National Institute of Technology Silchar Entrance Exam Set

The question probes the understanding of the fundamental principles of electromagnetic wave propagation and their interaction with materials, a core concept in electrical engineering and physics programs at National Institute of Technology Silchar. The scenario describes a plane electromagnetic wave incident on a semi-infinite dielectric medium. The key to solving this lies in understanding the conditions for total internal reflection (TIR). For a plane wave incident from a medium with refractive index \(n_1\) to a medium with refractive index \(n_2\), total internal reflection occurs when the angle of incidence \(\theta_i\) is greater than the critical angle \(\theta_c\). The critical angle is defined by Snell’s Law when the angle of refraction \(\theta_r\) is 90 degrees: \(n_1 \sin(\theta_i) = n_2 \sin(\theta_r)\). When \(\theta_r = 90^\circ\), we have \(n_1 \sin(\theta_c) = n_2 \sin(90^\circ)\), which simplifies to \(\sin(\theta_c) = \frac{n_2}{n_1}\). For TIR to be possible, we must have \(n_1 > n_2\). In this problem, the wave is incident from free space (where the refractive index \(n_1 = 1\)) onto a dielectric medium with a relative permittivity \(\epsilon_r = 4\). The refractive index of the dielectric medium is \(n_2 = \sqrt{\epsilon_r \mu_r}\). Assuming the medium is non-magnetic, the relative permeability \(\mu_r = 1\). Therefore, \(n_2 = \sqrt{4 \times 1} = 2\). Since \(n_1 = 1\) and \(n_2 = 2\), we have \(n_1 < n_2\). In this case, the condition for total internal reflection (\(n_1 > n_2\)) is not met. Instead, when light travels from a medium of lower refractive index to a medium of higher refractive index, refraction always occurs, and there is no critical angle beyond which reflection is total. The wave will be partially reflected and partially transmitted into the dielectric medium. The reflection coefficient for normal incidence would be \(\frac{n_1 – n_2}{n_1 + n_2} = \frac{1-2}{1+2} = -\frac{1}{3}\), indicating a phase reversal upon reflection. However, the question asks about the condition for total internal reflection. Since \(n_1 < n_2\), TIR is impossible. The wave will always be partially reflected and transmitted. Therefore, the statement that total internal reflection will occur for any angle of incidence is incorrect. The wave will be partially reflected and transmitted at all angles of incidence. The concept of critical angle and TIR is fundamental to understanding wave behavior at interfaces, which is crucial for designing optical components and understanding signal propagation in various media, areas of significant research at NIT Silchar.

The question probes the understanding of the fundamental principles governing the behavior of semiconductors under varying conditions, specifically focusing on the impact of temperature on carrier concentration and conductivity. In intrinsic semiconductors, the number of electrons in the conduction band equals the number of holes in the valence band, denoted by \(n_i\). The intrinsic carrier concentration \(n_i\) is highly temperature-dependent, increasing exponentially with temperature. This relationship is often approximated by the equation \(n_i \approx A T^{3/2} e^{-E_g / (2kT)}\), where \(A\) is a material-dependent constant, \(T\) is the absolute temperature, \(E_g\) is the band gap energy, and \(k\) is the Boltzmann constant. As temperature increases, \(n_i\) increases significantly. In an intrinsic semiconductor, conductivity (\(\sigma\)) is directly proportional to the intrinsic carrier concentration and the sum of the mobilities of electrons (\(\mu_n\)) and holes (\(\mu_p\)): \(\sigma = q n_i (\mu_n + \mu_p)\), where \(q\) is the elementary charge. While carrier mobilities (\(\mu_n\) and \(\mu_p\)) generally decrease with increasing temperature due to increased lattice scattering, the exponential increase in \(n_i\) dominates the conductivity behavior. Therefore, the conductivity of an intrinsic semiconductor increases substantially with rising temperature. The question asks about the primary factor that causes this increase in conductivity. Among the given options, the most accurate explanation is the exponential rise in intrinsic carrier concentration. While increased thermal agitation contributes to the mobility of charge carriers, its effect on conductivity is typically a decrease due to scattering. The dominant mechanism for conductivity enhancement in intrinsic semiconductors with temperature is the generation of more electron-hole pairs. This increased availability of charge carriers, despite potentially reduced mobility, leads to a net increase in conductivity. The National Institute of Technology Silchar’s curriculum in solid-state physics and semiconductor devices emphasizes these fundamental relationships, requiring students to discern the primary drivers of material properties under different environmental conditions. Understanding this exponential dependence is crucial for designing and analyzing semiconductor devices operating across a range of temperatures, a core competency expected of graduates from NIT Silchar’s engineering programs.

The question probes understanding of the fundamental principles of digital logic design, specifically concerning the minimization of Boolean expressions and the implications for hardware implementation. The scenario describes a combinational logic circuit designed to control a traffic light system at an intersection near the National Institute of Technology Silchar campus. The circuit’s output is determined by three input sensors: \(A\) (car approaching from North), \(B\) (car approaching from East), and \(C\) (pedestrian button pressed). The desired output is a high signal (1) to activate the green light for the North-South direction under specific conditions. The truth table provided implicitly defines the logic function. Let’s analyze the conditions for the green light to be active (output = 1): – If \(A\) is 1 (car from North) and \(B\) is 0 (no car from East), the green light should be on regardless of \(C\). This covers minterms \(A\bar{B}\bar{C}\) and \(A\bar{B}C\). – If \(A\) is 0 (no car from North) and \(B\) is 1 (car from East), the green light should be on only if \(C\) is 1 (pedestrian pressed). This covers minterm \(\bar{A}BC\). – If both \(A\) and \(B\) are 1, the green light should not be active (output = 0), implying a conflict or a priority for other directions not detailed here. – If \(A\) is 0 and \(B\) is 0, the green light should not be active, regardless of \(C\). Therefore, the Boolean expression representing the conditions where the green light is ON is: \(F(A, B, C) = A\bar{B} + \bar{A}BC\) Now, let’s simplify this expression using Boolean algebra or Karnaugh maps. Using Boolean algebra: \(F(A, B, C) = A\bar{B} + \bar{A}BC\) We can factor out \(A\bar{B}\) from the first term. The second term \(\bar{A}BC\) can be expanded using the consensus theorem or by adding redundant terms. A common simplification technique is to add \(A\bar{B}C\) to the expression, as \(A\bar{B} + A\bar{B}C = A\bar{B}\). So, \(F(A, B, C) = A\bar{B} + A\bar{B}C + \bar{A}BC\) Now, group terms: \(F(A, B, C) = A\bar{B}(1+C) + \bar{A}BC\) Since \(1+C = 1\), this simplifies to: \(F(A, B, C) = A\bar{B} + \bar{A}BC\) This expression is already in its minimal Sum-of-Products (SOP) form. Let’s consider if further simplification is possible. We can use the consensus theorem: \(XY + \bar{X}Z + YZ = XY + \bar{X}Z\). In our expression \(A\bar{B} + \bar{A}BC\), let \(X = A\), \(Y = \bar{B}\), and \(Z = BC\). This doesn’t directly fit the theorem. Let’s try another approach. Consider the implicants \(A\bar{B}\) and \(\bar{A}BC\). The prime implicants are \(A\bar{B}\) (covers minterms 4 and 5, i.e., \(A\bar{B}\bar{C}\) and \(A\bar{B}C\)) and \(\bar{A}BC\) (covers minterm 3, i.e., \(\bar{A}BC\)). Both are essential prime implicants because \(A\bar{B}\) is the only implicant covering minterm 4, and \(\bar{A}BC\) is the only implicant covering minterm 3. Thus, the minimal SOP form is indeed \(A\bar{B} + \bar{A}BC\). This expression requires two AND gates (one 2-input for \(A\bar{B}\), one 3-input for \(\bar{A}BC\)) and one OR gate (2-input) to implement. The question asks for the most efficient implementation in terms of gate count and complexity, which is directly related to the minimal Boolean expression. The expression \(A\bar{B} + \bar{A}BC\) is the most simplified form. Let’s evaluate the options based on this minimal expression. Option a) \(A\bar{B} + \bar{A}BC\): This is the minimal SOP form derived. Option b) \(A\bar{B} + \bar{A}BC + AC\): Adding \(AC\) is redundant. If \(A=1, C=1\), then either \(B=1\) (making \(\bar{A}BC=0\)) or \(B=0\) (making \(A\bar{B}=1\)). If \(A=1, C=1, B=0\), then \(A\bar{B}=1\). If \(A=1, C=1, B=1\), then \(A\bar{B}=0\) and \(\bar{A}BC=0\). So \(AC\) is covered by \(A\bar{B}\) when \(B=0\). Option c) \(A\bar{B} + \bar{A}BC + \bar{A}\bar{B}\): \(\bar{A}\bar{B}\) is not required by the logic. If \(A=0, B=0\), the output should be 0. Option d) \(A\bar{B} + BC\): This is incorrect because it omits the condition \(\bar{A}\) for the \(BC\) term. If \(A=1, B=1, C=1\), this expression would yield 1, but the original logic implies 0 when \(A=1, B=1\). Therefore, the most simplified and correct expression is \(A\bar{B} + \bar{A}BC\). This minimal form translates to the most efficient hardware implementation, using fewer logic gates, which is a key consideration in digital circuit design, aligning with the engineering principles taught at National Institute of Technology Silchar. Efficient design reduces power consumption, cost, and propagation delay, all critical factors in real-world applications like traffic control systems.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the aliasing phenomenon and its mitigation through anti-aliasing filters. In the context of the National Institute of Technology Silchar’s curriculum, which often emphasizes practical applications in fields like telecommunications and control systems, understanding sampling theory is paramount. Aliasing occurs when the sampling frequency (\(f_s\)) is not sufficiently high relative to the highest frequency component (\(f_{max}\)) present in the analog signal. According to the Nyquist-Shannon sampling theorem, to perfectly reconstruct an analog signal from its samples, the sampling frequency must be at least twice the highest frequency component in the signal, i.e., \(f_s > 2f_{max}\). If this condition is violated, higher frequencies in the analog signal “fold back” into the lower frequency range, appearing as distinct, lower-frequency components in the sampled signal, which are not present in the original signal. This distortion is known as aliasing. An anti-aliasing filter is a low-pass filter placed before the sampler. Its purpose is to attenuate or remove frequency components in the analog signal that are above half the sampling frequency (\(f_s/2\)), also known as the Nyquist frequency. By ensuring that the signal entering the sampler contains no frequencies greater than \(f_s/2\), the anti-aliasing filter effectively prevents aliasing from occurring, allowing for accurate digital representation and processing of the signal. Consider an analog signal with a maximum frequency component of 15 kHz. If this signal is sampled at a rate of 20 kHz, aliasing will occur because the sampling frequency (20 kHz) is less than twice the maximum frequency (2 * 15 kHz = 30 kHz). Specifically, frequencies above \(f_s/2 = 20 \text{ kHz} / 2 = 10 \text{ kHz}\) will alias. For instance, the 15 kHz component will appear as \(|15 \text{ kHz} – 20 \text{ kHz}| = 5 \text{ kHz}\) in the sampled data. To prevent this, an anti-aliasing filter with a cutoff frequency below 15 kHz but above the desired signal bandwidth (e.g., below 10 kHz) would be necessary to attenuate frequencies above 10 kHz before sampling. If the sampling rate were increased to 40 kHz, then the Nyquist frequency would be 20 kHz. In this case, the 15 kHz component would be below the Nyquist frequency, and aliasing would not occur for this specific frequency component, assuming no other components above 20 kHz exist. The anti-aliasing filter’s role is to guarantee that no frequencies above \(f_s/2\) are present at the sampler’s input, regardless of the original signal’s bandwidth, thereby ensuring faithful digital conversion.

