Laplace-Transform Response of a Damped Mass-Spring System to a Square-Wave Force: 04-BS-1, December 2017, Q6

Question 6: Laplace-Transform Response of a Damped Mass-Spring System 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. Overdamped mass-spring system $y''+3y'+2y=r(t)$, $y(0)=0,\ y'(0)=0$, with a unit-height rectangular force pulse $r(t)=u(t-1)-u(t-2)$.

Find. $y(t)$ for all $t\geq0$.

Approach. Laplace-transform the ODE (zero ICs eliminate all boundary terms), express $R(s)$ using unit-step functions, and use the second shifting theorem to invert term by term without ever needing an explicit convolution integral.

Transform the ODE. With zero initial conditions, $\mathcal L\{y''+3y'+2y\}=(s^2+3s+2)Y(s)$, so $$Y(s)=\frac{R(s)}{(s+1)(s+2)}.$$ The forcing $r(t)=u(t-1)-u(t-2)$ has $R(s)=\dfrac{e^{-s}-e^{-2s}}{s}$, so $$Y(s)=\bigl(e^{-s}-e^{-2s}\bigr)\,G(s),\qquad G(s)=\frac1{s(s+1)(s+2)}.$$
Partial fractions of $G(s)$ and its inverse transform. $$G(s)=\frac1{s(s+1)(s+2)}=\frac{1/2}{s}-\frac1{s+1}+\frac{1/2}{s+2}\quad\Rightarrow\quad g(t)=\mathcal L^{-1}\{G(s)\}=\frac12-e^{-t}+\frac12e^{-2t}.$$
Apply the second shifting theorem. Since $\mathcal L^{-1}\{e^{-as}G(s)\}=g(t-a)u(t-a)$, $$y(t)=g(t-1)u(t-1)-g(t-2)u(t-2).$$
Write the piecewise response. With $g(\tau)=\tfrac12-e^{-\tau}+\tfrac12e^{-2\tau}$ for $\tau\ge0$, $$y(t)=\begin{cases}0,&t<1,\\[4pt]\tfrac12-e^{-(t-1)}+\tfrac12e^{-2(t-1)},&1\le t<2,\\[4pt]-e^{-(t-1)}+\tfrac12e^{-2(t-1)}+e^{-(t-2)}-\tfrac12e^{-2(t-2)},&t\ge2.\end{cases}$$ (The middle and third lines agree at $t=2$ from the left/right by continuity of $y$, as required for a bounded forcing with no impulses.)

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

Quantity	Result
Step response $g(t)$	$\tfrac12-e^{-t}+\tfrac12e^{-2t}$
$y(t)$, $t<1$	$0$
$y(t)$, $1\le t<2$	$\tfrac12-e^{-(t-1)}+\tfrac12e^{-2(t-1)}$
$y(t)$, $t\ge2$	$g(t-1)-g(t-2)$ (formula above)

Concept & Theory

Topic. Laplace-transform solution of a linear ODE forced by a unit-step/rectangular-pulse input.

Key relations. Second shifting theorem: $\mathcal L^{-1}\{e^{-as}F(s)\}=f(t-a)u(t-a)$. A rectangular pulse active on $[1,2)$ is $u(t-1)-u(t-2)$, so its Laplace transform is $(e^{-s}-e^{-2s})/s$.

Why this works. Writing $R(s)=(e^{-s}-e^{-2s})/s$ and multiplying by the transfer function $1/[(s+1)(s+2)]$ gives $Y(s)$ as a sum of two shifted copies of the single step response $G(s)=1/[s(s+1)(s+2)]$ — this is exactly what the second shifting theorem is built for, and it avoids ever setting up an explicit convolution integral $\int r(\tau)h(t-\tau)d\tau$.

Common pitfall. Confusing the transfer function's impulse response $h(t)=e^{-t}-e^{-2t}$ (inverse of $1/[(s+1)(s+2)]$ alone) with the actual building block needed here, $g(t)=\mathcal L^{-1}\{1/[s(s+1)(s+2)]\}$ — the extra factor of $1/s$ from $R(s)$'s own step-function transform must not be dropped, since $r(t)$ is a finite-height pulse, not an impulse.

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

Question 6 of 8: Laplace-Transform Response of a Damped Mass-Spring System to a Square-Wave Force

Question 6: Laplace-Transform Response of a Damped Mass-Spring System to a Square-Wave Force (20 marks)