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PD controller principle and working

Short answer

A PD controller adds a derivative term Kd de/dt to the proportional action. It provides anticipatory phase lead, increasing damping ratio, reducing overshoot and settling time, and improving stability, while steady-state error remains unchanged. Noise amplification is a drawback.

Discuss principle and working performance of PD controller.

A PD (proportional–derivative) controller produces a control signal proportional to the error plus the rate of change of the error:

u(t)=Kpe(t)+Kdde(t)dt,Gc(s)=Kp+Kds=Kp(1+Tds), Td=KdKpu(t)=K_pe(t)+K_d\frac{de(t)}{dt},\qquad G_c(s)=K_p+K_ds=K_p(1+T_ds),\ T_d=\frac{K_d}{K_p}
          +-->[ Kp ]-----+
 e(t) ----|              (+)---> u(t)
          +-->[ Kd d/dt ]+

Principle

  • The P part acts on the present error.
  • The D part acts on the trend of the error. When the error is falling fast (output rushing toward the set point), de/dtde/dt is negative and the D term reduces the drive before the output overshoots. So D gives anticipatory action and extra damping.
  • It adds a zero at s=−Kp/Kds=-K_p/K_d to the open-loop transfer function, i.e. phase lead.

Working performance (second-order system)

For G(s)=ωn2s(s+2ζωn)G(s)=\dfrac{\omega_n^2}{s(s+2\zeta\omega_n)} with PD control (taking Kp=1K_p=1):

C(s)R(s)=ωn2(1+Tds)s2+(2ζωn+ωn2Td)s+ωn2\frac{C(s)}{R(s)}=\frac{\omega_n^2(1+T_ds)}{s^2+(2\zeta\omega_n+\omega_n^2T_d)s+\omega_n^2} ζ′=ζ+ωnTd2\zeta'=\zeta+\frac{\omega_nT_d}{2}
PropertyEffect of PD
Damping ratioincreases
Peak overshootdecreases
Rise time, settling timedecrease (faster)
Type of systemunchanged
Steady-state errorunchanged (for given KpK_p)
Stability marginimproves
Noiseamplified (drawback)

Example: with ωn=2\omega_n=2 rad/s and ζ=0.25\zeta=0.25, choosing Td=0.3T_d=0.3 s gives ζ′=0.25+0.3=0.55\zeta'=0.25+0.3=0.55, cutting the step overshoot from about 44% to about 16% (the added zero keeps it a little above the 12.6% of a plain ζ = 0.55 system).

✓ Answer checked by Claude Opus 5.5 (high effort)

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