PID control — proportional, integral, derivative terms: Question bank

3.5.38 · D5 · Physics › Guidance, Navigation & Control (GNC) › PID control — proportional, integral, derivative terms

True or false — justify

A larger

K_{p}

always makes the closed loop respond faster and is therefore always better.

False. Ye initial response ko speed up zaroor karta hai, lekin steady-state error ko sirf shrink karta hai — khatam nahi karta — aur ek point ke baad loop ko oscillation aur instability mein dhakel deta hai (dekho Stability & the Routh–Hurwitz criterion).

A pure proportional controller can achieve exactly zero steady-state error if you make

K_{p}

large enough.

False. Rest pe

u = K_{p} e_{ss}

ko disturbance

d

ke barabar hona chahiye, toh

e_{ss} = d / K_{p}

; yeh sirf tab zero jaata hai jab

K_{p} \to \infty

, jo physically impossible hai aur usse bahut pehle hi unstable ho jaata hai.

The integral term is what allows the loop to settle with zero steady-state error against a constant disturbance.

True. Steady state mein

u_{I}

ka change rukna zaroori hai, aur

\overset{u}{˙}_{I} = K_{i} e

, toh resting state sirf

e = 0

par hi possible hai — integrator tab tak push karta rehta hai jab tak error khatam na ho jaaye.

The derivative term reacts to how big the error is right now.

False. Yeh react karta hai is baat par ki error kitni tez badal rahi hai (

\overset{e}{˙}

), na ki uski magnitude par — ek bahut bada lekin steady error zero derivative action produce karta hai.

Adding a derivative term always improves stability.

Ideal maths mein mostly true, lekin practice mein false. Yeh damping add karta hai, phir bhi yeh high-frequency sensor noise ko amplify karta hai, toh ek unfiltered/bada

K_{d}

actuator ko chatter karwa sakta hai aur effectively real system ko destabilise kar sakta hai (dekho Sensor noise & filtering).

If the error signal is identically zero for all time, the integral term is also zero.

Generally false. Integral accumulated history hold karta hai; agar past error nonzero tha, toh

u_{I}

bada aur constant ho sakta hai even jab present error zero ho — woh stored value exactly wahi hai jo ek disturbance ko cancel karti hai.

Doubling

K_{i}

doubles how fast the accumulated offset is erased and carries no downside.

False. Zyada integral action offset ko jaldi mita deta hai lekin overshoot badhata hai aur windup ko aur bura kar deta hai jab actuator saturate ho jaata hai; stability/overshoot ka ek cost hota hai.

The discrete PID running on the flight computer computes the same control as the continuous law.

False, yeh sirf approximate karta hai. Integral ek rectangle sum ban jaata hai aur derivative ek backward finite difference; ye calculus versions ke kareeb aate hain sirf tabhi jab

Δ t \to 0

ho.

Spot the error

"At steady state with pure P,

e_{ss} = K_{p} d

, so a bigger gain gives a bigger leftover error."

Galat direction. Sahi relation hai

e_{ss} = d / K_{p}

: bada

K_{p}

error ko shrink karta hai. Student ne fraction ulta kar diya.

"The integral term removes offset because integrating a constant error gives zero."

Galat mechanism. Ek constant error ko integrate karne par growing ramp milti hai, zero nahi. Woh ramp

u

ko badhati rehti hai jab tak error khud zero nahi ho jaata — offset isliye khatam hota hai kyunki output sirf tab ruk sakti hai jab

e = 0

ho.

"Derivative action helps because

\overset{e}{˙}

tells us the current position error."

Value aur rate ko confuse kar raha hai.

\overset{e}{˙}

error ki slope hai, yaani uski trend. Yeh kehta hai "error tezi se gir raha hai, abhi brake karo," jo prediction hai — yeh present magnitude ke baare mein kuch nahi kehta.

"To fix integrator windup, just increase

K_{p}

so the actuator saturates less."

Symptom ko treat kar raha hai, cause ko nahi. Windup tab hota hai jab integrator accumulate karta rehta hai jabki actuator apni limit par clamped hota hai; ilaaj anti-windup hai (saturation ke dauran integration roko/clamp karo), alag

K_{p}

nahi.

"A noisy sensor is a problem for the integral term because summing noise blows up."

Galat term. Zero-mean noise ek sum mein average out ho jaata hai, toh integral relatively robust hota hai. Derivative noise ko amplify karta hai, kyunki differencing fast wiggles ko multiply karta hai (dekho Sensor noise & filtering).

"Since D 'sees the future,' a system with only a D term can hold a setpoint."

Impossible. Pure D zero output produce karta hai jab bhi error constant ho (constant nonzero error bhi include karke), toh yeh ek target hold karne ke liye zaroori steady push supply nahi kar sakta — yeh sirf transients ko shape karta hai.

