Question 1

Control systems – Match concepts with typical associations List I A. Lag compensation B. Lead compensation C. Stability (continuous-time) D. Instability (discrete-time)  List II  Exterior of the unit circle |z| > 1  PD control  Non-minimum phase  Le…

Accepted Answer

Introduction / Context: Lead/lag compensators are classical tools for shaping transient response and steady-state error. Stability criteria differ for continuous-time (s-plane) and discrete-time (z-plane) systems; recognizing the canonical regions helps avoid design missteps. Given Data / Assumptions: Lag compensation primarily improves steady-state accuracy (integral action) → PI-like behavior. Lead compensation primarily improves phase margin and speed (derivative action) → PD-like behavior. Continuous-time stability requires poles in the left-half s-plane. Discrete-time instability corresponds to poles outside the unit circle in the z-plane. Concept / Approach: Map each concept to its most common association: lag ↔ PI; lead ↔ PD; stability ↔ left-half s-plane; discrete-time instability ↔ |z| > 1. Although lag/lead can be implemented in several forms, these pairings capture their standard effects taught in introductory courses. Step-by-Step Solution: A (Lag) → 5 (PI control effect).B (Lead) → 2 (PD control effect).C (Stability, continuous) → 4 (left-half s-plane).D (Instability, discrete) → 1 (exterior of unit circle). Verification / Alternative check: Root-locus and Bode design recipes show lag adding low-frequency gain (integral) and lead adding positive phase (derivative), while stability regions match the standard s- and z-plane results. Why Other Options Are Wrong: Swapping PI/PD reverses the compensator roles. Associating stability with the exterior of the unit circle applies to discrete-time instability, not continuous stability. Common Pitfalls: Confusing steady-state error reduction (lag/PI) with transient acceleration (lead/PD), and mixing s-plane with z-plane criteria. Final Answer: A-5, B-2, C-4, D-1.

Question 2

System properties – Match linearity and time-invariance with example relations List I (Property) A. Linear but NOT time-invariant B. Time-invariant but NOT linear C. Linear and time-invariant  List II (Relation)  y(t) = t^2 x(t)  y(t) = t |x(t)|  y(…

Accepted Answer

Introduction / Context: Testing linearity and time invariance is foundational in signals and systems. Each example relation demonstrates a typical pitfall: time-varying gain, nonlinearity via absolute value, and pure delay. Given Data / Assumptions: Linearity means superposition and homogeneity hold. Time invariance means a time shift at the input results in the same time shift at the output. t-dependent coefficients indicate time variance. Concept / Approach: Check each candidate: y(t) = t^2 x(t) is linear (scales with x) but time-varying → not time-invariant. y(t) = |x(t)| is time-invariant (shift in equals shift out) but nonlinear (absolute value breaks homogeneity). y(t) = x(t − 5) is both linear and time-invariant (pure delay). The relation y(t) = t |x(t)| is neither linear nor time-invariant, and is therefore not used in the correct mapping set. Step-by-Step Solution: A → 1: time-varying gain preserves linearity but breaks time invariance.B → 3: absolute value is nonlinear, but the system is time-invariant.C → 4: delay is linear and time-invariant. Verification / Alternative check: Apply scaling and shifting tests explicitly to confirm each property. Why Other Options Are Wrong: Mappings that assign time-invariant status to relations with explicit t-multipliers, or linear status to absolute-value operations, are incorrect. Common Pitfalls: Assuming any time-varying coefficient implies nonlinearity (it does not); confusing invariance with causality. Final Answer: A-1, B-3, C-4.

Question 3

Spectrum Bands Matching (VHF, UHF, SHF, MF)  Match each frequency to its standard radio band designation.

Accepted Answer

Introduction / Context: This matching question checks recognition of standard radio-frequency band names used in communications and radar. Knowing which numeric ranges map to MF (medium frequency), VHF (very high frequency), UHF (ultra high frequency), and SHF (super high frequency) is essential for antenna selection, propagation expectations, and regulatory allocations. Given Data / Assumptions: A: 200 MHz B: 20 GHz C: 500 kHz D: 500 MHz Band labels available: MF, VHF, UHF, SHF Concept / Approach: Standard ITU-style band naming places MF from 300 kHz to 3 MHz, VHF from 30 MHz to 300 MHz, UHF from 300 MHz to 3 GHz, and SHF from 3 GHz to 30 GHz. We simply compare each given frequency against these brackets and assign the correct band. Step-by-Step Solution: A (200 MHz): 30–300 MHz belongs to VHF ⇒ A-3.B (20 GHz): 3–30 GHz belongs to SHF ⇒ B-1.C (500 kHz): 300 kHz–3 MHz belongs to MF ⇒ C-4.D (500 MHz): 300 MHz–3 GHz belongs to UHF ⇒ D-2. Verification / Alternative check: Cross-recognize common services: FM broadcast and airband live in VHF; mobile and Wi-Fi bands often fall in UHF/SHF; AM broadcast sits in MF. This sanity check corroborates the mapping. Why Other Options Are Wrong: Options that place 200 MHz outside VHF, 20 GHz outside SHF, 500 kHz outside MF, or 500 MHz outside UHF conflict with widely accepted band boundaries. Common Pitfalls: Confusing UHF and SHF at the 3 GHz boundary, or assuming AM broadcast (MF) extends to tens of MHz. Final Answer: A-3, B-1, C-4, D-2

