The constant induced EMF arises from the constant rate of change of magnetic flux, which is directly proportional to the constant \(k\). This understanding is crucial for students pursuing electrical engineering or physics at NIT Delhi, as it underpins the operation of generators, transformers, and various other electromagnetic devices. The question tests the ability to apply Faraday’s Law to a specific, yet common, scenario of a linearly changing magnetic field, requiring careful differentiation and understanding of the relationship between magnetic field strength, area, number of turns, and induced voltage.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its implications in the context of real-world signal acquisition for advanced engineering applications at institutions like the National Institute of Technology Delhi. The Nyquist-Shannon sampling theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be greater than twice the highest frequency component (\(f_{max}\)) present in the signal. Mathematically, this is expressed as \(f_s > 2f_{max}\). This minimum sampling rate is known as the Nyquist rate. In the given scenario, a sensor designed for monitoring subtle atmospheric pressure fluctuations is being considered. Such fluctuations, while potentially complex, are generally characterized by relatively low-frequency components. However, the system is also intended to capture transient, high-frequency anomalies that might indicate rapid meteorological events. If the sampling rate is set too low, these high-frequency components will be undersampled, leading to aliasing. Aliasing occurs when the undersampled high-frequency signal is misinterpreted as a lower frequency, corrupting the data and rendering it useless for accurate analysis or predictive modeling, which are crucial for research at the National Institute of Technology Delhi. The core of the question lies in identifying the most critical factor that dictates the appropriate sampling rate for such a sensor. While the *type* of sensor and the *desired accuracy* are important considerations in system design, they are secondary to the fundamental physical limitation imposed by the signal’s frequency content. The *bandwidth* of the signal directly encapsulates the range of frequencies present, and it is this bandwidth, specifically its maximum frequency component, that determines the theoretical minimum sampling rate required to avoid aliasing according to the Nyquist-Shannon theorem. Therefore, understanding and accurately characterizing the maximum frequency component of the atmospheric pressure signal, including potential transient anomalies, is paramount. This directly informs the selection of a sampling frequency that satisfies the Nyquist criterion, ensuring the integrity of the acquired data for subsequent advanced analysis and research at the National Institute of Technology Delhi.

The question probes the understanding of the fundamental principles governing the operation of a basic diode circuit, specifically focusing on forward bias and the concept of a cut-in voltage. A silicon diode, commonly used in electronics, has a typical cut-in voltage (also known as the threshold voltage or turn-on voltage) of approximately \(0.7 \text{ V}\). This is the minimum forward voltage required for the diode to begin conducting significant current. Consider a simple series circuit with a voltage source \(V_S\), a resistor \(R\), and a diode \(D\). When the voltage source \(V_S\) is applied such that the anode of the diode is at a higher potential than the cathode (forward bias), current will flow through the circuit. However, this current is negligible until the voltage across the diode, \(V_D\), reaches its cut-in voltage. If the applied source voltage \(V_S\) is less than the cut-in voltage of the diode, the diode remains in the “off” state, and effectively no current flows through the circuit. The voltage across the resistor \(R\) would be approximately zero, and the voltage across the diode would be equal to the source voltage \(V_S\). In the scenario presented, the voltage source is \(1.5 \text{ V}\) and the diode is a silicon diode with a cut-in voltage of \(0.7 \text{ V}\). The resistor is \(100 \Omega\). For the diode to conduct, the voltage across it must be at least \(0.7 \text{ V}\). In a series circuit, the sum of the voltage across the resistor (\(V_R\)) and the voltage across the diode (\(V_D\)) equals the source voltage (\(V_S\)): \(V_S = V_R + V_D\). If \(V_S = 1.5 \text{ V}\) and the diode is forward-biased, the voltage across the diode will attempt to reach \(0.7 \text{ V}\). Once \(V_D \approx 0.7 \text{ V}\), the diode starts conducting. The remaining voltage across the resistor would be \(V_R = V_S – V_D = 1.5 \text{ V} – 0.7 \text{ V} = 0.8 \text{ V}\). The current through the circuit would then be \(I = V_R / R = 0.8 \text{ V} / 100 \Omega = 0.008 \text{ A} = 8 \text{ mA}\). The question asks about the state of the diode when the source voltage is \(0.5 \text{ V}\). Since \(0.5 \text{ V}\) is less than the cut-in voltage of \(0.7 \text{ V}\), the diode will not conduct any significant current. In this condition, the diode is considered to be in the “off” state or reverse-biased (even though the polarity is correct for forward bias, the voltage is insufficient to overcome the barrier potential). The voltage across the diode will essentially be equal to the source voltage, \(0.5 \text{ V}\), and the current through the circuit will be negligible, close to zero. Therefore, the diode is not conducting. The core concept tested here is the understanding of the non-linear behavior of a semiconductor diode and the significance of its cut-in voltage in determining its operational state. For students aspiring to excel at the National Institute of Technology Delhi, a firm grasp of such fundamental semiconductor device characteristics is crucial for advanced studies in electronics and electrical engineering. This understanding forms the bedrock for analyzing more complex circuits and systems, including those involving integrated circuits and power electronics, which are areas of significant research and academic focus at NIT Delhi. The ability to predict a diode’s behavior based on applied voltage and its intrinsic properties is a foundational skill.

