Q: On–off control is a special case of proportional control with extremely high gain. What is the approximate proportional band (band width) expressed as a percentage for ideal on–off control?

Introduction / Context: On–off control is the simplest feedback strategy used in thermostats, level switches, and many binary actuators. It is often described as an extreme case of proportional control where the gain tends to infinity. Understanding its effective proportional band helps in comparing it with true proportional, PI, and PID controllers. Given Data / Assumptions: Ideal on–off behavior with no intentional hysteresis or differential gap. Control output takes only two states: fully on or fully off. Proportional band is defined as the error range over which the output linearly changes from 0% to 100%. Concept / Approach: For proportional control, output varies linearly with error within the proportional band. In on–off control, there is no linear region: the output switches instantaneously between extremes at a threshold. Therefore, the effective proportional band is 0 percent. Practical systems may add hysteresis to avoid chattering, but the idealized theoretical band remains zero. Step-by-Step Solution: Define proportional band PB as the span of error yielding a continuous output change.For on–off, output is discrete; continuous span is nonexistent.Hence, PB ≈ 0% for the ideal case. Verification / Alternative check: Plotting controller output vs error reveals a step function for on–off, versus a sloped line for proportional control—confirming a zero-width band for on–off. Why Other Options Are Wrong: 100%, 75%, 25%: These imply a nonzero linear region, which contradicts the binary nature of on–off control. Common Pitfalls: Confusing practical hysteresis (added for stability) with proportional band. Assuming on–off is a low-gain proportional controller; it is effectively infinite gain at the switching threshold. Overlooking cycling and wear issues when no hysteresis is used. Final Answer: 0

Question 1

Low-temperature measurement with thermocouples Which thermocouple type in this list is generally unsuitable for measuring sub-zero (below 0 °C) temperatures in practice?

Accepted Answer

Introduction / Context: Different thermocouple types are optimized for different temperature ranges and environments. Selection errors can cause poor sensitivity, noise, or drift—especially near the limits of a sensor's practical range. The question asks which listed type is not typically used for sub-zero measurements. Given Data / Assumptions: “Sub-zero” means below 0 °C. Chromel–alumel (Type K), Iron–constantan (Type J), and Copper–constantan (Type T) are base-metal thermocouples. Platinum–platinum + rhodium (Types S/R) are noble-metal thermocouples intended for high temperatures. Concept / Approach: Types S/R (Pt–PtRh) are designed for high-temperature service (furnaces, kilns) with excellent stability at elevated temperatures, but they are expensive and not favored at low temperatures, where their sensitivity and practicality are poor. By contrast, Types K, J, and especially T perform well at sub-zero temperatures and are commonly used in cryogenic and refrigeration applications. Step-by-Step Solution: Match each type to its practical range: T and K include sub-zero service; J covers moderate low temperatures; S/R focus on high temperatures.Identify the pair generally unsuitable below 0 °C: Pt–PtRh.Select “Platinum–platinum + rhodium.” Verification / Alternative check: Instrument catalogs and standards list S/R lower limits typically at or above 0 °C, with far better performance at high temperatures. Why Other Options Are Wrong: Chromel–alumel (K), Iron–constantan (J), Copper–constantan (T): Frequently used at sub-zero temperatures in industry. Common Pitfalls: Choosing noble-metal thermocouples for low-temperature duties leads to unnecessary cost and inferior sensitivity compared to Type T or K. Final Answer: Platinum–platinum + rhodium

Question 2

First-order element under sinusoidal forcing The steady-state sinusoidal response of an ideal first-order system exhibits what phase lag relative to the input?

Accepted Answer

Introduction / Context: Understanding frequency response is essential for controller tuning and stability analysis. A first-order process (with transfer function 1/(τs + 1)) has characteristic gain and phase behaviors that determine how it reacts to oscillatory inputs and noise. The phase lag is especially important for phase margin and closed-loop robustness. Given Data / Assumptions: Standard first-order transfer function: G(jω) = 1 / (1 + jωτ). We examine the steady-state sinusoidal response. Phase lag φ(ω) = arctan(ωτ) (expressed as a positive lag magnitude). Concept / Approach: The phase lag of a first-order system increases monotonically with frequency from 0° at ω → 0 to a limiting value as ω → ∞. The asymptotic maximum lag is 90°. At the corner frequency (ω = 1/τ), the phase lag equals 45°. Hence, the best single descriptive value for the maximum phase lag is 90° (noting this is the limit as frequency becomes very high). Step-by-Step Solution: Write φ(ω) = arctan(ωτ).Evaluate limits: φ(0) = 0°, φ(∞) → 90°.Therefore the maximal (asymptotic) lag is 90°. Verification / Alternative check: Bode plot of a first-order lag shows a straight-line phase characteristic approaching −90° at high frequency, confirming the result. Why Other Options Are Wrong: 30°, 120°, 180°: Do not match first-order behavior; 180° is characteristic of sign reversal or second-order resonant conditions, not a simple first-order lag. Common Pitfalls: Interpreting 90° as an exact constant; for a first-order element it is an upper limit reached asymptotically with increasing frequency. Final Answer: 90°

Question 3

Temperature range suitability of common instruments Which instrument from the list can reliably measure temperatures from about −20 °C up to approximately 300 °C in routine practice?

