Transient-response terms under proportional control: 22-Mec-A3 System Analysis and Control, May 2014, Q4

Question 4: Transient-response terms under proportional control (25 marks)

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.

Given. The Figure-1 loop: reference $R(s)$, error into a proportional controller $G_c(s)=K_p$, plant $200/[(s+1)(s+2)]$, and a feedback element of gain 0.02 (a non-unity, sensor-scaling feedback).

Element	Transfer function
Controller	$G_c(s)=K_p$
Plant	$G_p(s)=\dfrac{200}{(s+1)(s+2)}$
Feedback (sensor)	$H(s)=0.02$
Cases required	$K_p = 0.05$; $K_p = 0.5$; then the minimum $K_p$ for fastest decay

Find. The transient (natural-response) terms — i.e. the exponential and damped-sinusoidal modes fixed by the closed-loop poles — for each of the two gains, and the smallest proportional gain that makes the transient decay as fast as the loop allows.

[Figure not reproduced: Figure 1 redrawn: proportional controller, plant 200/[(s+1)(s+2)] and a 0.02 sensor gain in the feedback path. See the official exam paper.]

Approach. Reduce the loop to a single second-order transfer function, so that its characteristic polynomial gives the closed-loop poles as a function of $K_p$; the transient terms are then read directly from the pole pattern, and the "fastest decay" question becomes a question about where the two real poles coalesce.

Reduce the block diagram. The forward path is $G_cG_p$ and the feedback path is $H = 0.02$, so $$\frac{C(s)}{R(s)}=\frac{G_cG_p}{1+G_cG_pH}=\frac{200K_p}{(s+1)(s+2)+200K_p(0.02)} =\frac{200K_p}{s^{2}+3s+2+4K_p}$$ Note the useful simplification: the loop gain is $200\times0.02 = 4$ times $K_p$, so the characteristic equation depends on $K_p$ only through the product $4K_p$.
Write the characteristic equation. The transient modes are governed by $$\boxed{\;s^{2}+3s+(2+4K_p)=0\;}\qquad\Longrightarrow\qquad s_{1,2}=\frac{-3\pm\sqrt{9-4(2+4K_p)}}{2}=\frac{-3\pm\sqrt{1-16K_p}}{2}$$ The damping term $-3/2$ is fixed by the plant; only the discriminant moves with gain.
Part (a): $K_p = 0.05$. The constant term is $2+4(0.05)=2.2$ and the discriminant is $9-8.8=0.2>0$, so the poles are real and distinct: $$s_{1,2}=\frac{-3\pm 0.4472}{2}=-1.2764,\ -1.7236$$ The transient response therefore consists of two decaying exponentials, $$\boxed{\;c_{tr}(t)=A_1e^{-1.2764t}+A_2e^{-1.7236t}\;}$$ with time constants $\tau_1 = 1/1.2764 = 0.783$ s and $\tau_2 = 1/1.7236 = 0.580$ s. The response is over-damped and settles in roughly $4\tau_1 \approx 3.1$ s, dominated by the slower mode. For reference, the closed-loop dc gain is $200(0.05)/2.2 = 4.545$.
Part (b): $K_p = 0.5$. Now the constant term is $2+4(0.5)=4$ and the discriminant is $9-16=-7<0$, so the poles form a complex-conjugate pair: $$s_{1,2}=\frac{-3\pm j\sqrt{7}}{2}=-1.5\pm j1.3229$$ and the transient terms are a decaying sine and cosine at the damped natural frequency, $$\boxed{\;c_{tr}(t)=e^{-1.5t}\left(B_1\cos 1.3229t+B_2\sin 1.3229t\right)\;}$$ Comparing with the standard form $s^{2}+2\zeta\omega_n s+\omega_n^{2}$ gives $\omega_n=\sqrt{4}=2$ rad/s and $\zeta=1.5/2=0.75$, i.e. a well-damped oscillatory response with about 2.8 % overshoot and a settling time near $4/1.5 = 2.7$ s. The dc gain has risen to $200(0.5)/4 = 25$.
Part (c): the gain that gives the fastest decay. The decay rate of the transient is set by the least negative real part among the closed-loop poles. While the poles are real (small $K_p$) the slower pole is $s_1=(-3+\sqrt{1-16K_p})/2$, which becomes more negative as $K_p$ rises. Once the discriminant turns negative both poles sit at $\sigma=-1.5$ and the decay rate stops improving — further gain only adds oscillation. The transition is at the repeated root: $$9-4(2+4K_p)=0\quad\Longrightarrow\quad 2+4K_p=2.25\quad\Longrightarrow\quad \boxed{\;K_{p,\min}=0.0625\;}$$ at which the characteristic equation is $(s+1.5)^{2}=0$, giving a critically damped response with transient terms $c_{tr}(t)=(C_1+C_2t)e^{-1.5t}$.
Confirm the claim numerically. Sweeping the gain: at $K_p=0.03$ the slower pole is $-1.139$; at $K_p=0.05$ it is $-1.276$; at $K_p=0.0625$ it is exactly $-1.5$; and for every larger gain (0.1, 1, 10) the real part stays pinned at $-1.5$. Hence 0.0625 is the minimum gain achieving the fastest available decay envelope $e^{-1.5t}$ — which is precisely why the question asks for a minimum rather than an optimum.

