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Extremadura PAU Technology and Engineering II - independent engineering foundations study materials
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Extremadura PAU Technology and Engineering II - independent engineering foundations practice questions

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Question 1Materials, testing and selection

A cylindrical steel test specimen with an initial diameter of 10 mm is subjected to a static axial tensile load of 15 kN. What is the normal tensile stress (σ) induced in the cross-section of the specimen?

  1. 190.99 MPa Correct answer.
  2. 47.75 MPa
  3. 150.00 MPa
  4. 477.46 MPa

Explanation

Normal tensile stress is given by σ = F / A0. The initial cross-sectional area is A0 = π · d² / 4 = π · (10 mm)² / 4 = 78.54 mm². Converting 15 kN to 15,000 N: σ = 15,000 N / 78.54 mm² = 190.99 N/mm² = 190.99 MPa.

Question 2Materials, testing and selection

A metal structural rod with an original gauge length of 200 mm elongates by 0.4 mm under an axial tensile load within its elastic region. What is the engineering unit strain (ε) of the rod?

  1. 0.02 (2.0%)
  2. 0.002 (0.2%) Correct answer.
  3. 0.0004 (0.04%)
  4. 0.004 (0.4%)

Explanation

Engineering unit strain is defined as ε = ΔL / L0. Dividing elongation by original length: ε = 0.4 mm / 200 mm = 0.002, which equals 0.2%.

Question 3Materials, testing and selection

During a tensile test within the elastic limit, a metallic alloy exhibits a tensile stress of 210 MPa and an engineering strain of 0.001. What is the Young's modulus (modulus of elasticity, E) of the alloy?

  1. 21 GPa
  2. 105 GPa
  3. 210 GPa Correct answer.
  4. 420 GPa

Explanation

According to Hooke's Law (σ = E · ε), Young's modulus is E = σ / ε = 210 MPa / 0.001 = 210,000 MPa = 210 GPa.

Question 4Materials, testing and selection

A standard Charpy impact test uses a pendulum hammer with an initial potential energy of 300 J. After fracturing a notched specimen of cross-sectional area 0.8 cm², the hammer rises to a residual energy height of 180 J. What is the impact toughness (resilience, KCU) of the material?

  1. 150 J/cm² Correct answer.
  2. 375 J/cm²
  3. 225 J/cm²
  4. 120 J/cm²

Explanation

Absorbed energy ΔE = E_initial - E_residual = 300 J - 180 J = 120 J. The impact toughness KCU is ΔE / A = 120 J / 0.8 cm² = 150 J/cm².

Question 5Materials, testing and selection

Which heat treatment process involves heating steel above its critical transformation temperature (austenite range), holding it, and rapidly cooling (quenching) it in water or oil to achieve high hardness?

  1. Annealing (Recocido)
  2. Quenching (Temple) Correct answer.
  3. Tempering (Revenido)
  4. Normalizing (Normalizado)

Explanation

Quenching (temple) rapidly cools austenite to prevent equilibrium diffusion, trapping carbon in a supersaturated body-centered tetragonal structure known as martensite, which gives maximum hardness.

Question 6Materials, testing and selection

A steel bar of diameter 12 mm and length 300 mm is pulled by a tensile force F = 24 kN. Given Young's modulus E = 200 GPa, what is the total elongation (ΔL) of the bar?

  1. 1.273 mm
  2. 0.159 mm
  3. 0.318 mm Correct answer.
  4. 0.637 mm

Explanation

Area A0 = π · (12)² / 4 = 113.10 mm². Normal stress σ = 24,000 N / 113.10 mm² = 212.21 MPa. Strain ε = σ / E = 212.21 MPa / 200,000 MPa = 0.001061. Elongation ΔL = ε · L0 = 0.001061 · 300 mm = 0.318 mm.

Question 7Materials, testing and selection

During an axial tensile test, a cylindrical metal rod experiences a longitudinal strain ε_z = +0.0015 and a transverse diametral strain ε_x = -0.00045. What is the Poisson's ratio (ν) of the material?

