Vibration of Mass-Spring-Damper System (2) – Step Input Responses
Using Laplace Transform to find the time response without solving the differential equation
Consider the time responses of the mass-spring-damper system (2nd order system).
Use Laplace Transform to Equation (1), \[ m \ddot{x} + c \dot{x} + kx = F(t) \] \[ m \{ s^2 X(s) -sx(0) -\dot{x}(0) \} + c \{ sX(s) -x(0) \} +kX(s) = F(s) \\ \] \[ (ms^2 + cs + k)X(s) + (-ms -c)x(0) -m\dot{x}(0) = F(s) \] \[ \therefore X(s) = \dfrac{F(s)}{ms^2 + cs + k} + \dfrac{(ms+c)x(0) + m\dot{x}(0)}{ms^2 + cs + k} \] The first term represents the forced response and the second term the response due to the initial conditions.
Considering the initial condition \(x(0) = \dot{x}(0) = 0\), the transfer function is \[ \dfrac{X(s)}{F(s)} = \dfrac{1}{ms^2 + cs + k} = \dfrac{1}{m} \cdot \dfrac{1}{s^2 + 2\zeta\omega_{n}s + \omega_{n}^2} \]
For a unit step input \(f(t) = 1 \) (constant), the corresponding Laplace Transform is \(F(s) = \dfrac{1}{s} \)
Then the time response \(x(t)\) can be obtained from the inverse Laplace Transform of \(X(s) =\dfrac{1}{ms^2 + cs + k} \cdot \dfrac{1}{s} \)
Normalised second-order system
Applying a step input of \(f(t) = k\) to the above system leads to a unit steady-state response of 1 m.
Alternatively, let the normalised second-order system be such one that has the same steady-state response of 1 m with the unit step input \(f(t) = 1\). Then the normalised system has the transfer function of \[ \dfrac{X(s)}{F(s)} = \dfrac{k}{ms^2 + cs + k} = \dfrac{\omega_{n}^2}{s^2 + 2\zeta\omega_{n}s + \omega_{n}^2}\]
(i) if \(0<\zeta <1\) (underdamped)
Find the time response \(x(t)\) of the normalised second-order system applied with the unit step input \(f(t) = 1 \), which is \(F(s) = \dfrac{1}{s} \). \[ \begin{align} X(s) &= \dfrac{1}{s} \cdot \dfrac{\omega_{n}^2}{s^2 + 2\zeta\omega_{n}s + \omega_{n}^2} \\ &= \dfrac{1}{s} \cdot \dfrac{\omega_{n}^2}{s^2 + 2\zeta\omega_{n}s + \zeta^2\omega_{n}^2 + \omega_{n}^2 -\zeta^2\omega_{n}^2} \\ &= \dfrac{1}{s} \cdot \dfrac{\omega_{n}^2}{ (s + \zeta\omega_{n})^2 + \omega_{n}^2(1-\zeta^2)} \\ &= \dfrac{1}{s} \cdot \dfrac{a^2+b^2}{ (s + a)^2 + b^2} \\ \end{align} \] where \[ \left\{ \begin{array}{l} \begin{align} a &= \zeta\omega_{n} \\ b &=\omega_{n}\sqrt{1-\zeta^2} = \omega_{d} \end{align} \end{array} \right. \] Partial fraction expansion gives \[ \begin{align} X(s) &= \dfrac{1}{s} + \dfrac{-s-2a}{ (s + a)^2 + b^2} \\ &= \dfrac{1}{s} \; – \dfrac{s+a}{ (s + a)^2 + b^2} \; – \dfrac{a}{b} \cdot \dfrac{b}{(s + a)^2 + b^2}\\ \end{align} \] The time response is \[ \begin{align} x(t) &= 1 \;- e^{-at}\cos{bt} \;- \dfrac{a}{b}e^{-at}\sin{bt} \\ &= 1 \;- \dfrac{1}{b} e^{-at} ( b\cos{bt} + a\sin{bt} ) \\ &= 1 \;- \dfrac{e^{-\zeta\omega_{n}t}}{\omega_{n}\sqrt{1-\zeta^2}} ( \omega_{n}\sqrt{1-\zeta^2}\cos{\omega_{d}t} + \omega_{n}\zeta\sin{\omega_{d}t} ) \\ &= 1 \;- \dfrac{e^{-\zeta\omega_{n}t}}{\sqrt{1-\zeta^2}} ( \sqrt{1-\zeta^2}\cos{\omega_{d}t} + \zeta\sin{\omega_{d}t} ) \\ &= 1 \;- \dfrac{e^{-\zeta\omega_{n}t}}{\sqrt{1-\zeta^2}} \sin{(\omega_{d}t + \theta)} \\ \end{align} \] where \[ \theta = \tan^{-1}{\dfrac{\omega_{n}\sqrt{1-\zeta^2}}{\zeta\omega_{n}}} \]
(ii) if \(\zeta =1\) (critically damped)
Find the time response \(x(t)\) of the normalised second-order system applied with a unit step input \(f(t) = 1 \). \[ \begin{align} X(s) &= \dfrac{1}{s} \cdot \dfrac{\omega_{n}^2}{s^2 + 2\zeta\omega_{n}s + \omega_{n}^2} \\ &= \dfrac{1}{s} \cdot \dfrac{a^2+b^2}{ (s + a)^2 + b^2} \\ &= \dfrac{1}{s} \cdot \dfrac{a^2}{ (s + a)^2} \quad (\because b = 0) \\ \end{align} \] where \[ \left\{ \begin{array}{l} \begin{align} a &= \zeta\omega_{n} = \omega_{n} \\ b &=\omega_{n}\sqrt{1-\zeta^2} = \omega_{d} = 0 \end{align} \end{array} \right. \] Partial fraction expansion gives \[ \begin{align} X(s) &= \dfrac{1}{s} \;- \dfrac{s+2a}{ (s + a)^2} \\ &= \dfrac{1}{s} \;- \dfrac{1}{s+a} – \dfrac{a}{(s+a)^2} \\ \end{align} \] The time response is \[ \begin{align} x(t) &= 1 \;- e^{-at} \;-ate^{-at} \\ &= 1 \;- (1+at)e^{-at} \\ &= 1 \;- (1+\omega_{n}t)e^{-\omega_{n}t} \\ \end{align} \]
(iii) if \(\zeta >1\) ( overdamped)
Find the time response \(x(t)\) of the original second-order system applied with a unit step input \(f(t) = 1 \).
