Question 1 of 6: Unit step response, and the time constants of a transient

Question 1: Unit step response, and the time constants of a transient

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) — unit step response

Given. An open-loop plant $G(s) = 54 / [(2s+6)(s^2+3s+9)]$, driven by a unit step $r(t) = u_s(t)$, i.e. $R(s) = 1/s$. The system starts from rest.

Find. The closed-form time function $c(t)$ for $t \ge 0$, obtained by inverting $C(s) = G(s)\,R(s)$.

Approach. Normalise the denominator to monic form, expand $C(s)=G(s)/s$ in partial fractions with the complex pair kept together as a real quadratic, then invert term by term using the standard pairs $1/s$, $1/(s+\alpha)$ and $\omega/[(s+\sigma)^2+\omega^2]$ from the appended transform table.

1. Normalise the transfer function. The first factor carries a leading coefficient of 2, which must be divided out before any pole is read off: $$G(s)=\frac{54}{(2s+6)(s^2+3s+9)}=\frac{54}{2(s+3)(s^2+3s+9)}=\frac{27}{(s+3)(s^2+3s+9)}$$ The real pole is therefore at $s=-3$, not at $s=-6$.
2. Identify the second-order pair. Comparing $s^2+3s+9$ with the standard form $s^2+2\zeta\omega_n s+\omega_n^2$ gives $\omega_n=\sqrt{9}=3\ \text{rad/s}$ and $2\zeta\omega_n=3$, hence $\zeta = 0.5$. The pair is under-damped, with $$\sigma=\zeta\omega_n=1.5\ \text{s}^{-1},\qquad \omega_d=\omega_n\sqrt{1-\zeta^2}=\tfrac{3\sqrt{3}}{2}=2.598\ \text{rad/s}$$ so the complex poles sit at $s=-1.5\pm j2.598$.
3. Check the d.c. gain. Setting $s=0$, $G(0)=54/(6\times 9)=1$. The step response must therefore approach unity, which is the final-value check on the answer.
4. Expand the step response in partial fractions. With $R(s)=1/s$, $$C(s)=\frac{27}{s(s+3)(s^2+3s+9)}=\frac{A}{s}+\frac{B}{s+3}+\frac{Cs+D}{s^2+3s+9}$$ The two real residues follow by covering up their own factors: $$A=\left.\frac{27}{(s+3)(s^2+3s+9)}\right|_{s=0}=\frac{27}{3\times 9}=1,\qquad B=\left.\frac{27}{s(s^2+3s+9)}\right|_{s=-3}=\frac{27}{(-3)(9)}=-1$$
5. Match the remaining coefficients. Clearing denominators, $A(s+3)(s^2+3s+9)+Bs(s^2+3s+9)+(Cs+D)s(s+3)\equiv 27$. Collecting powers of $s$ with $A=1$ and $B=-1$ gives $s^3:\;A+B+C=0$ and $s^1:\;18A+9B+3D=0$, so that $$C=0,\qquad D=-3$$ The $s^2$ coefficient, $6A+3B+3C+D = 6-3+0-3 = 0$, closes the identity and confirms the expansion.
6. Invert term by term. The quadratic term is completed to $s^2+3s+9=(s+1.5)^2+(2.598)^2$, and since $C=0$ it contributes a pure sine: $$\mathcal{L}^{-1}\!\left\{\frac{-3}{(s+1.5)^2+2.598^2}\right\} =-\frac{3}{2.598}\,e^{-1.5t}\sin(2.598\,t)=-\frac{2}{\sqrt3}\,e^{-1.5t}\sin(2.598\,t)$$ Adding the two real terms gives the unit step response $$\boxed{\;c(t)=1-e^{-3t}-1.1547\,e^{-1.5t}\sin\!\left(2.598\,t\right),\qquad t\ge 0\;}$$
7. Verify the end points. At $t=0$ the expression gives $1-1-0=0$, as it must for a strictly proper plant starting from rest; as $t\to\infty$ both exponentials vanish and $c\to 1$, matching the d.c. gain. Numerically the response peaks at $c_{\mathrm{max}}=1.081$ at $t=1.64\ \text{s}$, i.e. about 8.1 % overshoot, and stays inside a 2 % band from $t=2.21\ \text{s}$ onwards. (The usual $4/\zeta\omega_n=4/1.5=2.7\ \text{s}$ estimate is conservative here because the real pole at $-3$ pulls the tail in faster than the pair's envelope alone.)

Part (b) — time constants and transient duration

Given. $G(s)=5(1-0.4s)/[(s+1)(0.2s+1)]$.

Find. The time constants of the two transient components and the time for the transient to die away almost completely.

Approach. The transient components are set by the poles alone. Each first-order factor written in time-constant form $(\tau s+1)$ contributes a mode $e^{-t/\tau}$, and the slowest mode governs how long the transient lasts. The zero changes the shape of the response, not its decay rates.

1. Read the time constants from the denominator. The factor $(s+1)$ is $(1.0\,s+1)$ in time-constant form and $(0.2s+1)$ is already there, so $$\tau_1=1.0\ \text{s}\;(\text{pole at }s=-1),\qquad \tau_2=0.2\ \text{s}\;(\text{pole at }s=-5)$$ The transient is the sum of $e^{-t}$ and $e^{-5t}$ terms.
2. Identify the dominant mode. The larger time constant dominates: $\tau_{\text{dom}}=\tau_1=1.0\ \text{s}$. The fast mode has decayed to less than 1 % of its initial value by $t=5\tau_2=1.0\ \text{s}$, by which time the slow mode has only fallen to $e^{-1}=37\,\%$.
3. Convert to a settling time. Using the usual engineering rule that a first-order mode is "practically gone" after four to five time constants, $$e^{-4}=0.0183\;(98.2\,\%\ \text{decayed}),\qquad e^{-5}=0.0067\;(99.3\,\%\ \text{decayed})$$ so $$\boxed{\;\tau_1=1.0\ \text{s},\quad \tau_2=0.2\ \text{s},\quad t_{\text{transient}}\approx 4\tau_1\ \text{to}\ 5\tau_1 = 4\ \text{to}\ 5\ \text{s}\;}$$
4. Note the effect of the zero. The numerator factor $(1-0.4s)$ places a zero at $s=+2.5$, in the right half plane, so this is a non-minimum-phase plant. Its step response initially moves the wrong way (undershoot) before recovering to the d.c. value $G(0)=5$. The undershoot is a shape effect only — it does not alter the two decay rates computed above.