Velocity feedback increases ζ and steady-state error

Velocity (derivative output) feedback adds a term proportional to $\dot{c}$ to the feedback signal. This increases the coefficient of $s$ in the characteristic equation.

Without velocity feedback ($k_d = 0$)

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

Damping ratio $= \zeta$. Velocity error constant and ramp error:

$$K_v = \lim_{s\to 0}sG(s) = \frac{\omega_n}{2\zeta}, \qquad e_{ss} = \frac{1}{K_v} = \frac{2\zeta}{\omega_n}$$

With velocity feedback

The minor loop has forward gain $\dfrac{\omega_n^2}{s(s+2\zeta\omega_n)}$ and feedback $sk_d$:

$$G'(s) = \frac{\frac{\omega_n^2}{s(s+2\zeta\omega_n)}}{1 + \frac{\omega_n^2 sk_d}{s(s+2\zeta\omega_n)}} = \frac{\omega_n^2}{s(s + 2\zeta\omega_n + k_d\omega_n^2)}$$

Closed loop (unity outer feedback):

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

New damping ratio

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

$$2\zeta'\omega_n = 2\zeta\omega_n + k_d\omega_n^2 \;\Rightarrow\; \zeta' = \zeta + \frac{k_d\omega_n}{2}$$

Since $k_d > 0$, $\zeta' > \zeta$. Damping increases, so overshoot falls.

New steady state error (unit ramp)

$$K_v' = \lim_{s\to 0}sG'(s) = \frac{\omega_n^2}{2\zeta\omega_n + k_d\omega_n^2} = \frac{\omega_n}{2\zeta + k_d\omega_n}$$ $$e_{ss}' = \frac{1}{K_v'} = \frac{2\zeta + k_d\omega_n}{\omega_n} = \frac{2\zeta}{\omega_n} + k_d$$

Since $k_d > 0$, $e_{ss}' > e_{ss}$. The steady state error to a ramp increases by $k_d$. (The step error stays zero, as the system is still Type 1.)

So both the damping ratio and the steady state error increase. This error can be reduced again by raising the forward gain.