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Working of PI controller

Short answer

A PI controller combines proportional action with integral action. The integral term accumulates error over time, eliminating steady-state error. It adds a pole at the origin and a zero; this may increase overshoot and reduce relative stability due to phase lag.

Discuss working of PI controller.

A PI controller combines proportional action with integral action. Its output is

u(t)=Kpe(t)+Ki∫0te(τ) dτ=Kp[e(t)+1Ti∫0te dτ]u(t)=K_pe(t)+K_i\int_0^te(\tau)\,d\tau=K_p\left[e(t)+\frac{1}{T_i}\int_0^te\,d\tau\right] Gc(s)=Kp+Kis=Kp(s+1Ti)s,Ti=KpKiG_c(s)=K_p+\frac{K_i}{s}=\frac{K_p\left(s+\frac{1}{T_i}\right)}{s},\quad T_i=\frac{K_p}{K_i}
          +--->[  Kp  ]---+
 e(t) ----|               (+)---> u(t)
          +--->[ Ki/s ]---+

Working

  1. P part reacts at once to the present error, giving fast correction.
  2. I part adds up (integrates) the error over time. As long as any error remains, the integral keeps growing and keeps changing the output.
  3. Therefore the system can settle only when the error is exactly zero. A constant controller output is then held by the integrator alone; this removes the offset that a pure P controller leaves.

For a step error EE, the output is u(t)=KpE+KiEtu(t)=K_pE+K_iEt: an instant jump followed by a ramp.

Effects

  • Adds a pole at the origin (type +1) and a zero at −1/Ti-1/T_i.
  • Steady-state error is eliminated for a step (type 0 plant) or ramp (type 1 plant).
  • Adds phase lag, so overshoot increases and relative stability decreases; the zero is placed near the origin to keep this small.
  • Acts like a low-pass filter (similar to a lag compensator); noise is not amplified.
  • Problem: integral wind-up when the actuator saturates.

Example: in a water-level control with P only, the level settles 2 cm below the set point; adding integral action slowly raises the valve opening until the level reaches the set point exactly.

✓ Answer checked by Claude Opus 5.5 (high effort)

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