--- title: "Derivative feedback controller: block diagram and effect" subject: "Control Systems" chapter: 7 key: "semester_3/control_systems:derivative-feedback-bd-effect" source: https://www.bctengineer.com/q/semester_3/control_systems/derivative-feedback-bd-effect site: BCT Engineering --- # Derivative feedback controller: block diagram and effect Control Systems · Chapter 7: Performance Specifications and Compensation Design - Asked in: 2080 Baisakh, 4 marks; 2079 Bhadra, 6 marks (asked 2 times) - Video: [Derivative feedback controller ,PI,PID controller](https://www.youtube.com/watch?v=d1LaXuj4wLY) — Yashwin Creations - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/derivative-feedback-bd-effect - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-7 ## Question What is derivative feedback controller? Draw block diagram and find the transfer function and hence show its effect on transient performance of the system. ## 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. ## Answer **Derivative (rate or tachometer) feedback control** feeds back a signal proportional to the **rate of change of the output**, $K_t\,dc/dt$, 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 ```text R(s) E(s) +-------------------+ C(s) --->(+)---->(+)---->| wn^2/(s(s+2z wn)) |--+----> ^- ^- +-------------------+ | | | | | +-------[ Kt s ]<-------------+ | | +-------------------------------------+ ``` #### Transfer function Inner loop (forward $G(s)=\dfrac{\omega_n^2}{s(s+2\zeta\omega_n)}$, feedback $K_ts$): $$ \begin{aligned} G_1(s)&=\frac{G}{1+GK_ts}=\frac{\omega_n^2}{s(s+2\zeta\omega_n)+K_t\omega_n^2s}=\frac{\omega_n^2}{s\left(s+2\zeta\omega_n+K_t\omega_n^2\right)} \end{aligned} $$ Closing the unity outer loop: $$ \frac{C(s)}{R(s)}=\frac{\omega_n^2}{s^2+\left(2\zeta\omega_n+K_t\omega_n^2\right)s+\omega_n^2} $$ Comparing with $s^2+2\zeta'\omega_ns+\omega_n^2$: $$ \zeta'=\zeta+\frac{K_t\omega_n}{2},\qquad \omega_n'=\omega_n $$ #### Effect on transient performance - **Damping increases** ($\zeta'>\zeta$) while $\omega_n$ is unchanged. - **Peak overshoot reduces**, since $M_p=e^{-\pi\zeta'/\sqrt{1-\zeta'^2}}$ falls as $\zeta'$ rises. - **Settling time reduces**: $t_s=4/(\zeta'\omega_n)$. - 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 $$ K_v=\lim_{s\to0}sG_1(s)=\frac{\omega_n}{2\zeta+K_t\omega_n},\qquad e_{ss}(\text{ramp})=\frac{2\zeta+K_t\omega_n}{\omega_n} $$ So the ramp error increases; this is usually offset by raising the forward gain. **Example:** $\omega_n=4$ rad/s, $\zeta=0.25$ ($M_p=44\%$). With $K_t=0.15$: $\zeta'=0.25+0.15\times4/2=0.55$, and $M_p$ falls to about 12.6%. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What signal does derivative (tachometer) feedback feed back? A. A signal proportional to the integral of the error B. A signal proportional to the rate of change of the output, Kt dc/dt C. A signal proportional to the rate of change of the error D. A signal proportional to the reference Answer: B. Rate feedback takes Kt dc/dt from the output (e.g. a tachogenerator) into an inner loop. (Key point: Definition) 2. For G = wn^2/[s(s+2 zeta wn)] with feedback Kt s in the inner loop, the closed-loop denominator becomes A. s^2 + 2 zeta wn s + wn^2 (unchanged) B. s^2 + (2 zeta wn + Kt wn^2)s + wn^2 C. s^2 + 2 zeta wn s + Kt wn^2 D. s^2 + (2 zeta wn + Kt)s + wn^2 Answer: B. The inner loop gives wn^2/[s(s + 2 zeta wn + Kt wn^2)], and unity outer feedback yields this denominator. (Key point: Closed-loop transfer function) 3. What happens to damping and natural frequency with derivative feedback? A. zeta' = zeta + Kt wn/2, wn unchanged B. zeta unchanged, wn increases C. both zeta and wn increase D. zeta' = zeta - Kt wn/2 Answer: A. Comparing coefficients gives zeta' = zeta + Kt wn/2 with wn' = wn, so overshoot and settling time fall. (Key point: Effect on damping) 4. What is the drawback of derivative feedback? A. It adds a closed-loop zero causing extra overshoot B. It reduces Kv, so the ramp steady-state error increases C. It reduces damping D. It makes the system type zero Answer: B. Kv = wn/(2 zeta + Kt wn) falls, so the ramp error (2 zeta + Kt wn)/wn rises; it adds no zero. 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