The question probes the understanding of the fundamental principles governing the operation of a basic semiconductor diode in a forward-biased configuration, specifically focusing on the relationship between applied voltage and current. In a silicon PN junction diode, the ideal forward voltage drop, often referred to as the turn-on voltage or knee voltage, is approximately \(0.7\) V. This is the voltage at which the diode begins to conduct significant current. Before this voltage, the current is negligible. After this voltage, the current increases exponentially with a small increase in voltage. Consider a scenario where a silicon diode is connected in series with a \(1 \text{ k}\Omega\) resistor and a variable voltage source. The goal is to determine the voltage across the resistor when the diode is just beginning to conduct. When the diode is just beginning to conduct, the voltage across it is approximately its forward voltage drop. For silicon, this is \(V_{diode} \approx 0.7\) V. The total voltage supplied by the source, \(V_{source}\), is distributed across the diode and the resistor. Therefore, \(V_{source} = V_{diode} + V_{resistor}\). The voltage across the resistor is given by Ohm’s Law: \(V_{resistor} = I \times R\), where \(I\) is the current flowing through the circuit and \(R\) is the resistance. When the diode is just beginning to conduct, the current \(I\) is still relatively small but is increasing rapidly. However, the question asks for the voltage across the resistor *when the diode is just beginning to conduct*. This implies we are at the threshold of significant current flow. At this point, the voltage across the diode is approximately \(0.7\) V. If we assume the voltage source is set to a value slightly above \(0.7\) V, say \(1.0\) V, then the voltage across the resistor would be \(V_{resistor} = V_{source} – V_{diode} = 1.0 \text{ V} – 0.7 \text{ V} = 0.3 \text{ V}\). The current would then be \(I = V_{resistor} / R = 0.3 \text{ V} / 1000 \Omega = 0.3 \text{ mA}\). The question is designed to test the understanding of the voltage division in a simple series circuit containing a diode. The critical concept is that the diode itself has a voltage drop, and the remaining voltage from the source is dropped across the series resistor. The voltage across the resistor is directly proportional to the current, and the current is dictated by the diode’s forward bias characteristics. The question focuses on the state where the diode is *just beginning* to conduct, implying the voltage across the diode is at its characteristic forward voltage drop. The voltage across the resistor is then the total source voltage minus this diode drop. The key is to recognize that the voltage across the resistor is not zero when the diode is beginning to conduct, but rather it is the source voltage minus the diode’s forward voltage drop. The specific value of the resistor (\(1 \text{ k}\Omega\)) and the source voltage are parameters that, when combined with the diode’s forward voltage drop, determine the current. The question implicitly asks for the voltage across the resistor when the diode’s voltage is at its threshold. Let’s re-evaluate the question’s intent. It asks about the voltage across the resistor *when the diode is just beginning to conduct*. This means the voltage across the diode is approximately \(0.7\) V. If the source voltage is, for example, \(1.0\) V, then the voltage across the resistor is \(1.0 \text{ V} – 0.7 \text{ V} = 0.3 \text{ V}\). The current would be \(0.3 \text{ V} / 1000 \Omega = 0.3 \text{ mA}\). The question is not asking for the current, but the voltage across the resistor. The crucial point is that the voltage across the resistor is the difference between the source voltage and the diode’s forward voltage drop. The question is framed to assess if the candidate understands that the diode “consumes” a certain voltage, and the remainder is available for the resistor. The specific value of the source voltage is not given, but the options are specific voltage values. This suggests we need to infer the most appropriate scenario. Consider the options provided. If the voltage across the resistor were \(0.7\) V, it would imply the diode voltage is \(0\) V, which is incorrect for conduction. If it were \(0\) V, it would mean the source voltage is equal to the diode voltage, and no current flows through the resistor, which contradicts “beginning to conduct.” If it were \(1.0\) V, it would imply the diode voltage is \(0\) V, which is also incorrect. Therefore, the voltage across the resistor must be less than the source voltage and dependent on the diode’s forward voltage drop. The question is subtly asking about the voltage across the resistor when the diode’s voltage is at its turn-on threshold. If we assume a source voltage that allows for this threshold to be met, the voltage across the resistor will be the source voltage minus \(0.7\) V. The options provided are specific voltage values. The most plausible scenario that aligns with the concept of a diode “just beginning to conduct” in a circuit with a \(1 \text{ k}\Omega\) resistor and a variable source, where the diode’s forward voltage is \(0.7\) V, is that the voltage across the resistor is the difference between the source voltage and the diode’s forward voltage. If we consider a source voltage of \(1.0\) V, then the voltage across the resistor is \(0.3\) V. This represents a state where the diode is indeed starting to conduct. The calculation is as follows: When a silicon diode is just beginning to conduct, the voltage across it is approximately \(V_{diode} \approx 0.7\) V. In a series circuit with a resistor \(R\) and a voltage source \(V_{source}\), the total voltage is divided: \(V_{source} = V_{diode} + V_{resistor}\). The voltage across the resistor is \(V_{resistor} = V_{source} – V_{diode}\). If we consider a source voltage of \(1.0\) V, then \(V_{resistor} = 1.0 \text{ V} – 0.7 \text{ V} = 0.3 \text{ V}\). The current through the resistor would be \(I = V_{resistor} / R = 0.3 \text{ V} / 1000 \Omega = 0.3 \text{ mA}\). The question asks for the voltage across the resistor. The most appropriate answer, reflecting the state where the diode is just beginning to conduct, is the voltage across the resistor when the source voltage is slightly above the diode’s forward voltage. Assuming a source voltage of \(1.0\) V, the voltage across the resistor is \(0.3\) V. This value is derived from the fundamental understanding of voltage division in a diode circuit at its conduction threshold. Final Answer is \(0.3\) V.