The question probes the understanding of fundamental principles in digital logic design, specifically concerning the minimization of Boolean expressions and the implications of using different logic gates. The scenario describes a situation where a designer at National Institute of Technology Silchar is tasked with implementing a specific logic function using only NAND gates. The core concept being tested is the universality of NAND gates, meaning any Boolean function can be implemented using only NAND gates. To solve this, one must understand how to convert a given logic expression into an equivalent form using only NAND operations. Let’s consider a hypothetical logic function \(F(A, B, C) = (A \cdot \bar{B}) + (\bar{A} \cdot C)\). To implement this using only NAND gates, we first need to express it in a form that can be directly translated. Using De Morgan’s laws: \(F = (A \cdot \bar{B}) + (\bar{A} \cdot C)\) \(F = \overline{\overline{(A \cdot \bar{B}) + (\bar{A} \cdot C)}}\) \(F = \overline{(\overline{A \cdot \bar{B}}) \cdot (\overline{\bar{A} \cdot C})}\) Now, we need to express \(\bar{B}\) and \(\bar{A}\) using NAND gates. \(\bar{B} = B \cdot B\) (AND with itself) \(\bar{B} = \overline{\overline{B \cdot B}}\) \(\bar{B} = \overline{B \text{ NAND } B}\) Similarly, \(\bar{A} = \overline{A \text{ NAND } A}\). Substituting these back into the expression for F: \(F = \overline{(\overline{A \cdot \overline{B}}) \cdot (\overline{\bar{A} \cdot C})}\) \(F = \overline{(\overline{A \cdot (\overline{B \text{ NAND } B})}) \cdot (\overline{(\overline{A \text{ NAND } A}) \cdot C})}\) This expression can be directly implemented using NAND gates. The key insight is that any Boolean function can be realized using only NAND gates. The question asks about the *most efficient* implementation in terms of gate count, which often relates to simplifying the expression before conversion or using specific NAND gate configurations. However, without a specific target expression or a minimization goal beyond universality, the most fundamental answer relates to the capability of NAND gates. The question, however, is framed around a scenario where a student at NIT Silchar is evaluating different approaches for implementing a complex logic circuit. The options presented are not about a specific calculation but about the *principle* of universal gates and their implications for design. The most accurate statement regarding the universal nature of NAND gates is that they can indeed construct any logic function, which is a foundational concept taught in digital electronics courses at institutions like NIT Silchar. The efficiency (gate count) is a secondary optimization, but the primary capability is universality. Therefore, the statement that accurately reflects this fundamental property is the correct choice.

The question probes the understanding of the fundamental principles governing the operation of a basic semiconductor diode in forward bias. When a diode is forward biased, current flows through it. The voltage across the diode, when conducting, is not zero but a small, relatively constant value known as the cut-in voltage or threshold voltage. This voltage is required to overcome the potential barrier at the P-N junction. For silicon diodes, this value is typically around \(0.7V\), and for germanium diodes, it’s around \(0.3V\). The question describes a scenario where a silicon diode is connected in series with a resistor and a voltage source. The voltage source is set to \(5V\), and the resistor has a resistance of \(1k\Omega\). The diode is forward-biased. To determine the voltage across the resistor, we first acknowledge that in a forward-biased silicon diode, the voltage drop across the diode itself is approximately \(0.7V\). Therefore, the remaining voltage from the source must be dropped across the resistor. The voltage across the resistor, \(V_R\), can be calculated using Kirchhoff’s Voltage Law (KVL) for the series circuit: \(V_{source} = V_D + V_R\). Substituting the known values, we get \(5V = 0.7V + V_R\). Solving for \(V_R\), we find \(V_R = 5V – 0.7V = 4.3V\). The current flowing through the circuit, \(I\), can then be calculated using Ohm’s Law for the resistor: \(I = \frac{V_R}{R}\). Thus, \(I = \frac{4.3V}{1k\Omega} = \frac{4.3V}{1000\Omega} = 0.0043A = 4.3mA\). The question asks for the voltage across the resistor. The calculation clearly shows this to be \(4.3V\). This concept is foundational for understanding basic electronic circuits and is crucial for students at the National Institute of Technology Silchar, particularly in disciplines like Electrical Engineering and Electronics and Communication Engineering, where circuit analysis is a core component. Understanding the non-ideal behavior of semiconductor devices, such as the voltage drop across a forward-biased diode, is essential for accurate circuit design and analysis.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning aliasing and the Nyquist-Shannon sampling theorem. Aliasing occurs when a signal is sampled at a rate lower than twice its highest frequency component. This results in high-frequency components being misrepresented as lower frequencies, leading to distortion. The Nyquist-Shannon sampling theorem states that to perfectly reconstruct a signal, the sampling frequency (\(f_s\)) must be strictly greater than twice the maximum frequency (\(f_{max}\)) present in the signal, i.e., \(f_s > 2f_{max}\). In the given scenario, a continuous-time signal with a maximum frequency of 15 kHz is sampled at a rate of 20 kHz. According to the Nyquist criterion, the minimum sampling rate required for perfect reconstruction is \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). Since the actual sampling rate (20 kHz) is less than the required minimum rate (30 kHz), aliasing will occur. The frequencies that will be aliased are those above \(f_s/2\), which in this case is \(20 \text{ kHz} / 2 = 10 \text{ kHz}\). Specifically, frequencies between 10 kHz and 15 kHz will fold back into the range of 0 to 10 kHz. For instance, a frequency of 12 kHz would appear as \(20 \text{ kHz} – 12 \text{ kHz} = 8 \text{ kHz}\), and a frequency of 15 kHz would appear as \(20 \text{ kHz} – 15 \text{ kHz} = 5 \text{ kHz}\). This phenomenon corrupts the original signal’s spectral content, making accurate reconstruction impossible without prior filtering. Therefore, the primary consequence of sampling below the Nyquist rate is the introduction of aliasing artifacts.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the aliasing phenomenon and its mitigation. Aliasing occurs when a continuous-time signal is sampled at a rate lower than twice its highest frequency component (Nyquist rate). This leads to the higher frequencies being misrepresented as lower frequencies in the sampled signal. To prevent aliasing, an anti-aliasing filter, which is a low-pass filter, is applied *before* sampling. This filter attenuates frequencies above the Nyquist frequency, ensuring that only frequencies below half the sampling rate are present in the signal being sampled. Consider a scenario where a signal contains frequencies up to \(f_{max}\). If this signal is sampled at a rate \(f_s\), the Nyquist frequency is \(f_s/2\). For accurate reconstruction, all frequency components of the original signal must be below \(f_s/2\). If the signal contains frequencies \(f > f_s/2\), these will alias. An anti-aliasing filter is designed to have a cutoff frequency \(f_c\) such that \(f_c < f_s/2\). This filter removes or significantly reduces the amplitude of frequencies above \(f_c\), thus ensuring that any remaining components are below the Nyquist frequency for the given sampling rate. Therefore, the primary role of the anti-aliasing filter is to limit the bandwidth of the analog signal before it is digitized.

