22-Mec-A3 System Analysis and Control · December 2017

Question 1 of 6: Impulse and Step Response; Free Response of a Second-Order ODE

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, December 2017 — 16-Mec-A3, System Analysis and Control. Three hours, closed book; a Casio or Sharp approved calculator and semi-log graph paper are the only permitted aids. Six questions are printed; any four constitute a complete paper and all questions are of equal value (25 marks each). A table of Laplace transforms is supplied as page 5. All six questions are solved here.

Reference texts. K. Ogata, Modern Control Engineering, 5th ed. (Prentice Hall) — the source of this paper's problem style; N. S. Nise, Control Systems Engineering, 7th ed. (Wiley); R. C. Dorf and R. H. Bishop, Modern Control Systems, 13th ed. (Pearson); G. F. Franklin, J. D. Powell and A. Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed. (Pearson).

Check: conventions used throughout. All transfer functions are written in the Laplace variable s with zero initial conditions unless a question states otherwise. “Unity feedback” means $H(s)=1$, so the closed-loop transfer function is $T(s)=G(s)/[1+G(s)]$ and the error signal is $E(s)=R(s)-C(s)$. Root-locus gains are the gain $K$ exactly as it appears in the printed $G(s)$.

Question 1: Impulse and Step Response; Free Response of a Second-Order ODE (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. Part (a): a unity-feedback loop with open-loop transfer function $G(s)=(2s+1)/s^2$ — a double integrator stabilised by a single proportional-plus-derivative zero at $s=-1/2$. Part (b): the homogeneous second-order equation $y''+3y'+2y=0$ with initial conditions $y(0)=0.1$ and $y'(0)=0.05$, and no forcing input.

Find. (a) the closed-loop unit-impulse response $c_\delta(t)$ and unit-step response $c_u(t)$ in closed form; (b) the free response $y(t)$ satisfying both initial conditions.

Approach. Close the loop algebraically to get $T(s)=G/(1+G)$, invert it directly for the impulse response and invert $T(s)/s$ for the step response; for part (b) transform the ODE retaining the initial-condition terms of the derivative theorem, then invert by partial fractions.

Form the closed-loop transfer function. With unity feedback, $$T(s)=\frac{C(s)}{R(s)}=\frac{G(s)}{1+G(s)}=\frac{\dfrac{2s+1}{s^2}}{1+\dfrac{2s+1}{s^2}}=\frac{2s+1}{s^2+2s+1}$$ The denominator factors perfectly: $s^2+2s+1=(s+1)^2$, so $$\boxed{\;T(s)=\frac{2s+1}{(s+1)^2}\;}$$ The closed loop has a repeated real pole at $s=-1$ and a zero at $s=-1/2$. Note that although the plant is a marginally-unstable double integrator, the derivative action supplied by the numerator zero has produced a critically damped closed loop.
Invert for the unit-impulse response. A unit impulse has $R(s)=1$, so $C(s)=T(s)$ and no partial-fraction work beyond splitting the numerator is required. Writing the numerator in terms of $(s+1)$, $$2s+1=2(s+1)-1$$ so that $$C_\delta(s)=\frac{2(s+1)-1}{(s+1)^2}=\frac{2}{s+1}-\frac{1}{(s+1)^2}$$ Using transform pairs (15) and (20) of the supplied table, $$\boxed{\;c_\delta(t)=2e^{-t}-t\,e^{-t}=(2-t)e^{-t},\qquad t\ge 0\;}$$
Invert for the unit-step response. Now $R(s)=1/s$, giving $$C_u(s)=\frac{2s+1}{s(s+1)^2}=\frac{A}{s}+\frac{B}{s+1}+\frac{C}{(s+1)^2}$$ The residues at the simple pole and at the repeated pole follow by cover-up: $$A=\left.\frac{2s+1}{(s+1)^2}\right|_{s=0}=\frac{1}{1}=1,\qquad C=\left.\frac{2s+1}{s}\right|_{s=-1}=\frac{-1}{-1}=1$$ and matching the coefficient of $s^2$ in $2s+1=A(s+1)^2+Bs(s+1)+Cs$ gives $A+B=0$, hence $B=-1$. Therefore $$\boxed{\;c_u(t)=1-e^{-t}+t\,e^{-t},\qquad t\ge 0\;}$$
Check the two answers against each other. The impulse response must be the time derivative of the step response, and it is: $$\frac{d}{dt}\!\left(1-e^{-t}+te^{-t}\right)=e^{-t}+e^{-t}-te^{-t}=(2-t)e^{-t}=c_\delta(t)$$ The boundary values are also consistent: $c_u(0)=1-1+0=0$ as required for a strictly proper plant starting from rest, and $c_u(\infty)=1$, confirming the zero steady-state step error that the loop's free integrators guarantee.
Characterise the step response. Although both poles are real, the response overshoots — a consequence of the closed-loop zero, not of complex poles. The peak occurs where $c_\delta(t)=(2-t)e^{-t}=0$, i.e. at $t_p=2\ \text{s}$, where $$c_u(2)=1-e^{-2}+2e^{-2}=1+e^{-2}=1.1353$$ so the overshoot is $13.53\%$. This is a useful reminder that the familiar “overshoot implies $\zeta<1$” rule applies only to a pole pair with no finite zero.