  1. 0.45
  2. 0.33
  3. 0.25
  4. 0.30 Correct answer.

Explanation

Poisson's ratio is defined as the negative ratio of transverse strain to axial strain: ν = - ε_transverse / ε_axial = - (-0.00045) / 0.0015 = 0.30.

Question 8Materials, testing and selection

A steel railway track segment of initial length L0 = 20 m with coefficient of thermal expansion α = 12 × 10⁻⁶ /°C is fixed rigidly between unyielding abutments. If the temperature increases by ΔT = 50°C, what thermal compressive stress (σ) develops in the steel (E = 200 GPa)?

  1. 120 MPa Correct answer.
  2. 240 MPa
  3. 60 MPa
  4. 12 MPa

Explanation

Thermal strain prevented by rigid constraints is ε_th = α · ΔT = (12 × 10⁻⁶ /°C) · 50°C = 0.0006. Thermal stress is σ = E · ε_th = 200,000 MPa · 0.0006 = 120 MPa.

Question 9Materials, testing and selection

An isotropic metal alloy has a Young's modulus E = 208 GPa and Poisson's ratio ν = 0.30. Using the elastic relationship G = E / [2(1 + ν)], what is the shear modulus (modulus of rigidity, G) of the alloy?

  1. 104 GPa
  2. 80 GPa Correct answer.
  3. 160 GPa
  4. 95 GPa

Explanation

Shear modulus G = E / [2(1 + ν)] = 208 GPa / [2 · (1 + 0.30)] = 208 / 2.60 = 80 GPa.

Question 10Materials, testing and selection

Which metallurgical heat treatment is applied immediately after quenching steel to relieve internal stresses, reduce extreme brittleness, and achieve a desired balance of ductility and toughness?

  1. Tempering (Revenido) Correct answer.
  2. Full Annealing (Recocido total)
  3. Nitriding (Nitruración)
  4. Carburizing (Cementación)

Explanation

Tempering (revenido) consists of reheating quenched martensitic steel to a temperature below the lower critical line (A1, ~727°C) and holding it, allowing carbon diffusion to transform brittle martensite into tempered martensite.

Question 11Materials, testing and selection

A flat composite tie bar of length 500 mm with rectangular cross-section 20 mm × 5 mm is loaded with F = 10 kN. If the maximum allowable strain is ε_max = 0.0012, what is the minimum required Young's modulus (E_min) of the material?

  1. 200.0 GPa
  2. 83.3 GPa Correct answer.
  3. 166.7 GPa
  4. 100.0 GPa

Explanation

Area A0 = 20 mm × 5 mm = 100 mm² = 10⁻⁴ m². Normal stress σ = 10,000 N / 100 mm² = 100 MPa. Minimum E = σ / ε_max = 100 MPa / 0.0012 = 83,333.33 MPa = 83.33 GPa ≈ 83.3 GPa.

Question 12Materials, testing and selection

What type of polymeric material consists of linear or branched polymer chains held together by weak van der Waals forces, allowing it to be repeatedly melted, reshaped, and recycled upon heating?

  1. Cross-linked vulcanized rubbers
  2. Thermosetting polymers (Termoestables)
  3. Thermoplastic polymers (Termoplásticos) Correct answer.
  4. Elastomers (Elastómeros)

Explanation

Thermoplastics (e.g., polyethylene, PVC, PET, polypropylene) consist of flexible linear or branched molecular chains. Because interchain bonds are physical van der Waals forces, heating softens them reversibly without thermal degradation.

Question 13Materials, testing and selection

A solid circular drive shaft of diameter d = 40 mm is subjected to a pure twisting torque T = 628.3 N·m. What is the maximum torsional shear stress (τ_max = 16 T / [π d³]) generated at the outer surface of the shaft?

  1. 50.0 MPa Correct answer.
  2. 100.0 MPa
  3. 12.5 MPa
  4. 25.0 MPa

Explanation

Maximum torsional shear stress is τ_max = T · r / J = 16 T / (π · d³). Substituting: τ_max = 16 · (628.3 N·m) / (π · 0.040³ m³) = 10,052.8 / (0.00020106) = 50,000,000 Pa = 50.0 MPa.