The question probes understanding of the fundamental principles governing the operation of a basic diode circuit, specifically in the context of rectification. When a sinusoidal AC voltage \( V_{in}(t) = V_p \sin(\omega t) \) is applied to a series circuit containing a diode and a resistor \( R \), the diode will conduct only when the anode is at a higher potential than the cathode. Assuming an ideal diode with zero forward voltage drop and infinite reverse resistance, the output voltage across the resistor \( V_{out}(t) \) will be equal to the input voltage during the positive half-cycles of the input waveform and zero during the negative half-cycles. The average DC voltage across the resistor is calculated by integrating the output voltage over one period and dividing by the period. For a sinusoidal input, the period is \( T = \frac{2\pi}{\omega} \). The output voltage is \( V_{out}(t) = V_{in}(t) \) for \( V_{in}(t) > 0 \) and \( V_{out}(t) = 0 \) for \( V_{in}(t) \le 0 \). So, \( V_{out}(t) = \begin{cases} V_p \sin(\omega t) & \text{if } \sin(\omega t) > 0 \\ 0 & \text{if } \sin(\omega t) \le 0 \end{cases} \) The average DC voltage \( V_{avg} \) is given by: \[ V_{avg} = \frac{1}{T} \int_{0}^{T} V_{out}(t) dt \] Since the output is zero during the negative half-cycle, we only integrate over the positive half-cycle, from \( t=0 \) to \( t=T/2 \). \[ V_{avg} = \frac{1}{T} \int_{0}^{T/2} V_p \sin(\omega t) dt \] Substitute \( T = \frac{2\pi}{\omega} \): \[ V_{avg} = \frac{\omega}{2\pi} \int_{0}^{\pi/\omega} V_p \sin(\omega t) dt \] \[ V_{avg} = \frac{\omega V_p}{2\pi} \left[ -\frac{\cos(\omega t)}{\omega} \right]_{0}^{\pi/\omega} \] \[ V_{avg} = \frac{V_p}{2\pi} [-\cos(\omega \frac{\pi}{\omega}) – (-\cos(0))] \] \[ V_{avg} = \frac{V_p}{2\pi} [-\cos(\pi) – (-1)] \] \[ V_{avg} = \frac{V_p}{2\pi} [-(-1) + 1] \] \[ V_{avg} = \frac{V_p}{2\pi} [1 + 1] \] \[ V_{avg} = \frac{2V_p}{2\pi} \] \[ V_{avg} = \frac{V_p}{\pi} \] If the peak input voltage \( V_p \) is 10V, then: \[ V_{avg} = \frac{10V}{\pi} \] This calculation demonstrates that the average DC voltage across the load resistor in a half-wave rectifier circuit is approximately \( \frac{V_p}{\pi} \). This is a fundamental concept in electronics, particularly relevant to power supply design and signal processing, areas of study within the electrical engineering curriculum at institutions like the National Institute of Technology Silchar. Understanding this average value is crucial for determining the DC component of the rectified waveform, which is essential for subsequent filtering stages and for powering DC-dependent loads. The efficiency and performance of rectification circuits are directly tied to this calculated average voltage.

The question probes the understanding of the fundamental principles of electromagnetic induction and Faraday’s Law, specifically as applied to a scenario involving a changing magnetic flux through a coil. The core concept is that a changing magnetic flux induces an electromotive force (EMF). Faraday’s Law quantifies this relationship: \( \mathcal{E} = -\frac{d\Phi_B}{dt} \), where \( \mathcal{E} \) is the induced EMF and \( \Phi_B \) is the magnetic flux. Lenz’s Law, incorporated in the negative sign, dictates the direction of the induced current, opposing the change in flux. In the given scenario, a circular coil with \( N \) turns and radius \( r \) is placed in a uniform magnetic field \( B \) that is perpendicular to the plane of the coil. The magnetic field strength is changing linearly with time, described by \( B(t) = kt \), where \( k \) is a positive constant. The magnetic flux through a single turn of the coil is given by \( \Phi_{B, \text{single}} = B \cdot A \cdot \cos(\theta) \), where \( A \) is the area of the coil and \( \theta \) is the angle between the magnetic field and the normal to the coil’s plane. Since the field is perpendicular to the plane, \( \theta = 0^\circ \) and \( \cos(\theta) = 1 \). The area of the circular coil is \( A = \pi r^2 \). Therefore, the magnetic flux through a single turn is \( \Phi_{B, \text{single}}(t) = (kt)(\pi r^2) = k\pi r^2 t \). The total magnetic flux through the coil with \( N \) turns is \( \Phi_B(t) = N \cdot \Phi_{B, \text{single}}(t) = Nk\pi r^2 t \). According to Faraday’s Law, the induced EMF is: \( \mathcal{E} = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}(Nk\pi r^2 t) \) 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 induced EMF will drive a current in the coil. The question asks about the nature of the induced EMF. Since the magnetic field strength \( B \) is increasing linearly with time (\( k > 0 \)), the magnetic flux is also increasing linearly with time. This continuous change in flux will induce a constant EMF in the coil as long as the magnetic field continues to increase linearly. The direction of the induced current will oppose this increase in flux, meaning it will create a magnetic field in the opposite direction to the applied field. The underlying concepts tested here are the direct application of Faraday’s Law of Induction and the understanding of how a linearly changing magnetic field affects the induced EMF in a coil. This is crucial for understanding principles in electrical engineering and physics, such as the operation of transformers and generators, which are foundational to many specializations offered at the National Institute of Technology Silchar. The ability to derive and interpret the induced EMF from a changing magnetic field is a core competency for students pursuing degrees in electrical, electronics, and computer science engineering at NIT Silchar, reflecting the institute’s emphasis on strong theoretical grounding.

The question probes the understanding of the fundamental principles of electromagnetic induction and Lenz’s Law, specifically in the context of a changing magnetic flux through a coil. Lenz’s Law states that the direction of an induced current is such that it opposes the change in magnetic flux that produced it. In this scenario, a bar magnet is moved towards a stationary coil. As the magnet approaches, the magnetic field lines passing through the coil increase. According to Lenz’s Law, the induced current in the coil will generate its own magnetic field that opposes this increase. This opposing field will be directed in a way that repels the approaching magnet. Therefore, the coil will exert a repulsive force on the magnet. This repulsion is a direct consequence of the induced current creating a magnetic field that opposes the motion causing the induction. The strength of this induced magnetic field, and thus the repulsive force, depends on the rate of change of magnetic flux, which is related to the speed of the magnet and the magnetic field strength. The core concept tested is the qualitative understanding of how induced currents counteract the very change that creates them, a cornerstone of Faraday’s Law of Induction. This principle is crucial in many applications studied at institutions like the National Institute of Technology Silchar, including electrical generators, transformers, and magnetic levitation systems, where controlling induced currents to produce desired forces is paramount.

The question probes the understanding of fundamental principles in materials science and engineering, specifically concerning the behavior of crystalline solids under stress, a core area for students entering programs at the National Institute of Technology Silchar. The scenario involves a tensile test on a polycrystalline metal, a common experimental technique. The key concept here is the relationship between macroscopic mechanical properties and microscopic structural features. In a polycrystalline material, deformation occurs through the movement of dislocations within individual grains. The yield strength is the stress at which plastic deformation begins. For polycrystalline metals, the Hall-Petch relationship describes how the yield strength increases with decreasing grain size. This is because smaller grains present more grain boundaries, which act as barriers to dislocation motion. Therefore, a finer grain structure leads to higher yield strength and increased hardness. The question asks about the primary mechanism responsible for the observed increase in yield strength when grain size is reduced. This directly relates to how dislocations interact with grain boundaries. Dislocations can either pile up at grain boundaries, requiring higher stress to initiate slip in the adjacent grain, or they can be absorbed by the boundary. The increased density of boundaries in finer-grained materials significantly impedes dislocation movement, thus increasing the overall resistance to plastic deformation. While other factors like work hardening (dislocation-dislocation interactions) and solid solution strengthening (solute atoms impeding dislocation motion) also contribute to yield strength, the question specifically focuses on the effect of grain size reduction. Elastic deformation occurs before yielding and is reversible. Ductility refers to the ability to deform plastically without fracture, which is related to but distinct from yield strength. Therefore, the impediment of dislocation motion by grain boundaries is the most direct and significant mechanism explaining the increase in yield strength with decreasing grain size.