Figure 1.1 — Closed-loop unit-step and unit-impulse responses of $T(s)=(2s+1)/(s+1)^2$. The step response overshoots to 1.1353 at $t=2$ s despite the critically damped (repeated real) pole pair, because of the closed-loop zero at $s=-1/2$.

Turning to part (b), the transform must now carry the initial conditions, since the system starts from a non-zero state rather than from rest.

Transform the ODE, retaining initial conditions. Pairs (8) and (9) of the supplied table give $\mathcal{L}[y']=sY-y(0)$ and $\mathcal{L}[y'']=s^2Y-sy(0)-y'(0)$. Substituting into $y''+3y'+2y=0$, $$\left[s^2Y-s(0.1)-0.05\right]+3\left[sY-0.1\right]+2Y=0$$ Collecting the terms in $Y$ and moving the known quantities to the right, $$\left(s^2+3s+2\right)Y(s)=0.1s+0.05+0.3=0.1s+0.35$$
Solve for Y(s) and factor. Since $s^2+3s+2=(s+1)(s+2)$, $$Y(s)=\frac{0.1s+0.35}{(s+1)(s+2)}$$ Both poles are simple and real, so a two-term partial fraction suffices.
Evaluate the residues. By the cover-up rule, $$k_1=\left.\frac{0.1s+0.35}{s+2}\right|_{s=-1}=\frac{-0.1+0.35}{1}=0.25,\qquad k_2=\left.\frac{0.1s+0.35}{s+1}\right|_{s=-2}=\frac{-0.2+0.35}{-1}=-0.15$$ Hence $Y(s)=\dfrac{0.25}{s+1}-\dfrac{0.15}{s+2}$, and inverting with pair (15), $$\boxed{\;y(t)=0.25\,e^{-t}-0.15\,e^{-2t},\qquad t\ge 0\;}$$
Verify against both initial conditions. This is the free check that the initial-condition bookkeeping was done correctly: $$y(0)=0.25-0.15=0.10\ \checkmark,\qquad y'(t)=-0.25e^{-t}+0.30e^{-2t}\;\Rightarrow\;y'(0)=-0.25+0.30=0.05\ \checkmark$$ Both match the data exactly, and $y(t)\to 0$ as $t\to\infty$ as expected for a stable homogeneous system with poles at $-1$ and $-2$.

Quantity	Result
(a) Closed-loop transfer function	$T(s)=(2s+1)/(s+1)^2$
(a) Unit-impulse response	$c_\delta(t)=(2-t)e^{-t}$
(a) Unit-step response	$c_u(t)=1-e^{-t}+t\,e^{-t}$
(a) Peak of step response	1.1353 at $t_p=2$ s (13.53 % overshoot)
(b) Free response	$y(t)=0.25e^{-t}-0.15e^{-2t}$

Topic. Laplace-domain solution of linear time-invariant systems: closed-loop reduction, partial-fraction inversion, and the effect of a closed-loop zero on transient shape.

Key relations. Unity-feedback closure $T=G/(1+G)$; derivative theorem $\mathcal{L}[y^{(n)}]=s^nY-s^{n-1}y(0)-\cdots-y^{(n-1)}(0)$; cover-up residues $k_i=[(s-p_i)F(s)]_{s=p_i}$; for a repeated pole the extra residue is $\frac{d}{ds}[(s-p)^2F(s)]$; impulse response $=\frac{d}{dt}$(step response).

Why this works. The Laplace transform converts both the loop algebra and the differential equation into ordinary algebra, and the inversion is then just pattern-matching against the supplied table. Part (a) is the zero-initial-condition case, in which the transfer function alone carries all the dynamics, so the impulse response is the inverse transform of $T(s)$ and the step response is the inverse transform of $T(s)/s$. Part (b) is the complementary case: there is no input at all, and the entire response is generated by the initial-condition terms that the derivative theorem injects on the right-hand side. The same characteristic polynomial governs both — the poles fix the modes $e^{-t}$, $e^{-2t}$, while the input or the initial state fixes only how much of each mode is excited.

Common pitfall. Two mistakes dominate. First, students often invert the open-loop $G(s)$ in part (a) and report an unstable ramp-like response, forgetting that the question asks for the response of the closed-loop system. Second, in part (b) it is easy to drop the $3y(0)=0.3$ contribution from the first-derivative term; that single omission changes the numerator from $0.1s+0.35$ to $0.1s+0.05$ and produces residues that fail the $y(0)=0.1$ check. Always substitute $t=0$ back into the final answer — it costs ten seconds and catches the error.

Source. Ogata, Modern Control Engineering, 5th ed., §2–5 (Laplace transform solution of LTI systems) and §5–3 (transient response of higher-order systems); transform pairs as supplied on page 5 of the examination.

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Quantity	Result
(a) Closed-loop transfer function	\(T(s)=(2s+1)/(s+1)^2\)
(a) Unit-impulse response	\(c_\delta(t)=(2-t)e^{-t}\)
(a) Unit-step response	\(c_u(t)=1-e^{-t}+t\,e^{-t}\)
(a) Peak of step response	1.1353 at \(t_p=2\) s (13.53 % overshoot)
(b) Free response	\(y(t)=0.25e^{-t}-0.15e^{-2t}\)