The question probes the understanding of the fundamental principles governing the operation of a basic electronic circuit, specifically focusing on the behavior of a diode in a forward-biased configuration. When a diode is forward-biased, it allows current to flow through it, but not without a certain voltage drop across it, known as the cut-in voltage or threshold voltage. For a silicon diode, this voltage is typically around 0.7V. The circuit described consists of a voltage source, a resistor, and a diode connected in series. The voltage source is given as 5V. The resistor is 1kΩ. The diode is a standard silicon diode. In a forward-biased diode circuit, the voltage across the diode will be approximately its cut-in voltage once the applied voltage exceeds this threshold. The remaining voltage from the source will be dropped across the resistor. The current flowing through the circuit is determined by Ohm’s Law applied to the resistor: \(I = \frac{V_{source} – V_{diode}}{R}\). Given \(V_{source} = 5V\), \(R = 1k\Omega = 1000\Omega\), and assuming \(V_{diode} \approx 0.7V\) for a silicon diode, the current is: \[ I = \frac{5V – 0.7V}{1000\Omega} = \frac{4.3V}{1000\Omega} = 0.0043A \] This current is equal to 4.3mA. The question asks about the behavior of the diode and the circuit when the input voltage is significantly higher than the diode’s forward voltage drop. The key concept here is that a forward-biased diode acts as a voltage regulator for its own terminals, maintaining a relatively constant voltage drop (the cut-in voltage) across it, provided the current through it is within its operating limits. The resistor in series limits the current to a safe level. Therefore, the diode will conduct, and the voltage across it will be approximately 0.7V, while the voltage across the resistor will be the source voltage minus the diode voltage. The current will be determined by the voltage across the resistor divided by its resistance. The National Institute of Technology Delhi Entrance Exam often tests such fundamental circuit analysis concepts, emphasizing the practical behavior of components in a circuit. Understanding this behavior is crucial for designing and analyzing various electronic systems, a core aspect of electrical engineering programs at NIT Delhi. This question assesses the candidate’s ability to apply basic circuit laws and component characteristics to predict circuit behavior, a skill vital for success in advanced engineering studies.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the aliasing phenomenon and its mitigation through sampling. When a continuous-time signal \(x(t)\) is sampled at a rate \(f_s\), the resulting discrete-time signal \(x[n] = x(nT)\), where \(T = 1/f_s\) is the sampling period, can exhibit aliasing if the signal’s bandwidth exceeds \(f_s/2\). Aliasing occurs when higher frequencies in the analog signal are misrepresented as lower frequencies in the sampled signal. The Nyquist-Shannon sampling theorem states that to perfectly reconstruct an analog signal from its samples, the sampling frequency \(f_s\) must be at least twice the highest frequency component present in the signal, i.e., \(f_s \ge 2f_{max}\). This minimum sampling frequency is known as the Nyquist rate. In the given scenario, the analog signal has frequency components up to \(15 \text{ kHz}\). Therefore, \(f_{max} = 15 \text{ kHz}\). To avoid aliasing, the sampling frequency \(f_s\) must satisfy \(f_s \ge 2 \times 15 \text{ kHz} = 30 \text{ kHz}\). The question asks for the minimum sampling frequency that prevents aliasing. This minimum frequency is precisely the Nyquist rate. Thus, the minimum sampling frequency required is \(30 \text{ kHz}\). Understanding this concept is crucial for students at the National Institute of Technology Delhi, particularly in programs like Electrical Engineering and Computer Science, where signal processing is a core subject. Proper sampling ensures the integrity of data acquisition and processing in various applications, from telecommunications to medical imaging. Failure to adhere to the Nyquist criterion can lead to distorted signals and erroneous results, highlighting the importance of this fundamental principle in practical engineering. The ability to identify the Nyquist rate and its implications for signal reconstruction is a key indicator of a student’s grasp of digital signal processing fundamentals, a vital skill for innovation and problem-solving within the technological landscape that NIT Delhi aims to foster.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its implications for aliasing. The Nyquist-Shannon sampling theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the signal. This minimum sampling frequency is known as the Nyquist rate, \(f_{Nyquist} = 2f_{max}\). In this scenario, the analog signal has a maximum frequency component of \(15 \text{ kHz}\). Therefore, the minimum sampling frequency required to avoid aliasing, according to the Nyquist-Shannon theorem, is \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). The question asks about the consequence of sampling at \(25 \text{ kHz}\), which is below the Nyquist rate. When a signal is sampled below its Nyquist rate, higher frequency components in the original analog signal are misrepresented as lower frequencies in the sampled digital signal. This phenomenon is called aliasing. Specifically, frequencies above \(f_s/2\) will be aliased. The folding frequency is \(f_s/2 = 25 \text{ kHz} / 2 = 12.5 \text{ kHz}\). Any frequency component in the original signal that is greater than \(12.5 \text{ kHz}\) will appear as a different, lower frequency after sampling at \(25 \text{ kHz}\). The component at \(15 \text{ kHz}\) is greater than \(12.5 \text{ kHz}\). The aliased frequency (\(f_{alias}\)) can be calculated using the formula \(f_{alias} = |f – n \cdot f_s|\), where \(f\) is the original frequency and \(n\) is an integer chosen such that \(0 \le f_{alias} \le f_s/2\). For \(f = 15 \text{ kHz}\) and \(f_s = 25 \text{ kHz}\), we can find \(n\). If \(n=1\), \(f_{alias} = |15 \text{ kHz} – 1 \cdot 25 \text{ kHz}| = |-10 \text{ kHz}| = 10 \text{ kHz}\). Since \(10 \text{ kHz}\) is within the range \(0 \le f_{alias} \le 12.5 \text{ kHz}\), the \(15 \text{ kHz}\) component will be aliased to \(10 \text{ kHz}\). Therefore, sampling an analog signal with a maximum frequency of \(15 \text{ kHz}\) at a rate of \(25 \text{ kHz}\) will result in the \(15 \text{ kHz}\) component being aliased to \(10 \text{ kHz}\), leading to distortion and loss of information. This is a fundamental concept in digital signal processing taught in courses at institutions like the National Institute of Technology Delhi, emphasizing the critical importance of adhering to sampling rate requirements for accurate signal representation. Understanding aliasing is crucial for designing effective analog-to-digital converters and for interpreting digital signal processing results across various engineering disciplines.