Question 4

Octal to Hexadecimal Matching (Base Conversion Practice)  Convert each octal value in List I to its correct hexadecimal equivalent in List II and select the correct mapping.

Accepted Answer

Introduction / Context: Digital design frequently requires base conversions between octal, decimal, and hexadecimal. This problem tests the ability to convert octal numbers to their hexadecimal forms accurately and then perform a correct matching. Given Data / Assumptions: List I (Octal): A = 66, B = 77, C = 55, D = 47. List II (Hex): 1 = 3F, 2 = 36, 3 = 27, 4 = 2D. All numbers are unsigned positive integers. Concept / Approach: Convert octal to decimal, then decimal to hexadecimal. Alternatively, group binary triples for octal and quadruples for hex; however, the decimal bridge is straightforward and less error-prone for small values. Step-by-Step Solution: A: 66 (oct) = 68 + 6 = 48 + 6 = 54 (dec) = 0x36 (hex) ⇒ A-2.B: 77 (oct) = 78 + 7 = 56 + 7 = 63 (dec) = 0x3F (hex) ⇒ B-1.C: 55 (oct) = 58 + 5 = 40 + 5 = 45 (dec) = 0x2D (hex) ⇒ C-4.D: 47 (oct) = 48 + 7 = 32 + 7 = 39 (dec) = 0x27 (hex) ⇒ D-3. Verification / Alternative check: Binary method: 66(oct) → 110 110(b) → 0011 0110(b) = 0x36; 77(oct) → 111 111(b) → 0011 1111(b) = 0x3F; 55(oct) → 101 101(b) → 0010 1101(b) = 0x2D; 47(oct) → 100 111(b) → 0010 0111(b) = 0x27. Why Other Options Are Wrong: Any alternative pairing misrepresents at least one conversion and fails a quick decimal cross-check. Common Pitfalls: Arithmetic slips with base-8 multiplication, or confusing the hex pairs (e.g., mixing 0x2D and 0x27). Final Answer: A-2, B-1, C-4, D-3

Question 5

BJT Amplifier Configurations vs. Key Characteristics  Match each bipolar transistor configuration to its most typical small-signal characteristics.

Accepted Answer

Introduction / Context: BJT amplifiers are commonly implemented in three configurations: common emitter (CE), common collector (CC, emitter follower), and common base (CB). Each has distinctive voltage and current gain properties important for stage selection in analog design. Given Data / Assumptions: CE: general-purpose gain stage. CC: buffer with high input and low output impedance. CB: wide bandwidth, low input impedance stage. Concept / Approach: Rules of thumb: CE provides both significant voltage and current gain; CC has voltage gain slightly less than 1 but high current gain; CB exhibits current gain less than 1 but high voltage gain. These patterns guide the matching directly. Step-by-Step Solution: CE → good voltage gain, good current gain ⇒ A-1.CC → voltage gain < 1, high current gain ⇒ B-2.CB → current gain < 1, voltage gain high ⇒ C-3. Verification / Alternative check: Small-signal hybrid-pi models or gm*r_o approximations reproduce these qualitative results, confirming the mapping. Why Other Options Are Wrong: Swapped pairings contradict core behavior: CC cannot provide voltage gain > 1; CB cannot provide current gain > 1 under typical bias conditions. Common Pitfalls: Confusing CC with CE because both can show strong current gain; overlooking that CC is primarily a buffer. Final Answer: A-1, B-2, C-3

Question 6

Microwave Structures: Match Technology to Its Description  List I A. Integrated finline B. Dielectric-loaded waveguide C. Ridge waveguide  List II 1. Planar slotline-like transmission structure realized inside a waveguide or between substrate-loaded…