Question 14Materials, testing and selection

A steel tension tie rod (E = 200 GPa, yield strength σ_y = 350 MPa) has an initial length of 2.0 m. If a safety factor SF = 2.0 is mandated against yield, what is the maximum allowable elongation (ΔL_max) within the design working stress?

  1. 3.50 mm
  2. 1.75 mm Correct answer.
  3. 7.00 mm
  4. 0.875 mm

Explanation

Allowable working stress σ_adm = σ_y / SF = 350 MPa / 2.0 = 175 MPa. Corresponding working strain ε_adm = σ_adm / E = 175 MPa / 200,000 MPa = 0.000875. Maximum allowable elongation ΔL_max = ε_adm · L0 = 0.000875 · 2000 mm = 1.75 mm.

Question 15Materials, testing and selection

A composite structural column consists of a concrete core reinforced with steel rods. If the steel (E_s = 200 GPa) and concrete (E_c = 25 GPa) undergo equal axial strain (ε_s = ε_c), what is the ratio of normal stress carried by the steel to that by the concrete (σ_s / σ_c)?

  1. 0.125
  2. 4.0
  3. 8.0 Correct answer.
  4. 16.0

Explanation

Under equal strain ε, Hooke's Law gives σ_s = E_s · ε and σ_c = E_c · ε. The stress ratio modular ratio n = σ_s / σ_c = E_s / E_c = 200 GPa / 25 GPa = 8.0.

Question 16Materials, testing and selection

During a Vickers hardness test (HV), a square-based diamond pyramid indenter with face angle 136° creates an indentation with mean diagonal length d = 0.50 mm under a load P = 30 kgf. Using HV = 1.8544 P / d², what is the Vickers hardness value?

  1. 445.1 HV
  2. 890.1 HV
  3. 111.3 HV
  4. 222.5 HV Correct answer.

Explanation

Vickers hardness formula is HV = 1.8544 · P / d² = 1.8544 · 30 kgf / (0.50 mm)² = 55.632 / 0.25 = 222.53 ≈ 222.5 HV.

Question 17Statics, strength and mechanical transmission

A simple gear drive consists of a driver gear with z1 = 15 teeth turning at n1 = 1200 rpm driving a follower gear with z2 = 60 teeth. What is the rotational speed (n2) of the follower gear?

  1. 4800 rpm
  2. 240 rpm
  3. 600 rpm
  4. 300 rpm Correct answer.

Explanation

The gear speed ratio is inversely proportional to teeth count: n2 / n1 = z1 / z2. Follower speed n2 = n1 · (z1 / z2) = 1200 rpm · (15 / 60) = 1200 · 0.25 = 300 rpm.

Question 18Statics, strength and mechanical transmission

A compound gear train consists of two pairs of gears: shaft 1 carries gear 1 (z1 = 20) driving gear 2 (z2 = 60) mounted on shaft 2. Shaft 2 also carries gear 3 (z3 = 15) driving gear 4 (z4 = 75) on shaft 3. What is the total speed ratio (i_total = n3 / n1) and input shaft speed if output shaft 3 turns at n3 = 100 rpm?

  1. i = 1/9; n1 = 900 rpm
  2. i = 1/5; n1 = 500 rpm
  3. i = 1/3; n1 = 300 rpm
  4. i = 1/15; n1 = 1500 rpm Correct answer.

Explanation

Individual ratios: i1 = z1 / z2 = 20 / 60 = 1/3; i2 = z3 / z4 = 15 / 75 = 1/5. Overall ratio i_total = i1 · i2 = (1/3) · (1/5) = 1/15. Input speed n1 = n3 / i_total = 100 rpm / (1/15) = 1500 rpm.

Question 19Statics, strength and mechanical transmission

A V-belt drive connects an electric motor running at n1 = 1450 rpm to an industrial compressor running at n2 = 580 rpm. If the driver pulley diameter is d1 = 120 mm, what must be the diameter (d2) of the driven compressor pulley?