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
\[
\left\{
\begin{array}{l} \begin{align}
\alpha_{1} &= \zeta\omega_{n} + \omega_{n} \sqrt{\zeta^2-1} \\
\alpha_{2} &= \zeta\omega_{n} \;- \omega_{n} \sqrt{\zeta^2-1} \\
\end{align} \end{array}
\right.
\]
Then
\[
\left\{
\begin{array}{l}
\alpha_{1}\alpha_{2} &= -\zeta^2\omega_{n}^2 \; – \omega_{n}^2 (\zeta^2-1) = \omega_{n}^2\\
\alpha_{1}+\alpha_{2}&= 2\zeta\omega_{n} \\
\end{array}
\right.
\]
\[
\begin{align}
X(s) &= \dfrac{1}{s} \cdot \dfrac{1}{s^2 + 2\zeta\omega_{n}s + \omega_{n}^2} \\
&= \dfrac{1}{s} \cdot \dfrac{1}{ (s + \alpha_{1})(s + \alpha_{2})} \\
\end{align}
\]
From the Inverse Laplace Transform of the ‘asymptotic double exponential’, the time response is
\[
\begin{align}
x(t) &= \dfrac{1}{\alpha_{1}\alpha_{2}} \Bigl(1- \dfrac{\alpha_{1}e^{-\alpha_{2}t} – \alpha_{2}e^{-\alpha_{1}t} }{\alpha_{1}-\alpha_{2}} \Bigr) \\
&= \dfrac{1}{\omega_{n}^2} \Bigl(1- \dfrac{(\zeta\omega_{n} + \omega_{n} \sqrt{\zeta^2-1})e^{-(\zeta\omega_{n} – \omega_{n} \sqrt{\zeta^2-1})t} – (\zeta\omega_{n} – \omega_{n} \sqrt{\zeta^2-1})e^{-(\zeta\omega_{n} + \omega_{n} \sqrt{\zeta^2-1})t} }{(\zeta\omega_{n} + \omega_{n} \sqrt{\zeta^2-1})-(\zeta\omega_{n} – \omega_{n} \sqrt{\zeta^2-1})} \Bigr) \\
& = \dfrac{1}{\omega_{n}^2} \Bigl(1 \; – \dfrac{ (\zeta+\sqrt{\zeta^2-1})e^{-\omega_{n}(\zeta-\sqrt{\zeta^2-1})t} – (\zeta-\sqrt{\zeta^2-1})e^{-\omega_{n}(\zeta+\sqrt{\zeta^2-1})t} }{2\sqrt{\zeta^2-1}} \Bigr)
\end{align}
\]
Stability analysis – Root Locus
The stability of a system is the ability to recover to its original operating condition after the disturbance disappears.
The characteristic equation is the denominator of the transfer function. The location of the roots \(s_{1}\), \(s_{2}\) of the characteristic equation in the complex s-plane indicates the stability of the system.
For the stability, all the roots need to have a negative real part, i.e. all the roots are in the left half of the s-plane so that the corresponding transient responses are decaying exponentials \(e^{-t} \), \( e^{-2t}\) etc.
The more negative the real part of the root, the quicker the transient decay of the response.
The larger the imaginary part of the root, the higher the frequency of the oscillation.
For the mass spring damper system, the characteristic equation is \[ ms^2+cs+k = m(s^2 + 2\zeta\omega_{n}s + \omega_{n}^2) =0 \] \[ s^2 + 2\zeta\omega_{n}s + \omega_{n}^2 = 0 \] The roots are \[ s = -\zeta \omega_{n} \pm \omega_{n} \sqrt{\zeta^2 -1} \quad = -\zeta \omega_{n} \pm i\omega_{n} \sqrt{1 – \zeta^2} \\ \]
| (i) if \(0<\zeta<1\) (underdamped) | the roots are two complex values | \(s = -\zeta \omega_{n} \pm i\omega_{n} \sqrt{1 – \zeta^2} \) |
| (ii) if \(\zeta = 1\) (critically damped) | the roots are repeated | \(s = -\zeta \omega_{n} \) |
| (iii) if \(\zeta > 1\) (overdamped) | the roots are two real values | \(s = -\zeta \omega_{n} \pm \omega_{n} \sqrt{\zeta^2 -1} \) |
For underdamping
The distance from the origin to the root is \(\omega_n\).
\(\cos\theta = \zeta\) i.e. small \(\theta\) means high damping \(\zeta\) close to 1. large \(\theta\) means small damping \(\zeta\) close to 0.
Reference
[1] Thomson, Theory of Vibration with Applications.
[2] Bolton, Control Systems.
[3] Cheever, The Unit Step Response. https://lpsa.swarthmore.edu/Transient/TransInputs/TransStep.html