Accepted Answer

Introduction / Context: Selecting a temperature instrument requires matching the sensor’s useful range to the application. Many processes operate near ambient yet occasionally reach several hundred degrees Celsius. An understanding of the typical coverage of classic instruments helps avoid over-specification and ensures dependable readings. Given Data / Assumptions: Target span is roughly −20 °C to 300 °C. Instruments listed: mercury-in-glass, vapor-pressure thermometer, and resistance thermometer (RTD). We assume standard, commonly available industrial or lab-grade devices. Concept / Approach: Mercury-in-glass thermometers commonly cover from about −39 °C to roughly 357 °C (boiling point of mercury), with many commercial scales spanning −20 to 300 °C comfortably. Vapor-pressure thermometers have ranges that depend on the fill fluid; some ranges may not simultaneously cover −20 to 300 °C without changing bulbs/fills. RTDs can indeed cover this span, but the question aims at a single, conventional instrument “from the list” that is routinely available pre-scaled for −20 to 300 °C—this is very typical for mercury-in-glass devices. Step-by-Step Solution: Check standard mercury ranges: include −20 to 300 °C scales.Note vapor-pressure ranges are fluid-specific and may not cover both ends in one fill.Recognize RTDs can measure this span but are not a single “glass thermometer” style; choose the most standard fit: mercury in glass. Verification / Alternative check: Supplier catalogs show many mercury thermometers directly spanning −20 to 300 °C with appropriate capillary and scale design. Why Other Options Are Wrong: Vapor pressure thermometer: Range depends on fill; a single instrument may not span −20 to 300 °C. Resistance thermometer: Technically capable, but the question context favors a traditional, self-contained thermometer; the mercury device is the canonical match. None of these: Incorrect because mercury-in-glass fits well. Common Pitfalls: Confusing “can be designed to measure” with “commonly supplied in a single off-the-shelf range.” Final Answer: Mercury in glass thermometer

Question 4

Root locus basics for closed-loop design For a unity-feedback system, the number of branches (loci) in the root locus equals the number of open-loop poles. If the open-loop transfer function has four finite poles, how many root-locus branches will b…

Accepted Answer

Introduction / Context: The root locus is a graphical method that shows how the closed-loop poles of a feedback system move in the complex plane as a real scalar gain K varies. It is a cornerstone tool for controller design and stability assessment in classical control theory. Given Data / Assumptions: Unity-feedback configuration is implied. Open-loop transfer function G(s)H(s) possesses four finite poles. Zeros may be finite or at infinity; asymptotes accommodate any excess poles. Concept / Approach: A fundamental rule of root locus construction states: the number of branches (also called loci) equals the number of open-loop poles, because each closed-loop pole (for any K) originates from a distinct open-loop pole at K = 0 and traces a continuous path as K increases. If there are fewer finite zeros than poles, some branches terminate at zeros at infinity along asymptotes determined by standard angle and centroid formulas. Step-by-Step Solution: Count open-loop poles: 4.Apply the rule: number of root-locus branches = number of open-loop poles.Therefore, four branches will appear on the root locus. Verification / Alternative check: Any textbook example with n poles and m zeros shows n branches; when m < n, (n − m) branches head to infinity, confirming the count. Why Other Options Are Wrong: 1, 2, 3: Do not match the pole count; each open-loop pole must seed a branch. Common Pitfalls: Confusing the number of asymptotes (n − m) with the number of branches (n). Asymptotes are directions for branches that head to infinity; they do not reduce branch count. Final Answer: 4

Question 5

For two non-interacting first-order systems connected in series in process control, the overall transfer function of the combined system is obtained by what operation on the individual transfer functions? Provide the most appropriate relationship an…