The question probes the understanding of the fundamental principles of data integrity and error detection in digital communication, a core concept in computer science and electrical engineering programs at institutions like the National Institute of Technology Silchar. The scenario describes a sender transmitting a sequence of bits and a receiver detecting an error. The core of the problem lies in identifying which error detection mechanism would *fail* to detect a specific type of error, namely a burst error where multiple consecutive bits are flipped. Let’s analyze the options: 1. **Parity Check:** A simple parity check adds a single bit to make the total number of ‘1’s either even or odd. If two bits flip within a block, the parity remains unchanged, and the error goes undetected. A burst error of an even number of flipped bits would also be missed. 2. **Longitudinal Redundancy Check (LRC):** LRC is similar to parity but applied across multiple bits (e.g., a column in a block of data). If an even number of bits flip in a column, the LRC for that column will not change. A burst error affecting an even number of bits in each of several consecutive rows, or an even number of bits within a single row, could go undetected. 3. **Cyclic Redundancy Check (CRC):** CRC uses polynomial division to generate a checksum. It is significantly more robust than parity or LRC and is designed to detect most burst errors, especially those of a length less than the degree of the generator polynomial. While not foolproof for *all* burst errors (e.g., a burst error exactly matching the generator polynomial might be missed), it is generally considered the most effective among these simple methods for burst error detection. 4. **Vertical Redundancy Check (VRC):** VRC is essentially the same as a simple parity check applied to each character or block. Like a simple parity check, it can fail to detect errors if an even number of bits are flipped within the checked unit. The question asks which method would *fail* to detect a burst error. While both parity (VRC) and LRC can fail, CRC is specifically designed to be more resilient to burst errors. Therefore, a burst error is *most likely* to be detected by CRC compared to the other methods. The question implies a scenario where an error *is* detected. If we consider a burst error of, say, 3 consecutive bits flipped, VRC and LRC would likely fail to detect it if the number of flipped bits within their respective checks is even. However, CRC is designed to catch such errors with high probability. The question is phrased to test the understanding of the *relative strengths* of these error detection methods. CRC is known for its superior ability to detect burst errors compared to simple parity (VRC) or LRC. Therefore, if a burst error *occurs*, it is less likely to be detected by VRC or LRC than by CRC. The question asks which method would *fail* to detect a burst error. This implies a scenario where the error *is* present. Let’s re-evaluate the question’s intent: “Which of the following error detection mechanisms, when implemented, would be *least likely* to detect a burst error of moderate length (e.g., 3-5 consecutive bits)?” * **VRC (Vertical Redundancy Check):** A single bit per block. If 2 or 4 bits flip in the block, it’s undetected. A burst of 3 bits would likely flip 3 bits in the block, changing the parity, so it *would* be detected. However, if the burst error affects bits such that an even number of bits flip in the block, it’s missed. * **LRC (Longitudinal Redundancy Check):** Similar to VRC but across columns. If a burst error affects an even number of bits in a column, it’s missed. * **CRC (Cyclic Redundancy Check):** Designed to detect burst errors up to the degree of the polynomial. It is highly effective. * **Simple Parity (which VRC is a form of):** Can miss errors if an even number of bits flip. The question is subtle. It asks which would *fail*. All simple methods can fail. However, CRC is specifically designed to *minimize* such failures for burst errors. Therefore, the methods that are *less sophisticated* in handling burst errors are more prone to failure. VRC (simple parity) is the most basic. If a burst error flips an odd number of bits within a VRC block, it *will* be detected. If it flips an even number, it won’t. The question is about *failure to detect*. A burst error of length \(k\) can be detected by CRC if \(k\) is less than the degree of the generator polynomial. VRC and LRC are more susceptible. If a burst error of 3 bits flips bits at positions 1, 2, and 3 in a block, and the VRC is calculated for the entire block, the parity will change, and the error will be detected. Let’s consider the *most likely* failure. A burst error of length \(k\) will flip \(k\) consecutive bits. * VRC: If \(k\) is odd, the parity will change. If \(k\) is even, the parity will not change. So, VRC fails for even-length burst errors. * LRC: Similar to VRC, depends on the number of flipped bits in the checked unit. * CRC: Highly effective for burst errors up to a certain length. The question asks which would *fail*. This implies a scenario where the error is *not* detected. VRC is the most basic and has the highest probability of failing to detect a burst error compared to CRC. If a burst error of length 4 occurs, and it affects bits within a single VRC block, the parity will remain unchanged, and the error will be missed. CRC, with a suitable generator polynomial, would likely detect this. Therefore, the mechanism that is *least sophisticated* in handling burst errors, and thus most prone to *failure* in detecting them, is the Vertical Redundancy Check (VRC), which is a form of simple parity. Final Answer Calculation: The question asks which method would *fail* to detect a burst error. – VRC (Vertical Redundancy Check) is a single parity bit per block. If a burst error flips an even number of bits within that block, VRC will fail to detect it. – LRC (Longitudinal Redundancy Check) is similar, checking bits across rows. An even number of flips in a column leads to failure. – CRC (Cyclic Redundancy Check) is designed to detect burst errors and is significantly more robust. It is less likely to fail. Comparing VRC and LRC, both are susceptible. However, VRC is the most fundamental and simplest form of error detection, making it the most likely to fail for a general burst error scenario without specific constraints on the burst length relative to the block structure. The question asks which would *fail*, implying a scenario where detection is missed. VRC’s failure mode (even number of bit flips) is a direct consequence of its simplicity. The correct answer is VRC because it is the most basic and has the highest probability of failing to detect a burst error compared to more sophisticated methods like CRC.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning aliasing and the Nyquist-Shannon sampling theorem. Aliasing occurs when a signal is sampled at a rate lower than twice its highest frequency component. This leads to the misinterpretation of higher frequencies as lower frequencies in the sampled data. The Nyquist frequency is defined as half the sampling rate, and for perfect reconstruction, the signal’s bandwidth must be less than the Nyquist frequency. Consider a signal with a maximum frequency component of \(f_{max}\). According to the Nyquist-Shannon sampling theorem, the minimum sampling rate \(f_s\) required to avoid aliasing and ensure perfect reconstruction is \(f_s \ge 2 f_{max}\). If a signal with a maximum frequency of \(f_{max} = 15 \text{ kHz}\) is sampled at a rate of \(f_s = 25 \text{ kHz}\), the Nyquist frequency is \(f_{Nyquist} = f_s / 2 = 25 \text{ kHz} / 2 = 12.5 \text{ kHz}\). Since \(f_{max} > f_{Nyquist}\) (15 kHz > 12.5 kHz), aliasing will occur. The higher frequencies in the original signal will be folded back into the lower frequency range, specifically appearing as frequencies within the range of \(0\) to \(f_{Nyquist}\). A frequency \(f\) in the original signal, where \(f > f_{Nyquist}\), will be aliased to a frequency \(f_{alias}\) given by \(f_{alias} = |f – k \cdot f_s|\), where \(k\) is an integer chosen such that \(0 \le f_{alias} < f_{Nyquist}\). For a frequency of \(15 \text{ kHz}\) sampled at \(25 \text{ kHz}\), we can find the aliased frequency: Let \(f = 15 \text{ kHz}\) and \(f_s = 25 \text{ kHz}\). We need to find an integer \(k\) such that \(0 \le |15000 – k \cdot 25000| < 12500\). If \(k=1\), \(|15000 – 1 \cdot 25000| = |-10000| = 10000\). Since \(10000 < 12500\), the aliased frequency is \(10 \text{ kHz}\). Therefore, a component at \(15 \text{ kHz}\) in the original signal will be indistinguishable from a component at \(10 \text{ kHz}\) in the sampled signal. This is a core concept in understanding signal reconstruction and the limitations imposed by sampling rates, crucial for fields like telecommunications and digital audio processing, which are relevant to the interdisciplinary studies at NIT Silchar.

The question probes the understanding of the fundamental principles of digital signal processing, specifically focusing on the Nyquist-Shannon sampling theorem and its implications in the context of analog-to-digital conversion. The 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 (\(f_{max}\)) present in the analog signal. This minimum sampling frequency is known as the Nyquist rate, \(f_s \ge 2f_{max}\). In this scenario, the analog signal contains frequency components up to \(15 \text{ kHz}\). Therefore, \(f_{max} = 15 \text{ kHz}\). According to the Nyquist-Shannon sampling theorem, the minimum sampling frequency required to avoid aliasing and ensure perfect reconstruction is \(2 \times f_{max}\). Calculation: Minimum sampling frequency = \(2 \times f_{max}\) Minimum sampling frequency = \(2 \times 15 \text{ kHz}\) Minimum sampling frequency = \(30 \text{ kHz}\) The question asks for the minimum sampling frequency required for faithful reconstruction. This directly corresponds to the Nyquist rate. The National Institute of Technology Silchar, with its strong emphasis on electronics and communication engineering, would expect students to grasp this foundational concept. Understanding aliasing and the necessity of adequate sampling is crucial for designing effective digital systems, from audio processing to telecommunications. The ability to apply the Nyquist criterion demonstrates a grasp of the trade-offs between sampling rate, data storage, and signal fidelity, which are vital considerations in any digital engineering project undertaken at NIT Silchar. The question is designed to test this core understanding without requiring complex calculations, focusing instead on the conceptual application of the theorem.

The question probes the understanding of the fundamental principles governing the operation of a basic semiconductor diode in forward bias, specifically relating to its voltage-current characteristics. In forward bias, a diode exhibits a non-linear relationship between voltage and current. The key concept here is the “knee voltage” or “cut-in voltage,” which is the minimum forward voltage required for the diode to conduct significant current. For silicon diodes, this is typically around 0.6V to 0.7V, and for germanium diodes, it’s around 0.2V to 0.3V. The question describes a scenario where a diode is subjected to a forward bias voltage of 0.5V. Given that the typical knee voltage for a silicon diode (which is the most common type unless otherwise specified) is approximately 0.7V, a forward bias of 0.5V is insufficient to overcome the potential barrier within the semiconductor junction. Consequently, the diode will be in a state of very low conductivity, allowing only a negligible leakage current to flow. This negligible current is often approximated as zero for practical purposes in introductory circuit analysis when the applied voltage is below the knee voltage. Therefore, the current through the diode will be extremely small, effectively close to zero.

The question probes the understanding of fundamental principles in materials science and engineering, particularly concerning the behavior of crystalline solids under stress, a core area of study at institutions like the National Institute of Technology Silchar. The scenario describes a tensile test on a polycrystalline metallic specimen. The yield strength is defined as the stress at which plastic deformation begins. In metals, plastic deformation primarily occurs through the movement of dislocations. The critical resolved shear stress (\(\tau_{CRSS}\)) is the minimum shear stress required to initiate dislocation motion on a specific slip system. The relationship between the applied tensile stress (\(\sigma\)) and the resolved shear stress (\(\tau\)) on a slip system is given by Schmid’s Law: \(\tau = \sigma \cos\phi \cos\lambda\), where \(\phi\) is the angle between the tensile axis and the normal to the slip plane, and \(\lambda\) is the angle between the tensile axis and the slip direction. Yielding occurs when \(\tau\) reaches \(\tau_{CRSS}\). Therefore, the applied tensile stress at yielding is \(\sigma_y = \frac{\tau_{CRSS}}{\cos\phi \cos\lambda}\). The term \(\cos\phi \cos\lambda\) is known as the orientation factor. For plastic deformation to occur, there must be at least one slip system oriented favorably for dislocation motion. In a polycrystalline material, yielding is governed by the slip system with the most favorable orientation factor, which is the one that requires the lowest applied tensile stress to reach \(\tau_{CRSS}\). This occurs when \(\cos\phi \cos\lambda\) is maximized. The maximum value of \(\cos\phi \cos\lambda\) is 0.5, which happens when \(\phi = 45^\circ\) and \(\lambda = 45^\circ\). Thus, the minimum yield stress for a single crystal is \(\sigma_y = \frac{\tau_{CRSS}}{0.5} = 2\tau_{CRSS}\). However, the question asks about a polycrystalline material and the initiation of plastic deformation. While the critical resolved shear stress is a fundamental material property, the macroscopic yield strength of a polycrystalline metal is influenced by grain boundaries, work hardening, and other microstructural features. The question, however, focuses on the *initiation* of plastic deformation, which is fundamentally tied to the ease with which dislocations can move. The most favorable orientation factor for yielding in a single crystal is 0.5. If a polycrystalline material contains grains with orientations close to this ideal, yielding will initiate in those grains. The question is designed to test the understanding that plastic deformation begins when the resolved shear stress on *any* slip system reaches the critical resolved shear stress. The most easily activated slip system will be the one with the highest orientation factor. Therefore, the yield strength is directly related to the critical resolved shear stress and inversely related to the orientation factor. The minimum stress required for yielding will occur when the orientation factor is maximized. The maximum value of the orientation factor (\(\cos\phi \cos\lambda\)) is 0.5. Thus, the yield strength is \(\sigma_y = \frac{\tau_{CRSS}}{0.5} = 2\tau_{CRSS}\). This represents the theoretical lower bound for yielding in a single crystal. For polycrystalline materials, grain boundaries impede dislocation motion, often leading to higher yield strengths than predicted by single-crystal behavior. However, the question is about the *onset* of plastic deformation, which is initiated by the most favorably oriented grains. The concept of critical resolved shear stress and its relationship to applied stress via Schmid’s Law is paramount. The question implicitly asks for the condition under which plastic deformation *begins* in a polycrystalline aggregate, which is dictated by the most favorably oriented grains. The critical resolved shear stress is the fundamental material property that dictates the stress required for dislocation motion. The applied stress must be sufficient to generate this resolved shear stress on at least one slip system. The most favorable orientation factor (0.5) dictates the minimum applied stress needed to achieve this. Therefore, the yield strength is directly proportional to the critical resolved shear stress.

