Will now determine the maximum overshoot called \(OS_{\max }\) as shown in this diagram
We will use \(\frac{dy}{dt}=0\) to find \(y_{\max }\). Since \(\frac{dy}{dt}=0\) will generate many solutions, we will take the first one.\[ y\left ( t\right ) =1-\frac{e^{-\xi \omega _{n}t}}{\sqrt{1-\xi ^{2}}}\sin \left ( \omega _{d}t+\phi \right ) \] Then\begin{align} \frac{dy}{dt} & =0=\xi \omega _{n}\frac{e^{-\xi \omega _{n}t}}{\sqrt{1-\xi ^{2}}}\sin \left ( \omega _{d}t+\phi \right ) -\frac{e^{-\xi \omega _{n}t}}{\sqrt{1-\xi ^{2}}}\omega _{d}\cos \left ( \omega _{d}t+\phi \right ) \nonumber \\ 0 & =\xi \sin \left ( \omega _{d}+\phi \right ) -\sqrt{1-\xi ^{2}}\cos \left ( \omega _{d}t+\phi \right ) \tag{1} \end{align}
To solve this, since \(\cos \phi =\xi \) and \(\sin \phi =\sqrt{1-\xi ^{2}}\)
then (1) becomes\[ 0=\cos \phi \sin \left ( \omega _{d}t+\phi \right ) -\sin \phi \cos \left ( \omega _{d}t+\phi \right ) \] Using \(\sin \left ( A-B\right ) =\cos A\sin B-\sin A\cos B\) the above can be written as (using \(A=\phi \))\[ 0=\sin \left ( \phi -\left ( \omega _{d}t+\phi \right ) \right ) \]
Hence
\[ \sin \left ( \omega _{d}t\right ) =0 \]
The solution is \(\omega _{d}=\omega _{n}\sqrt{1-\xi ^{2}}t=k\pi \) for \(k=0,1,2,\cdots \). We pick \(k=1\) since this is the first one after \(t=0\), hence\begin{align*} \omega _{n}\sqrt{1-\xi ^{2}}t_{\max } & =\pi \\ t_{\max } & =\frac{\pi }{\omega _{n}\sqrt{1-\xi ^{2}}}\\ & =\frac{\pi }{\omega _{d}} \end{align*}
To find \(y_{\max }\left ( t\right ) =y\left ( t_{\max }\right ) \), we plug the above \(t_{\max }\) back in the original solution which is \(y\left ( t\right ) =1-\frac{e^{-\xi \omega _{n}t}}{\sqrt{1-\xi ^{2}}}\sin \left ( \omega _{d}t+\phi \right ) \). Hence\begin{align} OS_{\max } & =y\left ( t_{\max }\right ) -1\nonumber \\ & =\left ( 1-\frac{e^{-\xi \omega _{n}t_{\max }}}{\sqrt{1-\xi ^{2}}}\sin \left ( \omega _{d}t_{\max }+\phi \right ) \right ) -1 \tag{2} \end{align}
Reader: Show that the above reduces to\[ OS_{\max }=e^{\frac{-\pi \xi }{\sqrt{1-\xi ^{2}}}}\]
Reader solution: substitute \(t_{\max }=\frac{\pi }{\omega _{n}\sqrt{1-\xi ^{2}}}\) in (2) gives
\begin{align*} OS_{\max } & =-\frac{e^{-\xi \omega _{n}\left ( \frac{\pi }{\omega _{n}\sqrt{1-\xi ^{2}}}\right ) }}{\sqrt{1-\xi ^{2}}}\sin \left ( \omega _{d}\left ( \frac{\pi }{\omega _{d}}\right ) +\phi \right ) \\ & =-\frac{e^{\frac{-\pi \xi }{\sqrt{1-\xi ^{2}}}}}{\sqrt{1-\xi ^{2}}}\sin \left ( \pi +\phi \right ) \end{align*}
But \(\sin \left ( \pi +\phi \right ) =-\sin \phi \) which is \(-\sqrt{1-\xi ^{2}}\), hence the above becomes
\begin{align*} OS_{\max } & =-\frac{e^{\frac{-\pi \xi }{\sqrt{1-\xi ^{2}}}}}{\sqrt{1-\xi ^{2}}}\left ( -\sqrt{1-\xi ^{2}}\right ) \\ & =e^{\frac{-\pi \xi }{\sqrt{1-\xi ^{2}}}} \end{align*}
Notice the overshoot do not depend on \(\omega _{n}\). It only depends on damping. There are two ways to change damping. Either change the system itself, or add a controller to compensate.
Second main property of second order system is resonance. This arises in the frequency context. When the frequency the system is operating at is close to the natural frequency of the system. We are now interested in \(\left \vert G\left ( j\omega \right ) \right \vert \) vs. \(\omega \). We will call the resonance frequency \(\omega _{r}\) and \(\left \vert G\left ( j\omega _{r}\right ) \right \vert =M_{r}\). From\begin{align*} G\left ( s\right ) & =\frac{\omega _{n}^{2}}{s^{2}+2\xi \omega _{n}s+\omega _{n}^{2}}\\ \left \vert G\left ( j\omega \right ) \right \vert & =\frac{\omega _{n}^{2}}{\sqrt{\left ( \omega _{n}^{2}-\omega ^{2}\right ) ^{2}+4\xi ^{2}\omega _{n}^{2}\omega ^{2}}} \end{align*}
To find where this is maximum, \[ \frac{d}{d\omega }\left \vert G\left ( j\omega \right ) \right \vert =0 \] To simplify, we will instead use \(\left \vert G\left ( j\omega \right ) \right \vert ^{2}\) to get rid of the square root of the denominator giving
\[ \left \vert G\left ( j\omega \right ) \right \vert =\frac{\omega _{n}^{4}}{\left ( \omega _{n}^{2}-\omega ^{2}\right ) ^{2}+4\xi ^{2}\omega _{n}^{2}\omega ^{2}}\]
Then the maximum is where the denominator is minimum. Hence \begin{align*} \frac{d}{d\omega }\left ( \left ( \omega _{n}^{2}-\omega ^{2}\right ) ^{2}+4\xi ^{2}\omega _{n}^{2}\omega ^{2}\right ) & =0\\ 2\left ( \omega _{n}^{2}-\omega ^{2}\right ) 2\omega +8\xi ^{2}\omega _{n}^{2}\omega & =0\\ \omega _{r} & =\omega _{n}\sqrt{1-2\xi ^{2}} \end{align*}
So the above \(\omega _{r}\) is where \(G\left ( j\omega \right ) \) is maximum. To find \(M_{r}\) we plug-in is \(\omega _{r}\) in place of \(\omega \) in \(\left \vert G\left ( j\omega \right ) \right \vert \).
Reader:
Show that \[ M_{r}=\left \vert G\left ( j\omega _{r}\right ) \right \vert =\frac{1}{2\xi \sqrt{1-\xi ^{2}}}\]
Reader answer:
From\[ \left \vert G\left ( j\omega \right ) \right \vert =\frac{\omega _{n}^{2}}{\sqrt{\left ( \omega _{n}^{2}-\omega ^{2}\right ) ^{2}+4\xi ^{2}\omega _{n}^{2}\omega ^{2}}}\]
Replacing \(\omega \) in the above by \(\omega _{r}=\omega _{n}\sqrt{1-2\xi ^{2}}\) and working out the algebra gives \(\left \vert G\left ( j\omega =\omega _{r}\right ) \right \vert =\frac{1}{2\xi \sqrt{1-\xi ^{2}}}\). To verify, here is small Matlab code