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Measures to track the reference input

Short answer

Use a closed-loop system with integral action (PI or PID controller) to raise the system type. An integrator makes the position error constant infinite for step inputs, eliminating the steady-state error permanently because the integral term accumulates any error and forces the output to match the reference.

There exists a permanent mismatch to track the reference by the actual system. Suggest a way out to follow your system as per reference and justify.

A permanent mismatch between the reference and the output is a steady-state error esse_{ss}. It can be removed by using a closed-loop system with integral action (PI or PID controller), i.e. by raising the type number of the loop.

Justification

For unity feedback, E(s)=R(s)1+G(s)E(s) = \dfrac{R(s)}{1+G(s)} and

ess=lim⁡s→0sE(s)e_{ss} = \lim_{s\to0} sE(s)

For a unit step, ess=11+Kpe_{ss} = \dfrac{1}{1+K_p}, where the position error constant is Kp=lim⁡s→0G(s)K_p = \lim_{s\to0}G(s).

  • Type-0 system, e.g. G(s)=Kτs+1G(s) = \dfrac{K}{\tau s+1}: Kp=KK_p = K, so ess=11+K≠0e_{ss} = \dfrac{1}{1+K} \ne 0. The mismatch is permanent.
  • Raising KK reduces the error (e.g. K=99K = 99 gives 1%) but never makes it zero, and high gain can make higher-order systems oscillatory or unstable.
  • Add an integrator (PI controller Gc(s)=Kc+Ki/sG_c(s) = K_c + K_i/s): the loop becomes type 1 and the position constant Kp→∞K_p \to \infty, so
ess=11+∞=0e_{ss} = \frac{1}{1+\infty} = 0

The integrator keeps changing its output as long as any error exists, so the output is forced to follow the reference exactly.

Other measures

  • If the system is open loop, convert it to closed loop with a sensor so that the error is measured and corrected.
  • For ramp references, use a type-2 loop (double integration) or a lag compensator to raise KvK_v.
  • Calibrate the sensor and feedback element: an error in HH appears directly in the output, since C/R≈1/HC/R \approx 1/H.

✓ Answer checked by Claude Opus 5.5 (high effort)

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