Question 1

Computing Basics: Match Term to Definition/Role  List I A. Firmware B. Compiler C. Assembler D. Virtual addressing  List II  Converts a program in a high-level language to machine language  Converts an assembly-language program into machine language…

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Introduction / Context: Knowing the roles of firmware, compilers, assemblers, and virtual addressing is foundational for systems programming and embedded design. These terms describe how software is built and how memory is abstracted by operating systems and hardware. Given Data / Assumptions: Firmware resides in ROM/flash and tightly couples to hardware. Compilers translate C/C++/Java-like source to machine code (often via assembly/object code). Assemblers convert symbolic assembly mnemonics to opcodes. Virtual addressing maps process-visible addresses to physical memory via MMUs/page tables. Concept / Approach: Each List I term has a canonical definition that unambiguously matches a List II description, focusing on the principal function rather than implementation details such as intermediate representations or linkers. Step-by-Step Solution: A (Firmware) → device-resident software in non-volatile memory ⇒ 3.B (Compiler) → high-level language to machine code ⇒ 1.C (Assembler) → assembly mnemonics to machine code ⇒ 2.D (Virtual addressing) → maps logical addresses across memories ⇒ 4. Verification / Alternative check: Toolchains: source → compiler → object → linker → executable; low-level .asm → assembler → object. Operating systems expose large virtual memory spaces; MMU translates to physical frames. Why Other Options Are Wrong: Swapping compiler and assembler misunderstands abstraction levels. Firmware is not a translator; it is deployed software integral to hardware. Virtual addressing is not compilation; it is memory indirection and protection. Common Pitfalls: Conflating firmware with microcode (internal CPU control store) or drivers (host OS software). Also, assuming assemblers optimize like compilers; assemblers mostly map mnemonics directly to opcodes. Final Answer: A-3, B-1, C-2, D-4

Question 2

Control Components: Match Each Device to Its Function in a Control System  List I A. Servomotor B. Amplidyne C. Potentiometer D. Flapper valve (pneumatic)  List II  Error detector  Transducer  Actuator  Power amplifier

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Introduction / Context: Classical control systems combine sensing, power amplification, and actuation. Correctly identifying each component’s role helps in designing and troubleshooting servo loops, especially in electro-mechanical and pneumatic systems. Given Data / Assumptions: Servomotor provides mechanical actuation under command signal control. Amplidyne is an electromechanical power amplifier capable of large gain. Potentiometer commonly serves as an error detector by comparing reference and feedback positions via voltage division. Flapper-nozzle pneumatic element converts displacement to pressure change, serving as a transducer in pneumatic controllers. Concept / Approach: Map each device to its canonical function: actuator for motion, power amplifier for control power, error detection for command-feedback difference, and transduction for physical-to-signal conversion (here, displacement to pressure). Step-by-Step Solution: A (Servomotor) → actuator converting control signals to motion/torque ⇒ 3.B (Amplidyne) → power amplifier with high gain for heavy loads ⇒ 4.C (Potentiometer) → error detector via differential voltage ⇒ 1.D (Flapper valve) → pneumatic transducer (displacement→pressure) ⇒ 2. Verification / Alternative check: Block diagrams of servo mechanisms place potentiometers in the error-summing junction, amplidynes before servomotors for high-power drive, and flapper-nozzle stages at pneumatic controller inputs. Why Other Options Are Wrong: Assigning the servomotor as a power amplifier is incorrect; it is the load actuator. The flapper valve is not an error comparator; it transduces mechanical displacement into pneumatic pressure. Common Pitfalls: Confusing transducers (sensor/pressure conversion) with actuators (motion/force). Also, mixing an electronic power amplifier with an amplidyne’s electromechanical amplification concept. Final Answer: A-3, B-4, C-1, D-2

Question 3

Information Units: Match Unit Name to Its Associated Logarithmic Base  List I A. bit B. nat C. decit  List II  10  e  2

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Introduction / Context: Information measures use logarithms with different bases, leading to familiar units: bits, nats, and decimal digits (decits). Knowing the base associated with each unit is key when converting entropy or information rates between unit systems. Given Data / Assumptions: bit uses base 2 logarithms. nat uses base e logarithms. decit (also called Hartley) uses base 10 logarithms. Concept / Approach: Map each unit to its defining logarithm base. Conversion follows change-of-base formulas: log_b(x) = log_k(x) / log_k(b). Thus 1 nat = log_2(e) bits ≈ 1.4427 bits; 1 decit = log_2(10) bits ≈ 3.3219 bits. Step-by-Step Solution: A (bit) → base 2 ⇒ 3.B (nat) → base e ⇒ 2.C (decit) → base 10 ⇒ 1. Verification / Alternative check: Entropy conversions: H_bits = H_nats / log_e(2). For decimal digits, H_bits = H_decits * log_2(10). These confirm the base associations. Why Other Options Are Wrong: Swapping bases produces incorrect conversion factors; for instance, treating a nat as base 10 would wildly misstate information content. Common Pitfalls: Confusing “decade” (frequency factor of 10) with decit (information unit) and mixing up natural logarithms with base-10 common logarithms. Final Answer: A-3, B-2, C-1