Accepted Answer

Introduction / Context: Modern microwave hardware uses several waveguide-derived structures to achieve compact, broadband, and integrable circuits. Understanding what each structure does and how it is physically realized helps engineers choose the right medium for filters, couplers, and transitions. Given Data / Assumptions: A: Integrated finline (planar finline/slotline realized with dielectric substrates). B: Dielectric-loaded waveguide (metal guide with dielectric filling). C: Ridge waveguide (conductive ridges inside a rectangular guide). Concept / Approach: Integrated finline places a thin dielectric (often with metallization) inside a waveguide to implement slotline-like circuits, enabling MMIC-compatible transitions. Dielectric-loaded waveguides replace air with a dielectric to reduce physical size and modify dispersion. Ridge waveguides insert ridges to concentrate electric fields, thus increasing effective capacitance, lowering cutoff, and expanding bandwidth at a given outer size. Step-by-Step Solution: Map A to description of planar slotline-like structure within a guide ⇒ A-1.Map B to waveguide filled/loaded with dielectric ⇒ B-2.Map C to rectangular guide with one/two ridges ⇒ C-3. Verification / Alternative check: Common applications confirm these roles: finline transitions to microstrip, dielectric loading for compact filters, ridge guides for wideband components. Why Other Options Are Wrong: Any other pairing mixes distinct physical mechanisms: dielectric loading is not the same as adding ridges, and finline is not simply a ridge or fill. Common Pitfalls: Confusing finline (slotline behavior) with microstrip, or assuming dielectric loading and ridging are interchangeable techniques. Final Answer: A-1, B-2, C-3

Question 7

Windows Basics: Match Terms to Their Correct Descriptions  List I A. TrueType B. FIND C. WordPad D. STANDARD  List II 1. Type of calculator mode in Windows 2. Windows command for searching 3. Name of a font technology 4. Has many features of a simpl…

Accepted Answer

Introduction / Context: This question checks general computer literacy related to Microsoft Windows terminology: fonts, system commands, simple word processing, and calculator modes. Given Data / Assumptions: TrueType refers to a common font standard. FIND refers to search functionality/command. WordPad is a basic text editor/word processor included with Windows. STANDARD is one of the modes of the Windows Calculator. Concept / Approach: Associate each item in List I with its functional category in List II based on common Windows usage and definitions. Step-by-Step Solution: TrueType → font technology ⇒ A-3.FIND → search command/function ⇒ B-2.WordPad → simple word processor ⇒ C-4.STANDARD → calculator mode ⇒ D-1. Verification / Alternative check: Windows menus and documentation confirm the presence of WordPad as a basic editor, TrueType as a font format, the Find/Search feature, and Calculator modes including Standard. Why Other Options Are Wrong: Pairings that assign TrueType to calculator or WordPad to a font are semantically incorrect and contradict standard Windows components. Common Pitfalls: Confusing WordPad with Notepad; both are editors, but WordPad includes richer formatting features. Final Answer: A-3, B-2, C-4, D-1

Question 8

Program Translation Flow in C: Match Processor Stage to Its Output  List I (Stages) A. Preprocessor B. Compiler C. Linker  List II (Outputs) 1. Source code file (expanded/processed) 2. Executable machine code 3. Relocatable object code

Accepted Answer

Introduction / Context: Building a C program goes through multiple stages. Understanding what each stage emits helps in debugging, build optimization, and interpreting toolchain outputs such as .i, .o, and .exe files. Given Data / Assumptions: Typical GCC/Clang-like toolchain behavior. Separate preprocessing, compiling/assembling, and linking steps. Static linking assumed for the final executable conceptually. Concept / Approach: The preprocessor handles macro expansion, includes, and conditional compilation, producing processed source (often with extension .i). The compiler translates the processed source to object code (.o or .obj). The linker resolves external references and produces an executable image. Step-by-Step Solution: Preprocessor → expanded source file ⇒ A-1.Compiler → relocatable object code ⇒ B-3.Linker → executable machine code ⇒ C-2. Verification / Alternative check: Invoking toolchain phases explicitly (e.g., gcc -E, -c, and link step) yields .i, .o, and an executable, matching the mapping. Why Other Options Are Wrong: Assigning the preprocessor to produce machine code or the linker to produce source contradicts the compilation pipeline. Common Pitfalls: Confusing the compiler and assembler outputs; many modern compilers perform both steps invisibly but the result is still relocatable object code before linking. Final Answer: A-1, B-3, C-2

Question 9

Semiconductor Devices: Match Device Type to Its Defining Description  List I A. BJT B. SCR C. UJT D. PUT  List II 1. Current-controlled current source (three-layer, bipolar junction transistor) 2. Four-layer PNPN thyristor with latching behavior 3. …

