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Derivative feedback controller: block diagram and effect

Short answer

Derivative feedback uses an inner rate loop (tachometer) to feed back Kt s C(s). The closed-loop transfer function remains second-order with ωn unchanged, but damping increases to ζ + Kt ωn/2, reducing overshoot and settling time while leaving the natural frequency unchanged.

What is derivative feedback controller? Draw block diagram and find the transfer function and hence show its effect on transient performance of the system.

Derivative (rate or tachometer) feedback control feeds back a signal proportional to the rate of change of the output, Kt dc/dtK_t\,dc/dt, through an inner (minor) loop, in addition to the usual unity feedback. In a position servo this signal comes from a tachogenerator on the motor shaft.

Block diagram

R(s)    E(s)        +-------------------+     C(s)
--->(+)---->(+)---->| wn^2/(s(s+2z wn)) |--+---->
     ^-      ^-     +-------------------+  |
     |       |                             |
     |       +-------[ Kt s ]<-------------+
     |                                     |
     +-------------------------------------+

Transfer function

Inner loop (forward G(s)=ωn2s(s+2ζωn)G(s)=\dfrac{\omega_n^2}{s(s+2\zeta\omega_n)}, feedback KtsK_ts):

G1(s)=G1+GKts=ωn2s(s+2ζωn)+Ktωn2s=ωn2s(s+2ζωn+Ktωn2)\begin{aligned} G_1(s)&=\frac{G}{1+GK_ts}=\frac{\omega_n^2}{s(s+2\zeta\omega_n)+K_t\omega_n^2s}=\frac{\omega_n^2}{s\left(s+2\zeta\omega_n+K_t\omega_n^2\right)} \end{aligned}

Closing the unity outer loop:

C(s)R(s)=ωn2s2+(2ζωn+Ktωn2)s+ωn2\frac{C(s)}{R(s)}=\frac{\omega_n^2}{s^2+\left(2\zeta\omega_n+K_t\omega_n^2\right)s+\omega_n^2}

Comparing with s2+2ζ′ωns+ωn2s^2+2\zeta'\omega_ns+\omega_n^2:

ζ′=ζ+Ktωn2,ωn′=ωn\zeta'=\zeta+\frac{K_t\omega_n}{2},\qquad \omega_n'=\omega_n

Effect on transient performance

  • Damping increases (ζ′>ζ\zeta'>\zeta) while ωn\omega_n is unchanged.
  • Peak overshoot reduces, since Mp=e−πζ′/1−ζ′2M_p=e^{-\pi\zeta'/\sqrt{1-\zeta'^2}} falls as ζ′\zeta' rises.
  • Settling time reduces: ts=4/(ζ′ωn)t_s=4/(\zeta'\omega_n).
  • Rise time increases slightly (slower initial rise).
  • Unlike derivative error (PD) control, it adds no zero to the closed loop, so it does not cause the extra overshoot a zero can produce.

Drawback: steady-state error

Kv=lim⁡s→0sG1(s)=ωn2ζ+Ktωn,ess(ramp)=2ζ+KtωnωnK_v=\lim_{s\to0}sG_1(s)=\frac{\omega_n}{2\zeta+K_t\omega_n},\qquad e_{ss}(\text{ramp})=\frac{2\zeta+K_t\omega_n}{\omega_n}

So the ramp error increases; this is usually offset by raising the forward gain.

Example: ωn=4\omega_n=4 rad/s, ζ=0.25\zeta=0.25 (Mp=44%M_p=44\%). With Kt=0.15K_t=0.15: ζ′=0.25+0.15×4/2=0.55\zeta'=0.25+0.15\times4/2=0.55, and MpM_p falls to about 12.6%.

✓ Answer checked by Claude Opus 5.5 (high effort)

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