Question 4

Solid-state materials and band theory — match each item to its band-structure characteristic  List I (Material / dopant) A. Copper (a good conductor) B. Rubber (an electrical insulator) C. Antimony doped into Si/Ge (Group V donor) D. Boron doped int…

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Introduction / Context: This matching question tests fundamental solid-state physics as used in electronics: how intrinsic materials and dopants shape band structures and hence conductivity. Conductors, insulators, and doped semiconductors can be recognized by simple band diagrams and the placement of donor/acceptor energy levels relative to the conduction and valence bands. Given Data / Assumptions: Copper represents a metallic conductor. Rubber represents an insulating polymer with a large band gap. Antimony (Sb, Group V) acts as a donor in Si/Ge. Boron (B, Group III) acts as an acceptor in Si/Ge. Concept / Approach: In metals, the valence and conduction bands overlap or a partially filled band exists, so electrons move freely. Insulators show a large forbidden gap between valence and conduction bands, limiting carrier excitation. In semiconductors, doping introduces discrete energy levels: donors (Group V) add levels just below the conduction band; acceptors (Group III) add levels just above the valence band, easing promotion of carriers with modest thermal energy. Step-by-Step Solution: Copper → overlapping bands → maps to statement 4.Rubber → large forbidden gap → maps to statement 3.Antimony (donor) → discrete level just below E_c → maps to statement 2.Boron (acceptor) → discrete level just above E_v → maps to statement 1. Verification / Alternative check: Introductory band diagrams for metals, insulators, and doped semiconductors consistently show these placements. Device physics texts use the same donor/acceptor level locations to explain n-type and p-type behavior and the ease of thermal ionization at room temperature. Why Other Options Are Wrong: Assigning rubber to donor/acceptor levels confuses intrinsic insulation with doped semiconductor behavior. Placing “overlap” on rubber or dopants contradicts their basic roles. Swapping donor/acceptor positions in energy band diagrams reverses n-type vs p-type physics. Common Pitfalls: Mixing up “discrete level near E_c” (donor) and “near E_v” (acceptor); assuming all good conductors simply have tiny gaps rather than true band overlap; forgetting that polymers like rubber are wide-gap insulators. Final Answer: A-4, B-3, C-2, D-1

Question 5

Digital logic families — match each family to its hallmark characteristic  List I (Logic family) A. TTL (Transistor–Transistor Logic) B. ECL (Emitter-Coupled Logic) C. MOS (PMOS/NMOS families) D. CMOS (Complementary MOS)  List II (Trait) 1. Low prop…

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Introduction / Context: Comparing logic families is a staple of digital electronics. Each technology balances speed, power, and integration density differently. Recognizing these trade-offs guides family selection for timing-critical vs. low-power applications. Given Data / Assumptions: TTL uses saturated bipolar transistors. ECL avoids saturation via emitter-coupled differential pairs. MOS (pre-CMOS) emphasizes integration density compared with bipolar. CMOS leverages complementary devices for very low static power. Concept / Approach: Map families to their hallmark: TTL → saturated bipolar (4); ECL → fastest, low propagation delay (1); MOS → high packing density (3); CMOS → low power consumption (2). While modern variants blur lines, the classical properties remain correct for fundamentals. Step-by-Step Solution: Match A (TTL) → 4.Match B (ECL) → 1.Match C (MOS) → 3.Match D (CMOS) → 2. Verification / Alternative check: Family datasheets and textbooks list typical t_pd and power numbers: ECL shows sub-nanosecond speeds at higher static power; CMOS shows nanojoule-level energy per transition and nearly zero static power. Why Other Options Are Wrong: Swapping MOS and CMOS traits confuses density with static power advantages. Calling TTL “fastest” ignores ECL’s unsaturated design. Common Pitfalls: Assuming a single “best” family—real designs weigh speed, noise margins, and power together. Final Answer: A-4, B-1, C-3, D-2

Question 6

Fourier tools and their signal/spectrum properties — match each to the appropriate description  List I A. Fourier Series (FS) B. Fourier Transform (FT) C. Discrete-Time Fourier Transform (DTFT) D. Discrete Fourier Transform (DFT)  List II (Property)…

