Damped Mass–Spring Response to a Square-Wave Force: 04-BS-1, May 2017, Q8

Question 8: Damped Mass–Spring Response to a Square-Wave Force 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. An overdamped second-order system at rest, forced by a unit-height square pulse active only on $1\le t<2$.

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

Approach. Write the pulse as a difference of unit step functions, $r(t)=u(t-1)-u(t-2)$, transform with zero initial conditions, and use the second shifting theorem to invert.

Laplace-transform the equation. With $y(0)=y'(0)=0$ and $\mathcal L\{u(t-a)\}=e^{-as}/s$: $$\big(s^2+3s+2\big)Y(s)=\frac{e^{-s}-e^{-2s}}{s}\ \Rightarrow\ Y(s)=\frac{e^{-s}-e^{-2s}}{s(s+1)(s+2)}.$$ Note $s^2+3s+2=(s+1)(s+2)$.
Partial fractions of the transfer factor. $$\frac{1}{s(s+1)(s+2)}=\frac{1/2}{s}-\frac{1}{s+1}+\frac{1/2}{s+2}.$$
Invert the unit-step response $h(\tau)$. This is the response to a single unit step at $t=0$: $$h(\tau)=\frac12-e^{-\tau}+\frac12e^{-2\tau},\qquad\tau\ge0.$$
Apply the second shifting theorem. Since $Y(s)=\big(e^{-s}-e^{-2s}\big)H(s)$, the time response is $y(t)=h(t-1)u(t-1)-h(t-2)u(t-2)$: $$\boxed{y(t)=\begin{cases}0,&0\le t<1\\[2pt]\dfrac12-e^{-(t-1)}+\dfrac12e^{-2(t-1)},&1\le t<2\\[4pt]\left[e^{-(t-2)}-e^{-(t-1)}\right]+\dfrac12\left[e^{-2(t-1)}-e^{-2(t-2)}\right],&t\ge2\end{cases}}$$ (the $t\ge2$ branch is $h(t-1)-h(t-2)$ expanded and simplified; the constant $\tfrac12-\tfrac12=0$ terms cancel). (Check: substituting the middle branch back into $y''+3y'+2y$ on $1

Interval	$y(t)$
$0\le t<1$	$0$
$1\le t<2$	$\tfrac12-e^{-(t-1)}+\tfrac12e^{-2(t-1)}$
$t\ge2$	$\left[e^{-(t-2)}-e^{-(t-1)}\right]+\tfrac12\left[e^{-2(t-1)}-e^{-2(t-2)}\right]$

Concept & Theory

Topic. Laplace-transform solution of a forced linear ODE with a piecewise (square-wave) input, via unit step functions.

Key relations. Second shifting theorem: if $\mathcal L\{f(t)\}=F(s)$ then $\mathcal L\{f(t-a)u(t-a)\}=e^{-as}F(s)$. A rectangular pulse active on $[a,b)$ is $u(t-a)-u(t-b)$.

Why this works. The response to any shifted, scaled step is just the unit-step response $h$ shifted and gated by the same $u(t-a)$ — so once $h(\tau)$ is found from the transfer function alone (no forcing specifics needed), both step edges of the square wave reuse it directly with no fresh partial-fraction work.

Common pitfall. Forgetting the $u(t-a)$ gating factor and writing $h(t-a)$ as valid for all $t$ (it must be zero before $t=a$); sign errors subtracting the two shifted copies of $h$ for $t\ge2$; not simplifying the constant terms that cancel in the $t\ge2$ branch.

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

Question 8 of 8: Damped Mass–Spring Response to a Square-Wave Force

Question 8: Damped Mass–Spring Response to a Square-Wave Force 20 marks