Q1 Past Paper · PPSC/FPSC/NTS easy
Magnitude in Bode plot for gain K is
A K dB without log scaling incorrectly stated ✓ B 20 log10(K) dB horizontal line ✓ C log10 K without factor 20 only as incorrect variant ✓ D 20 log10(s) always without K ✓ Show Answer 💡 Explanation: Constant gain contributes flat magnitude in dB.
Q2 Past Paper · PPSC/FPSC/NTS easy
Pole at origin (integrator) contributes to Bode magnitude slope of
A −20 dB/decade ✓ B +20 dB/decade ✓ C −40 dB/decade always for single pole at origin without other poles ✓ D 0 dB/decade always ✓ Show Answer 💡 Explanation: Each integrator adds −20 dB/decade slope.
Q3 Past Paper · PPSC/FPSC/NTS easy
Zero at origin (differentiator) contributes magnitude slope of
A +20 dB/decade ✓ B −20 dB/decade for zero at origin—positive slope for differentiator ✓ C 0 dB/decade always ✓ D −6 dB/octave only without sign clarity—still +20 dB/decade ✓ Show Answer 💡 Explanation: Differentiator boosts high-frequency gain.
Q4 Past Paper · PPSC/FPSC/NTS medium
Corner frequency of first-order pole at 1/τ rad/s is
A ω = τ only incorrectly ✓ B ω = 1/τ ✓ C ω = τ² only ✓ D ω = 0 always ✓ Show Answer 💡 Explanation: Break frequency where asymptote bends for real pole.
Q5 Past Paper · PPSC/FPSC/NTS medium
Phase of first-order lag at corner frequency is approximately
A 0° always ✓ B −90° at corner exactly for asymptotic mid—approx −45° at ω=1/τ ✓ C −45° ✓ D +45° always ✓ Show Answer 💡 Explanation: At break frequency, first-order lag phase is −45°.
Q6 Past Paper · PPSC/FPSC/NTS medium
Gain crossover frequency is where
A phase equals −180° only without magnitude condition for gain crossover definition ✓ B output equals input in time domain step only without frequency definition ✓ C open-loop magnitude equals 0 dB (unity gain) ✓ D PID integral time constant only ✓ Show Answer 💡 Explanation: ωgc is used with phase margin for stability assessment.
Q7 Past Paper · PPSC/FPSC/NTS medium
Phase crossover frequency is where
A magnitude equals 0 dB only without phase −180° condition for phase crossover ✓ B system becomes type 2 always ✓ C open-loop phase equals −180° ✓ D derivative action is maximum always ✓ Show Answer 💡 Explanation: ωpc used with gain margin definition.
Q8 Past Paper · PPSC/FPSC/NTS medium
Gain margin is
A phase increase in degrees at gain crossover only without gain definition ✓ B time delay only without dB definition ✓ C distance relay reach in ohms only ✓ D amount gain can increase before instability at phase crossover ✓ Show Answer 💡 Explanation: GM = −|G(jωpc)| in dB if stable; positive GM means stable.
Q9 Past Paper · PPSC/FPSC/NTS medium
Phase margin is
A gain increase in dB at phase crossover only ✓ B steady-state error to ramp only without phase definition ✓ C motor slip at full load only ✓ D additional phase lag before −180° at gain crossover frequency ✓ Show Answer 💡 Explanation: PM indicates relative stability; typical design target 30°–60°.
Q10 hard
Resonant peak in closed-loop frequency response relates to
A only DC gain without dynamics ✓ B damping ratio of dominant second-order poles ✓ C only transport delay only without resonance ✓ D only fuse cut-off current only ✓ Show Answer 💡 Explanation: Low damping gives high resonant peak Mr.
Q11 hard
Asymptotic Bode magnitude of second-order underdamped pair near ωn shows
A flat 0 dB always without roll-off ✓ B −40 dB/decade slope far above ωn with possible peak near ωn ✓ C +40 dB/decade always above ωn without resonance possibility ✓ D undefined without zeros always ✓ Show Answer 💡 Explanation: Complex pole pair rolls off at −40 dB/decade at high ω.
Q12 easy
Decade on Bode frequency axis means
A twofold change only ✓ B tenfold change in frequency ✓ C phase change of 45° only without frequency meaning ✓ D one period of oscillation only without factor ten ✓ Show Answer 💡 Explanation: Slope quoted as dB per decade of frequency.