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 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 \(15 \text{ kHz}\). Therefore, 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 problem states that the signal is sampled at \(25 \text{ kHz}\). Since \(25 \text{ kHz} < 30 \text{ kHz}\), the sampling frequency is below the Nyquist rate. When the sampling frequency is less than twice the maximum frequency of the signal, higher frequency components in the original analog signal will appear as lower frequencies in the sampled digital signal. This phenomenon is known as aliasing. Specifically, a frequency \(f\) in the analog signal will be aliased to a frequency \(f_{alias}\) in the sampled signal, where \(f_{alias} = |f – k \cdot f_s|\) for some integer \(k\), such that \(0 \le f_{alias} \le f_s/2\). For the \(15 \text{ kHz}\) component, with a sampling frequency of \(25 \text{ kHz}\), the aliased frequency would be calculated as follows: \(f_{alias} = |15 \text{ kHz} – k \cdot 25 \text{ kHz}|\). To find the aliased frequency within the range \(0\) to \(f_s/2 = 12.5 \text{ kHz}\), we can test values of \(k\). If \(k=1\), \(f_{alias} = |15 \text{ kHz} – 1 \cdot 25 \text{ kHz}| = |-10 \text{ kHz}| = 10 \text{ kHz}\). Since \(10 \text{ kHz}\) is within the range \(0 \le f_{alias} \le 12.5 \text{ kHz}\), the \(15 \text{ kHz}\) component will appear as \(10 \text{ kHz}\) in the sampled signal. This understanding is crucial in fields like telecommunications and audio processing, areas of significant research and application at institutions like the National Institute of Technology Silchar. Proper sampling ensures data integrity and prevents distortion, which is a core principle in designing efficient and accurate digital systems. The ability to identify and mitigate aliasing is a fundamental skill for engineers and researchers working with signals.

The question probes the understanding of fundamental principles in material science and engineering, particularly concerning the behavior of crystalline structures under stress, a core area of study at institutions like the National Institute of Technology Silchar. The scenario describes a metallic alloy exhibiting anisotropic elastic properties, meaning its stiffness varies with direction. This anisotropy arises from the underlying crystal lattice structure. When a tensile stress is applied along a specific crystallographic direction, the strain experienced by the material will be a complex function of the elastic constants in different directions. The key concept here is the compliance tensor, denoted by \(S_{ijkl}\), which relates stress and strain. For an anisotropic material, the strain \(\epsilon_{ij}\) is related to the stress \(\sigma_{kl}\) by \(\epsilon_{ij} = S_{ijkl} \sigma_{kl}\). In a simplified 2D representation or for specific stress states, this can be reduced. However, the question asks about the *most likely* consequence of applying stress to an anisotropic material without specifying the exact stress tensor or the full elastic tensor. The options present different potential outcomes. Option a) suggests that the material will deform uniformly in all directions. This is characteristic of isotropic materials, where the elastic properties are the same in every direction. Anisotropic materials, by definition, do not deform uniformly. Option b) proposes that the material will exhibit localized yielding at grain boundaries. While grain boundaries can influence deformation, the primary characteristic of anisotropy is directional dependence of elastic properties, not necessarily preferential yielding at boundaries due to the applied stress direction itself. Option c) states that the material will experience strain that is dependent on the crystallographic orientation of the applied stress. This directly aligns with the definition of anisotropy. The elastic response (strain) is dictated by how the crystal lattice is oriented relative to the applied stress. For instance, applying stress along a close-packed direction in a metal might result in less strain than applying it along a more open direction, assuming different stiffnesses. Option d) posits that the material will undergo phase transformation. While stress can induce phase transformations in some materials, this is a specific phenomenon and not a general consequence of applying stress to an anisotropic material; it’s not the *most likely* or defining characteristic of anisotropy itself. Therefore, the most accurate and fundamental consequence of applying stress to an anisotropic material is that the resulting strain will be orientation-dependent.

The question probes understanding of the fundamental principles of digital signal processing, specifically concerning the Discrete Fourier Transform (DFT) and its properties. The scenario describes a real-valued discrete-time signal \(x[n]\) of finite duration \(N\). The DFT of this signal is denoted as \(X[k]\). A key property of the DFT for real-valued signals is that its spectrum exhibits conjugate symmetry, meaning \(X[k] = X^*[N-k]\), where \(X^*\) denotes the complex conjugate. The question asks about the nature of the DFT of a signal \(y[n] = x[n] \cos(\frac{2\pi}{N} n)\). Let’s analyze the components: \(x[n]\) is real-valued. The term \(\cos(\frac{2\pi}{N} n)\) can be expressed using Euler’s formula as \(\frac{e^{j\frac{2\pi}{N} n} + e^{-j\frac{2\pi}{N} n}}{2}\). Therefore, \(y[n] = x[n] \left( \frac{e^{j\frac{2\pi}{N} n} + e^{-j\frac{2\pi}{N} n}}{2} \right) = \frac{1}{2} x[n] e^{j\frac{2\pi}{N} n} + \frac{1}{2} x[n] e^{-j\frac{2\pi}{N} n}\). The DFT of a product of two sequences is not simply the product of their DFTs. However, we can consider the effect of multiplying by a complex exponential on the DFT. Multiplying a signal \(x[n]\) by \(e^{j\omega_0 n}\) corresponds to a frequency shift in its DFT. Specifically, if \(x[n]\) has DFT \(X[k]\), then the DFT of \(x[n] e^{j\frac{2\pi}{N} n}\) is \(X[k-1]\) (with indices taken modulo \(N\)), and the DFT of \(x[n] e^{-j\frac{2\pi}{N} n}\) is \(X[k+1]\) (with indices taken modulo \(N\)). So, the DFT of \(y[n]\), let’s call it \(Y[k]\), will be related to the DFT of \(x[n]\), \(X[k]\), as follows: \(Y[k] = \frac{1}{2} \text{DFT}\{x[n] e^{j\frac{2\pi}{N} n}\} + \frac{1}{2} \text{DFT}\{x[n] e^{-j\frac{2\pi}{N} n}\}\) \(Y[k] = \frac{1}{2} X[k-1] + \frac{1}{2} X[k+1]\) (indices modulo \(N\)). Now, let’s consider the properties of \(Y[k]\). Since \(x[n]\) is real, \(X[k]\) has conjugate symmetry: \(X[k] = X^*[N-k]\). We need to check if \(Y[k]\) exhibits conjugate symmetry. Let’s evaluate \(Y^*[N-k]\): \(Y^*[N-k] = \left( \frac{1}{2} X[N-k-1] + \frac{1}{2} X[N-k+1] \right)^*\) \(Y^*[N-k] = \frac{1}{2} X^*[N-k-1] + \frac{1}{2} X^*[N-k+1]\) Using the conjugate symmetry of \(X[k]\): \(X^*[N-k-1] = X[N – (N-k-1)] = X[k+1]\) \(X^*[N-k+1] = X[N – (N-k+1)] = X[k-1]\) Substituting these back into the expression for \(Y^*[N-k]\): \(Y^*[N-k] = \frac{1}{2} X[k+1] + \frac{1}{2} X[k-1]\) This is exactly the expression for \(Y[k]\). Therefore, \(Y[k] = Y^*[N-k]\), which means the DFT of \(y[n]\) is also conjugate symmetric. This property is characteristic of real-valued signals. Since \(y[n]\) is a product of two real signals, \(y[n]\) itself is real-valued. The DFT of any real-valued discrete-time signal must exhibit conjugate symmetry. The question is designed to test this understanding by applying a frequency-domain operation (multiplication by cosine, which is a sum of two complex exponentials) to a signal whose DFT has known properties. The resulting signal \(y[n]\) is real, and thus its DFT must be conjugate symmetric.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the aliasing phenomenon and its mitigation. Aliasing occurs when a continuous-time signal is sampled at a rate lower than twice its highest frequency component (Nyquist rate). This leads to the misrepresentation of higher frequencies as lower frequencies in the sampled signal. To prevent aliasing, an anti-aliasing filter is employed before sampling. This filter is a low-pass filter designed to attenuate frequencies above half the sampling rate. Consider a scenario where a signal contains frequency components up to \(f_{max}\). If this signal is sampled at a rate \(f_s\), and \(f_{max} > f_s/2\), aliasing will occur. The anti-aliasing filter’s cutoff frequency, \(f_c\), must be set such that \(f_c < f_s/2\) to ensure that all frequencies above \(f_s/2\) are sufficiently attenuated before sampling. If the filter's cutoff frequency is set too high, for instance, at \(f_c > f_s/2\), then frequencies between \(f_s/2\) and \(f_c\) will still pass through and, upon sampling, will be aliased into the lower frequency band, corrupting the desired signal information. Therefore, to guarantee that no aliasing occurs, the cutoff frequency of the anti-aliasing filter must be strictly less than half the sampling frequency.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its practical implications in the context of audio signal acquisition for a hypothetical advanced research project at the National Institute of Technology Silchar. The theorem states that to perfectly reconstruct a signal, the sampling frequency must be at least twice the highest frequency component of the signal. In this scenario, the audio signal has a maximum frequency component of 20 kHz. Therefore, the minimum sampling rate required is \(2 \times 20 \text{ kHz} = 40 \text{ kHz}\). However, practical considerations and the need for a guard band to account for non-ideal filters and potential aliasing introduce a requirement for a higher sampling rate. A common practice in digital audio, especially for high-fidelity applications, is to use sampling rates significantly above the theoretical minimum to ensure robust reconstruction and to accommodate a wider dynamic range and potential for future signal processing. A sampling rate of 44.1 kHz, as used in CDs, is a well-established standard that provides a good balance between fidelity and data storage. For an advanced research project aiming for superior audio quality and flexibility, a sampling rate of 96 kHz offers a substantial improvement, allowing for more precise representation of the original analog signal, better performance of anti-aliasing filters, and greater headroom for digital signal manipulation without introducing artifacts. This higher rate captures more detail in the audio spectrum, which is crucial for sophisticated analysis and synthesis tasks often undertaken in advanced research environments like those at NIT Silchar. The other options represent sampling rates that are either too low to adequately capture the full audio spectrum (22.05 kHz, 32 kHz) or are less common for high-fidelity audio research compared to 96 kHz, potentially leading to aliasing or reduced signal fidelity.

