Question 6 of 6: Effect of Loop Gain on Step Response and Steady-State Error

Question 6: Effect of Loop Gain on Step Response and Steady-State Error (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. Unity-feedback loop of Fig. 3 with forward gain $K$ and plant $G(s)=\dfrac{1}{(s+2)(s+10)}$; unit-step input; two gain settings, $K = 7$ and $K = 20$.

Find. (a) the closed-loop step responses $c(t)$ at both gains; (b) the steady-state errors, verified both from the time responses and from the static error constant; (c) a comparison of settling time and response character.

Approach. Form the closed-loop transfer function once in terms of $K$, factor its denominator at each gain, invert by partial fractions, then check each final value against $e_{ss}=1/(1+K_p)$ and compare the dominant time constants.

1. Form the closed-loop transfer function. With unity feedback, $$\frac{C(s)}{R(s)}=\frac{KG(s)}{1+KG(s)}=\frac{K}{(s+2)(s+10)+K} =\frac{K}{s^2+12s+(20+K)}$$ The damping term $12s$ is fixed by the plant; only the constant term moves with $K$, so raising the gain raises $\omega_n=\sqrt{20+K}$ and lowers $\zeta = 6/\sqrt{20+K}$.
2. Case $K = 7$: factor and invert. The denominator becomes $s^2+12s+27=(s+3)(s+9)$ — two real poles, so the response is overdamped. With $C(s)=7/[s(s+3)(s+9)]$ the residues are $7/27$, $-7/18$ and $7/54$, giving $$\boxed{c_1(t)=\frac{7}{27}-\frac{7}{18}e^{-3t}+\frac{7}{54}e^{-9t}}$$ The coefficients sum to zero at $t = 0$ as required, and $c_1(\infty)=7/27=0.2593$.
3. Case $K = 20$: factor and invert. Now $s^2+12s+40$, whose roots are $s=-6\pm j2$: complex, so the response is underdamped with $\omega_n=\sqrt{40}=6.325\ \text{rad/s}$ and $\zeta = 6/6.325 = 0.949$. Expanding $C(s)=20/[s(s^2+12s+40)]$ as $0.5/s + (Bs+D)/(s^2+12s+40)$ gives $B=-0.5$, $D=-6$, and completing the square as $(s+6)^2+2^2$ yields $$\boxed{c_2(t)=0.5-e^{-6t}\left(0.5\cos 2t+1.5\sin 2t\right)}$$ with $c_2(0)=0$ and $c_2(\infty)=0.5$.

4. Part (b): verify the errors directly from the responses. Because $E = R - C$ and $R$ is a unit step, the steady-state error is simply $1-c(\infty)$: $$e_{ss}\big|_{K=7}=1-\frac{7}{27}=\frac{20}{27}=0.7407,\qquad e_{ss}\big|_{K=20}=1-0.5=0.5000$$
5. Part (b): confirm against the static error constant. The loop is type 0, with $$K_p=\lim_{s\to0}\frac{K}{(s+2)(s+10)}=\frac{K}{20}$$ so $K_p = 0.35$ at $K=7$ and $K_p = 1.0$ at $K=20$, giving $$\frac{1}{1+0.35}=0.7407,\qquad \frac{1}{1+1.0}=0.5000$$ Both agree exactly with the time-domain values, which is the "direct verification" the question asks for. Note that even at $K = 20$ the error is still 50 % — a type-0 loop cannot be made accurate by gain alone without eventually destroying its damping.
6. Part (c): compare settling times. At $K=7$ the dominant pole is $s=-3$, so the $2\ \%$ settling time is approximately $4/3 = 1.33\ \text{s}$; evaluating the exact response gives $1.44\ \text{s}$. At $K=20$ both poles have real part $-6$, giving $4/6 = 0.67\ \text{s}$ by the envelope estimate and $0.83\ \text{s}$ exactly (the sine term inflates the envelope slightly). The higher gain therefore settles about 1.7 times faster.
7. Part (c): compare the nature of the responses. At $K=7$ the poles are real and distinct, so the response rises monotonically to $0.2593$ with no overshoot and no oscillation. At $K=20$ the poles are complex with $\zeta = 0.949$; the response is nominally underdamped but so heavily damped that its overshoot is only $M_p = e^{-\pi\zeta/\sqrt{1-\zeta^2}} = 0.008\ \%$ — a peak of $0.50004$ at $t = \pi/2\ \text{s}$, entirely invisible in practice. In summary, raising $K$ from 7 to 20 halves the steady-state error, nearly halves the settling time, and moves the system from overdamped to just inside the underdamped region without producing any practically significant overshoot. Continuing past $K=20$ would keep improving accuracy and speed of response only until the overshoot became objectionable; since $\zeta$ falls as $6/\sqrt{20+K}$, a common design limit of $\zeta = 0.7$ would already be reached at $K \approx 53$.