--- title: "P, I, D controllers and role of PI, PD" subject: "Control Systems" chapter: 7 key: "semester_3/control_systems:pid-pi-pd-role" source: https://www.bctengineer.com/q/semester_3/control_systems/pid-pi-pd-role site: BCT Engineering --- # P, I, D controllers and role of PI, PD Control Systems · Chapter 7: Performance Specifications and Compensation Design - Asked in: 2066 Bhadra (old course), 8 marks - Video: [P, PI, PD, and PID Controllers: A Comparative Analysis in Control System Engineering](https://www.youtube.com/watch?v=OYPXiC7QjUM) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/pid-pi-pd-role - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7 ## Question Mention P, I and D controllers. Also explain the role of PI and PD controllers on transient and steady state performance specification. ## Short answer P controller provides output proportional to error. I eliminates steady-state error but slows response. D adds damping. PI increases system type, making steady-state error zero but increasing overshoot. PD increases damping, reduces overshoot and settling time, improving transient performance. ## Answer #### P, I and D controllers - **Proportional (P):** $u(t) = K_p e(t)$, $G_c(s) = K_p$. Output is proportional to the present error. Raising $K_p$ speeds up the response and reduces (but does not remove) steady-state error; too much gain gives overshoot and instability. - **Integral (I):** $u(t) = K_i \int e\,dt$, $G_c(s) = K_i/s$. Output depends on the accumulated past error. It adds a pole at the origin and removes steady-state offset, but adds 90° of phase lag and slows the response. - **Derivative (D):** $u(t) = K_d\, de/dt$, $G_c(s) = K_d s$. Output depends on the rate of change of error (future trend). It adds damping, but is never used alone because it gives no output for a constant error and amplifies noise. #### Role of the PI controller $G_c(s) = K_p + \dfrac{K_i}{s}$, a pole at the origin and a zero at $-K_i/K_p$. | Specification | Effect of PI | |---|---| | System type | Increased by one | | Steady-state error | Step error becomes zero (type-0 plant); ramp error falls | | Rise time | Slightly reduced | | Overshoot | Increases | | Settling time | Increases (slower) | | Stability | Relative stability decreases | | Bandwidth | Decreases, so noise is filtered | #### Role of the PD controller $G_c(s) = K_p + K_d s$, a zero at $-K_p/K_d$. For $G = \dfrac{\omega_n^2}{s(s+2\zeta\omega_n)}$ with PD, the characteristic equation becomes $s^2 + (2\zeta\omega_n + K_d\omega_n^2)s + K_p\omega_n^2 = 0$, so the effective damping ratio increases. | Specification | Effect of PD | |---|---| | Damping | Increased | | Peak overshoot | Reduced | | Rise time and settling time | Reduced (faster) | | Stability | Improved | | Steady-state error | Type unchanged; error depends only on $K_p$ | | Bandwidth | Increased; noise is amplified | So PI is chosen to improve **steady-state** performance and PD to improve **transient** performance. A PID controller combines both. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the I controller do? A. Removes steady-state offset, but adds 90 degrees lag and slows the response B. Adds damping by reacting to the rate of change of error C. Increases bandwidth and amplifies noise D. Gives output proportional to present error only Answer: A. Ki/s accumulates past error, adding a pole at the origin; it eliminates offset at the cost of phase lag. (Key point: Integral action) 2. Why is D never used alone? A. It always destabilizes the plant B. It gives no output for a constant error and amplifies noise C. It makes the system type zero D. It adds a pole at the origin Answer: B. Kd*de/dt is zero for constant error and boosts noise, so it only works together with P or PI. (Key point: Derivative alone) 3. Which describes the effect of a PI controller? A. Leaves type unchanged B. Improves steady-state accuracy but increases overshoot and settling time C. Improves transient speed and reduces overshoot D. Raises bandwidth and reduces rise time greatly Answer: B. PI raises the type, removing step error, but relative stability falls and the response slows. (Key point: PI effect) 4. With PD control on G = wn^2/[s(s+2*zeta*wn)], what changes? A. Natural frequency drops to zero B. The system type rises C. Damping rises because the s-coefficient grows by Kd*wn^2 D. Steady-state error vanishes Answer: C. The characteristic equation has s coefficient 2*zeta*wn + Kd*wn^2, so effective damping increases and overshoot falls. 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