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 analog-to-digital conversion. The 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 (\(f_{max}\)) present in the signal. This minimum sampling rate is known as the Nyquist rate, \(f_{Nyquist} = 2f_{max}\). In the given 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 rate *below* this minimum requirement. When the sampling frequency (\(f_s\)) is less than \(2f_{max}\), higher frequency components in the analog signal are misrepresented as lower frequencies in the sampled digital signal. This phenomenon is called aliasing. Aliasing distorts the reconstructed analog signal, making it impossible to recover the original information accurately. The lower frequency components of the original signal are preserved, but the higher frequency components are folded back into the lower frequency spectrum, creating spurious signals. This is a critical concept in digital signal processing, particularly relevant in fields like telecommunications, audio processing, and image processing, all of which are areas of study at institutions like the National Institute of Technology Silchar. Understanding aliasing is crucial for designing effective sampling systems and for interpreting digital signal processing results correctly.

The question probes the understanding of fundamental principles in materials science and engineering, particularly concerning the behavior of crystalline solids under stress, a core area for aspiring engineers at the National Institute of Technology Silchar. The scenario describes a metal alloy exhibiting a specific stress-strain curve. The key to answering lies in identifying the point where the material transitions from elastic to plastic deformation. Elastic deformation is characterized by a reversible change in shape, where the material returns to its original form upon removal of the applied stress. This region of the stress-strain curve is typically linear, following Hooke’s Law. Plastic deformation, on the other hand, is permanent; the material undergoes irreversible changes in its atomic structure, such as the movement of dislocations. The yield strength is defined as the stress at which plastic deformation begins. In the provided stress-strain curve, the initial portion is linear, indicating elastic behavior. Beyond a certain stress level, the curve deviates from linearity and begins to exhibit a permanent deformation. This deviation point, where the material starts to yield, is the crucial parameter. While the ultimate tensile strength represents the maximum stress the material can withstand before necking, and the fracture strength is the stress at which the material breaks, neither of these directly signifies the onset of permanent deformation. The elastic limit is often used interchangeably with the yield strength, representing the maximum stress that can be applied without causing permanent deformation. Therefore, identifying the stress value at the point of deviation from linearity on the stress-strain graph is paramount. For the purpose of this question, let’s assume the stress-strain curve shows a clear transition from a linear elastic region to a non-linear plastic region at a stress of 250 MPa. This 250 MPa marks the yield strength, the point at which permanent deformation begins.

The question probes the understanding of fundamental principles in digital logic design, specifically concerning the minimization of Boolean expressions and the implications of using different logic gates. The core concept here is Karnaugh maps (K-maps) and the direct implementation of a given Boolean function using NAND gates. The given Boolean function is \( F(A, B, C) = \Sigma m(1, 3, 6, 7) \). This represents the minterms for which the function is true. The minterms are: \( m_1 = A’B’C \) \( m_3 = A’BC \) \( m_6 = ABC’ \) \( m_7 = ABC \) We can represent this function using a 3-variable K-map. “` BC 00 01 11 10 A 0 0 1 1 0 1 0 0 1 1 “` Grouping the 1s in the K-map: 1. Group of two 1s at \( m_1 \) and \( m_3 \): \( A’C \) 2. Group of two 1s at \( m_6 \) and \( m_7 \): \( AB \) So, the minimized Sum of Products (SOP) form is \( F(A, B, C) = A’C + AB \). Now, we need to implement this using only NAND gates. A key principle for NAND-only implementation is that: – Any SOP expression can be implemented using NAND gates. – The final output of the SOP expression is inverted. – To get the original SOP expression, the final output needs to be inverted again. Let’s convert the SOP expression \( F = A’C + AB \) into a form suitable for NAND implementation. First, we can use De Morgan’s Law to express the function in a way that facilitates NAND gate implementation. \( F = A’C + AB \) Apply double negation: \( F = \overline{\overline{A’C + AB}} \) Using De Morgan’s Law \( \overline{X+Y} = \overline{X} \cdot \overline{Y} \): \( F = \overline{\overline{A’C}} \cdot \overline{\overline{AB}} \) This doesn’t directly lead to a NAND implementation. A more direct approach for SOP to NAND conversion: 1. Implement each product term using NAND gates. 2. Feed the outputs of these NAND gates into a final NAND gate. For the term \( A’C \), we need to invert A first. An inverter can be made from a NAND gate by connecting its inputs together. So, \( A’ \) can be obtained by \( \overline{A \cdot A} \). Then, \( A’C \) is \( \overline{A’ \cdot C} \). This requires two NAND gates: one for \( A’ \) and one for \( A’C \). For the term \( AB \), we can directly implement it as \( \overline{A \cdot B} \). This requires one NAND gate. Now, we have \( \overline{A’C} \) and \( \overline{AB} \). To get \( F = A’C + AB \), we need to perform an OR operation. However, we are restricted to NAND gates. The expression \( F = A’C + AB \) can be rewritten using De Morgan’s law as \( F = \overline{\overline{A’C} \cdot \overline{AB}} \). This structure \( \overline{X \cdot Y} \) is directly implemented by a NAND gate. So, we need to implement \( \overline{A’C} \) and \( \overline{AB} \) and then feed them into a final NAND gate. Let’s break down the implementation: 1. Invert A: \( A’ = \overline{A \cdot A} \) (1 NAND gate) 2. Implement \( \overline{A’C} \): \( \overline{A’ \cdot C} \) (1 NAND gate, using the output of step 1) 3. Implement \( \overline{AB} \): \( \overline{A \cdot B} \) (1 NAND gate) 4. Implement \( F = \overline{\overline{A’C} \cdot \overline{AB}} \): \( F = \overline{(\overline{A’ \cdot C}) \cdot (\overline{A \cdot B})} \) (1 NAND gate, taking outputs from step 2 and step 3) Total NAND gates required: 1 (for \( A’ \)) + 1 (for \( \overline{A’C} \)) + 1 (for \( \overline{AB} \)) + 1 (for the final output) = 4 NAND gates. Let’s verify the logic: Gate 1: Inputs A, A. Output \( A’ \). Gate 2: Inputs \( A’ \), C. Output \( \overline{A’C} \). Gate 3: Inputs A, B. Output \( \overline{AB} \). Gate 4: Inputs \( \overline{A’C} \), \( \overline{AB} \). Output \( \overline{\overline{A’C} \cdot \overline{AB}} \). Using De Morgan’s Law: \( \overline{\overline{A’C} \cdot \overline{AB}} = \overline{\overline{A’C}} + \overline{\overline{AB}} = A’C + AB \). This matches the minimized SOP expression. Therefore, the minimum number of NAND gates required is 4. The question asks for the minimum number of NAND gates to implement the given function. The process outlined above provides a direct and minimal implementation of the SOP form using NAND gates. Any other approach would likely involve more gates or not be a direct conversion. For instance, converting to NOR gates first and then to NAND gates would be an indirect and potentially less efficient method. The direct SOP to NAND conversion is standard and yields the minimum required gates for this type of expression.

The question probes the understanding of the fundamental principles governing the behavior of semiconductors under varying environmental conditions, specifically focusing on the impact of temperature on intrinsic carrier concentration and conductivity. For an intrinsic semiconductor, the intrinsic carrier concentration \(n_i\) is a function of temperature \(T\) and the band gap energy \(E_g\). A commonly used approximation for \(n_i\) is given by \(n_i \approx AT^{3/2}e^{-E_g/(2kT)}\), where \(A\) is a material-dependent constant and \(k\) is the Boltzmann constant. The conductivity \(\sigma\) of an intrinsic semiconductor is directly proportional to the intrinsic carrier concentration and the sum of the electron and hole mobilities (\(\mu_n\) and \(\mu_p\)): \(\sigma = q n_i (\mu_n + \mu_p)\), where \(q\) is the elementary charge. Both electron and hole mobilities are generally temperature-dependent, typically decreasing with increasing temperature, often approximated by \(\mu \propto T^{-m}\) where \(m\) is a positive constant (e.g., \(m \approx 2.5\) for lattice scattering). Let’s consider the temperature dependence of conductivity. The term \(n_i\) increases exponentially with temperature due to the \(e^{-E_g/(2kT)}\) factor. The mobility terms (\(\mu_n + \mu_p\)) decrease with temperature. The overall conductivity \(\sigma\) is the product of \(n_i\) and \((\mu_n + \mu_p)\). If we consider the dominant exponential increase in \(n_i\) due to the band gap, and a power-law decrease in mobility, the overall conductivity will increase with temperature. However, the rate of increase is not linear. The exponential term in \(n_i\) dominates the power-law decrease in mobility at typical operating temperatures for many semiconductors. The question asks about the *rate of increase* of conductivity with temperature. The conductivity \(\sigma\) is approximately proportional to \(T^{3/2}e^{-E_g/(2kT)} \cdot T^{-m}\), which simplifies to \(\sigma \propto T^{(3/2 – m)}e^{-E_g/(2kT)}\). Since \(E_g\) is a positive value and \(k\) is a positive constant, the exponential term \(e^{-E_g/(2kT)}\) increases rapidly as \(T\) increases. The term \(T^{(3/2 – m)}\) will decrease if \(m > 3/2\), which is common. However, the exponential increase in carrier concentration due to thermal excitation across the band gap is the most significant factor driving the increase in conductivity for intrinsic semiconductors. This exponential relationship means that the conductivity does not increase linearly; rather, its rate of increase itself increases with temperature, exhibiting a non-linear, accelerating growth pattern. This is characteristic of an exponential function’s behavior. Therefore, the rate of increase of conductivity with temperature is itself increasing.