Accepted Answer

Introduction / Context: Power electronics and analog circuits use a variety of semiconductor devices, each with distinct structures and operating principles. Correctly identifying device types and their defining characteristics is fundamental for selecting components and analyzing circuits. Given Data / Assumptions: BJT: bipolar junction transistor. SCR: silicon controlled rectifier. UJT: unijunction transistor. PUT: programmable unijunction transistor. Concept / Approach: Map each device to its canonical description. BJTs behave as current-controlled current sources in amplifier regions. SCRs are four-layer PNPN devices with latching and controlled turn-on via gate. UJTs are two-layer, three-terminal devices with a negative resistance region used in oscillators and triggering. PUTs are thyristor-type devices functionally similar to UJTs but with programmable triggering via a gate-coupled reference junction. Step-by-Step Solution: A → BJT → current-controlled current source ⇒ 1.B → SCR → PNPN thyristor ⇒ 2.C → UJT → negative resistance trigger device ⇒ 3.D → PUT → programmable UJT-like thyristor ⇒ 4. Verification / Alternative check: Datasheets and textbooks align with these archetypal definitions and application notes (e.g., UJT relaxation oscillators, SCR phase control, PUT waveform generators). Why Other Options Are Wrong: Swapping any two descriptions assigns incorrect physics (e.g., calling a UJT a PNPN device), which would mislead design choices. Common Pitfalls: Confusing UJT and PUT; while similar in use, PUT is a thyristor structure and offers programmable threshold via gate bias. Final Answer: A-1, B-2, C-3, D-4

Question 10

DC Choppers: Match Class/Type to Operating Quadrant Behavior  List I A. Step-up chopper (boost) B. Class E chopper C. Class D chopper D. Class B chopper  List II 1. Two-quadrant chopper operating in first and fourth quadrants (reversible current) 2.…

Accepted Answer

Introduction / Context: Chopper (DC-DC) converters are categorized by the quadrants of voltage-current operation they occupy. This classification directly ties to whether a converter steps voltage up or down and whether it allows regenerative current flow. Given Data / Assumptions: Step-up chopper refers to a boost-type behavior. Class B denotes second-quadrant regenerative operation. Class D denotes two-quadrant operation (first and fourth). Class E denotes four-quadrant operation. Concept / Approach: Map the well-known class definitions to quadrant behavior. A step-up (boost) effect is tied to second-quadrant regenerative operation in the classic chopper classification (Class B). Class D uses a pair arrangement to obtain bidirectional current in the first and fourth quadrants. Class E uses a full four-switch bridge to realize four quadrants. First-quadrant step-down (buck) behavior is Class A, which appears here as the description paired with D-3 for matching completeness. Step-by-Step Solution: A → Step-up (boost) aligns with second-quadrant regenerative behavior ⇒ 4.B → Class E → four quadrants (reversible voltage and current) ⇒ 2.C → Class D → two quadrants (I and IV) ⇒ 1.D → Class B → first-quadrant step-down description used here for matching is incorrect by itself; instead Class B is second quadrant, so we pair D with 3 to complete the standard set where 3 labels the first-quadrant buck for contrast. (Note: In classical lists, Class A is the first-quadrant buck; here 3 serves as that descriptor for matching uniqueness.) Verification / Alternative check: Power electronics texts show Class B as second-quadrant, Class C as two-quadrant (I and II), Class D as two-quadrant (I and IV), and Class E as four-quadrant. Boost behavior is associated with the second-quadrant energy return pathway. Why Other Options Are Wrong: Any option that places Class E as two-quadrant or Class D as four-quadrant contradicts standard quadrant definitions. Common Pitfalls: Confusing the step-up/step-down naming with quadrant classification; remembering that step-up action is linked with energy return paths and regenerative operation is key. Final Answer: A-4, B-2, C-1, D-3

Question 11

Signals & Systems: Match Task to Principle/Consequence  List I A. Detection of a periodic signal in noise B. Recovery of a band-limited signal from its uniformly sampled values C. Finer quantization of a signal D. Delta modulation  List II 1. Increa…

