--- title: "Application of PI controller with example" subject: "Control Systems" chapter: 7 key: "semester_3/control_systems:pi-application-example" source: https://www.bctengineer.com/q/semester_3/control_systems/pi-application-example site: BCT Engineering --- # Application of PI controller with example Control Systems · Chapter 7: Performance Specifications and Compensation Design - Asked in: 2070 Chaitra, 4 marks - Video: [Proportional Integral PI Controller](https://www.youtube.com/watch?v=lVQxEwxVqgk) — TutorialsPoint - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/pi-application-example - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7 ## Question Discuss the application of a PI controller with suitable example. ## Short answer PI controller Gc(s)=Kp+Ki/s adds a pole at the origin and a zero, raising system type by one to eliminate steady-state error. Example: DC motor G(s)=10/(s+2) with PI 4+8/s yields type-1, zero offset, and fast first-order response time constant 0.025 s. ## Answer A **PI controller** produces a control signal proportional to both the error and the integral of the error: $$ u(t) = K_p e(t) + K_i \int_0^t e(t)\,dt, \qquad G_c(s) = K_p + \frac{K_i}{s} = \frac{K_p\left(s + K_i/K_p\right)}{s} $$ It adds a pole at the origin and a zero at $s = -K_i/K_p$. #### Why it is applied - The pole at the origin **raises the system type by one**. A type-0 plant gets zero steady-state error to a step; a type-1 plant gets zero error to a ramp. - It removes the steady-state offset that a pure P controller always leaves. - The zero, placed close to the origin, limits the extra phase lag, so stability is not badly affected. - It acts like a lag compensator (low-pass), so it also filters high-frequency noise. - Drawbacks: slightly slower response, possibly more overshoot, and integral wind-up when the actuator saturates. #### Example: speed control of a DC motor Let the motor (speed output) be $G(s) = \dfrac{10}{s+2}$ with unity feedback (type 0). - With **P control** $K_p = 4$: $K_P = \lim_{s\to0} \dfrac{40}{s+2} = 20$, so the step error $e_{ss} = \dfrac{1}{1+20} = 0.048$, i.e. a 4.8% speed offset that grows when load torque is applied. - With **PI control** $G_c = 4 + \dfrac{8}{s}$: $$ G_c G = \frac{4(s+2)}{s}\cdot\frac{10}{s+2} = \frac{40}{s} $$ The system becomes type 1, $K_P = \infty$ and $e_{ss} = 0$. The closed loop is $\dfrac{40}{s+40}$, a fast first-order response with time constant 0.025 s and no offset, even under constant load disturbances. Other common uses: liquid-level and flow control, temperature control of furnaces, pressure control in process plants and voltage regulators, wherever zero steady-state offset is essential. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does a PI controller Gc(s) = Kp + Ki/s add to the open-loop transfer function? A. Two zeros and no pole B. A pole at the origin and a zero at s = -Ki/Kp C. A zero at the origin and a pole at -Ki/Kp D. Only a pole at s = -Kp/Ki Answer: B. Gc = Kp(s + Ki/Kp)/s has a pole at s = 0 and a zero at -Ki/Kp. (Key point: PI structure) 2. Why does a PI controller remove steady-state offset that a P controller leaves? A. It reduces the plant gain at high frequency B. It adds damping through derivative action C. The pole at the origin raises the system type by one D. It increases the bandwidth Answer: C. Raising the type makes a type-0 plant give zero step error and a type-1 plant zero ramp error. (Key point: Why PI removes offset) 3. For G(s) = 10/(s+2) with Kp = 4, the P-control step error is 1/(1+20) = 0.048. With PI control Gc = 4 + 8/s, what happens? A. Gc*G = 80/s, ess = 0.5 B. Gc*G = 40/s, the system becomes type 1 and ess = 0 C. Gc*G = 40/(s+2), ess stays 0.048 D. Gc*G = 40/s but ess = 1/41 Answer: B. The PI zero at -2 cancels the plant pole, giving 40/s; the integrator makes the step error zero. (Key point: DC motor example) 4. Which is a known drawback of PI control mentioned in the answer? A. It always makes the system unstable B. It amplifies high-frequency noise C. Integral wind-up when the actuator saturates D. It permanently leaves a steady-state offset Answer: C. The integral keeps accumulating during saturation, causing wind-up; PI also slows response and may raise overshoot, but it filters noise. 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