The question probes the understanding of fundamental principles in material science and engineering, particularly concerning the behavior of materials under stress and the role of microstructural features. The scenario describes a tensile test on a metallic alloy, a common experimental procedure in materials engineering programs at institutions like the National Institute of Technology Silchar. The key observation is the significant increase in ductility after a specific heat treatment. This phenomenon is directly linked to changes in the material’s microstructure. The heat treatment described, likely an annealing or tempering process, aims to reduce internal stresses and promote grain growth or recrystallization. In a metallic alloy, grain boundaries act as barriers to dislocation movement, which is the primary mechanism for plastic deformation. Smaller grains, while increasing yield strength (Hall-Petch effect), can limit overall ductility if they lead to a higher density of grain boundaries that impede dislocation slip. Conversely, a controlled increase in grain size, or a reduction in the density of internal defects and precipitates through annealing, can facilitate easier dislocation motion across larger regions of the crystal lattice. This enhanced mobility allows the material to undergo more extensive plastic deformation before fracture, thus increasing its ductility. The question requires distinguishing between factors that primarily affect strength versus those that influence ductility. Increased dislocation density, for instance, would increase strength but potentially decrease ductility. Alloying elements can have complex effects, often increasing strength by impeding dislocation motion, but their impact on ductility depends on how they interact with the lattice and form precipitates. Work hardening, a process of plastic deformation that increases dislocation density, also increases strength at the expense of ductility. Therefore, the observed increase in ductility points towards a microstructural refinement that facilitates slip, rather than hindering it. The most plausible explanation for enhanced ductility after heat treatment is the reduction of internal defects and stresses, and potentially a more favorable grain structure for slip propagation, which is achieved through processes like annealing or tempering. This aligns with the core concepts taught in materials science and engineering at NIT Silchar, emphasizing the structure-property relationships.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its implications in reconstructing analog signals. The theorem states that to perfectly reconstruct a continuous-time 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, \(f_N = 2f_{max}\). In the given 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 \(f_s \ge 2 \times 15 \text{ kHz} = 30 \text{ kHz}\). The question asks for the *most appropriate* sampling frequency that guarantees accurate reconstruction. While any frequency above 30 kHz would technically satisfy the theorem, practical considerations and the desire for a robust reconstruction often lead to sampling at a frequency somewhat higher than the absolute minimum. This provides a margin of safety against imperfections in the sampling and reconstruction filters and allows for easier design of these filters. Considering the options: – 15 kHz is less than the Nyquist rate, so it would lead to aliasing and loss of information. – 25 kHz is also less than the Nyquist rate (30 kHz), thus insufficient for accurate reconstruction. – 40 kHz is greater than the Nyquist rate of 30 kHz. This frequency is commonly used in digital audio (e.g., CD quality is 44.1 kHz) and provides a good balance between data rate and reconstruction fidelity. It allows for a wider transition band in anti-aliasing and reconstruction filters, making them easier to implement. – 60 kHz is also greater than the Nyquist rate. While it would also allow for accurate reconstruction, it results in a higher data rate and increased processing requirements without a significant commensurate improvement in reconstruction quality for a signal with a maximum frequency of 15 kHz. The benefits of sampling at 60 kHz over 40 kHz for this specific signal are marginal and may not justify the increased computational and storage overhead, which is a key consideration in practical system design at institutions like the National Institute of Technology Silchar. Therefore, 40 kHz represents the most appropriate and commonly adopted sampling frequency in such scenarios, balancing theoretical requirements with practical engineering considerations for accurate signal reconstruction. This reflects the pragmatic approach to signal processing taught and researched at the National Institute of Technology Silchar, where efficiency and effectiveness are paramount.

The question probes the understanding of the fundamental principles governing the operation of a basic semiconductor diode in a circuit, specifically focusing on its behavior under forward bias. When a diode is forward-biased, the applied voltage overcomes the built-in potential barrier of the p-n junction. This allows majority carriers (electrons in the n-type material and holes in the p-type material) to diffuse across the junction. The current flow is primarily due to the movement of these majority carriers. The voltage drop across a silicon diode under forward bias is typically around 0.7V, and for germanium diodes, it’s around 0.3V. This voltage is often referred to as the “turn-on voltage” or “forward voltage drop.” Once this threshold is reached, the diode exhibits a very low resistance, allowing significant current to flow. The relationship between voltage and current in the forward-biased region is exponential, as described by the Shockley diode equation, but for practical purposes in introductory circuit analysis, the constant voltage drop model is often used. The question requires identifying the primary mechanism responsible for current conduction in this state. The movement of minority carriers is significantly less under forward bias compared to majority carriers. The depletion region, which is wide under reverse bias, narrows considerably under forward bias, facilitating carrier movement. Therefore, the dominant factor enabling current flow is the diffusion of majority charge carriers across the junction, driven by the applied forward voltage exceeding the barrier potential.

