Question 7 of 7: Steady-State Response of a Damped Mass-Spring System to a Periodic Square Wave

Question 7: Steady-State Response of a Damped Mass-Spring System to a Periodic Square Wave (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 underdamped-parameter (actually overdamped: see roots below) second-order system driven by a period-2 square wave of height 2 on the first half-period.

Find. The response $y(t)$.

Approach. Expand the periodic forcing $r(t)$ as a Fourier series, then superpose the particular (steady-state) response to each harmonic using the standard frequency-response formula for a damped linear oscillator; the transient homogeneous part decays and does not affect the long-time response.

1. Homogeneous roots (confirm decay). $r^2+5r+3=0\Rightarrow r=\dfrac{-5\pm\sqrt{13}}{2}\approx-0.697,\,-4.303$, both negative, so any transient $C_1e^{r_1t}+C_2e^{r_2t}\to0$ and the long-time response is exactly the periodic particular solution found below.
2. Fourier series of $r(t)$. Period $T=2$, fundamental frequency $\omega_n=n\pi$. The average is $a_0/2=1$. Since $r$ is a rectangular pulse of height 2 on the first half of each period, $$a_n=\frac1{1}\int_0^2r(t)\cos(n\pi t)\,dt=0\ \text{(all }n\text{)},\qquad b_n=\int_0^2 r(t)\sin(n\pi t)\,dt=\frac{2}{n\pi}\big(1-(-1)^n\big)=\begin{cases}\dfrac{4}{n\pi}&n\text{ odd}\\0&n\text{ even}\end{cases}$$ $$r(t)=1+\sum_{n\ \text{odd}}\frac{4}{n\pi}\sin(n\pi t).$$
3. Response to the constant term. For $y''+5y'+3y=1$, the constant particular solution solves $3y_p=1\Rightarrow y_p=\tfrac13$.
4. Response to each harmonic. For $y''+5y'+3y=B_n\sin(\omega_nt)$ with $\omega_n=n\pi$, try $y_p=M_n\sin(\omega_nt)+N_n\cos(\omega_nt)$. Writing $a_n^\ast=3-\omega_n^2$, matching coefficients gives the linear system $a_n^\ast M_n-5\omega_nN_n=B_n$, $5\omega_nM_n+a_n^\ast N_n=0$, whose solution is $$M_n=\frac{B_n\,a_n^\ast}{(a_n^\ast)^2+25\omega_n^2},\qquad N_n=\frac{-5\omega_nB_n}{(a_n^\ast)^2+25\omega_n^2}.$$
5. Assemble the steady-state response. Summing the constant term and all odd harmonics ($B_n=4/(n\pi)$, $\omega_n=n\pi$, $a_n^\ast=3-n^2\pi^2$):

$$y_{ss}(t)=\boxed{\frac13+\sum_{\substack{n=1\\n\ \text{odd}}}^{\infty}\frac{4/(n\pi)}{(3-n^2\pi^2)^2+25n^2\pi^2}\Big[(3-n^2\pi^2)\sin(n\pi t)-5n\pi\cos(n\pi t)\Big]} $$