The core principle tested here is the understanding of how different types of feedback mechanisms influence the stability and responsiveness of a system, particularly in the context of engineering design and control systems, which are fundamental to many programs at the National Institute of Technology Delhi. A positive feedback loop, by its nature, amplifies deviations from a set point. If a system is designed with a positive feedback mechanism for error correction, any initial error will be magnified, leading to instability and potentially runaway behavior. For instance, if a thermostat were designed with positive feedback for temperature control, a slight increase in temperature would cause the heating system to turn on more intensely, further increasing the temperature, creating a cycle of escalating deviation. Conversely, negative feedback acts to counteract deviations, driving the system back towards its equilibrium. This is crucial for maintaining stable operation in devices and processes. Therefore, a system designed with positive feedback for error correction would inherently be unstable and prone to oscillations or divergence, making it unsuitable for precise control or predictable operation. The question probes the candidate’s ability to discern the fundamental impact of feedback types on system dynamics, a concept vital for advanced coursework in areas like control theory, electronics, and mechanical systems at NIT Delhi.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its implications in real-world scenarios. The Nyquist-Shannon sampling theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the signal. This minimum sampling frequency is known as the Nyquist rate, \(f_{Nyquist} = 2f_{max}\). In this scenario, the analog signal has a maximum frequency component of 15 kHz. Therefore, according to the Nyquist-Shannon sampling theorem, the minimum sampling frequency required to avoid aliasing and ensure perfect reconstruction is \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). The question asks about the consequence of sampling at a frequency *lower* than this minimum requirement. When the sampling frequency (\(f_s\)) is less than \(2f_{max}\), higher frequency components in the original signal are misrepresented as lower frequencies in the sampled signal. This phenomenon is called aliasing. Aliasing distorts the reconstructed signal, making it impossible to recover the original waveform accurately. The higher frequencies “fold back” into the lower frequency spectrum, creating spurious signals that were not present in the original analog input. This is a critical concept in digital signal processing, as it dictates the design of analog-to-digital converters (ADCs) and the preprocessing steps (like anti-aliasing filters) required before sampling. For students at the National Institute of Technology Delhi, understanding aliasing is crucial for courses in communications, control systems, and embedded systems, where analog signals are frequently converted to digital for processing. The ability to identify the cause and effect of undersampling is a key indicator of foundational knowledge in these areas.

The question probes the understanding of the fundamental principles governing the operation of a basic semiconductor diode in a forward-biased configuration, specifically focusing on the voltage drop across it. In a forward-biased diode, the applied voltage overcomes the built-in potential barrier, allowing current to flow. For a silicon diode, this barrier potential is approximately \(0.7\) V. When a voltage of \(5\) V is applied across a series combination of a silicon diode and a \(1\) kΩ resistor, and the diode is forward-biased, the voltage across the diode will be its forward voltage drop. The remaining voltage will be dropped across the resistor. Therefore, the voltage across the resistor is \(V_R = V_{applied} – V_{diode}\). Assuming ideal forward bias conditions where the diode voltage is stable at its threshold, the voltage across the resistor is \(5 \text{ V} – 0.7 \text{ V} = 4.3 \text{ V}\). The current flowing through the circuit can then be calculated using Ohm’s Law: \(I = \frac{V_R}{R} = \frac{4.3 \text{ V}}{1 \text{ k}\Omega} = \frac{4.3 \text{ V}}{1000 \Omega} = 4.3 \text{ mA}\). The question asks for the voltage drop across the resistor. This voltage drop is the applied voltage minus the forward voltage drop of the diode. Thus, the voltage across the resistor is \(5 \text{ V} – 0.7 \text{ V} = 4.3 \text{ V}\). This scenario is fundamental to understanding basic circuit analysis involving semiconductor devices, a core concept for aspiring engineers at the National Institute of Technology Delhi. Understanding this forward voltage drop is crucial for designing circuits that utilize diodes for rectification, switching, or voltage regulation, ensuring proper current and voltage levels are maintained. The ability to predict these values is a hallmark of strong foundational knowledge in electronics engineering, a key area of focus at NIT Delhi.

