Derivative feedback in second order time response

In derivative (rate/tachometer) feedback control, a signal proportional to the rate of change of the output, $K_t\,\dot c(t)$, is fed back negatively in an inner loop, besides the main unity feedback. In a motor position system, a tachogenerator on the shaft provides this velocity signal.

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

Analysis

Inner loop:

$$G_1(s)=\frac{\omega_n^2}{s(s+2\zeta\omega_n+K_t\omega_n^2)}$$

Closed loop:

$$\frac{C(s)}{R(s)}=\frac{\omega_n^2}{s^2+(2\zeta\omega_n+K_t\omega_n^2)s+\omega_n^2}$$ $$\zeta_{eff}=\zeta+\frac{K_t\omega_n}{2},\qquad\omega_n\ \text{unchanged}$$

Working / effect on time response

1. When the output moves fast towards the reference, the feedback signal $K_t\dot c$ is large and subtracts from the actuating signal, braking the system early. This is extra damping.
2. Overshoot falls as $\zeta_{eff}$ rises; $t_s=4/(\zeta_{eff}\omega_n)$ falls.
3. $\omega_n$ is unchanged and no zero is added, so the response is smooth (unlike PD, whose zero can add overshoot).
4. Ramp error increases: $K_v=\dfrac{\omega_n}{2\zeta+K_t\omega_n}$; this is offset by raising the amplifier gain.

Example: $\omega_n=5$ rad/s, $\zeta=0.2$ ($M_p=52.7\%$). With $K_t=0.12$: $\zeta_{eff}=0.2+0.3=0.5$, $M_p=16.3\%$.