  1. 240 mm
  2. 480 mm
  3. 150 mm
  4. 300 mm Correct answer.

Explanation

Assuming negligible belt slip, speed ratio is n1 / n2 = d2 / d1. Driven pulley diameter d2 = d1 · (n1 / n2) = 120 mm · (1450 rpm / 580 rpm) = 120 mm · 2.5 = 300 mm.

Question 20Statics, strength and mechanical transmission

A Class 1 lever has a load arm length L_load = 0.4 m and an effort arm length L_effort = 1.6 m. What effort force (F_effort) is required to balance a heavy load load weight W = 800 N?

  1. 100 N
  2. 200 N Correct answer.
  3. 3200 N
  4. 400 N

Explanation

By the principle of moments for a lever in static equilibrium: F_effort · L_effort = W · L_load. Effort force F_effort = W · (L_load / L_effort) = 800 N · (0.4 m / 1.6 m) = 800 N · 0.25 = 200 N.

Question 21Statics, strength and mechanical transmission

A planetary (epicyclic) gear set has a sun gear with z_s = 20 teeth and a fixed outer ring (annulus) gear with z_r = 80 teeth. If the sun gear is driven at n_s = 1000 rpm, what is the speed (n_c) of the planet carrier output arm (n_c = n_s · z_s / [z_s + z_r])?

  1. 166.7 rpm
  2. 500 rpm
  3. 200 rpm Correct answer.
  4. 250 rpm

Explanation

For a fixed ring gear (n_r = 0), planetary carrier speed is n_c = n_s · z_s / (z_s + z_r) = 1000 rpm · 20 / (20 + 80) = 1000 · (20 / 100) = 200 rpm.

Question 22Statics, strength and mechanical transmission

A power screw jack has a single-start trapezoidal thread with pitch p = 6 mm and mean pitch diameter d_m = 30 mm. Taking coefficient of thread friction μ = 0.10, what is the thread lead angle (λ = arctan(p / [π d_m])) and efficiency (η = tan(λ) / tan(λ + φ)) when raising a load?

  1. λ = 5.00°, η = 45.0%
  2. λ = 3.64°, η = 39.0% Correct answer.
  3. λ = 6.42°, η = 52.0%
  4. λ = 2.18°, η = 28.5%

Explanation

Lead angle λ = arctan[p / (π · d_m)] = arctan[6 / (π · 30)] = arctan(6 / 94.248) = arctan(0.06366) = 3.642°. Friction angle φ = arctan(μ) = arctan(0.10) = 5.711°. Efficiency η = tan(λ) / tan(λ + φ) = tan(3.642°) / tan(3.642° + 5.711°) = 0.06366 / tan(9.353°) = 0.06366 / 0.1647 = 0.3865 ≈ 39.0%.

Question 23Electrical circuits, machines and digital logic

Which basic logic gate produces a HIGH (1) output ONLY when all of its inputs are LOW (0)?

  1. AND gate
  2. NOR gate Correct answer.
  3. NAND gate
  4. XOR gate

Explanation

A NOR gate implements NOT(A + B). When both inputs A = 0 and B = 0, output Y = NOT(0 + 0) = NOT(0) = 1. If any input is 1, output is 0.

Question 24Electrical circuits, machines and digital logic

According to De Morgan's Laws in Boolean algebra, what is the simplified equivalent expression for NOT(A · B)?

  1. NOT(A + B)
  2. NOT A · NOT B
  3. NOT A + NOT B Correct answer.
  4. A + B

Explanation

De Morgan's First Theorem states that the complement of a product equals the sum of the individual complements: (A · B)' = A' + B' (or NOT A + NOT B).

Question 25Electrical circuits, machines and digital logic

What digital combinational component selects one of N data input channels and routes it to a single output line based on a set of binary select lines?

  1. Demultiplexer (Demultiplexor / DEMUX)
  2. Priority Encoder (Codificador de prioridad)
  3. Decoder (Decodificador)
  4. Multiplexer (Multiplexor / MUX) Correct answer.