The question probes the understanding of the foundational principles of digital signal processing, specifically concerning the Discrete Fourier Transform (DFT) and its properties. The scenario describes a signal \(x[n]\) and its DFT \(X[k]\). The core concept being tested is the linearity property of the DFT, which states that if \(y[n] = ax[n] + bz[n]\), then \(Y[k] = aX[k] + bZ[k]\). In this problem, we are given \(x[n]\) and its DFT \(X[k]\). We are then introduced to a new signal \(y[n] = 3x[n-2]\). To find the DFT of \(y[n]\), denoted as \(Y[k]\), we need to apply the time-shifting property of the DFT. The time-shifting property states that if \(y[n] = x[n-n_0]\), then \(Y[k] = X[k]e^{-j\frac{2\pi kn_0}{N}}\), where \(N\) is the length of the DFT. In this case, \(n_0 = 2\). Therefore, the DFT of \(x[n-2]\) is \(X[k]e^{-j\frac{2\pi k(2)}{N}}\). Since \(y[n] = 3x[n-2]\), by the linearity property, \(Y[k] = 3 \times (X[k]e^{-j\frac{4\pi k}{N}})\). Thus, \(Y[k] = 3e^{-j\frac{4\pi k}{N}}X[k]\). This demonstrates how a linear transformation (scaling by 3) and a time shift (by 2 samples) affect the DFT of a signal. Understanding these properties is crucial for various applications in signal analysis and processing, which are core to the curriculum at NIT Delhi, particularly in fields like Electrical Engineering and Computer Science. The ability to manipulate signals in the frequency domain by understanding these transformations is a fundamental skill for any aspiring engineer.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its implications for aliasing. The Nyquist-Shannon sampling theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the original analog signal. Mathematically, this is expressed as \(f_s \ge 2f_{max}\). Aliasing occurs when this condition is violated, meaning \(f_s < 2f_{max}\). In such a scenario, higher frequencies in the analog signal are misrepresented as lower frequencies in the sampled digital signal, leading to distortion. Consider an analog signal with a maximum frequency component of 15 kHz. According to the Nyquist-Shannon sampling theorem, the minimum sampling frequency required to avoid aliasing is \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). If the sampling frequency used is 25 kHz, which is less than the required minimum of 30 kHz, aliasing will occur. The frequencies that will be aliased into the baseband (0 to \(f_s/2\), which is 0 to 12.5 kHz in this case) are those above \(f_s/2\). Specifically, a frequency \(f\) greater than \(f_s/2\) will appear as \(|f – n f_s|\) for some integer \(n\) such that the aliased frequency is within the range \([0, f_s/2]\). For a sampling frequency of 25 kHz, the folding frequency is 12.5 kHz. A frequency of 15 kHz, which is above 12.5 kHz, will be aliased. The aliased frequency can be calculated as \(f_{alias} = |f – n f_s|\) where \(n\) is chosen such that \(0 \le f_{alias} \le f_s/2\). For \(f = 15 \text{ kHz}\) and \(f_s = 25 \text{ kHz}\), we can choose \(n=1\). Then, \(f_{alias} = |15 \text{ kHz} – 1 \times 25 \text{ kHz}| = |-10 \text{ kHz}| = 10 \text{ kHz}\). This 10 kHz frequency falls within the desired baseband range of 0 to 12.5 kHz. Therefore, the 15 kHz component will be incorrectly represented as 10 kHz in the sampled digital signal. This phenomenon is a critical consideration in the design of analog-to-digital converters (ADCs) and signal acquisition systems at institutions like the National Institute of Technology Delhi, where precise data representation is paramount for research and development in areas such as telecommunications and control systems. Understanding aliasing is fundamental to ensuring the integrity of sampled data and preventing erroneous conclusions in experimental analysis.

The question probes the understanding of the fundamental principles governing the operation of a basic semiconductor diode in a forward-biased configuration, specifically focusing on the voltage drop across it. In a forward-biased diode, the applied voltage overcomes the built-in potential barrier of the p-n junction. For silicon diodes, this barrier potential, often referred to as the turn-on voltage or knee voltage, is approximately \(0.7\) volts. Once this threshold is reached, the diode conducts current readily. However, the voltage across the diode does not increase linearly with current; it tends to stabilize around this characteristic voltage. Even with a significant increase in current, the voltage drop across a silicon diode in forward bias remains relatively constant, typically within the range of \(0.6\) to \(0.8\) volts, with \(0.7\) volts being a common approximation. The question describes a scenario where a silicon diode is forward-biased by a \(5\)V source through a \(1\)kΩ resistor. The current flowing through the diode is given as \(2\)mA. To find the voltage across the diode, we first calculate the voltage drop across the resistor. Using Ohm’s Law, \(V_R = I \times R\). Here, \(I = 2\)mA = \(2 \times 10^{-3}\) A, and \(R = 1\)kΩ = \(1000\) Ω. Therefore, \(V_R = (2 \times 10^{-3} \text{ A}) \times (1000 \text{ Ω}) = 2\)V. The total voltage from the source is \(5\)V, and this voltage is distributed across the resistor and the diode. So, \(V_{\text{source}} = V_R + V_D\), where \(V_D\) is the voltage across the diode. Rearranging the equation, \(V_D = V_{\text{source}} – V_R\). Substituting the values, \(V_D = 5 \text{ V} – 2 \text{ V} = 3\)V. This result, \(3\)V, is significantly higher than the typical forward voltage drop of a silicon diode. This discrepancy indicates that the premise of \(2\)mA current with a \(5\)V source and \(1\)kΩ resistor is inconsistent with the expected behavior of a silicon diode. If the diode were operating at its typical forward voltage of \(0.7\)V, the voltage across the resistor would be \(5\)V – \(0.7\)V = \(4.3\)V. The current through the resistor (and thus the diode) would then be \(I = V_R / R = 4.3 \text{ V} / 1000 \text{ Ω} = 0.0043\) A or \(4.3\)mA. The question, however, states a current of \(2\)mA. This implies that either the diode is not silicon, or the given current value is not representative of a typical forward-biased silicon diode operating under these conditions. Given the context of an entrance exam for a technology institute like NIT Delhi, questions often test the understanding of ideal or near-ideal component behavior and the ability to identify inconsistencies or apply fundamental laws. The most plausible interpretation, assuming a standard silicon diode, is that the stated current of \(2\)mA is a hypothetical value that, when combined with the circuit parameters, leads to a voltage across the diode that deviates from the expected characteristic. However, the question asks for the voltage across the diode *given* these conditions. If we strictly adhere to the given values and Ohm’s Law, the calculation yields \(3\)V. But this contradicts the fundamental property of a silicon diode. The question is designed to test if the candidate recognizes this contradiction and understands the typical operating voltage. The most appropriate answer, reflecting the fundamental characteristic of a silicon diode in forward bias, is its approximate turn-on voltage. The calculation \(V_D = 