Q13 easy
Octave corresponds to
A doubling of frequency ✓ B tenfold frequency change which is decade not octave ✓ C tripling frequency always ✓ D zero frequency span only ✓ Show Answer 💡 Explanation: Some slopes expressed as dB/octave (6 dB/oct for pole).
Q14 hard
Minimum phase system Bode plot is uniquely determined by
A phase curve alone without magnitude for minimum phase systems—magnitude alone suffices ✓ B time delay only without magnitude ✓ C magnitude curve alone ✓ D PID settings only without plant dynamics ✓ Show Answer 💡 Explanation: Hilbert transform relationship links magnitude and phase for minimum phase.
Q15 Past Paper · PPSC/FPSC/NTS hard
Transport delay adds phase
A constant −90° at all frequencies always ✓ B −ωT radians linearly with frequency ✓ C +ωT always positive linear phase which is wrong sign for delay ✓ D zero phase always regardless of delay ✓ Show Answer 💡 Explanation: e^(−jωT) contributes −ωT phase lag.
Q16 Past Paper · PPSC/FPSC/NTS medium
Bode plot is useful for designing
A only DC machine commutation timing only without frequency design ✓ B lead-lag compensators and assessing closed-loop bandwidth ✓ C only cable joint stress cones only without control design ✓ D only insulator fog profile only without controls ✓ Show Answer 💡 Explanation: Frequency shaping with compensators uses Bode insights.
Q17 medium
Slope of −20 dB/decade on Bode magnitude often indicates
A one pole or combination equivalent to first-order roll-off ✓ B two integrators always without checking net order ✓ C zero at origin always without sign ✓ D pure time delay only without magnitude slope contribution like integrator ✓ Show Answer 💡 Explanation: Net pole excess of one gives −20 dB/decade asymptote.
Q18 Past Paper · PPSC/FPSC/NTS easy
Linear system is BIBO stable if
A bounded input produces bounded output ✓ B any input produces zero output always ✓ C output always unbounded for step input always ✓ D poles need not be considered ever ✓ Show Answer 💡 Explanation: BIBO stability requires impulse response absolutely integrable for LTI.
Q19 Past Paper · PPSC/FPSC/NTS easy
For continuous-time LTI system, internal stability (asymptotic) requires poles to lie
A strictly in left half of s-plane ✓ B on imaginary axis always without left half plane requirement for asymptotic stability ✓ C right half plane for stability incorrectly ✓ D at origin only always for all stable systems ✓ Show Answer 💡 Explanation: Re(s)<0 for all poles ensures decaying modes.
Q20 Past Paper · PPSC/FPSC/NTS medium
Marginally stable system has
A all poles in RHP always as marginal definition incorrectly ✓ B only zeros on jω axis without pole condition ✓ C non-repeated poles on jω axis and none in RHP ✓ D always exponentially decaying modes only ✓ Show Answer 💡 Explanation: Undamped sustained oscillations occur for poles on imaginary axis.
Q21 Past Paper · PPSC/FPSC/NTS hard
Routh-Hurwitz criterion determines stability using
A only Bode phase at DC only without polynomial test ✓ B only Nyquist plot drawing without polynomial ✓ C characteristic equation coefficients without finding roots explicitly ✓ D only motor nameplate data only ✓ Show Answer 💡 Explanation: R-H tabulation signs of first column indicate stability.
Q22 Past Paper · PPSC/FPSC/NTS medium
Tuning PID by Ziegler-Nichols ultimate gain method uses
A only step response rise time without ultimate gain experiment incorrectly ✓ B only Bode zero placement without experiment incorrectly ✓ C only motor slip measurement incorrectly ✓ D critical gain Ku and oscillation period Pu at stability limit ✓ Show Answer 💡 Explanation: Closed-loop or open-loop Z-N rules based on Ku, Pu.
Q23 Past Paper · PPSC/FPSC/NTS easy
Increasing Kp too much can cause
A guaranteed stability always incorrectly ✓ B zero overshoot always incorrectly ✓ C instability and excessive oscillations ✓ D elimination of noise always incorrectly ✓ Show Answer 💡 Explanation: Excessive proportional gain reduces margins.