Explanation

A Multiplexer (MUX) acts as a digitally controlled multi-position switch, selecting one of 2ⁿ data input signals based on n select control lines and forwarding it to one output line.

Question 26Electrical circuits, machines and digital logic

In an edge-triggered D flip-flop, what is the output state Q(t+1) immediately following an active clock transition?

  1. Q(t+1) = D (Data input state) Correct answer.
  2. Q(t+1) = 0 (Always reset)
  3. Q(t+1) = Q(t) (Always hold)
  4. Q(t+1) = NOT Q(t) (Toggle)

Explanation

The characteristic equation of a D flip-flop is Q(t+1) = D. The next state Q(t+1) simply captures and stores whatever logic value (0 or 1) is present at the D data input line at the active clock edge.

Question 27Electrical circuits, machines and digital logic

What is the logic function realized by a 1-bit Half-Adder circuit for the Sum (S) and Carry-out (C) outputs from inputs A and B?

  1. S = A OR B; C = A AND B
  2. S = A XOR B; C = A AND B Correct answer.
  3. S = A AND B; C = A XOR B
  4. S = A XNOR B; C = A OR B

Explanation

A Half-Adder adds two single binary bits. The Sum output S is 1 when inputs differ: S = A ⊕ B (A XOR B). The Carry-out C is 1 only when both inputs are 1: C = A · B (A AND B).

Question 28Electrical circuits, machines and digital logic

Simplify the Boolean function F(A, B, C) = A'B'C + A'BC + AB'C + ABC to its minimal Sum-of-Products form.

  1. F = A · C
  2. F = A
  3. F = C Correct answer.
  4. F = B + C

Explanation

Factor out C from all four minterms: F = C · (A'B' + A'B + AB' + AB). Notice that A'B' + A'B + AB' + AB = A'(B' + B) + A(B' + B) = A'(1) + A(1) = A' + A = 1. Therefore, F = C · 1 = C.

Question 29Electrical circuits, machines and digital logic

Using a 3-variable Karnaugh map, simplify the logic function F(A,B,C) defined by minterms Σm(1, 3, 4, 5, 7).

  1. F = A + B C
  2. F = B' + A C
  3. F = A' C + B
  4. F = C + A B' Correct answer.

Explanation

Minterms in binary: m1(001), m3(011), m5(101), m7(111) form a 4-cell group covering all cells where C=1, yielding term 'C'. Minterms m4(100) and m5(101) combine to yield term 'A B''. Thus, minimized F = C + A B'.

Question 30Electrical circuits, machines and digital logic

What is the characteristic input condition that causes an invalid / indeterminate state (Q = 1 and Q' = 1) in a basic active-HIGH NOR-based SR latch?

  1. S = 1, R = 1 Correct answer.
  2. S = 1, R = 0
  3. S = 0, R = 1
  4. S = 0, R = 0

Explanation

In a NOR-based SR latch, setting both Set (S = 1) and Reset (R = 1) forces both NOR outputs Q and Q' to 0 simultaneously. When inputs return to 00, the final state becomes unpredictable (forbidden race condition).

Question 31Electrical circuits, machines and digital logic

A Master-Slave JK flip-flop has inputs J = 1 and K = 1. What happens to output state Q at each active clock pulse?

  1. Output Q resets to 0 regardless of previous state.
  2. Output Q toggles to its inverse state Q(t+1) = NOT Q(t). Correct answer.
  3. Output Q sets to 1 regardless of previous state.
  4. Output Q remains latched in its previous state Q(t).

Explanation

The characteristic equation of a JK flip-flop is Q(t+1) = J · Q' + K' · Q. When J = 1 and K = 1: Q(t+1) = 1 · Q' + 0 · Q = Q'. Thus, the flip-flop toggles output state on every clock pulse.

Question 32Electrical circuits, machines and digital logic

How many edge-triggered flip-flops are required to construct a 4-bit MOD-16 synchronous binary counter?