5 \text{ V} – (2 \text{ mA} \times 1 \text{ k}\Omega) = 5 \text{ V} – 2 \text{ V} = 3 \text{ V}\) is mathematically correct based on the provided numbers but physically unrealistic for a silicon diode. Therefore, the question implicitly tests the candidate’s knowledge of the diode’s intrinsic properties. The expected voltage drop for a forward-biased silicon diode is approximately \(0.7\)V. This is the voltage required to initiate significant current flow by overcoming the depletion region’s built-in potential. While the actual voltage can vary slightly with current and temperature, \(0.7\)V is the standard approximation taught and used. The scenario presented, with a \(5\)V source and a \(1\)kΩ resistor, would typically result in a current much higher than \(2\)mA if the diode were operating at its characteristic \(0.7\)V. The question’s premise is thus a distractor, pushing the candidate to perform a calculation that leads to a physically improbable result for the component in question. The correct approach is to recognize the inherent voltage drop of a silicon diode in forward bias, which is approximately \(0.7\)V, regardless of the specific (and potentially inconsistent) circuit parameters provided, as long as the circuit is indeed forward-biased. This tests a deeper understanding of semiconductor device physics beyond simple circuit analysis.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its implications for aliasing. The Nyquist-Shannon sampling theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the signal. This minimum sampling rate is known as the Nyquist rate, \(f_{Nyquist} = 2f_{max}\). In this scenario, the analog signal has a maximum frequency component of 15 kHz. Therefore, the minimum sampling frequency required to avoid aliasing is \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). The question states that the signal is sampled at 25 kHz. Since 25 kHz is less than the required Nyquist rate of 30 kHz, aliasing will occur. Aliasing is the phenomenon where higher frequencies in the analog signal are misrepresented as lower frequencies in the sampled digital signal, leading to distortion and loss of information. When the sampling frequency (\(f_s\)) is less than \(2f_{max}\), frequencies above \(f_s/2\) will be folded back into the lower frequency range, appearing as spurious signals. Specifically, a frequency \(f\) in the original signal will appear as \(|f – n f_s|\) for some integer \(n\), where \(f_s/2\) is the folding frequency. For a frequency of 15 kHz sampled at 25 kHz, the folding frequency is \(25 \text{ kHz} / 2 = 12.5 \text{ kHz}\). The 15 kHz component will be aliased to \(|15 \text{ kHz} – 1 \times 25 \text{ kHz}| = |-10 \text{ kHz}| = 10 \text{ kHz}\). This demonstrates that the original 15 kHz signal is incorrectly represented as a 10 kHz signal in the digital domain. This concept is crucial for students at the National Institute of Technology Delhi, particularly in programs like Electrical Engineering and Electronics and Communication Engineering, where understanding signal integrity and digital representation is paramount for designing and analyzing communication systems, control systems, and various other electronic applications. The ability to identify and mitigate aliasing is a core skill for ensuring accurate data acquisition and processing.

The question probes the understanding of the ethical considerations in scientific research, particularly concerning the dissemination of findings. The core principle being tested is the responsibility of researchers to ensure that their work, when published or presented, does not inadvertently lead to harmful applications or misinterpretations that could jeopardize public safety or societal well-being. While transparency and open communication are vital, they are not absolute when balanced against potential immediate and severe risks. Consider a scenario where a research team at the National Institute of Technology Delhi, specializing in advanced materials science, discovers a novel compound with exceptional structural integrity and energy absorption properties. This compound, if improperly synthesized or handled, could be unstable and prone to explosive decomposition under specific environmental stimuli. The team is preparing to present their findings at an international conference and submit a paper for peer review. The ethical dilemma lies in how to communicate the discovery without providing a “recipe” for a dangerous substance. Simply withholding the synthesis details would violate the principle of open scientific inquiry and hinder further legitimate research. However, publishing the exact synthesis parameters, including precise precursor ratios, reaction temperatures, and purification methods, could be misused by individuals or groups with malicious intent, potentially leading to the creation of an explosive device. Therefore, the most ethically sound approach, aligning with the responsible conduct of research emphasized at institutions like NIT Delhi, is to present the findings in a manner that highlights the compound’s properties and potential beneficial applications (e.g., in aerospace or protective gear) while deliberately omitting or obscuring the specific, critical synthesis parameters that could enable its dangerous misuse. This allows for scientific progress and peer validation without directly facilitating harm. The research team should, however, clearly document these omitted details internally and be prepared to share them under controlled circumstances with vetted researchers or regulatory bodies if a legitimate need arises for further safety or application-specific research. This nuanced approach balances scientific integrity with paramount ethical obligations.

The core principle being tested here is the understanding of the fundamental theorem of calculus and its application in evaluating definite integrals. The question asks for the value of a definite integral, which represents the net area under a curve between two specified limits. The integral to be evaluated is \[\int_{0}^{\pi/2} \cos(x) e^{\sin(x)} dx\]. To solve this, we can use a substitution method. Let \(u = \sin(x)\). Then, the differential \(du\) is \(du = \cos(x) dx\). We also need to change the limits of integration according to the substitution: When \(x = 0\), \(u = \sin(0) = 0\). When \(x = \pi/2\), \(u = \sin(\pi/2) = 1\). Substituting \(u\) and \(du\) into the integral, and changing the limits, we get: \[\int_{0}^{1} e^u du\] Now, we find the antiderivative of \(e^u\), which is \(e^u\). Evaluating the definite integral: \[[e^u]_{0}^{1} = e^1 – e^0\] \[= e – 1\] Therefore, the exact value of the definite integral is \(e – 1\). This result signifies the net accumulation of the function \(f(x) = \cos(x) e^{\sin(x)}\) over the interval \([0, \pi/2]\). Understanding such definite integrals is crucial in various fields of engineering and applied sciences, including signal processing, fluid dynamics, and thermodynamics, which are core areas of study at the National Institute of Technology Delhi. The ability to perform these calculations demonstrates a grasp of foundational calculus concepts essential for advanced coursework and research at the institution.