Q24 Past Paper · PPSC/FPSC/NTS medium
Too much derivative gain amplifies
A DC error only without high-frequency noise concern incorrectly ✓ B measurement noise ✓ C integral windup primarily incorrectly though D does not integrate noise same way ✓ D steady-state reference only incorrectly ✓ Show Answer 💡 Explanation: High Kd boosts high-frequency noise in error signal.
Q25 hard
PID in parallel form differs from ideal series form in
A being unable to implement derivative ever incorrectly ✓ B requiring no tuning ever incorrectly ✓ C eliminating steady-state error without I incorrectly ✓ D algebraic arrangement but can be made equivalent with conversion ✓ Show Answer 💡 Explanation: Parallel and series PID forms relate by equivalent parameter conversion.
Q26 hard
Set-point weighting on PID separates
A response to reference changes versus load disturbances ✓ B only fuse blowing from overload incorrectly ✓ C only distance zones incorrectly ✓ D only symmetrical sequences incorrectly ✓ Show Answer 💡 Explanation: Two-degree-of-freedom PID reduces overshoot to set-point steps.
Q27 Past Paper · PPSC/FPSC/NTS medium
Sampled-data PID implementation must consider
A only analog Bode plot of continuous plant without sampling effects incorrectly ✓ B discretization, aliasing and integral approximation ✓ C infinite sampling rate without implementation constraints incorrectly ✓ D elimination of derivative always incorrectly as requirement ✓ Show Answer 💡 Explanation: Digital control uses difference equations and Tustin etc.
Q28 Past Paper · PPSC/FPSC/NTS medium
Derivative filter in practical PID limits
A low-frequency gain to zero always incorrectly as filter purpose ✓ B high-frequency noise amplification by filtering D path ✓ C integral action entirely incorrectly ✓ D proportional gain to zero incorrectly ✓ Show Answer 💡 Explanation: First-order filter on derivative term is standard industrial practice.
Q29 hard
Feedforward control added to PID can
A improve disturbance rejection for measurable disturbances ✓ B replace stability analysis always incorrectly ✓ C guarantee RHP poles always incorrectly ✓ D eliminate need for feedback entirely incorrectly ✓ Show Answer 💡 Explanation: Feedforward anticipates known disturbance effects.
Q30 Past Paper · PPSC/FPSC/NTS medium
Cascade control uses
A only single P controller always incorrectly ✓ B only open-loop control without feedback incorrectly ✓ C inner fast loop and outer slow loop for improved disturbance rejection ✓ D only fuse in series without controllers incorrectly ✓ Show Answer 💡 Explanation: Inner loop handles fast dynamics; outer sets inner reference.
Q31 Past Paper · PPSC/FPSC/NTS easy
PI controller is commonly used when
A derivative noise is desired to increase incorrectly ✓ B zero proportional gain is required incorrectly ✓ C system must be type 0 open-loop without closed-loop integrator effect incorrectly stated ✓ D steady-state accuracy needed without excessive noise sensitivity from D ✓ Show Answer 💡 Explanation: PI gives integral action for SSE with simpler tuning than full PID.
Q32 medium
Closed-loop bandwidth with PID tuning is often traded against
A only cable charging current magnitude incorrectly ✓ B robustness margins and noise rejection ✓ C only tower sag at minimum temperature incorrectly ✓ D only SCR latching current only incorrectly ✓ Show Answer 💡 Explanation: Aggressive tuning raises bandwidth but may reduce PM/GM.
Q33 Past Paper · PPSC/FPSC/NTS medium
Integral windup occurs when
A derivative term exceeds proportional always incorrectly as windup definition ✓ B actuator saturates while integrator continues accumulating error ✓ C system is always stable during saturation incorrectly ✓ D gain margin is infinite incorrectly ✓ Show Answer 💡 Explanation: Anti-windup schemes freeze or back-calculate integral during saturation.
Q34 Past Paper · PPSC/FPSC/NTS easy
Transfer function of a linear time-invariant system is defined as
A ratio of time-domain outputs without transformation ✓ B ratio of Laplace transform of output to input with zero initial conditions ✓ C Fourier series coefficients only ✓ D PID tuning constants only ✓ Show Answer 💡 Explanation: G(s) = Y(s)/U(s) with zero initial conditions.
Q35 Past Paper · PPSC/FPSC/NTS easy
Poles of a transfer function are values of s where
A numerator equals zero only ✓ B output is maximum always ✓ C denominator equals zero ✓ D gain margin is infinite always ✓ Show Answer 💡 Explanation: Poles determine natural modes and stability.