  1. 16 flip-flops
  2. 2 flip-flops
  3. 4 flip-flops Correct answer.
  4. 8 flip-flops

Explanation

A counter with N flip-flops can count 2ᴺ unique states (MOD-2ᴺ). For MOD-16, 2ᴺ = 16 -> N = 4 flip-flops (states 0000 to 1111).

Question 33Electrical circuits, machines and digital logic

Which logic gate family is considered 'universal' because any Boolean logic function can be constructed using only that single gate type?

  1. NAND gates (or NOR gates) Correct answer.
  2. XOR gates and NOT gates
  3. XNOR gates
  4. AND gates and OR gates

Explanation

NAND gates and NOR gates are universal gates because AND, OR, and NOT operations can all be synthesized purely from combinations of NAND (or NOR) gates alone.

Question 34Electrical circuits, machines and digital logic

A 4-to-1 Multiplexer (MUX) has data inputs D0, D1, D2, D3 and select lines S1, S0. If S1 = 1 and S0 = 0, which data input is connected to output Y?

  1. D0
  2. D2 Correct answer.
  3. D1
  4. D3

Explanation

The select binary combination (S1 S0)₂ = (1 0)₂ equals decimal 2. Therefore, data input D2 is selected and routed to output Y.

Question 35Electrical circuits, machines and digital logic

What is the primary operational difference between a combinational logic circuit and a sequential logic circuit?

  1. Combinational circuits are built with flip-flops; sequential circuits are built only with logic gates.
  2. Combinational circuits use feedback loops; sequential circuits do not.
  3. Combinational outputs depend solely on present inputs; sequential outputs depend on present inputs AND past state history (memory). Correct answer.
  4. Combinational circuits require a clock signal; sequential circuits operate asynchronously only.

Explanation

Combinational circuits (e.g., adders, MUXes) have memoryless outputs determined instantly by current inputs. Sequential circuits (e.g., counters, registers) contain internal memory elements (flip-flops) so outputs depend on input history.

Question 36Electrical circuits, machines and digital logic

Minimize the 4-variable logic function F(A, B, C, D) given by minterms Σm(0, 2, 8, 10, 14, 15) with don't-care conditions d(A,B,C,D) = Σd(5, 7, 13).

  1. F = B' D' + A B C Correct answer.
  2. F = A' C + B D
  3. F = A D' + B' C
  4. F = C' D + A B

Explanation

Group minterms on a 4-variable K-map: m0(0000), m2(0010), m8(1000), m10(1010) form a 4-corner quad yielding term 'B' D''. Minterms m14(1110) and m15(1111) group together to yield 'A B C'. Minimal expression is F = B' D' + A B C.

Question 37Electrical circuits, machines and digital logic

A 10-bit R-2R ladder Digital-to-Analog Converter (DAC) has a reference voltage V_ref = 5.0 V. What analog output voltage (V_out) corresponds to the digital input code (1000000000)₂ (MSB = 1, all other bits 0)?

  1. 4.995 V
  2. 2.50 V Correct answer.
  3. 5.00 V
  4. 1.25 V

Explanation

For an R-2R DAC, digital input code (1000000000)₂ represents decimal 512 out of 2¹⁰ = 1024 total counts. Analog output voltage V_out = V_ref · (512 / 1024) = 5.0 V · 0.50 = 2.50 V.

Question 38Electrical circuits, machines and digital logic

In a 4-bit Asynchronous (Ripple) Binary Up-Counter constructed with negative-edge-triggered JK flip-flops (J=1, K=1), where is the clock input of each succeeding flip-flop (FF1, FF2, FF3) connected?

  1. The clock input of FF_n is connected to ground.
  2. All clock inputs are tied together to a master system clock.
  3. The clock input of FF_n is driven by the Q output of preceding flip-flop FF_{n-1}. Correct answer.
  4. The clock input of FF_n is driven by an external AND gate.

Explanation

In an asynchronous ripple counter, only the first flip-flop (FF0) receives the external clock signal. Each subsequent stage FF_n is clocked by the Q (or Q') output transition of the previous stage FF_{n-1}, causing a cascading ripple delay.

Question 39Electrical circuits, machines and digital logic

How many 2-to-1 Multiplexers are required to construct a 8-to-1 Multiplexer?