The question probes the understanding of fundamental principles in digital logic design, specifically related to combinational circuits and their implementation using universal gates. The core concept tested is the ability to synthesize a given Boolean function using only NAND gates, a common exercise in digital electronics. Let the given Boolean function be \( F(A, B, C) = \sum m(1, 3, 6, 7) \). First, we represent this function in Sum of Products (SOP) form using the minterms: \( F(A, B, C) = A’B’C + A’BC + AB’C’ + ABC’ \) To implement this using only NAND gates, we can follow these steps: 1. **Convert SOP to Product of Sums (POS) by finding the maxterms:** The missing minterms are 0, 2, 4, 5. \( F'(A, B, C) = \sum m(0, 2, 4, 5) = A’B’C’ + A’BC’ + AB’C’ + ABC \) Using De Morgan’s theorem, \( F(A, B, C) = (F'(A, B, C))’ = (A’B’C’ + A’BC’ + AB’C’ + ABC)’ \) \( F(A, B, C) = (A’B’C’)’ \cdot (A’BC’)’ \cdot (AB’C’)’ \cdot (ABC)’ \) This is the POS form: \( F(A, B, C) = (A+B+C’) \cdot (A+B’+C’) \cdot (A’+B+C’) \cdot (A’+B’+C) \) 2. **Implement POS using NAND gates:** A POS expression can be directly implemented using NAND gates. Each term \( (X+Y+Z’) \) is a sum term. To implement \( (X+Y+Z’) \) using NAND gates, we need to express it in a form suitable for NAND implementation. \( (X+Y+Z’) = ((X+Y+Z’)’)’ \) Using De Morgan’s theorem: \( (X+Y+Z’)’ = X’ \cdot Y’ \cdot (Z’)’ = X’ \cdot Y’ \cdot Z \) So, \( (X+Y+Z’) = (X’ \cdot Y’ \cdot Z)’ \) This means each sum term can be implemented as a NAND gate if we have the inverted inputs. However, a more direct approach for POS to NAND conversion is to realize that each sum term \( (P+Q+R) \) can be implemented by a NAND gate with inverted inputs. \( (A+B+C’) \) can be implemented by a NAND gate with inputs A, B, and C (since C’ is the input). \( (A+B’+C’) \) can be implemented by a NAND gate with inputs A, B’, and C. \( (A’+B+C’) \) can be implemented by a NAND gate with inputs A’, B, and C. \( (A’+B’+C) \) can be implemented by a NAND gate with inputs A’, B’, and C. The final output is the product of these terms: \( F = Term1 \cdot Term2 \cdot Term3 \cdot Term4 \). To implement a product using NAND gates, we need to NAND the results of the individual terms. \( F = (Term1 \cdot Term2 \cdot Term3 \cdot Term4) \) \( F = ((Term1 \cdot Term2 \cdot Term3 \cdot Term4)’)’ \) \( F = ((Term1)’ \cdot (Term2)’ \cdot (Term3)’ \cdot (Term4)’)’ \) This requires a final NAND gate whose inputs are the outputs of the first level of NAND gates. Let’s re-evaluate the direct conversion of SOP to NAND. \( F(A, B, C) = A’B’C + A’BC + AB’C’ + ABC’ \) To implement this with NAND gates, we can use the following procedure: 1. Implement the SOP expression directly using NAND gates. 2. For each AND gate in the SOP, replace it with a NAND gate. 3. For each OR gate in the SOP, replace it with a NAND gate followed by an inverter (which is another NAND gate with inputs tied together). 4. If the final output is an OR gate, replace it with a NAND gate. If the final output is an AND gate, it needs to be converted to NAND. A more systematic way for SOP to NAND: \( F = A’B’C + A’BC + AB’C’ + ABC’ \) \( F = (A’B’C)’ \cdot (A’BC)’ \cdot (AB’C’)’ \cdot (ABC’)’ \) — This is incorrect, this is for NOR implementation of AND. Let’s use the standard SOP to NAND conversion: \( F = A’B’C + A’BC + AB’C’ + ABC’ \) \( F = (A’B’C)’ \cdot (A’BC)’ \cdot (AB’C’)’ \cdot (ABC’)’ \) — This is incorrect. The correct approach for SOP to NAND: \( F = A’B’C + A’BC + AB’C’ + ABC’ \) We need to perform double negation: \( F = ((A’B’C + A’BC + AB’C’ + ABC’)’)’ \) Using De Morgan’s: \( F = ((A’B’C)’ \cdot (A’BC)’ \cdot (AB’C’)’ \cdot (ABC’)’)’ \) This expression is now in a form that can be directly implemented with NAND gates. The terms \( (A’B’C)’, (A’BC)’, (AB’C’)’, (ABC’)’ \) are the outputs of the first level of NAND gates. The final output is the NAND of these four terms. Let’s verify the terms: \( (A’B’C)’ \) requires NAND gates for \( A’, B’, C \) and then a NAND gate. \( (A’BC)’ \) requires NAND gates for \( A’, B, C \) and then a NAND gate. \( (AB’C’)’ \) requires NAND gates for \( A, B’, C’ \) and then a NAND gate. \( (ABC’)’ \) requires NAND gates for \( A, B, C’ \) and then a NAND gate. The number of NAND gates required is: – Inverters for A’, B’, C’: 3 NAND gates (each input tied to one input, output is the inversion). – For the first term \( (A’B’C)’ \): 1 NAND gate with inputs A’, B’, C. – For the second term \( (A’BC)’ \): 1 NAND gate with inputs A’, B, C. – For the third term \( (AB’C’)’ \): 1 NAND gate with inputs A, B’, C’. – For the fourth term \( (ABC’)’ \): 1 NAND gate with inputs A, B, C’. – Final output NAND gate: 1 NAND gate with the outputs of the previous four NAND gates. Total NAND gates = 3 (for inversions) + 4 (for product terms) + 1 (final output) = 8 NAND gates. Let’s re-examine the direct conversion of SOP to NAND without explicit inversion gates first. \( F = A’B’C + A’BC + AB’C’ + ABC’ \) We can group terms: \( F = A’C(B’ + B) + AC'(B’ + B) \) \( F = A’C(1) + AC'(1) \) \( F = A’C + AC’ \) This is the XOR function: \( F = A \oplus C \) Now, implement \( F = A \oplus C \) using only NAND gates. The XOR function \( A \oplus C \) can be expressed as \( (A+C)(A’+C’) \). Using De Morgan’s, \( A \oplus C = (A+C)(A’+C’) = ((A+C)’)’ \cdot ((A’+C’)’)’ \) This is not directly helpful for NAND. A standard NAND implementation of XOR: \( A \oplus C = (A \cdot C’) + (A’ \cdot C) \) \( A \oplus C = ((A \cdot C’) + (A’ \cdot C))” \) \( A \oplus C = ((A \cdot C’)’) \cdot ((A’ \cdot C)’)’ \) This requires: 1. \( (A \cdot C’)’ \): 1 NAND gate with inputs A and C’. 2. \( (A’ \cdot C)’ \): 1 NAND gate with inputs A’ and C. 3. Final NAND gate with the outputs of the above two. To get A’ and C’: – \( A’ \) requires 1 NAND gate (inputs tied). – \( C’ \) requires 1 NAND gate (inputs tied). So, the total NAND gates for \( A \oplus C \) are: – 2 NAND gates for inversions (A’, C’). – 2 NAND gates for the product terms \( (A \cdot C’)’ \) and \( (A’ \cdot C)’ \). – 1 NAND gate for the final output. Total = 2 + 2 + 1 = 5 NAND gates. Let’s verify the original SOP simplification again. \( F(A, B, C) = \sum m(1, 3, 6, 7) \) Minterms: 001 (1) = A’B’C 011 (3) = A’BC 110 (6) = ABC’ 111 (7) = ABC Karnaugh Map: BC A 00 01 11 10 0 0 1 1 0 1 0 0 1 1 Grouping: – Group of two: A’BC and ABC (covers minterms 3 and 7) -> BC – Group of two: ABC’ and ABC (covers minterms 6 and 7) -> AB – Group of two: A’B’C and A’BC (covers minterms 1 and 3) -> A’C This is incorrect. Let’s redo the K-map. BC A 00 01 11 10 0 0 1 1 0 (A=0) 1 0 0 1 1 (A=1) Minterms: m1 (001) = A’B’C m3 (011) = A’BC m6 (110) = ABC’ m7 (111) = ABC K-map filling: BC A 00 01 11 10 0 0 1 1 0 (m0, m1, m3, m2) 1 0 0 1 1 (m4, m5, m7, m6) Correct K-map: BC A 00 01 11 10 0 0 1 1 0 1 0 0 1 1 Groups: 1. Group of four: m1, m3, m7, m6. This is not a valid group of four. 2. Group of two: m1 (A’B’C) and m3 (A’BC) -> A’C 3. Group of two: m6 (ABC’) and m7 (ABC) -> AB This simplification is still incorrect. Let’s look at the K-map again. m1: 001 m3: 011 m6: 110 m7: 111 BC A 00 01 11 10 0 0 1 1 0 1 0 0 1 1 The ‘1’s are at (0,01), (0,11), (1,11), (1,10). – Group of two: (0,01) and (0,11) -> A’C – Group of two: (0,11) and (1,11) -> C – Group of two: (1,11) and (1,10) -> AB Let’s try to cover all ‘1’s with minimum groups. – Group 1: m1 (001) and m3 (011) -> A’C – Group 2: m6 (110) and m7 (111) -> AB – Group 3: m3 (011) and m7 (111) -> C The minimal SOP is \( F = A’C + AB + C \). We can simplify further: \( F = A’C + AB + C = A’C + C + AB = C(A’+1) + AB = C(1) + AB = C + AB \). So, \( F(A, B, C) = C + AB \). Now, implement \( F = C + AB \) using only NAND gates. 1. Convert to NAND form: \( F = (C + AB)” \) \( F = ((C + AB)’)’ \) \( F = ((C)’ \cdot (AB)’)’ \) This requires: – \( C’ \): 1 NAND gate (inputs tied). – \( (AB)’ \): 1 NAND gate with inputs A and B. – Final NAND gate with inputs \( C’ \) and \( (AB)’ \). Total NAND gates = 1 (for C’) + 1 (for AB) + 1 (final output) = 3 NAND gates. Let’s verify the simplification \( F = C + AB \). m1 (001): C=1, AB=0. F = 1+0 = 1. Correct. m3 (011): C=1, AB=0. F = 1+0 = 1. Correct. m6 (110): C=0, AB=1. F = 0+1 = 1. Correct. m7 (111): C=1, AB=1. F = 1+1 = 1. Correct. The simplified function is indeed \( F = C + AB \). Implementation of \( F = C + AB \) using NAND gates: \( F = ((C)’ \cdot (AB)’)’ \) Gate 1: NAND gate with inputs A, B. Output is \( (AB)’ \). Gate 2: NAND gate with inputs C, C. Output is \( C’ \). Gate 3: NAND gate with inputs \( (AB)’ \) and \( C’ \). Output is \( ((AB)’)’ \cdot (C’)’ )’ = (AB \cdot C)” = AB + C \). This implementation uses 3 NAND gates. Let’s consider the options provided in a typical exam scenario. The question asks for the minimum number of NAND gates. The question is about implementing a Boolean function using only NAND gates, a fundamental concept in digital logic design taught at institutions like NIT Silchar. The ability to simplify Boolean expressions and then efficiently implement them using universal gates like NAND is crucial for designing digital circuits. This skill is foundational for advanced topics in computer architecture, VLSI design, and embedded systems. The challenge lies in correctly simplifying the expression and then applying the rules for NAND gate implementation, ensuring the minimum number of gates is used. The provided scenario requires careful application of Boolean algebra and De Morgan’s theorems. The correct answer is 3. Let’s consider why other options might be plausible but incorrect. – 4 gates: This might arise from an incomplete simplification or an inefficient NAND implementation strategy, perhaps by not fully utilizing the universal gate property or by introducing unnecessary inversions. For example, if one were to implement \( C + AB \) as \( (C’ \cdot (AB)’)’ \) but then implement \( C’ \) as \( (C \cdot C)’ \) and \( (AB)’ \) as \( (A \cdot B)’ \), and then NAND these two, it would be 3 gates. If one tried to implement \( C + AB \) as \( (C + AB)” \) and then \( C” \) as \( (C’)’ \) and \( AB” \) as \( (AB)’ \), and then NAND them, it would still be 3. An implementation that uses 4 gates might involve a less optimal conversion, perhaps by not simplifying \( C + AB \) to its minimal form first. For instance, if one tried to implement the original SOP \( A’B’C + A’BC + AB’C’ + ABC’ \) directly without simplification, it would require significantly more gates. A common mistake might be to implement each product term with a NAND gate and then OR them, which would require an additional NAND gate for the OR operation, leading to more gates. – 5 gates: As calculated earlier, a direct implementation of \( A \oplus C \) (which was an incorrect simplification of the original SOP) requires 5 gates. This option tests if the candidate can correctly simplify the initial expression. Another way to get 5 gates might be if the simplification \( C + AB \) is implemented as \( ((C)’ \cdot (AB)’)’ \) but the \( C’ \) is implemented as \( (C \cdot C)’ \) and \( (AB)’ \) as \( (A \cdot B)’ \), and then the final output is \( ((C \cdot C)’ \cdot (A \cdot B)’)’ \). This is still 3 gates. A 5-gate implementation might arise from a misunderstanding of how to implement the OR operation using NAND gates. For example, \( X+Y = ((X’) \cdot (Y’))’ \). If \( X=C \) and \( Y=AB \), then \( C+AB = ((C’) \cdot (AB)’)’ \). This requires \( C’ \) (1 gate), \( (AB)’ \) (1 gate), and the final NAND (1 gate), totaling 3 gates. A 5-gate implementation might occur if one implements \( C \) and \( AB \) separately, then inverts them, and then NANDs them, which is not the correct way. – 6 gates: This could result from a more complex, non-minimal implementation. For example, if the simplification \( C + AB \) was not fully utilized, and one attempted to implement \( C \) and \( AB \) separately and then combine them in a way that requires more gates. Perhaps by implementing \( C \) as \( (C \cdot 1)’ \) and \( AB \) as \( (A \cdot B)’ \), and then NANDing them, but this doesn’t naturally lead to 6 gates. A 6-gate implementation might arise from an incorrect conversion of the SOP to NAND, where intermediate terms are unnecessarily inverted or combined. For instance, if the SOP was converted to \( ((A’B’C)’ \cdot (A’BC)’ \cdot (AB’C’)’ \cdot (ABC’)’)’ \) and then the terms \( A’ \), \( B’ \), \( C’ \) were generated using separate NAND gates for each input, and then these were fed into the product NAND gates, and finally the output NAND gate. This would be \( 3 \times 1 \) (for A’, B’, C’) + \( 4 \times 1 \) (for the product terms) + \( 1 \) (final output) = 8 gates. So 6 gates is unlikely from this path. A 6-gate implementation might be a result of a specific, less common conversion method or a misapplication of De Morgan’s theorem. The key to achieving the minimum number of gates is the simplification of the Boolean function to its most reduced form and then applying the standard NAND implementation techniques. For \( F = C + AB \), the direct NAND implementation is \( ((C)’ \cdot (AB)’)’ \), which uses 3 gates.