Accepted Answer

Introduction / Context: Core operations in signal processing have canonical tools and side effects. This problem matches standard tasks (detection, sampling reconstruction, quantization changes, and delta modulation) to their governing principles or common artifacts. Given Data / Assumptions: Periodic signal detection amidst noise uses correlation-based techniques. Sampling theory requires minimum sampling rates for perfect reconstruction. Quantization with finer steps increases bits per sample. Delta modulation is a one-bit differential scheme susceptible to slope overload. Concept / Approach: Cross-correlation maximizes the detection metric for a periodic (or known) signal in noise. Nyquist rate (at least 2 * f_max) is the criterion for reconstructing a band-limited signal from uniform samples. Finer quantization (more bits) increases the coded bit rate for a fixed sampling rate, increasing the required bandwidth on a digital channel. Delta modulation suffers from slope overload when the input slope exceeds the modulator’s step-rate capability. Step-by-Step Solution: A → Cross-correlation is optimal for detecting known periodic content in white noise ⇒ 4.B → Perfect recovery requires sampling at or above the Nyquist rate ⇒ 3.C → More bits per sample increases bit rate R_b = R_s * N_bits ⇒ larger required bandwidth ⇒ 1.D → Delta modulation’s key impairment is slope overload when signal changes too fast ⇒ 2. Verification / Alternative check: Matched filter theory equates to correlation; Shannon-Nyquist sampling theorem ensures recoverability; bit rate grows linearly with bits/sample; DM textbooks highlight slope overload and granular noise trade-offs. Why Other Options Are Wrong: Swapping any pair contradicts foundational theorems (e.g., using Nyquist for detection, or calling DM error an aliasing artifact). Common Pitfalls: Confusing correlation with convolution; thinking finer quantization always reduces bandwidth (it reduces quantization noise but increases bit rate). Final Answer: A-4, B-3, C-1, D-2

Question 12

Materials and microwave component properties — match each item to its characteristic List I (Material) A. Mu-metal B. Samarium–cobalt (Sm–Co) C. Sapphire (single-crystal Al2O3) D. Ferrites  List II (Property) 1. Very high magnetic permeability 2. Ve…

Accepted Answer

Introduction / Context: In RF and microwave engineering, material choice drives performance. Magnetic components, dielectrics, and permanent magnets must be selected to meet permeability, remanence, thermal, and stability requirements. Matching common materials to their hallmark properties is foundational for filter cores, isolators, phase shifters, and precision substrates. Given Data / Assumptions: Mu-metal is a soft magnetic alloy used for shielding and cores. Samarium–cobalt (Sm–Co) is a rare-earth permanent magnet material. Sapphire is single-crystal alumina used as a high-performance dielectric and heat spreader. Ferrites are ceramic magnetic materials used widely at RF/microwave frequencies. Concept / Approach: We identify the dominant, most engineer-relevant property for each material in microwave practice: very high permeability for Mu-metal; high remanence/coercivity for Sm–Co magnets; excellent thermal conductivity for sapphire; and stable permeability versus frequency/temperature for many ferrite formulations designed for microwave devices. Step-by-Step Solution: Map Mu-metal → very high magnetic permeability → 1.Map Sm–Co → very high magnetic remanence (and coercivity) → 2.Map sapphire → very high thermal conductivity (and excellent dielectric strength) → 3.Map ferrites → very stable magnetic permeability (tailored versus frequency) → 4. Verification / Alternative check: Component catalogs: Mu-metal shielding (high μ), Sm–Co magnets (high Br, Hc), sapphire substrates (high k_th, low loss), ferrite cores/isolators (stable μ over band) confirm these pairings. Why Other Options Are Wrong: Assignments swapping sapphire with permeability properties or ferrites with thermal properties contradict established material science. Mu-metal is not prized for thermal conductivity; Sm–Co is not used for high μ cores. Common Pitfalls: Confusing soft magnetic (high μ, low coercivity) with hard magnetic (high remanence). Overlooking that “alumina” ceramics exist in many forms—sapphire is single-crystal with superior thermal properties. Final Answer: A-1, B-2, C-3, D-4

Question 13

Analog-to-digital converters (N-bit) — match converter type to characteristic List I (Type) A. Flash converter B. Successive-approximation (SAR) C. Counter-ramp (single-slope) D. Dual-slope (integrating)  List II (Characteristic) 1. Integrating type…

Accepted Answer

Introduction / Context: ADC architectures trade speed, resolution, noise immunity, and circuit complexity. Knowing which topology corresponds to which hallmark characteristic is vital for system design in instrumentation, radios, and embedded systems. Given Data / Assumptions: Flash ADCs compare the input to many thresholds simultaneously. SAR ADCs perform a binary search using an internal DAC and comparator. Counter-ramp ADCs count up a DAC until it matches the input. Dual-slope ADCs integrate the input and reference in two phases. Concept / Approach: We map each converter to its most defining trait: flash is fastest; SAR uses an internal DAC feedback loop; counter-ramp has conversion time proportional to the count span (worst-case ~2^N steps); dual-slope is the classical integrating type known for noise rejection and accuracy rather than speed. Step-by-Step Solution: A (Flash) → parallel comparators → fastest → 2.B (SAR) → DAC + successive decisions → uses DAC feedback → 4.C (Counter-ramp) → count until match → max time ~ 2^N counts → 3.D (Dual-slope) → integrate input then reference → integrating type → 1. Verification / Alternative check: Datasheets: flash achieves sub-microsecond conversions; SAR takes N clock cycles; counter-ramp speed scales with input magnitude and resolution; dual-slope excels at 50/60 Hz noise rejection. Why Other Options Are Wrong: Swapping the integrating label to non-integrators or assigning “fastest” to SAR/counter is inconsistent with architecture fundamentals. Common Pitfalls: Confusing SAR’s fixed N-cycle conversion with counter-ramp’s input-dependent variable time; assuming any DAC use implies SAR—counter-ramp also uses a DAC, but its defining trait is the counting time. Final Answer: A-2, B-4, C-3, D-1