  1. 8 multiplexers
  2. 4 multiplexers
  3. 3 multiplexers
  4. 7 multiplexers Correct answer.

Explanation

To build an 8-to-1 MUX tree using 2-to-1 MUXes: First stage uses 4 MUXes (reduces 8 inputs to 4). Second stage uses 2 MUXes (reduces 4 to 2). Third stage uses 1 MUX (reduces 2 to 1). Total required = 4 + 2 + 1 = 7 multiplexers.

Question 40Measurement, feedback, PLC and automation

What is the primary distinguishing feature of a Closed-Loop Automatic Control System compared to an Open-Loop Control System?

  1. Closed-loop systems incorporate a feedback sensor that measures output y(t) and compares it with reference input r(t) to generate an error signal e(t). Correct answer.
  2. Open-loop systems use digital microcontrollers; closed-loop systems are purely analog.
  3. Closed-loop systems cannot suffer from instability.
  4. Closed-loop systems use higher power electric motors.

Explanation

A closed-loop system continuously monitors output y(t) using feedback sensors, comparing it against setpoint reference r(t). The error e(t) = r(t) - y(t) drives the controller to compensate for disturbances.

Question 41Measurement, feedback, PLC and automation

In control engineering, what is the mathematical definition of a system Transfer Function G(s)?

  1. Ratio of time-domain output y(t) to input u(t).
  2. Ratio of Laplace transform of output Y(s) to Laplace transform of input U(s), assuming all initial conditions are zero. Correct answer.
  3. Product of system gain and input frequency.
  4. Derivative of output with respect to time.

Explanation

The Transfer Function G(s) = Y(s) / U(s) is defined in the s-domain as the ratio of Laplace transformed output Y(s) to input U(s) under zero initial conditions.

Question 42Measurement, feedback, PLC and automation

Two control blocks with individual transfer functions G1(s) = 4 / (s + 2) and G2(s) = 5 / (s + 3) are connected in series (cascade). What is the combined overall transfer function G_total(s)?

  1. 20 / (2s + 5)
  2. 9 / (s + 5)
  3. 20 / [(s + 2)(s + 3)] Correct answer.
  4. (9s + 22) / [(s + 2)(s + 3)]

Explanation

When blocks are connected in series (cascade), their individual transfer functions multiply: G_total(s) = G1(s) · G2(s) = [4 / (s + 2)] · [5 / (s + 3)] = 20 / [(s + 2)(s + 3)].

Question 43Measurement, feedback, PLC and automation

What type of industrial sensor utilizes a platinum wire resistor whose electrical resistance increases predictably with rising temperature (e.g., Pt100)?

  1. NTC Thermistor
  2. Optical Pyrometer
  3. Thermocouple (Termopar Type K)
  4. Resistance Temperature Detector (RTD / Termorresistencia Pt100) Correct answer.

Explanation

An RTD (Resistance Temperature Detector) such as Pt100 uses pure platinum metal with a positive temperature coefficient (PTC). Its resistance is exactly 100 Ω at 0 °C and increases linearly with temperature.

Question 44Measurement, feedback, PLC and automation

In a negative feedback control system with forward transfer function G(s) and feedback transfer function H(s) = 1 (unity feedback), what is the closed-loop transfer function T(s) = Y(s) / R(s)?

  1. T(s) = G(s) / (1 + G(s)) Correct answer.
  2. T(s) = G(s) / (1 - G(s))
  3. T(s) = 1 + G(s)
  4. T(s) = G(s) · H(s)

Explanation

For negative feedback control, closed-loop transfer function is T(s) = G(s) / [1 + G(s)H(s)]. When H(s) = 1, T(s) = G(s) / [1 + G(s)].

Question 45Measurement, feedback, PLC and automation

A first-order physical system has a transfer function G(s) = 5 / (2s + 1). What is the time constant (τ) of the system, and how long does it take to reach 98% of its final steady-state response to a step input (4 τ)?

