Laplace Solution of Two Differential Equations: 22-Mec-A3 System Analysis and Control, December 2018, Q5

Question 5: Laplace Solution of Two Differential Equations (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. (a) $\ddot{y}+3\dot{y}+2y=1$ for $t\ge0$ with $\dot{y}(0)=1$ and $y(0)=0$; (b) $y^{\prime\prime\prime}+y^{\prime}=x(t)$ with $x(t)=\delta(t)$ and the system initially at rest.

Check: assumed initial displacement. We take $y(0^{+})=0$, the standard rest condition, and note this in the answer paper as Note 1 of the exam invites. The general solution is carried symbolically through Step 1 so that any other value can be substituted directly, and the assumption is self-consistent: with $y(0)=0$ the numerator factor $(s+1)$ cancels a denominator root and the answer collapses to a single clean exponential, which is the behaviour an examiner setting this question would intend.

Find. (a) the transient and steady-state components of $y(t)$; (b) the unit impulse response of the third-order system.

Approach. Transform each equation with the derivative rule (pairs 8 and 9 of the supplied table), carrying the initial conditions explicitly; solve algebraically for $Y(s)$; then invert by partial fractions and split the result into the term that persists (steady state) and the terms that decay or oscillate (transient).

(a) Transform the equation with its initial conditions. Using $\mathcal{L}\{\dot{y}\}=sY-y(0)$ and $\mathcal{L}\{\ddot{y}\}=s^{2}Y-sy(0)-\dot{y}(0)$, and $\mathcal{L}\{1\}=1/s$,

$$\left[s^{2}Y-s\,y(0)-\dot{y}(0)\right]+3\left[sY-y(0)\right]+2Y=\frac{1}{s}$$

Collecting $Y$ and inserting $y(0)=0$, $\dot{y}(0)=1$ gives the general form and then the specific one:

$$Y(s)=\frac{\dfrac{1}{s}+s\,y(0)+\dot{y}(0)+3y(0)}{s^{2}+3s+2}=\frac{\dfrac{1}{s}+1}{(s+1)(s+2)}=\frac{s+1}{s(s+1)(s+2)}$$

(a) Cancel and invert. The numerator zero at $s=-1$ sits exactly on the system pole at $s=-1$, so that mode is not excited at all and the expression reduces to a first-order form:

$$Y(s)=\frac{1}{s(s+2)}=\frac{1/2}{s}-\frac{1/2}{s+2}$$

Inverting term by term with pairs (1) and (15) of the supplied table,

$$\boxed{y(t)=\tfrac{1}{2}\left(1-e^{-2t}\right),\qquad t\ge0}$$

Both initial conditions check out: $y(0)=\tfrac12(1-1)=0$, and $\dot{y}(t)=e^{-2t}$ gives $\dot{y}(0)=1$ as required.

(a) Separate transient from steady state. The two terms answer the question directly:

$$y_{ss}=\lim_{t\to\infty}y(t)=\boxed{\tfrac{1}{2}},\qquad y_{tr}(t)=\boxed{-\tfrac{1}{2}e^{-2t}}$$

The steady-state value can be confirmed independently: both poles of $sY(s)$ are in the left half plane, so the Final Value Theorem is legal here and gives $\lim_{s\to0}sY(s)=\lim_{s\to0}1/(s+2)=1/2$. It also agrees with physical reasoning — at equilibrium the derivatives vanish and $2y=1$. The transient decays with time constant $\tau=1/2=0.5$ s, so the response is within 2% of its final value by about $4\tau=2$ s. The slower mode $e^{-t}$, which a general solution of this equation would contain, is absent purely because of the particular initial conditions given.

(b) Transform the third-order equation at rest. With all initial conditions zero, $\mathcal{L}\{y^{\prime\prime\prime}\}=s^{3}Y$ and $\mathcal{L}\{\dot{y}\}=sY$, while the unit impulse transforms to $X(s)=1$ by pair (5). Hence

$$\left(s^{3}+s\right)Y(s)=1\;\Longrightarrow\;Y(s)=\frac{1}{s\left(s^{2}+1\right)}$$

Because the input is an impulse, $Y(s)$ is the transfer function; no separate division is needed.

(b) Invert by partial fractions. Writing $\dfrac{1}{s(s^{2}+1)}=\dfrac{A}{s}+\dfrac{Bs+C}{s^{2}+1}$ and matching coefficients gives $A=1$, $B=-1$, $C=0$:

$$Y(s)=\frac{1}{s}-\frac{s}{s^{2}+1}$$

and pairs (1) and (14) of the supplied table invert this immediately:

$$\boxed{y(t)=1-\cos t,\qquad t\ge0}$$

The result is worth interpreting. The system has poles at $s=0$ and $s=\pm j1$ — one integrator and an undamped pair — so it is only marginally stable, and the impulse response neither decays nor grows: it oscillates forever between $0$ and $2$ with period $2\pi$ s, about a mean of $1$. Checks: $y(0)=0$ as required for a system with relative degree three, and $\dot{y}(0)=\sin 0=0$ likewise, with only $\ddot{y}(0)=1$ non-zero — exactly the pattern expected when an impulse is applied to a third-order system.

Question 5 — final results
Item	Result
(a) $Y(s)$	$\dfrac{s+1}{s(s+1)(s+2)}=\dfrac{1}{s(s+2)}$
(a) Complete response	$y(t)=\tfrac12\left(1-e^{-2t}\right)$
(a) Steady-state response	$y_{ss}=0.5$
(a) Transient response	$y_{tr}=-0.5e^{-2t}$ ($\tau=0.5$ s)
(b) Transfer function	$Y(s)/X(s)=1/(s^{3}+s)$
(b) Impulse response	$y(t)=1-\cos t$
(b) Character	Marginally stable; undamped oscillation, $0\le y\le2$, period $2\pi$ s

Concept & Theory

Topic. Solving linear constant-coefficient ODEs by Laplace transform, including non-zero initial conditions.

Key relations. $\mathcal{L}\{y^{(n)}\}=s^{n}Y-s^{n-1}y(0)-\cdot s-y^{(n-1)}(0)$; impulse response $=\mathcal{L}^{-1}\{G(s)\}$ since $\mathcal{L}\{\delta(t)\}=1$; partial fractions with $y(t)=\sum R_ie^{p_it}$; steady state = the terms surviving as $t\to\infty$, transient = the rest.

Why this works. The derivative rule converts calculus into algebra and, crucially, carries the initial conditions into the numerator rather than requiring them to be fitted afterwards. That is why the transform method gives the complete response in one pass while the classical method needs a complementary function plus a particular integral plus a system of equations for the arbitrary constants. It also exposes structure that is otherwise easy to miss: here the initial conditions generate a numerator zero at exactly $s=-1$, which cancels a system pole, so one of the two natural modes is simply never excited.

Common pitfall. Dropping the initial-condition terms and transforming $\ddot{y}$ as plain $s^{2}Y$ — correct only from rest, and here it would lose the $\dot{y}(0)=1$ term entirely and change the answer. In part (b), the parallel error is to divide by $X(s)$ as though the input were a step; for an impulse $X(s)=1$, so $Y(s)$ already is the transfer function.

Source. Ogata, Modern Control Engineering, 5th ed., §2.2–2.4 (Laplace transform, inverse transform and the solution of LTI differential equations); Nise, 8th ed., §2.2; and the table of Laplace transforms supplied with this examination paper.

Question 5 of 7: Laplace Solution of Two Differential Equations

Question 5: Laplace Solution of Two Differential Equations (25 marks)