Q36 Past Paper · PPSC/FPSC/NTS easy
Zeros of a transfer function are values of s where
A denominator equals zero ✓ B system is always unstable ✓ C phase margin is zero only ✓ D numerator equals zero ✓ Show Answer 💡 Explanation: Zeros affect magnitude and phase but not poles of closed loop alone.
Q37 Past Paper · PPSC/FPSC/NTS easy
Standard first-order system G(s) = K/(τs+1) has time constant
A τ ✓ B K only ✓ C τ/K only ✓ D 1/K only without τ ✓ Show Answer 💡 Explanation: Time constant τ governs exponential response speed.
Q38 Past Paper · PPSC/FPSC/NTS easy
DC gain of transfer function G(s) is found by
A s approaching infinity only always for DC gain definition—DC is s=0 ✓ B only imaginary axis without s=0 ✓ C only at resonant frequency always ✓ D evaluating G(s) at s = 0 ✓ Show Answer 💡 Explanation: Steady-state gain for step input is G(0) if applicable.
Q39 Past Paper · PPSC/FPSC/NTS medium
Second-order underdamped system is characterized by
A real distinct poles only always ✓ B complex conjugate poles with damping ratio ζ < 1 ✓ C ζ greater than 1 always ✓ D no transient response ever ✓ Show Answer 💡 Explanation: Underdamped response oscillates with decay.
Q40 Past Paper · PPSC/FPSC/NTS medium
Natural frequency ωn of second-order system appears in standard form
A s² + 2ζωn s + ωn² ✓ B s + ωn only without quadratic term ✓ C ωn² s + 1 only ✓ D PID derivative term only ✓ Show Answer 💡 Explanation: ωn is undamped natural frequency in rad/s.
Q41 Past Paper · PPSC/FPSC/NTS easy
Block diagram reduction uses rules for
A series, parallel and feedback loop algebra ✓ B only Bode plotting without algebra ✓ C only symmetrical components only on AC faults ✓ D only fuse coordination curves only ✓ Show Answer 💡 Explanation: Transfer functions combine by multiplication, summation and feedback formula.
Q42 Past Paper · PPSC/FPSC/NTS medium
Closed-loop transfer function with unity feedback H=1 is
A G/(1-G) always ✓ B 1/G always ✓ C G only without feedback effect ✓ D G/(1+G) where G is open-loop transfer function ✓ Show Answer 💡 Explanation: Negative unity feedback gives T = G/(1+G).
Q43 Past Paper · PPSC/FPSC/NTS hard
Type of a system indicates number of
A zeros at origin only ✓ B delay elements counted twice always without definition ✓ C PID derivative paths only ✓ D integrators (poles at origin) in open-loop transfer function ✓ Show Answer 💡 Explanation: Type determines steady-state error to polynomial inputs.
Q44 hard
Steady-state error to step input for type 0 system with step input is
A finite and generally non-zero unless gain is infinite ✓ B always zero for any type 0 without integrator—generally non-zero ✓ C always infinite for step on type 0—actually finite non-zero typically ✓ D undefined without Laplace ✓ Show Answer 💡 Explanation: Type 0 cannot track step without error unless loop gain → ∞.
Q45 hard
Lead compensator transfer function typically has
A pole closer to origin than zero always for lead definition—lead has zero closer ✓ B equal zero and pole at origin only always ✓ C zero closer to origin than pole ✓ D no effect on phase ✓ Show Answer 💡 Explanation: Lead adds positive phase in mid frequencies.
Q46 hard
Lag compensator provides
A positive phase boost at all frequencies without attenuation ever ✓ B elimination of all poles ✓ C high-frequency gain reduction and improved steady-state accuracy ✓ D only derivative action without lag pole-zero pair ✓ Show Answer 💡 Explanation: Lag increases low-frequency gain while attenuating highs.
Q47 Past Paper · PPSC/FPSC/NTS hard
Transport delay e^(-Ts) in transfer function causes
A constant gain at all frequencies without phase effect ever ✓ B phase lag increasing linearly with frequency ✓ C elimination of stability issues always ✓ D infinite bandwidth always ✓ Show Answer 💡 Explanation: Pure delay reduces phase margin significantly.