Accepted Answer

Introduction / Context: In process control and instrumentation, complex plants are frequently modeled by connecting simpler dynamic blocks in series. Each block has a transfer function that relates its output to its input in the Laplace domain. This question tests your understanding of how to combine transfer functions for non-interacting first-order elements connected in series, which is a foundational concept for control-loop design and stability analysis. Given Data / Assumptions: Two first-order, linear, time-invariant systems are connected in series. Systems are non-interacting (the output of the first is the input to the second with no side coupling or feedback between them). Standard Laplace-transform transfer function representation is used. Concept / Approach: For linear time-invariant blocks connected in series, the overall transfer function G_total(s) equals the product of the individual transfer functions. If G1(s) and G2(s) are the two blocks, then G_total(s) = G2(s) * G1(s). This follows directly from block-diagram algebra and the cascade property of Laplace-domain models. Step-by-Step Solution: Let the input be U(s) and the intermediate variable be X(s), with output Y(s).X(s) = G1(s) * U(s)Y(s) = G2(s) * X(s) = G2(s) * G1(s) * U(s)Therefore, G_total(s) = Y(s)/U(s) = G2(s) * G1(s) Verification / Alternative check: If each element is a standard first-order lag, G1(s) = K1/(1 + τ1s) and G2(s) = K2/(1 + τ2s). Then G_total(s) = (K1K2)/[(1 + τ1s)(1 + τ2s)], which shows the expected second-order denominator formed by multiplication—consistent with series connection behavior. Why Other Options Are Wrong: Ratio: Ratios arise in feedback loops, not in simple series cascades.Sum: Summation corresponds to parallel paths, not series.Difference: Also corresponds to summing junctions with a minus sign, not series. Common Pitfalls: Confusing series combination (product) with parallel combination (sum). Assuming interaction or hidden feedback between stages, which would invalidate simple multiplication. Mixing time-domain convolution with Laplace-domain multiplication; in the Laplace domain, cascade implies multiplication. Final Answer: product

Question 6

In a standard Bode plot used for frequency-response analysis, the phase angle Φ is plotted against angular frequency ω on what kind of graph paper or axis scaling?

Accepted Answer

Introduction / Context: Bode plots are ubiquitous in control engineering for analyzing system stability and performance in the frequency domain. A complete Bode diagram consists of two plots: magnitude (in decibels) vs frequency and phase (in degrees) vs frequency. Understanding the axis scaling is essential to correctly interpret slopes, phase margins, and crossover frequencies. Given Data / Assumptions: We consider standard Bode diagrams for linear time-invariant systems. Frequency axis is represented by angular frequency ω (or frequency f), spanning several decades. Phase Φ is plotted in degrees. Concept / Approach: In Bode plots, the horizontal axis (frequency) is logarithmic to accommodate multiple decades conveniently. Both the magnitude and phase plots share this log frequency axis. The vertical axis is linear (decibels for magnitude, degrees for phase). Hence, the phase plot is made on semi-log axis: log scale horizontally and linear scale vertically. Step-by-Step Solution: Identify axes: horizontal = frequency ω, vertical = Φ (degrees).Apply Bode convention: frequency axis is log-scaled to compress decades.Vertical axis for phase is linear; thus the overall paper is semi-log. Verification / Alternative check: Typical plotting software and textbooks show phase curves (e.g., arctangent-type for first-order systems) laid out over log-spaced ω to clearly display low-, mid-, and high-frequency asymptotes and transitions—confirming semi-log usage. Why Other Options Are Wrong: Log-log: Used for magnitude (in some contexts like Nyquist slopes) but the phase ordinate is not logarithmic.Ordinary (linear-linear): Would not compress frequency decades and is not standard for Bode plots.Triangular: Not a recognized plotting format in control theory. Common Pitfalls: Confusing semi-log with log-log due to the decibel axis on the magnitude plot; decibels are linear in the vertical direction. Plotting frequency linearly, which obscures asymptotic behavior across decades. Mixing up Bode (semi-log) with Nichols or Nyquist which have different representations. Final Answer: semi-log

Question 7

Consider process dynamics statements about transfer lag and system interaction. Identify the incorrect statement among the following descriptions of first-order and higher-order systems and interacting tanks.

