Question

Which part of a PID controller is specifically responsible for eliminating the steady-state error?

• A. Proportional term
• B. Feedforward term
• C. Integral term
• D. Derivative term

Hint:

To remember PID controller actions: - P (Proportional) $\rightarrow$ Corrects the present error. - I (Integral) $\rightarrow$ Corrects past accumulated errors and eliminates steady-state error ($e_{ss} = 0$). - D (Derivative) $\rightarrow$ Anticipates future errors and improves transient stability.

Answer 1

The Correct Option is C

Concept: A Proportional-Integral-Derivative (PID) controller combines three distinct operational modes to manipulate a system's error signal $e(t) = r(t) - y(t)$. Its continuous-time control law is defined as: \[ u(t) = K_p e(t) + K_i \int_{0}^{t} e(\tau) \, d\tau + K_d \frac{de(t)}{dt} \] Each block serves a specific role in adjusting the system's response characteristics.

Step 1: Analyze the specific function of individual terms.
Let us break down the primary impact of each term on the system's behavior:
• Proportional Term ($K_p \cdot e(t)$): Generates an output proportional to the current error amplitude. While it reduces steady-state error, it cannot eliminate it completely for Type 0 systems without causing instability due to excessively high loop gains.
• Derivative Term ($K_d \cdot \frac{de}{dt}$): Senses the rate of change of the error signal to predict future behavior. This adds damping and improves transient stability, but has no effect on the steady-state error under constant DC tracking conditions.
• Integral Term ($K_i \cdot \int e\,dt$): Continuously integrates the error signal over time.

Step 2: Mechanism of error elimination by the Integral block.
In the frequency domain, the integral action adds a pole exactly at the origin ($s = 0$), which increases the overall system type by 1. The accumulated value in the integrator continues to grow as long as any non-zero error persists, driving the plant until the steady-state tracking error is reduced completely to zero ($e_{ss} = 0$).