Q48 hard
Impulse response of LTI system is inverse Laplace transform of
A input only without system dynamics ✓ B PID output only without plant ✓ C distance relay impedance only ✓ D transfer function G(s) ✓ Show Answer 💡 Explanation: Impulse response characterizes system completely with LTI assumption.
Q49 medium
Convolution in time domain corresponds to
A subtraction of transfer functions only ✓ B division of zeros only ✓ C multiplication in Laplace domain ✓ D Fourier series only without Laplace ✓ Show Answer 💡 Explanation: y(t)=u(t)*g(t) ⟺ Y(s)=U(s)G(s) with zero ICs.
Q50 hard
State-space model ẋ=Ax+Bu, y=Cx+Du relates to transfer function by
A only Bode magnitude plot without matrices ✓ B G(s)=C(sI−A)⁻¹B+D ✓ C only per unit impedance conversion ✓ D only symmetrical sequence networks only ✓ Show Answer 💡 Explanation: State-space and transfer function are equivalent for LTI SISO/MIMO.
Q51 hard
Minimum phase system has all zeros in
A left half of s-plane (or on jω axis) ✓ B right half plane always ✓ C origin only always ✓ D infinity only always without left half plane constraint ✓ Show Answer 💡 Explanation: Minimum phase systems have monotonic phase lag with frequency.
Q52 Past Paper · PPSC/FPSC/NTS easy
Bode plot displays
A only time response without frequency content ever ✓ B only pole-zero map on complex plane without frequency axis as Bode definition ✓ C only Nyquist real axis only without magnitude and phase plots ✓ D magnitude in dB and phase versus frequency on logarithmic scale ✓ Show Answer 💡 Explanation: Bode diagrams aid frequency-domain analysis and controller design.
Q53 Past Paper · PPSC/FPSC/NTS easy
Derivative (D) action responds to
A only steady-state error magnitude without rate sensitivity incorrectly ✓ B only DC offset always incorrectly ✓ C rate of change of error and adds damping ✓ D integral of error only incorrectly ✓ Show Answer 💡 Explanation: D anticipates error trend and reduces overshoot.
Q54 hard
Nyquist criterion uses
A only step response overshoot percent only without Nyquist ✓ B only PID integral windup only without frequency encirclement ✓ C only symmetrical components only without Nyquist ✓ D open-loop frequency response encirclements of −1 point ✓ Show Answer 💡 Explanation: N = Z − P relates encirclements to closed-loop RHP zeros.
Q55 medium
Negative feedback reduces sensitivity of closed-loop gain to
A increases sensitivity always at all frequencies without exception incorrectly as blanket statement ✓ B has no effect ever on sensitivity incorrectly ✓ C plant parameter variations at low frequencies often ✓ D only changes fuse rating without loop effect ✓ Show Answer 💡 Explanation: Feedback can desensitize system to certain perturbations.
Q56 Past Paper · PPSC/FPSC/NTS medium
Root locus shows
A open-loop zeros only without pole motion ✓ B only Bode magnitude without pole paths ✓ C only steady-state error constants only without locus ✓ D closed-loop pole trajectories as gain varies from 0 to ∞ ✓ Show Answer 💡 Explanation: Evans root locus guides gain selection for desired damping.
Q57 Past Paper · PPSC/FPSC/NTS medium
Adding a pole in forward path generally
A always improves phase margin without exception incorrectly ✓ B has no effect on phase ever incorrectly ✓ C eliminates need for feedback incorrectly ✓ D degrades stability by adding phase lag ✓ Show Answer 💡 Explanation: Extra lag reduces phase margin unless compensated.
Q58 Past Paper · PPSC/FPSC/NTS medium
Adding a zero in forward path near crossover can
A always destabilize without exception incorrectly ✓ B eliminate all dynamics incorrectly ✓ C improve phase margin ✓ D replace sensor entirely incorrectly ✓ Show Answer 💡 Explanation: Lead zero adds positive phase near crossover.
Q59 Past Paper · PPSC/FPSC/NTS easy
Stable closed-loop system with adequate gain and phase margins typically has
A guaranteed zero steady-state error to all inputs without type consideration incorrectly as blanket ✓ B infinite bandwidth always incorrectly ✓ C poles in RHP always incorrectly ✓ D well-damped transient response without excessive oscillation ✓ Show Answer 💡 Explanation: Adequate margins correlate with acceptable transient behaviour.