The question probes the understanding of the fundamental principles governing the operation of a basic transistor circuit, specifically focusing on the concept of biasing and its impact on transistor behavior. In a common-emitter configuration with a voltage divider bias, the quiescent collector current (\(I_C\)) and collector-emitter voltage (\(V_{CE}\)) are crucial for determining the operating point. For a silicon bipolar junction transistor (BJT) with a typical forward current gain (\(\beta\)) of 100, and given the biasing resistors \(R_1 = 100 \, \text{k}\Omega\), \(R_2 = 10 \, \text{k}\Omega\), \(R_C = 1 \, \text{k}\Omega\), and \(R_E = 470 \, \Omega\), with a supply voltage \(V_{CC} = 12 \, \text{V}\), we can determine the quiescent point. First, calculate the voltage at the base (\(V_B\)) using the voltage divider rule: \(V_B = V_{CC} \times \frac{R_2}{R_1 + R_2} = 12 \, \text{V} \times \frac{10 \, \text{k}\Omega}{100 \, \text{k}\Omega + 10 \, \text{k}\Omega} = 12 \, \text{V} \times \frac{10}{110} = \frac{120}{110} \, \text{V} \approx 1.09 \, \text{V}\) Next, determine the emitter voltage (\(V_E\)) by subtracting the base-emitter voltage drop (\(V_{BE}\)), which is approximately \(0.7 \, \text{V}\) for a silicon transistor: \(V_E = V_B – V_{BE} = 1.09 \, \text{V} – 0.7 \, \text{V} = 0.39 \, \text{V}\) Now, calculate the emitter current (\(I_E\)) using Ohm’s law: \(I_E = \frac{V_E}{R_E} = \frac{0.39 \, \text{V}}{470 \, \Omega} \approx 0.00083 \, \text{A} = 0.83 \, \text{mA}\) Assuming the transistor is in the active region, the collector current (\(I_C\)) is approximately equal to the emitter current (\(I_E\)), especially when \(\beta\) is large: \(I_C \approx I_E = 0.83 \, \text{mA}\) Finally, calculate the collector-emitter voltage (\(V_{CE}\)): \(V_{CE} = V_{CC} – I_C \times R_C – I_E \times R_E\) Since \(I_C \approx I_E\), we can simplify: \(V_{CE} = V_{CC} – (I_C \times R_C) – V_E\) \(V_{CE} = 12 \, \text{V} – (0.83 \, \text{mA} \times 1 \, \text{k}\Omega) – 0.39 \, \text{V}\) \(V_{CE} = 12 \, \text{V} – 0.83 \, \text{V} – 0.39 \, \text{V} = 10.78 \, \text{V}\) The operating point (\(Q\)-point) is therefore approximately \((I_C, V_{CE}) = (0.83 \, \text{mA}, 10.78 \, \text{V})\). This point is well within the active region, indicating stable amplification. The question assesses the candidate’s ability to apply fundamental circuit analysis techniques to determine the DC operating conditions of a BJT, a core concept in analog electronics taught at institutions like NIT Delhi. Understanding the \(Q\)-point is crucial for designing amplifiers, as it dictates the bias stability and the range of linear operation, directly impacting the fidelity and performance of electronic circuits. This analysis highlights the importance of proper biasing to ensure the transistor functions as an amplifier rather than a switch or being in cutoff, a foundational principle for any aspiring electronics engineer.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its implications for aliasing. The Nyquist-Shannon sampling theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency \(f_s\) must be at least twice the highest frequency component \(f_{max}\) present in the signal. This minimum sampling frequency is known as the Nyquist rate, \(f_{Nyquist} = 2f_{max}\). In this scenario, the analog signal has a maximum frequency component of 15 kHz. According to the Nyquist-Shannon sampling theorem, the minimum sampling frequency required to avoid aliasing is \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). The digital system samples the signal at 25 kHz. Since 25 kHz is less than the required minimum of 30 kHz, aliasing will occur. Aliasing is the phenomenon where high-frequency components in the analog signal are incorrectly interpreted as lower frequencies in the sampled digital signal. Specifically, when the sampling frequency \(f_s\) is less than \(2f_{max}\), frequencies above \(f_s/2\) will be aliased. The aliased frequency \(f_{alias}\) of a signal component with frequency \(f\) is given by \(f_{alias} = |f – k \cdot f_s|\), where \(k\) is an integer chosen such that \(f_{alias}\) falls within the range \([0, f_s/2]\). For a frequency component of 15 kHz in the original signal, and a sampling frequency of 25 kHz, the aliased frequency can be calculated. The folding frequency is \(f_s/2 = 25 \text{ kHz} / 2 = 12.5 \text{ kHz}\). The frequency 15 kHz is above the folding frequency. To find the aliased frequency, we can use the formula \(f_{alias} = |f – k \cdot f_s|\). Let’s try \(k=1\): \(f_{alias} = |15 \text{ kHz} – 1 \cdot 25 \text{ kHz}| = |-10 \text{ kHz}| = 10 \text{ kHz}\). Since 10 kHz is within the range \([0, 12.5 \text{ kHz}]\), this is the aliased frequency. Therefore, the 15 kHz component will appear as a 10 kHz component in the sampled signal. This demonstrates a fundamental concept in digital signal processing taught at institutions like NIT Delhi, emphasizing the importance of adhering to sampling rate requirements to preserve signal integrity. Understanding aliasing is crucial for designing effective digital systems in fields like telecommunications, audio processing, and control systems, all of which are areas of study at NIT Delhi.