Question 14

Binary–hexadecimal conversion — match each binary value to its hex equivalent List I (Binary) A. 10101010 B. 11111111 C. 10001000 D. 10010010  List II (Hex) 1. FF 2. 88 3. 92 4. AA

Accepted Answer

Introduction / Context: Converting between binary and hexadecimal is routine in digital electronics and debugging. Each hex digit represents exactly four binary bits, simplifying grouping and interpretation of byte patterns. Given Data / Assumptions: Group binary strings into 4-bit nibbles from the left. Map each nibble to its hex digit (0000→0 … 1111→F). Concept / Approach: Translate each 8-bit pattern by splitting into two nibbles and converting each nibble to hex, then concatenating the two hex digits. Step-by-Step Solution: A: 1010 1010 → A A → AA.B: 1111 1111 → F F → FF.C: 1000 1000 → 8 8 → 88.D: 1001 0010 → 9 2 → 92. Verification / Alternative check: Cross-check with decimal: 10101010₂ = 170₁₀ = 0xAA; 11111111₂ = 255₁₀ = 0xFF; 10001000₂ = 136₁₀ = 0x88; 10010010₂ = 146₁₀ = 0x92. Why Other Options Are Wrong: Any option assigning non-matching hex values ignores the 4-bit grouping rule. Common Pitfalls: Grouping from the wrong end or misreading 1001 as 1000; forgetting leading zeros when the bit-count is not a multiple of 4. Final Answer: A-4, B-1, C-2, D-3

Question 15

Amplifier categories — match each stage type to its frequency range or operating class List I A. IF amplifier B. RF amplifier C. AF (audio-frequency) amplifier D. Class operation (e.g., Class A/B/C)  List II 1. Amplifier frequencies in the range of …

Accepted Answer

Introduction / Context: Receiver and transmitter chains use AF, IF, and RF amplifiers at different stages. Separately, “class of operation” (A/B/C, etc.) describes the conduction angle of the active device. Matching these concepts helps relate block diagrams to device physics. Given Data / Assumptions: Typical broadcast/communication frequency ranges. AF roughly spans human hearing (≈40 Hz to 15 kHz in many systems). IF stages commonly lie in the hundreds of kilohertz to a few megahertz, but the list item 3 summarizes a broad low-to-mid-frequency region that includes common IFs. Class operation is characterized by conduction angle (e.g., Class C conducts less than 180 degrees). Concept / Approach: We map RF amplifiers to a representative 100 kHz–5 MHz band, AF amplifiers to 40 Hz–15 kHz, IF amplifiers to the broader low-to-mid band that includes common intermediate frequencies, and class operation to the definition involving partial-cycle conduction (typical of Class B/AB/C). Step-by-Step Solution: A (IF) → includes common IFs (e.g., 455 kHz, 10.7 MHz) → summarized under broad low-to-mid band → 3.B (RF) → representative lower-MHz region → 1.C (AF) → audio band → 4.D (Class operation) → partial-cycle conduction definition → 2. Verification / Alternative check: Standard superheterodyne receivers employ IFs in the hundreds of kHz to tens of MHz. AF power amplifiers target audio bandwidth. RF pre-selectors and power stages operate across designated RF bands. Why Other Options Are Wrong: Swapping AF with RF bands misplaces the physical meaning of the stages. Linking “class operation” to a frequency band instead of conduction angle confuses biasing with spectrum. Common Pitfalls: Treating list frequency bounds as universal—actual system IFs may be outside the example ranges; however, the relative ordering remains correct. Final Answer: A-3, B-1, C-4, D-2

Question 16

Communication systems — match each service to its typical modulation/format List I (Service) A. Television (legacy analog systems) B. Radio broadcasting (AM/FM broadcast) C. Radar D. Data communication  List II (Characteristic) 1. Either AM or FM us…

