Damped Mass-Spring System Driven by a Square Pulse (Laplace Transform): 04-BS-1, May 2014, Q8

Question 8: Damped Mass-Spring System Driven by a Square Pulse (Laplace Transform) (20 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. Overdamped mass-spring system $y''+3y'+2y=r(t)$, zero ICs, forced by a unit square pulse active on $[1,2)$: $r(t)=u(t-1)-u(t-2)$ in terms of the Heaviside step $u$.

Find. $y(t)$ for all $t\ge0$.

Approach. Use the Laplace transform. First find the system's unit-step response $h(t)$ (the solution to $y''+3y'+2y=u(t)$ with zero ICs) by inverting $\mathcal L\{h\}=\dfrac{1}{s(s+1)(s+2)}$ via partial fractions. Then, since $r(t)=u(t-1)-u(t-2)$ and the ICs are zero, the second shifting theorem gives $y(t)=h(t-1)u(t-1)-h(t-2)u(t-2)$ directly, with no new inverse transform needed.

Transform the ODE. With zero ICs, $\mathcal L\{y''+3y'+2y\}=(s^2+3s+2)Y(s)=(s+1)(s+2)Y(s)$, and $\mathcal L\{r(t)\}=\dfrac{e^{-s}-e^{-2s}}{s}$, so $$Y(s)=\frac{e^{-s}-e^{-2s}}{s(s+1)(s+2)}.$$
Partial fractions of the transfer function. $$\frac{1}{s(s+1)(s+2)}=\frac{A}{s}+\frac{B}{s+1}+\frac{C}{s+2},\qquad A=\tfrac12,\ B=-1,\ C=\tfrac12$$ (found by clearing denominators and evaluating at $s=0,-1,-2$).
Invert to get the unit-step response $h(t)$. $$h(t)=\tfrac12-e^{-t}+\tfrac12e^{-2t},\qquad t\ge0$$ (check: $h(0)=\tfrac12-1+\tfrac12=0$ and $h'(0)=e^{-t}-e^{-2t}\big|_{t=0}=0$, matching the zero ICs).
Apply the second shifting theorem. Since $Y(s)=(e^{-s}-e^{-2s})\cdot\mathcal L\{h\}(s)$, each exponential factor delays and gates $h$: $$y(t)=h(t-1)\,u(t-1)-h(t-2)\,u(t-2).$$ Written piecewise: $$y(t)=\begin{cases}0,&t<1,\\[4pt] \tfrac12-e^{-(t-1)}+\tfrac12e^{-2(t-1)},&1\le t<2,\\[4pt] \big[\tfrac12-e^{-(t-1)}+\tfrac12e^{-2(t-1)}\big]-\big[\tfrac12-e^{-(t-2)}+\tfrac12e^{-2(t-2)}\big],&t\ge2.\end{cases}$$

$$\boxed{y(t)=h(t-1)u(t-1)-h(t-2)u(t-2)},\quad h(\tau)=\tfrac12-e^{-\tau}+\tfrac12e^{-2\tau}$$

Quantity	Result
Unit-step response $h(t)$	$\tfrac12-e^{-t}+\tfrac12e^{-2t}$
$y(t)$, $t<1$	$0$
$y(t)$, $1\le t<2$	$h(t-1)$
$y(t)$, $t\ge2$	$h(t-1)-h(t-2)$
Sample value $y(1.5)$	$\approx0.0774$
Sample value $y(2.5)$	$\approx0.2244$ (peak region)

Concept & Theory

Topic. Laplace-transform solution of a linear ODE forced by a finite-duration pulse (second shifting/time-delay theorem).

Key relations. $\mathcal L\{u(t-a)f(t-a)\}=e^{-as}F(s)$ (second shifting theorem); a rectangular pulse active on $[a,b)$ is $u(t-a)-u(t-b)$, so its transform is $(e^{-as}-e^{-bs})/s$ and the response to it is the system's step response, delayed and differenced at $a$ and $b$.

Why this works. Because the system is linear and time-invariant with zero initial conditions, the response to $u(t-1)-u(t-2)$ is exactly the (linear) superposition of the delayed step responses $h(t-1)u(t-1)$ and $-h(t-2)u(t-2)$ — no new partial-fraction inversion is needed once $h(t)$ (the response to a single unit step at $t=0$) is known.

Common pitfall. Attempting to invert $\dfrac{e^{-s}-e^{-2s}}{s(s+1)(s+2)}$ directly term-by-term without first recognizing the shared factor $\dfrac{1}{s(s+1)(s+2)}=\mathcal L\{h\}$, leading to redundant partial-fraction work; also forgetting the $u(t-1)$ gate, which makes $y(t)\equiv0$ for $t<1$ rather than applying the formula for all $t$.

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 6 (§6.3 Unit Step Function, Second Shifting Theorem; §6.4 Short Impulses).

Question 8 of 8: Damped Mass-Spring System Driven by a Square Pulse (Laplace Transform)

Question 8: Damped Mass-Spring System Driven by a Square Pulse (Laplace Transform) (20 marks)