Q60 hard
Limit cycle in nonlinear system refers to
A exponential decay always in nonlinear system incorrectly for limit cycle definition ✓ B only linear system resonance at natural frequency only without nonlinearity requirement for limit cycle term usage ✓ C only fuse arcing only without control context though arcing is nonlinear oscillation colloquially ✓ D sustained oscillation amplitude determined by nonlinearity ✓ Show Answer 💡 Explanation: Nonlinearities can produce stable amplitude oscillations.
Q61 hard
Lyapunov stability analysis uses
A only Bode plot asymptotes only without Lyapunov function ✓ B energy-like Lyapunov function with negative semi-definite derivative ✓ C only per unit conversion only without state function ✓ D only distance relay mho circle only without Lyapunov ✓ Show Answer 💡 Explanation: Lyapunov method proves stability without solving differential equation.
Q62 medium
Phase margin of 45° roughly corresponds to
A unstable response always incorrectly ✓ B overdamped without oscillation always incorrectly for 45° PM typical correlation ✓ C moderately damped closed-loop step response ✓ D zero overshoot always incorrectly as blanket ✓ Show Answer 💡 Explanation: Rule of thumb links PM to damping of dominant second-order approximation.
Q63 Past Paper · PPSC/FPSC/NTS medium
Gain margin less than 0 dB indicates
A closed-loop instability for unity feedback minimum-phase assumptions typically ✓ B guaranteed stability always incorrectly ✓ C infinite phase margin always incorrectly ✓ D zero steady-state error always incorrectly ✓ Show Answer 💡 Explanation: GM<0 means open-loop gain exceeds unity at −180° phase.
Q64 hard
Hurwitz polynomial has
A roots on RHP always incorrectly ✓ B only real roots always as Hurwitz requirement incorrectly stated—complex allowed with negative real parts ✓ C all roots with negative real parts ✓ D roots at infinity only incorrectly ✓ Show Answer 💡 Explanation: Characteristic polynomial of stable LTI system is Hurwitz.
Q65 Past Paper · PPSC/FPSC/NTS medium
Delay in control loop tends to
A reduce phase margin and can destabilize system ✓ B always improve damping without exception incorrectly ✓ C eliminate steady-state error without integrator incorrectly as delay effect ✓ D increase gain margin always incorrectly ✓ Show Answer 💡 Explanation: Time delay adds lag proportional to frequency.
Q66 hard
Conditional stability may occur when
A system is stable only for limited gain range ✓ B system is stable for all gains always incorrectly ✓ C poles always in RHP incorrectly for conditional stability definition ✓ D only nonlinear systems without gain dependence incorrectly—linear can be conditionally stable with gain ✓ Show Answer 💡 Explanation: Some compensator configurations stable only between gain limits.
Q67 Past Paper · PPSC/FPSC/NTS easy
PID controller transfer function is
A Gc(s) = Kp + Ki/s + Kd s ✓ B Kp only without integral or derivative paths ✓ C Ki s only incorrectly ✓ D Kd/s only incorrectly ✓ Show Answer 💡 Explanation: Proportional, integral and derivative actions combine in parallel form.
Q68 Past Paper · PPSC/FPSC/NTS easy
Proportional (P) action primarily
A eliminates all steady-state error always without integrator incorrectly as blanket ✓ B reduces rise time and steady-state error partially but not entirely for type 0 plants typically ✓ C eliminates overshoot always incorrectly ✓ D provides infinite gain at DC always incorrectly—that is integral ✓ Show Answer 💡 Explanation: P gain improves response but often leaves SSE unless plant has integrator.
Q69 Past Paper · PPSC/FPSC/NTS easy
Integral (I) action eliminates
A all measurement noise always without amplification concern incorrectly ✓ B need for sensor ever incorrectly ✓ C stability margins always improves without limit incorrectly ✓ D steady-state error to step input for stable closed-loop systems when properly applied ✓ Show Answer 💡 Explanation: Integrator provides infinite gain at DC tracking constant references.
Q70 Past Paper · PPSC/FPSC/NTS medium
Anti-windup methods include
A stopping integration when output saturates or back-calculation ✓ B increasing integral gain during saturation incorrectly ✓ C removing proportional term always incorrectly ✓ D opening feedback loop permanently incorrectly ✓ Show Answer 💡 Explanation: Prevents large overshoot when leaving saturation.