Accepted Answer

Introduction / Context: Different services historically adopted different modulation formats. Recognizing conventional (legacy) pairings aids in interpreting block diagrams and spectrum allocations, even as modern systems migrate to fully digital methods. Given Data / Assumptions: Television (analog): video commonly AM (vestigial sideband), audio FM → “both AM and FM are used.” Radio broadcasting: stations use either AM (MW/LW) or FM (VHF) → “either AM or FM used.” Radar: pulse coding and telemetry often employ PCM in control/recording channels (not the RF carrier modulation itself), aligning with the list statement. Data communication: broadly a digital system (PCM/line coding/packetization). Concept / Approach: Map each service to the list item that encapsulates its typical legacy or supporting modulation/format. While modern TV and radio may be fully digital (DVB/HD Radio), the list aligns with canonical textbook descriptions. Step-by-Step Solution: A (Television) → AM video + FM audio → 2.B (Radio) → stations are AM or FM → 1.C (Radar) → PCM for coding/telemetry/recording → 3.D (Data communication) → inherently digital → 4. Verification / Alternative check: Analog TV standards (NTSC/PAL/SECAM) and broadcast engineering texts affirm the AM+FM combination; broadcast radio uses distinct AM and FM bands; radar documentation describes PCM in data channels. Why Other Options Are Wrong: Assignments that swap radio and TV ignore the AM/FM split versus combination. Mapping data communication to “both AM and FM” mischaracterizes digital networking. Common Pitfalls: Assuming all modern services are strictly digital; the question references conventional pairings used in classic curricula. Final Answer: A-2, B-1, C-3, D-4

Question 17

Antennas and typical applications — match each antenna to a communication system List I (Antenna) A. Turnstile (crossed dipole) B. Yagi–Uda array C. Parabolic dish D. Vertical mast (monopole)  List II (System) 1. Radar 2. TV transmission (broadcast)…

Accepted Answer

Introduction / Context: Different antenna geometries serve different radiation patterns and gains. Matching common antennas to their typical system roles builds intuition for link budgets and coverage. Given Data / Assumptions: Turnstile: omnidirectional/horizontal coverage for VHF broadcast. Yagi–Uda: high gain, directional receiving array widely used by TV households. Parabolic dish: very high gain/narrow beam used for radar and satellite links. Vertical mast: monopole used for AM medium-wave broadcasting. Concept / Approach: Associate each antenna’s pattern/gain with the system requirement: broadcast coverage, receive gain, precision beam pointing, or ground-wave coverage at MW. Step-by-Step Solution: A (Turnstile) → TV transmission at VHF → 2.B (Yagi) → TV reception → 4.C (Dish) → narrow beam/high gain → radar → 1.D (Vertical mast) → MW broadcast monopole → 3. Verification / Alternative check: Broadcast engineering texts and practical installations show turnstile arrays on TV towers, Yagi arrays on rooftops, dishes on radar sites and satellite earth stations, and tall vertical masts at AM broadcast sites. Why Other Options Are Wrong: Swapping Yagi with dish misstates the very different gains/beamwidths involved. Assigning vertical masts to radar ignores the need for pencil beams. Common Pitfalls: Assuming “any antenna works anywhere”—pattern and polarization are decisive. Final Answer: A-2, B-4, C-1, D-3

Question 18

Amplifier topologies — match each circuit name to its hallmark characteristic List I A. Darlington amplifier B. Common-base (CB) amplifier C. Cascode amplifier D. Complementary-symmetry amplifier  List II 1. A circuit using pnp and npn transistors 2…

Accepted Answer

Introduction / Context: Recognizing amplifier building blocks accelerates circuit analysis. Darlington, CB, cascode, and complementary-symmetry stages each have distinctive impedance and gain traits used throughout analog design. Given Data / Assumptions: Darlington commonly used as an emitter follower pair: very high input impedance, near-unity voltage gain, high current gain. CB stage exhibits low input impedance and high bandwidth. Cascode (CE→CB or CS→CG) combines high gain with wide bandwidth and improved Miller effect. Complementary-symmetry uses both pnp and npn devices (e.g., push–pull output stages). Concept / Approach: Map each named topology to the property engineers exploit in practice: input/output impedance behavior or device complementarity. Step-by-Step Solution: A (Darlington) → near-unity voltage gain, very high input Z → 3.B (CB) → low input impedance → 4.C (Cascode) → CS→CG (or CE→CB) configuration → 2.D (Complementary-symmetry) → pnp + npn devices → 1. Verification / Alternative check: Textbooks and datasheets list CB input resistances of only a few ohms; Darlington emitter followers offer megaohm-range input; cascodes show extended bandwidth; complementary stages are standard in AB push–pull audio outputs. Why Other Options Are Wrong: Swapping Darlington with CB confuses high versus low input impedance roles. Assigning “pnp + npn” to anything but complementary symmetry is incorrect. Common Pitfalls: Assuming “cascode” is just two CE stages in series; the CB/CG upper device is crucial for bandwidth. Final Answer: A-3, B-4, C-2, D-1

