Vibration of Mass-Spring-Damper System (3) – Impulse Responses
Similar to the unit input responses, the time response of the mass-spring-damper system with impulse input can be analytically found.
The Dirac delta function is
\[\delta(x)=\left\{\begin{array}{ll}
+\infty, & x=0 \\
0, & x \neq 0
\end{array}\right.
\]
which also satisfies
\[
\int_{-\infty}^{\infty} \delta(x) d x=1
\]
The Laplace Transform of the delta function is
\[
\mathcal{L}\{\delta(t)\} = 1
\]
(i) if \(0<\zeta <1\) (underdamped)
Find the time response \(x(t)\) of the normalised second-order system applied with the impulse input \(f(t)\), which is \(F(s) = 1\). \[ \begin{align} X(s) &= 1 \cdot \dfrac{\omega_{n}^2}{s^2 + 2\zeta\omega_{n}s + \omega_{n}^2} \\ &= \dfrac{\omega_{n}^2}{s^2 + 2\zeta\omega_{n}s + \zeta^2\omega_{n}^2 + \omega_{n}^2 -\zeta^2\omega_{n}^2} \\ &= \dfrac{\omega_{n}^2}{ (s + \zeta\omega_{n})^2 + \omega_{n}^2(1-\zeta^2)} \\ &= \dfrac{a^2+b^2}{ (s + a)^2 + b^2} \\ &= \dfrac{a^2+b^2}{b} \dfrac{b}{ (s + a)^2 + b^2} \\ \end{align} \] where \[ \left\{ \begin{array}{l} \begin{align} a &= \zeta\omega_{n} \\ b &=\omega_{n}\sqrt{1-\zeta^2} = \omega_{d} \end{align} \end{array} \right. \] The time response is \[ \begin{align} x(t) &= \dfrac{a^2+b^2}{b} e^{-at}\sin{bt} \\ &= \dfrac{\omega_{n}^2}{\omega_{n} \sqrt{1-\zeta^2}} e^{-\zeta\omega_{n}t}\sin{\omega_{d}t}\\ &= \dfrac{\omega_{n}}{\sqrt{1-\zeta^2}} e^{-\zeta\omega_{n}t}\sin{\omega_{d}t}\\ \end{align} \]
(ii) if \(\zeta =1\) (critically damped)
Find the time response \(x(t)\) of the normalised second-order system applied with the impulse input \(f(t)\). \[ \begin{align} X(s) &= 1 \cdot \dfrac{\omega_{n}^2}{s^2 + 2\zeta\omega_{n}s + \omega_{n}^2} \\ &= \dfrac{a^2+b^2}{ (s + a)^2 + b^2} \\ &= \dfrac{a^2}{ (s + a)^2} \quad (\because b = 0) \\ \end{align} \] where \[ \left\{ \begin{array}{l} \begin{align} a &= \zeta\omega_{n} = \omega_{n} \\ b &=\omega_{n}\sqrt{1-\zeta^2} = \omega_{d} = 0 \end{align} \end{array} \right. \] The time response is \[ \begin{align} x(t) &= a^2 te^{-at} \\ &= \omega_{n}^2 te^{-\omega_{n}t} \\ \end{align} \]
(iii) if \(\zeta >1\) ( overdamped)
Find the time response \(x(t)\) of the original second-order system applied with the impulse input \(f(t)\).
The characteristic equation \(s^2 + 2\zeta\omega_{n}s + \omega_{n}^2 = 0\) has the two real roots \(s = -\zeta\omega_{n} \pm \omega_{n} \sqrt{\zeta^2-1}\)
Let \(\alpha_{1}\) and \(\alpha_{2}\) be the absolute values of the two roots
\[
\left\{
\begin{array}{l} \begin{align}
\alpha_{1} &= \zeta\omega_{n} + \omega_{n} \sqrt{\zeta^2-1} \\
\alpha_{2} &= \zeta\omega_{n} \;- \omega_{n} \sqrt{\zeta^2-1} \\
\end{align} \end{array}
\right.
\]
\[
\begin{align}
X(s) &= 1 \cdot \dfrac{1}{s^2 + 2\zeta\omega_{n}s + \omega_{n}^2} \\
&= \dfrac{1}{ (s + \alpha_{1})(s + \alpha_{2})} \\
\end{align}
\]
From the Inverse Laplace Transform of the ‘dual exponential’, the time response is
\[
\begin{align}
x(t) &= \dfrac{1}{\alpha_{1} \;- \alpha_{2}} \Bigl( e^{-\alpha_{2}t} – e^{-\alpha_{1}t} \Bigr) \\
&= \dfrac{1}{2\omega_{n}\sqrt{\zeta^2-1}} \Bigl[ e^{-\omega_{n}(\zeta-\sqrt{\zeta^2-1})t} – e^{-\omega_{n}(\zeta+\sqrt{\zeta^2-1})t} \Bigr] \\
\end{align}
\]
Reference
[1] Cheever, Common Laplace Transform Pairs. https://lpsa.swarthmore.edu/LaplaceZTable/Common%20Laplace%20Transform%20Pairs.pdf