Normalised unit-step responses for the two gains, showing the over-damped (K_p = 0.05) and under-damped (K_p = 0.5) characters.

Check: part (c) is printed with the symbol $K$ while parts (a) and (b) use $K_p$; since the only adjustable gain in Figure 1 is the proportional controller gain, $K$ is read here as $K_p$. Under this reading the three parts form a coherent progression (over-damped → under-damped → the critically damped boundary between them), and the answer 0.0625 lies between the two gains examined in (a) and (b).

Case	Closed-loop poles	Transient terms	Character
(a) K_p = 0.05	−1.2764, −1.7236	$A_1e^{-1.2764t}+A_2e^{-1.7236t}$	over-damped, τ = 0.783 s and 0.580 s
(b) K_p = 0.5	−1.5 ± j1.3229	$e^{-1.5t}(B_1\cos 1.3229t+B_2\sin 1.3229t)$	under-damped, ω_n = 2 rad/s, ζ = 0.75
(c) K_p,min = 0.0625	−1.5 (double)	$(C_1+C_2t)e^{-1.5t}$	critically damped — fastest decay envelope

Topic. Second-order closed-loop transient modes under proportional control, and the critically damped gain.

Key relations. Closed-loop transfer function $G_cG_p/(1+G_cG_pH)$; characteristic equation $s^{2}+2\zeta\omega_ns+\omega_n^{2}=0$ with $\omega_n=\sqrt{2+4K_p}$ and $2\zeta\omega_n=3$; real poles for $9>4(2+4K_p)$, complex for $9<4(2+4K_p)$; decay envelope $e^{-\sigma t}$ with $\sigma=\min|\mathrm{Re}(s_i)|$.

Why this works. The transient (natural) response is the zero-input solution of the closed-loop differential equation, so its functional form depends only on the roots of the characteristic polynomial — not on the input or the numerator gain. Because the feedback element scales the loop gain by 0.02, the plant's large dc gain of 200 is tamed to an effective $4K_p$, and the sum of the poles ($-3$) is invariant with gain: raising $K_p$ can only slide the poles along the real axis toward each other and then vertically off it. That geometry is what makes the repeated-root gain the exact boundary between "faster with more gain" and "no faster, just more ringing".

Common pitfall. Forgetting the 0.02 feedback gain and using $s^{2}+3s+2+200K_p$ — which shifts the critical gain by a factor of 50. A second is confusing "fastest decay" with "fastest rise" or with minimum settling time including overshoot; the question asks about the decay of the transient terms, which is governed by the real part of the dominant pole. A third is reporting complex poles for case (a): the discriminant there is small but positive, so the poles are genuinely real and close together.

Source. Ogata, Modern Control Engineering, 5th ed., §5.3 (second-order systems and transient specifications); Franklin, Powell & Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed., Ch. 3.3–3.4.

Question 4 of 6: Transient-response terms under proportional control

Question 4: Transient-response terms under proportional control (25 marks)