The question probes the understanding of the fundamental principles governing the operation of a basic semiconductor diode in a forward-biased configuration, specifically focusing on the voltage drop across it. In a forward-biased diode, the applied voltage overcomes the built-in potential barrier of the p-n junction. For silicon diodes, this barrier potential, often referred to as the turn-on voltage or threshold voltage, is approximately \(0.7\) Volts. Once this threshold is surpassed, the diode conducts current. However, the voltage across the diode does not increase linearly with current; it remains relatively constant, close to the barrier potential, for a wide range of forward currents. This behavior is due to the exponential relationship between current and voltage in a diode, described by the Shockley diode equation: \(I = I_s (e^{V/(nV_T)} – 1)\), where \(I\) is the diode current, \(I_s\) is the reverse saturation current, \(V\) is the voltage across the diode, \(n\) is the ideality factor, and \(V_T\) is the thermal voltage. As \(V\) increases beyond the barrier potential, the exponential term dominates, leading to a rapid increase in current with only a small incremental increase in voltage. Therefore, in a practical forward-biased silicon diode circuit, the voltage drop across the diode will be approximately \(0.7\) V, assuming it is conducting. The question asks about the voltage across the diode when it is forward-biased and conducting. The most accurate representation of this voltage drop for a silicon diode is \(0.7\) V.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its implications for aliasing. The scenario describes a continuous-time signal \(x(t)\) with a maximum frequency component of \(f_{max} = 15\) kHz. The Nyquist-Shannon sampling theorem states that to perfectly reconstruct a continuous-time signal from its samples, the sampling frequency \(f_s\) must be at least twice the maximum frequency component of the signal, i.e., \(f_s \ge 2f_{max}\). This minimum sampling frequency is known as the Nyquist rate. In this case, the Nyquist rate for the signal \(x(t)\) is \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). The question states that the signal is sampled at a frequency \(f_s = 25 \text{ kHz}\). Since \(25 \text{ kHz} < 30 \text{ kHz}\), the sampling frequency is below the Nyquist rate. When a signal is sampled below its Nyquist rate, higher frequency components in the original signal "fold over" or masquerade as lower frequencies in the sampled signal. This phenomenon is called aliasing. The aliased frequency \(f_{alias}\) of a frequency \(f > f_s/2\) is given by \(f_{alias} = |f – k f_s|\), where \(k\) is an integer chosen such that \(0 \le f_{alias} < f_s/2\). In this scenario, the signal has components up to 15 kHz. The sampling frequency is 25 kHz, so the folding frequency is \(f_s/2 = 12.5\) kHz. Consider a frequency component at \(f = 15\) kHz. Since \(15 \text{ kHz} > 12.5 \text{ kHz}\), this component will be aliased. To find the aliased frequency, we use the formula. We need to find an integer \(k\) such that \(0 \le |15 \text{ kHz} – k \times 25 \text{ kHz}| < 12.5 \text{ kHz}\). If \(k=1\), \(|15 \text{ kHz} – 1 \times 25 \text{ kHz}| = |-10 \text{ kHz}| = 10 \text{ kHz}\). Since \(0 \le 10 \text{ kHz} < 12.5 \text{ kHz}\), the aliased frequency is 10 kHz. Therefore, the 15 kHz component of the original signal will appear as a 10 kHz component in the sampled signal, leading to distortion and an inability to perfectly reconstruct the original signal. This is a critical concept in digital signal processing and is fundamental to understanding the limitations and requirements of analog-to-digital conversion, a core area of study in electrical engineering and related disciplines at institutions like the National Institute of Technology Delhi. Understanding aliasing is crucial for designing effective anti-aliasing filters and selecting appropriate sampling rates to preserve signal integrity, ensuring accurate data acquisition and processing in various engineering applications.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its implications for signal reconstruction. The Nyquist-Shannon sampling theorem states that a band-limited signal can be perfectly reconstructed from its samples if the sampling frequency \(f_s\) is greater than twice the maximum frequency component \(f_{max}\) of the signal, i.e., \(f_s > 2f_{max}\). This minimum sampling rate is known as the Nyquist rate. In the given scenario, a continuous-time signal is sampled at a rate of \(10 \text{ kHz}\). The signal is known to contain frequency components up to \(6 \text{ kHz}\). To perfectly reconstruct this signal without aliasing, the sampling frequency must be strictly greater than twice the maximum frequency component. Calculation: Maximum frequency component, \(f_{max} = 6 \text{ kHz}\). Required minimum sampling frequency (Nyquist rate) = \(2 \times f_{max} = 2 \times 6 \text{ kHz} = 12 \text{ kHz}\). The actual sampling frequency is \(f_s = 10 \text{ kHz}\). Since the actual sampling frequency (\(10 \text{ kHz}\)) is less than the required Nyquist rate (\(12 \text{ kHz}\)), the signal cannot be perfectly reconstructed. Specifically, frequencies in the signal above \(f_s/2 = 10 \text{ kHz} / 2 = 5 \text{ kHz}\) will be subject to aliasing. The portion of the signal between \(5 \text{ kHz}\) and \(6 \text{ kHz}\) will be aliased to frequencies below \(5 \text{ kHz}\), leading to distortion and loss of information. Therefore, the signal cannot be faithfully reconstructed. This concept is crucial for students at the National Institute of Technology Delhi, particularly in programs like Electrical Engineering and Electronics and Communication Engineering, where understanding signal processing is foundational for areas such as telecommunications, control systems, and embedded systems design. Proper sampling ensures the integrity of information transmitted and processed digitally, preventing loss of data and ensuring accurate representation of the original analog signal. Failure to adhere to the Nyquist criterion can lead to significant errors in system performance and data interpretation.