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Introduction / Context: Each Fourier representation pairs a class of signals with characteristic spectra. Knowing whether the spectrum is discrete or continuous, and periodic or aperiodic, helps choose the right tool for analysis and implementation (e.g., spectral lines vs. continuous bands). Given Data / Assumptions: FS models continuous-time periodic signals. FT models continuous-time aperiodic signals. DTFT models discrete-time aperiodic sequences and yields periodic spectra. DFT computes sampled spectra of finite-length sequences, producing discrete spectra that repeat periodically. Concept / Approach: We map as follows: FS → discrete, periodic spectrum (sums of harmonics); FT → continuous, aperiodic spectrum; DTFT → continuous spectrum periodic in frequency (period 2π in rad/s); DFT → discrete spectrum that is periodic due to circularity. Step-by-Step Solution: A (FS) → 1.B (FT) → 4.C (DTFT) → 2.D (DFT) → 3. Verification / Alternative check: Standard signals texts depict FS with spectral lines at integer multiples; DTFT magnitude plots repeat every 2π; DFT bins are discrete and circularly periodic; FT spectra fill a continuum for aperiodic signals. Why Other Options Are Wrong: Assigning “TEM”-like periodicity to FT or discreteness to DTFT contradicts definitions. Common Pitfalls: Confusing “periodic in time” with “periodic in frequency”. FS: time periodic → discrete spectrum; DTFT: time discrete → frequency periodic. Final Answer: A-1, B-4, C-2, D-3

Question 7

Discrete-time system properties — match each input–output relation to linearity/causality  List I (I/P → O/P) A. y(n) = x(n) B. y(n) = x(n^2) C. y(n) = [x(-n)]^2 D. y(n) = [x(n)]^2  List II (System property) 1. Nonlinear, non-causal 2. Linear, non-c…

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Introduction / Context: Determining linearity and causality from an input–output equation is a core DSP skill. Linearity means superposition holds; causality means the output at index n depends only on inputs at indices ≤ n. Given Data / Assumptions: x(n) is a generic discrete-time sequence. Time index n ranges over all integers. Standard definitions of linearity and causality apply. Concept / Approach: Check linearity via additivity and homogeneity; check causality by whether the formula references future samples (k > n) when computing y(n). Time reversal x(-n) uses future values for n > 0; time warping x(n^2) references x at indices larger than n for n > 1, breaking causality (though linearity may still hold if no nonlinear functions like squaring are present). Squaring makes a system nonlinear. Step-by-Step Solution: A: y(n) = x(n) → linear (identity) and causal → 3.B: y(n) = x(n^2) → linear mapping of indices but non-causal (for n > 1, n^2 > n) → 2.C: y(n) = [x(-n)]^2 → squaring → nonlinear; time reversal → non-causal → 1.D: y(n) = [x(n)]^2 → squaring → nonlinear; depends only on x(n) → causal → 4. Verification / Alternative check: Test superposition: for C and D, [ax1 + bx2]^2 ≠ a[x1]^2 + b[x2]^2, failing linearity. For causality: choose an impulse at a future index and observe whether current y(n) changes. Why Other Options Are Wrong: Labeling B “causal” ignores that x(n^2) references future indices. Calling C “causal” or “linear” contradicts squaring and time reversal. Common Pitfalls: Confusing linear time-warps with linear systems; linearity refers to signals, not just index algebra. Final Answer: A-3, B-2, C-1, D-4

Question 8

Hybrid h-parameters — match each parameter to its unit  List I (h-parameter) A. h_ie (input impedance with output shorted) B. h_oe (output admittance with input open) C. h_fe (forward current gain with output shorted)  List II (Unit) 1. siemens (S) …

Accepted Answer

Introduction / Context: The hybrid (h-parameter) model is a popular small-signal representation of bipolar junction transistors. Correct units reinforce physical meaning: impedance in ohms, admittance in siemens, and current gain as dimensionless. Given Data / Assumptions: Common-emitter h-parameter definitions: h_ie, h_oe, h_fe. Small-signal (incremental) linearization applies around a bias point. Concept / Approach: By definition: h_ie is an input impedance (V/I) → ohms; h_oe is an output admittance (I/V) → siemens; h_fe is a ratio of currents (ΔI_c/ΔI_b) → dimensionless. Recognize that “e” subscript denotes the common-emitter configuration and the “i/o, e/f” letters indicate input/output and impedance/admittance or current/voltage relationships. Step-by-Step Solution: Match A (h_ie) → 3 (ohms).Match B (h_oe) → 1 (siemens).Match C (h_fe) → 2 (no units). Verification / Alternative check: Dimensional analysis of the hybrid equations v1 = h11 i1 + h12 v2 and i2 = h21 i1 + h22 v2 confirms the unit assignments for the corresponding elements. Why Other Options Are Wrong: Swapping ohms and siemens reverses impedance/admittance meanings. Assigning units to a pure gain ratio (h_fe) is incorrect. Common Pitfalls: Memorizing symbols without understanding that “ie” and “oe” encode which port and which type of quantity is being referenced. Final Answer: A-3, B-1, C-2

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