Why questions

Why must a pure-P controller output drop to zero exactly when the error is zero?

Kyunki

u_{P} = K_{p} e

directly present error se tied hai; agar

e = 0

toh

u_{P} = 0

, kuch bhi disturbance ko oppose karne ke liye nahi bachta — precisely isliye ek offset persist karna hi padta hai.

Why does the integrator "wind up" specifically during actuator saturation and not during normal operation?

Jab saturated hota hai, actuator commanded output deliver nahi kar sakta, toh error bada rehta hai aur integral ek aisi value accumulate karta rehta hai jis par plant kabhi actually act nahi karta — woh stored surplus phir buri tarah overshoot karta hai jab loop catch up karta hai.

Why is differentiation, rather than integration, the operation that amplifies noise?

Differentiation slope par respond karta hai; ek chhota fast wiggle tiny amplitude rakhta hai lekin bahut badi slope, toh uska derivative bada hota hai — high frequencies multiply ho jaati hain, jabki integration unhe smooth karta hai.

Why do we describe P as "present," I as "past," and D as "future"?

P error ki value abhi use karta hai, I uski poori history accumulate karta hai, aur D uski rate of change use karke extrapolate karta hai ki yeh kahan ja raha hai — value, integral, derivative exactly present/past/future information hain.

Why can raising

K_{p}

turn a stable loop unstable?

Ek bada gain plant ke apne lags aur delays ke through ek badi correction feed back karta hai; ek threshold ke baad correction out of phase aakar error ko cancel karne ki jagah reinforce karta hai — woh boundary wahi hai jo Stability & the Routh–Hurwitz criterion pin down karta hai.

Why does the derivative term reduce overshoot on the approach to target?

Jaise

y

target

r

ki taraf daudta hai, error girta hai toh

\overset{e}{˙} < 0

, jisse

u_{D} < 0

milta hai; yeh command se pehle subtract karta hai, system ke setpoint se aage nikal jaane se pehle approach ko brake karta hai.

Edge cases

What does a PID controller command at the instant the system first reaches the setpoint after approaching fast?

e = 0

toh P kuch contribute nahi karta, lekin

\overset{e}{˙} \neq = 0

(abhi bhi move kar raha hai) aur integral nonzero value hold kar sakta hai — toh

u

generally zero nahi hota; D aur I overshoot rokne aur disturbance cancel karne ke liye abhi bhi act karte hain.

If the disturbance

d

is zero, does the integral term ever do anything at all?

Sirf transients ke dauran. Jab error zero ho jaaye aur wahan ruke,

u_{I}

change karna band kar deta hai; koi disturbance nahi toh loop

u_{I} = 0

ke saath rest kar sakta hai, toh integral action sirf transient offset clean karne ke liye zaroori tha.

What happens to a pure-integral controller (

K_{p} = K_{d} = 0

) responding to a step?

Yeh sluggishly respond karta hai aur oscillate karne lagta hai — koi present-error ya damping term nahi hone se, accumulator overshoot karta hai aur dheere unwind karta hai, poor, ringing performance deta hai despite eventual zero offset ke.

What is the effect of the derivative term when the error is constant but nonzero (a stuck offset)?

Zero —

\overset{e}{˙} = 0

ka matlab hai

u_{D} = 0

, toh D standing offsets ke liye blind hai; sirf I unhe eliminate kar sakta hai. Isliye ek constant leftover error ek symptom hai jo integral ki taraf point karta hai, derivative ki taraf nahi.

t = 0

with a sudden step in setpoint, what does the ideal derivative term momentarily command?

Ek impulse —

e

mein ek step ek instant ke liye infinite slope rakhta hai, toh ek unfiltered derivative enormously spike karta hai. Yeh "derivative kick" isliye real controllers measurement ko differentiate karte hain ya derivative ko filter karte hain (dekho Sensor noise & filtering).

K_{i}

is set to exactly zero, what class of disturbance can the loop still reject with zero steady-state error?

Koi bhi jo constant hain — integral action ke bina ek constant disturbance hamesha

e_{ss} = d / K_{p} \neq = 0

chhod jaata hai; sirf woh disturbance jo khud zero tak decay ho jaaye fully reject ho sakti hai (dekho Steady-state error and system type).

Recall Har trap ki ek-line summary

Har PID term ka exactly ek kaam hai aur exactly ek failure mode: P fast hai lekin offset chhod jaata hai, I offset khatam karta hai lekin wind up ho jaata hai, D damp karta hai lekin noise amplify karta hai. Upar diye almost har trap mein ya toh galat term ko symptom ke liye blame kiya ja raha hai, ya yeh bhool jaate hain ki D rate padhta hai jabki P value padhta hai aur I history padhta hai.