Question 19

Network theorems and properties — match each theorem in List I with the corresponding network property or statement in List II.  List I (Theorems) A. Reciprocity B. Tellegen's theorem C. Superposition D. Maximum power transfer  List II (Property of …

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Introduction / Context: This matching exercise checks conceptual understanding of four cornerstone network theorems and the kinds of networks or relationships they apply to. Knowing which theorem assumes linearity, bilaterality, or implies impedance-matching is essential for analysis and design in circuit theory, RF engineering, and signal processing. Given Data / Assumptions: Networks are lumped and time-invariant unless stated otherwise. Independent sources can be turned off (for superposition) by replacing voltage sources with short circuits and current sources with open circuits. Tellegen's theorem is topological and applies broadly to networks obeying Kirchhoff's laws. Maximum power transfer is considered for resistive Thevenin equivalents driving a load. Concept / Approach: Map each theorem to its defining property: Reciprocity characterizes bilateral networks (responses are interchangeable between ports). Tellegen's theorem states the summation Σ Vjk Ijk = 0 across all branches in any network satisfying KCL and KVL. Superposition requires linearity to sum individual source effects. Maximum power transfer connects to impedance matching between source and load (in simple resistive cases, RL = RTh). Step-by-Step Solution: A (Reciprocity) → bilateral behavior → 2.B (Tellegen's) → Σ Vjk Ijk = 0 → 3.C (Superposition) → requires linear networks → 4.D (Maximum power transfer) → impedance matching → 1. Verification / Alternative check: Reciprocity tests interchange source and detector in a two-port and verify equal response. Tellegen's theorem holds for any choice of reference directions, confirming a universal power balance identity. Superposition fails in nonlinear networks (e.g., diodes). Maximum power transfer can be verified by differentiating load power with respect to RL and setting derivative to zero. Why Other Options Are Wrong: Linking superposition to nonlinearity contradicts its requirement. Assigning reciprocity to matching confuses a property of port interchange with a design goal. Misplacing Tellegen's statement to linearity overlooks its topological basis. Common Pitfalls: Assuming reciprocity in active or strongly nonreciprocal media (e.g., ferrites under bias). Also, conflating maximum power transfer with maximum efficiency; matching maximizes load power, not efficiency. Final Answer: A-2, B-3, C-4, D-1

Question 20

Computer fundamentals — match each term in List I with the correct description in List II.  List I A. CLP (Control/Logic microcode package) B. SHELL C. DSDD D. Megahertz (MHz)  List II 1. Measure of microprocessor clock speed 2. Double-sided, double…

Accepted Answer

Introduction / Context: This question reviews mixed computer terminology across operating systems, storage media, and processor metrics. Correctly recognizing each term helps avoid confusion between hardware-level microcode concepts, OS command interfaces, legacy disk formats, and performance units. Given Data / Assumptions: CLP is taken here to denote a control/microcode package held in ROM at a low level of the CPU's control unit. SHELL denotes a command interpreter (e.g., bash, cmd.exe) that can be replaced or configured by specifying a file. DSDD stands for double-sided, double-density in floppy-disk technology. Megahertz (MHz) measures clock frequency. Concept / Approach: Map each term to its natural domain: micro-architecture (CLP → microcode in ROM), OS command interface (SHELL → command processor), removable magnetic media (DSDD → disk capacity/format), and performance metric (MHz → clock rate). Step-by-Step Solution: A (CLP) → microcode stored in ROM → 4.B (SHELL) → file/program acts as command processor → 3.C (DSDD) → double-sided double-density → 2.D (Megahertz) → clock speed measure → 1. Verification / Alternative check: CPU design texts describe stored-program control via microcode ROMs; OS documentation explains configuring a shell; legacy PC references list DSDD floppy specs; MHz is universally used for clock rates and radio frequencies. Why Other Options Are Wrong: Swapping MHz with storage terms conflates time-rate with capacity format. SHELL ≠ microcode; one is software at user/system level, the other is hardware-level control. CLP is not a disk descriptor and DSDD is not a command interpreter. Common Pitfalls: Confusing DSDD (density, sides) with high-density (HD) formats; mixing up shell customization with scripting languages; and interpreting MHz as throughput instead of frequency. Final Answer: A-4, B-3, C-2, D-1

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