  1. τ = 0.5 s; settling time t_s = 2 s
  2. τ = 1 s; settling time t_s = 4 s
  3. τ = 2 s; settling time t_s = 8 s Correct answer.
  4. τ = 5 s; settling time t_s = 20 s

Explanation

Standard first-order form is G(s) = K / (τ s + 1). Comparing G(s) = 5 / (2s + 1), the time constant τ = 2 seconds. The 98% settling time is t_s ≈ 4 τ = 4 · 2 s = 8 seconds.

Question 46Measurement, feedback, PLC and automation

A closed-loop system has forward transfer function G(s) = 10 / (s + 2) and negative unity feedback H(s) = 1. What is the closed-loop transfer function T(s) = Y(s) / R(s)?

  1. T(s) = 10 / (s + 2)
  2. T(s) = 10 / (s - 8)
  3. T(s) = 1 / (s + 12)
  4. T(s) = 10 / (s + 12) Correct answer.

Explanation

Closed-loop transfer function T(s) = G(s) / [1 + G(s)]. Substitute G(s): T(s) = [10 / (s + 2)] / [1 + 10 / (s + 2)] = 10 / [(s + 2) + 10] = 10 / (s + 12).

Question 47Measurement, feedback, PLC and automation

The characteristic equation of a second-order feedback control system is s² + 6s + 25 = 0. What are the undamped natural frequency (ω_n) and damping ratio (ζ) of the system?

  1. ω_n = 5 rad/s, ζ = 0.60 (Underdamped) Correct answer.
  2. ω_n = 25 rad/s, ζ = 0.12 (Underdamped)
  3. ω_n = 5 rad/s, ζ = 1.00 (Critically damped)
  4. ω_n = 6 rad/s, ζ = 0.50 (Underdamped)

Explanation

Standard 2nd-order characteristic equation is s² + 2 ζ ω_n s + ω_n² = 0. Comparing terms: ω_n² = 25 -> ω_n = 5 rad/s. 2 ζ ω_n = 6 -> 2 · ζ · 5 = 6 -> 10 ζ = 6 -> ζ = 0.60. Since 0 < ζ < 1, the response is underdamped.

Question 48Measurement, feedback, PLC and automation

According to the Routh-Hurwitz stability criterion, what condition must the closed-loop characteristic polynomial P(s) = s³ + 3s² + 3s + K = 0 satisfy for the system to be BIBO stable?

  1. K < 0
  2. 0 < K < 9 Correct answer.
  3. K > 9
  4. 0 < K < 3

Explanation

Routh array for s³ + 3s² + 3s + K = 0: Row s³: 1 3 Row s²: 3 K Row s¹: (3·3 - 1·K) / 3 = (9 - K) / 3 Row s⁰: K For stability, all elements in column 1 must be positive: K > 0 and (9 - K)/3 > 0 -> 9 - K > 0 -> K < 9. Therefore, 0 < K < 9.

Question 49Measurement, feedback, PLC and automation

What type of sensor converts linear position displacement directly into a proportional AC electrical voltage signal using primary and secondary transformer coils?

  1. Strain Gauge (Galgas extensométricas)
  2. Capacitive Proximity Sensor
  3. Linear Variable Differential Transformer (LVDT) Correct answer.
  4. Optical Shaft Encoder

Explanation

An LVDT (Linear Variable Differential Transformer) consists of one primary winding, two secondary windings, and a movable ferromagnetic core. Core motion alters magnetic coupling, generating an AC voltage proportional to linear displacement.

Question 50Measurement, feedback, PLC and automation

A unity feedback control system with forward transfer function G(s) = 20 / [s(s + 5)] is subjected to a unit step input r(t) = 1(t). What is the steady-state position error e_ss?

  1. e_ss = 0.25
  2. e_ss = 0.05
  3. e_ss = ∞
  4. e_ss = 0 Correct answer.

Explanation

G(s) contains an integrator pole at s = 0, making it a Type 1 system. Position error constant K_p = lim_{s->0} G(s) = ∞. Steady-state error to a unit step input is e_ss = 1 / (1 + K_p) = 1 / (1 + ∞) = 0.