9.3: General Solution for Underdamped Second Order Systems
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We want to solve Equation 9.2.2 for output \(x(t)\) during positive time \(t>0\), given any input \(u(t)\), and given appropriate initial conditions at \(t = 0\). We will use Laplace transformation with application of the inverse convolution transform. To simplify the notation, let us denote \(X(s) \equiv L[x(t)]\). Transforming Equation 9.2.2 with use of Equation 2.2.11 gives
\[s^{2} X(s)-s x(0)-\dot{x}(0)+2 \zeta \omega_{n}[s X(s)-x(0)]+\omega_{n}^{2} X(s)=\omega_{n}^{2} L[u(t)] \nonumber \]
We denote the two initial conditions, \(x_{0} \equiv x(0)\) initial “position” and \(\dot{x}_{0} \equiv \dot{x}(0)\) initial “velocity”. Collecting terms algebraically and rearranging the equation gives
\[\left(s^{2}+2 \zeta \omega_{n} s+\omega_{n}^{2}\right) X(s)=\left(s+2 \zeta \omega_{n}\right) x_{0}+\dot{x}_{0}+\omega_{n}^{2} L[u(t)]\label{eqn:9.14} \]
Note that up to this point in the derivation, no restriction has been placed on the value of damping ratio \(\zeta\).
To cast Equation \(\ref{eqn:9.14}\) into an easily solvable form, we use two algebraic tricks that are not obvious a priori. The first trick is to re-write the left-hand-side quadratic term [which, essentially, is the same as the quadratic term in characteristic Equation 9.1.4]:
\[s^{2}+2 \zeta \omega_{n} s+\omega_{n}^{2}=\left(s+\zeta \omega_{n}\right)^{2}+\omega_{n}^{2}-\left(\zeta \omega_{n}\right)^{2}=\left(s+\zeta \omega_{n}\right)^{2}+\omega_{d}^{2}\label{eqn:9.15} \]
We shall regard \(\omega_{d}^{2}\) as a positive parameter in the following, so Equation \(\ref{eqn:9.15}\) is nominally valid only for an underdamped system (\(0 \leq \zeta<1\)). The second trick is to split into a particular form the IC terms on the right-hand side of Equation \(\ref{eqn:9.14}\):
\[\left(s+2 \zeta \omega_{n}\right) x_{0}+\dot{x}_{0}=\left(s+\zeta \omega_{n}\right) x_{0}+\left(\dot{x}_{0}+\zeta \omega_{n} x_{0}\right)\label{eqn:9.16} \]
Applying Equations \(\ref{eqn:9.15}\) and \(\ref{eqn:9.16}\) to Equation \(\ref{eqn:9.14}\), then solving for \(X(s)\) gives:
\[X(s)=\frac{s+\zeta \omega_{n}}{\left(s+\zeta \omega_{n}\right)^{2}+\omega_{d}^{2}} x_{0}+\frac{\omega_{d}}{\left(s+\zeta \omega_{n}\right)^{2}+\omega_{d}^{2}}\left(\frac{\dot{x}_{0}+\zeta \omega_{n} x_{0}}{\omega_{d}}\right)+\frac{\omega_{n}^{2}}{\omega_{d}}\overbrace{\frac{\omega_{d}}{\left(s+\zeta \omega_{n}\right)^{2}+\omega_{d}^{2}}}^{F_{1}(s)}\overbrace{L[u(t)]}^{F_{2}(s)}\label{eqn:9.17} \]
The motivation for casting the solution into form Equation \(\ref{eqn:9.17}\) is a relevant general Laplace transform pair that has not appeared previously in this book: given a function \(f(t)\), its Laplace transform \(F(s)\), and the exponential function \(e^{\sigma t}\), where \(\sigma\) is a constant, then the Laplace transform of the product \(e^{\sigma t} f(t)\) is
\[L\left[e^{\sigma t} f(t)\right]=\int_{t=0}^{t=\infty} e^{-s t} e^{\sigma t} f(t) d t=\int_{t=0}^{t=\infty} e^{-(s-\sigma) t} f(t) d t \equiv F(s-\sigma)\label{eqn:9.18} \]
The associated inverse transform is
\[L^{-1}[F(s-\sigma)]=e^{\sigma t} f(t)\label{eqn:9.19} \]
Returning to Equation \(\ref{eqn:9.17}\), we identify \(\sigma=-\zeta \omega_{n}\) and invert the two IC-response terms using Equation \(\ref{eqn:9.19}\) in conjunction with sine and cosine transforms Equation 2.4.7 and Equation 2.4.8. To invert the forced-response term, we apply both Equation \(\ref{eqn:9.19}\) and inverse convolution transform Equation 6.1.5. This leads to the two equivalent general equations for output \(x(t)\) of an underdamped 2nd order system:
\[x(t) = \overbrace{e^{-\zeta \omega_{n} t}\left[x_{0} \cos \omega_{d} t+\left(\frac{\dot{x}_{0}+\zeta \omega_{n} x_{0}}{\omega_{d}}\right) \sin \omega_{d} t\right.}^{\text{IC response}} + \overbrace{\frac{\omega_{n}^{2}}{\omega_{d}} \int_{\tau=0}^{\tau=t} e^{-\zeta \omega_{n} \tau} \sin \omega_{d} \tau \times u(t-\tau) d \tau}^{\text{forced response}}\label{eqn:9.20a} \]
\[x(t)=e^{-\zeta \omega_{n} t}\left[x_{0} \cos \omega_{d} t+\left(\frac{\dot{x}_{0}+\zeta \omega_{n} x_{0}}{\omega_{d}}\right) \sin \omega_{d} t\right]+\frac{\omega_{n}^{2}}{\omega_{d}} \int_{\tau=0}^{\tau=t} e^{-\zeta \omega_{n}(t-\tau)} \sin \omega_{d}(t-\tau) \times u(\tau) d \tau\label{eqn:9.20b} \]
Note the emphasis that Equations \(\ref{eqn:9.20a}\) and \(\ref{eqn:9.20b}\) are valid for underdamped systems. This solution is nominally not valid for overdamped systems, although we will see in Section 9.10 that it can be converted easily. The nominal restriction to underdamped systems stems from the use of sinusoidal transforms Equation 2.4.7 and Equation 2.4.8, which are valid in this case only for positive \(\omega_{d}^{2}\), which holds only if \(0 \leq \zeta<1\), from Equation 9.1.12. For example, we used the inverse transform \(L^{-1}\left[\frac{\omega}{s^{2}+\omega^{2}}\right]=\sin \omega t\), valid for \(\omega^{2}>0\), in order to find Equations \(\ref{eqn:9.20a}\) and \(\ref{eqn:9.20b}\), but the following transform holds for a negative term in the denominator: \(L^{-1}\left[\frac{\omega}{s^{2}-\omega^{2}}\right]= \sinh \omega t\), a hyperbolic sine.