Derivative feedback for ramp input

Derivative (rate) feedback feeds back $K_t\,\dfrac{dc}{dt}$ in addition to the output, for example with a tachogenerator on a position servo. It increases damping but, for a ramp input, it increases the steady-state error.

Take the standard second-order plant with derivative feedback in a minor loop:

$$G(s) = \frac{\omega_n^2}{s(s + 2\zeta\omega_n)}, \qquad H(s) = 1 + K_t s$$

Closed-loop transfer function:

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

Effect on damping

Comparing with $s^2 + 2\zeta'\omega_n s + \omega_n^2$:

$$\zeta' = \zeta + \frac{K_t\omega_n}{2}$$

The natural frequency is unchanged, but damping rises, so overshoot and oscillation fall and the system becomes more stable.

Effect on the unit ramp response

Error $E(s) = R(s) - C(s)$ with $R(s) = 1/s^2$:

$$\begin{aligned} E(s) &= \frac{s^2 + (2\zeta\omega_n + K_t\omega_n^2)s}{s^2 + (2\zeta\omega_n + K_t\omega_n^2)s + \omega_n^2}\cdot\frac{1}{s^2} \\ e_{ss} &= \lim_{s\to0} sE(s) = \frac{2\zeta\omega_n + K_t\omega_n^2}{\omega_n^2} = \frac{2\zeta}{\omega_n} + K_t \end{aligned}$$

Without rate feedback $e_{ss} = 2\zeta/\omega_n$. Rate feedback adds $K_t$ to the ramp error.

Example: $\omega_n = 4$, $\zeta = 0.25$, $K_t = 0.15$: $\zeta' = 0.25 + 0.3 = 0.55$ (overshoot drops from 44% to 12.6%) but $e_{ss}$ rises from 0.125 to 0.275.

Remedy: increase the forward gain (amplifier) to bring the ramp error back down; the extra damping from $K_t$ keeps the response well damped. In contrast, a PD controller in the forward path adds similar damping while leaving $K_v$, and so the ramp error $2\zeta/\omega_n$, unchanged. Derivative feedback therefore trades ramp accuracy for damping.