Second-order identification and rate-feedback design: 22-Elec-A2 Systems and Control, May 2013, Q6

Question 6 — Second-order identification and rate-feedback design [10 + 10]

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part A — Model from the step response [10]

Given (read from Figure Q6.1). Unit-step response of $G(s)$: steady-state $y_{ss}\approx0.86$; first peak $\approx1.15$ at $t\approx1\ \text{s}$; successive peaks about $2\ \text{s}$ apart (period $\approx2\ \text{s}$).

DC gain. $K_{dc}=y_{ss}=\boxed{0.86}$.
Damping ratio from overshoot. $\text{PO}=\dfrac{1.15-0.86}{0.86}=0.337$, so $\zeta=\dfrac{-\ln0.337}{\sqrt{\pi^2+\ln^2 0.337}}=\boxed{0.327}$.
Natural frequency from the oscillation period. Period $\approx2\ \text{s}\Rightarrow\omega_d=2\pi/2=\pi\ \text{rad/s}$; $\omega_n=\dfrac{\omega_d}{\sqrt{1-\zeta^2}}=\dfrac{\pi}{\sqrt{1-0.327^2}}=\boxed{3.32\ \text{rad/s}}$.
Model transfer function. $$G_m(s)=\frac{K_{dc}\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}=\frac{9.50}{s^2+2.17s+11.04}.$$ (Check $G_m(0)=9.50/11.04=0.86$.)

Part B — Proportional + rate-feedback design [10]

Figure Q6.2 — Proportional gain $K_p$ in the forward path, rate feedback $H(s)=K_d s+1$ in the feedback path.

Approach. The rate feedback adds damping without changing the DC loop gain (since $H(0)=1$). Use the DC gain to meet the error spec and the closed-loop characteristic polynomial to meet the overshoot spec.

Gain for the error spec. With $H(0)=1$, the loop is type 0 with $K_{pos}=K_pK_{dc}$. $e_{ss}=5\%\Rightarrow K_{pos}=19\Rightarrow K_p=\dfrac{19}{0.86}=\boxed{22.1}$.
Closed-loop characteristic polynomial. $1+K_pG(s)(K_d s+1)=0$ gives $$s^2+\big(2\zeta\omega_n+K_pK_{dc}\omega_n^2K_d\big)s+\omega_n^2\big(1+K_pK_{dc}\big)=0.$$
New natural frequency. $\omega_n'=\omega_n\sqrt{1+K_pK_{dc}}=3.32\sqrt{20}=14.87\ \text{rad/s}$.
Rate gain for the overshoot spec. $\text{PO}=10\%\Rightarrow\zeta'=0.591$, so $2\zeta'\omega_n'=2\zeta\omega_n+K_pK_{dc}\omega_n^2K_d$: $$K_d=\frac{2(0.591)(14.87)-2(0.327)(3.32)}{19(3.32)^2}=\boxed{0.073}.$$
Stability & specs. Both coefficients of the quadratic are positive, so the system is stable (YES). Compensated: $\zeta'=0.591$, $\omega_n'=14.87\Rightarrow\text{PO}=10\%$, $T_{settle}=4/(\zeta'\omega_n')=0.455\ \text{s}$, $e_{ss}=5\%$. Type 0: $K_{pos}=19,\ K_v=0,\ K_a=0$.

Quantity	Result
Process model	$G_m(s)=\dfrac{9.50}{s^2+2.17s+11.04}$ ($K_{dc}=0.86,\ \zeta=0.327,\ \omega_n=3.32$)
Controller gains	$K_p=22.1,\ K_d=0.073$
Stable?	YES (both quadratic coefficients $\gt 0$)
Compensated specs	$\text{PO}=10\%,\ T_{settle}=0.455\ \text{s},\ e_{ss}=5\%$
Type / error constants	$\boxed{\text{Type 0};\ K_{pos}=19,\ K_v=0,\ K_a=0}$

Concept & Theory

Topic. Time-domain identification of a second-order plant and minor-loop (rate) feedback design.

Key relations. $\text{PO}=e^{-\zeta\pi/\sqrt{1-\zeta^2}}$, $\omega_d=2\pi/T_{period}$; rate feedback $H(s)=K_d s+1$ raises the effective damping term without altering $H(0)$, so the DC gain and error are set by $K_p$ alone.

Why this works. Because the tachometer/rate term contributes zero at DC, the steady-state (error) design and the damping (overshoot) design decouple: $K_p$ fixes $K_{pos}$ and $e_{ss}$; $K_d$ then tunes the $s$-coefficient to hit $\zeta'$ without disturbing $\omega_n'$.

Common pitfall. Reading overshoot as (peak $-1$) instead of (peak $-y_{ss})/y_{ss}$ when $y_{ss}\ne1$, or forgetting the rate feedback also multiplies the closed-loop numerator (it does not here, since it is in the feedback path, not the forward path).

Source. Nise §4 (transient identification), §12 (rate/velocity feedback, minor-loop compensation).

Question 6 of 8: Second-order identification and rate-feedback design

Question 6 — Second-order identification and rate-feedback design [10 + 10]

Part A — Model from the step response [10]

Part B — Proportional + rate-feedback design [10]