Accepted Answer

Introduction / Context: Control engineers distinguish between capacity (dynamic) lags and transfer (transportation or dead-time) lags. This question probes your understanding of transfer lag versus system order and the notion of interacting versus non-interacting processes, such as jacketed vessels where energy exchange couples dynamics. Given Data / Assumptions: Single first-order system: one energy or material storage with no pure time delay. Higher-order systems: multiple storages in series or interacting. Transfer lag (dead time) refers to a pure time delay e^{-θs} component. Concept / Approach: A first-order system without transport effects exhibits capacity lag but not transfer lag. Higher order does not automatically imply transfer lag; dead time arises from transport phenomena (e.g., material flow in long pipes, sensing delays) or from approximations of many small lags in series. Interaction occurs when states influence each other, as in a jacketed stirred tank where process and jacket temperatures mutually affect dynamics. Step-by-Step Solution: Evaluate (a): A single first-order element has no pure dead time — statement is correct.Evaluate (b): Jacketed stirred tank is interacting due to coupled energy balances — statement is correct.Evaluate (c): Claiming all higher-order systems inherently have transfer lag is incorrect; higher order may still have zero dead time — this is the wrong statement.Evaluate (d): As the number of cascade stages decreases, the approximation to dead time weakens, hence apparent transfer lag decreases — statement is acceptable. Verification / Alternative check: Classical models show that N equal first-order lags in series approximate a dead-time element as N increases. Conversely, fewer stages reduce the approximation to dead time, supporting option (d). Why Other Options Are Wrong: (a) Correct: only capacity lag is present.(b) Correct: a canonical interacting example.(d) Correct trend for reduced stage count. Common Pitfalls: Equating higher order with dead time; they are distinct features. Confusing interacting with coupled feedback; interaction is internal coupling of states. Overlooking that dead time is often a transport or measurement artifact. Final Answer: Transfer lag is a characteristic of all higher order systems (other than first order systems).

Question 8

For measuring the temperature of a red-hot furnace (very high temperature, non-contact preferred), which temperature instrument is most appropriate in industrial practice?

Accepted Answer

Introduction / Context: Very high temperatures inside furnaces pose challenges for conventional contact thermometers. Instruments must withstand harsh environments, provide reliable readings, and often work without physical contact to avoid probe damage or measurement disturbance. This question compares standard instruments and identifies the best choice for a red-hot furnace. Given Data / Assumptions: Red-hot furnace temperatures commonly exceed the upper limits of many contact sensors. Non-contact measurement reduces probe degradation and measurement lag. Line-of-sight access is available for optical instruments. Concept / Approach: Optical pyrometers measure temperature by comparing the visible radiation from a hot target to a calibrated source. They are ideal for high temperatures where contact methods may fail or introduce errors. Thermocouples can measure high temperatures but require protection and may suffer drift; resistance thermometers and bimetallic devices have much lower upper limits. Step-by-Step Solution: Assess temperature range: red-hot suggests very high (well above typical RTD ranges).Compare instruments: RTD and bimetallic are limited; thermocouples are contact and may degrade; optical pyrometer is non-contact and suitable.Select the optical pyrometer as the most appropriate tool. Verification / Alternative check: Radiation pyrometers (including optical types) are standard for furnaces and kilns due to high-temperature capability, fast response, and non-intrusive operation. Why Other Options Are Wrong: Platinum resistance thermometer: Accurate but not rated for extreme furnace temperatures and requires contact.Thermocouple: Usable at high temperatures but suffers from sheath degradation and contact limitations.Bimetallic thermometer: Low upper range and slow response; unsuitable for furnaces. Common Pitfalls: Assuming contact sensors are always better because they are cheaper. Ignoring radiation corrections (emissivity) in pyrometry; correct setup is essential. Overlooking viewing window cleanliness and alignment, which affect optical readings. Final Answer: Optical pyrometer

Question 9

On–off control is a special case of proportional control with extremely high gain. What is the approximate proportional band (band width) expressed as a percentage for ideal on–off control?

Accepted Answer

Introduction / Context: On–off control is the simplest feedback strategy used in thermostats, level switches, and many binary actuators. It is often described as an extreme case of proportional control where the gain tends to infinity. Understanding its effective proportional band helps in comparing it with true proportional, PI, and PID controllers. Given Data / Assumptions: Ideal on–off behavior with no intentional hysteresis or differential gap. Control output takes only two states: fully on or fully off. Proportional band is defined as the error range over which the output linearly changes from 0% to 100%. Concept / Approach: For proportional control, output varies linearly with error within the proportional band. In on–off control, there is no linear region: the output switches instantaneously between extremes at a threshold. Therefore, the effective proportional band is 0 percent. Practical systems may add hysteresis to avoid chattering, but the idealized theoretical band remains zero. Step-by-Step Solution: Define proportional band PB as the span of error yielding a continuous output change.For on–off, output is discrete; continuous span is nonexistent.Hence, PB ≈ 0% for the ideal case. Verification / Alternative check: Plotting controller output vs error reveals a step function for on–off, versus a sloped line for proportional control—confirming a zero-width band for on–off. Why Other Options Are Wrong: 100%, 75%, 25%: These imply a nonzero linear region, which contradicts the binary nature of on–off control. Common Pitfalls: Confusing practical hysteresis (added for stability) with proportional band. Assuming on–off is a low-gain proportional controller; it is effectively infinite gain at the switching threshold. Overlooking cycling and wear issues when no hysteresis is used. Final Answer: 0

Process Control and Instrumentation Questions