We have the standard second order system
\[ G\left ( s\right ) =\frac{\omega _{n}^{2}}{s^{2}+2\xi \omega _{n}s+\omega _{n}^{2}}\]
The magnitude of the frequency response, for \(0<\zeta <1\) would look something like
As \(\zeta \) becomes smaller (system less damped) then at resonance, the amplitude of the steady state response will increase, since \(M_{r}=\left \vert G\left ( j\omega _{r}\right ) \right \vert =\frac{1}{2\xi \sqrt{1-\xi ^{2}}}\)
Second order system typical concerns are
What to do when these requirements (some or all) do not meet our requirements? We add a controller and use feedback. Today we will talk about adding a controller (pure gain \(k\)) and feedback to address the problem of designing to meet the requirement of specification made on amount of overshoot while keeping the final steady state value within some acceptable limit.
Lets take the open loop \(G\left ( s\right ) \) and say we have \(\zeta =0.01\), then the overshoot \begin{align*} y_{\max } & =1+OS_{\max }\\ & =1+e^{\frac{-\pi \zeta }{\sqrt{1-\zeta ^{2}}}}\\ & =1+e^{\frac{-\pi \left ( 0.01\right ) }{\sqrt{1-\left ( 0.01\right ) ^{2}}}}\\ & =1.969\,1 \end{align*}
This is almost two times the steady state final value which is one in this example. This is not good (it depends on the application as well). We will now start on the overshoot design problem. We add feedback and pure gain controller
We will create the user specs. We want \(y_{\max }^{new}\leq \gamma \). For example, \(\gamma =1.2\). Remember that when adding controller \(k\) and using feedback, the final value will no longer be one. Since with \(k\) present, \(y_{ss}^{new}\) will no longer be one (as mentioned above), so we now need another specification which says by how much the new steady state response can deviate from the original steady state response (which is one in this example). So we write\[ \left \vert y_{ss}^{new}-y_{ss}\right \vert \leq \varepsilon \] Or, since \(y_{ss}=1\) (for this example only, since the input is step), then the above becomes\[ \left \vert y_{ss}^{new}-1\right \vert \leq \varepsilon \] For example, if \(\varepsilon =0.1\), then the above says that the new steady state response (final value) should remain within \(10\%\) of the steady state response before adding the feedback and the controller.
So now we have two specifications to meet by using feedback with the controller in place. We need to see if we can find \(k\) which meets both the above design requirements. Again, the design requirements are
Once we add the feedback and the controller, we obtain a closed loop transfer function\begin{align} T\left ( s\right ) & =\frac{KG\left ( s\right ) }{1+KG\left ( s\right ) }\nonumber \\ & =\frac{K\frac{\omega _{n}^{2}}{s^{2}+2\xi \omega _{n}s+\omega _{n}^{2}}}{1+\frac{K\omega _{n}^{2}}{s^{2}+2\xi \omega _{n}s+\omega _{n}^{2}}}\nonumber \\ & =\frac{K\omega _{n}^{2}}{s^{2}+2\xi \omega _{n}s+\omega _{n}^{2}\left ( 1+K\right ) } \tag{1} \end{align}
So the new \(T\left ( s\right ) \) has different natural frequency, given by (will call \(\bar{\omega }_{n},\bar{\zeta }\) the new \(\omega _{n}\) and the new \(\zeta \) ). \[ \bar{\omega }_{n}^{2}=\omega _{n}^{2}\left ( 1+K\right ) \] or \begin{equation} \bar{\omega }_{n}=\omega _{n}\sqrt{\left ( 1+K\right ) } \tag{2} \end{equation} Re-scalling (1) gives\[ T\left ( s\right ) =\frac{\frac{K\omega _{n}^{2}}{\bar{\omega }_{n}^{2}}\bar{\omega }_{n}^{2}}{s^{2}+2\frac{\xi \omega _{n}}{\bar{\zeta }\bar{\omega }_{n}}\bar{\zeta }\bar{\omega }_{n}s+\bar{\omega }_{n}^{2}}\] We want \(\frac{\xi \omega _{n}}{\bar{\zeta }\bar{\omega }_{n}}=1\) to obtain the same form of the standard second order system. Hence, using (2)\begin{align} \frac{\xi \omega _{n}}{\bar{\zeta }\bar{\omega }_{n}} & =1\nonumber \\ \frac{\xi \omega _{n}}{\bar{\zeta }\left ( \omega _{n}\sqrt{\left ( 1+K\right ) }\right ) } & =1\\ \frac{\xi }{\bar{\zeta }\sqrt{\left ( 1+K\right ) }} & =1\nonumber \\ \bar{\zeta } & =\frac{\xi }{\sqrt{\left ( 1+K\right ) }} \tag{3} \end{align}
And we call \(\frac{K\omega _{n}^{2}}{\bar{\omega }_{n}^{2}}\) as \(\Delta \) which is scaling term. Hence \begin{equation} \Delta =\frac{K\omega _{n}^{2}}{\omega _{n}^{2}\left ( 1+K\right ) }=\frac{K}{1+K} \tag{4} \end{equation} Therefore, \(T\left ( s\right ) \) can now be written in standard second order system as\[ T\left ( s\right ) =\left ( \frac{K}{1+K}\right ) \frac{\bar{\omega }_{n}^{2}}{s^{2}+2\bar{\zeta }\bar{\omega }_{n}s+\bar{\omega }_{n}^{2}}\] Where \(\bar{\zeta }\) is given by (3) and \(\bar{\omega }_{n}\) is given by (2). Summary is below before we go to next stage and design for \(K\)
\begin{align*} T\left ( s\right ) & =\left ( \frac{K}{1+K}\right ) \frac{\bar{\omega }_{n}^{2}}{s^{2}+2\bar{\zeta }\bar{\omega }_{n}s+\bar{\omega }_{n}^{2}}\\ \bar{\zeta } & =\frac{\xi }{\sqrt{\left ( 1+K\right ) }}\\ \bar{\omega }_{n} & =\omega _{n}\sqrt{\left ( 1+K\right ) } \end{align*}
Now we find to find \(K\) which meets \(y_{\max }^{new}\leq \gamma \) and \(\left \vert y_{ss}^{new}-1\right \vert \leq \varepsilon \) at the same time. We start \(y_{\max }^{new}\leq \gamma \). Since we know that\[ y_{\max }^{new}=1+e^{\frac{-\pi \bar{\zeta }}{\sqrt{1-\bar{\zeta }^{2}}}}\]
Then, since system is linear, then \(y_{\max }^{new}\) is just just the constant \(\left ( \frac{K}{1+K}\right ) \) times the above, or \[ y_{\max }^{new}=\left ( \frac{K}{1+K}\right ) \left ( 1+e^{\frac{-\pi \bar{\zeta }}{\sqrt{1-\bar{\zeta }^{2}}}}\right ) \] So our requirement becomes\begin{align*} \frac{K}{1+K}\left ( 1+e^{\frac{-\pi \bar{\zeta }}{\sqrt{1-\bar{\zeta }^{2}}}}\right ) & \leq \gamma \\ e^{\frac{-\pi \bar{\zeta }}{\sqrt{1-\bar{\zeta }^{2}}}} & \leq \gamma \frac{1+K}{K}-1\\ & \leq \frac{\gamma \left ( 1+K\right ) -K}{K} \end{align*}
Taking natural logs gives\[ \frac{-\pi \bar{\zeta }}{\sqrt{1-\bar{\zeta }^{2}}}\leq \ln \left ( \frac{\gamma \left ( 1+K\right ) -K}{K}\right ) \] By multiplying both sides by \(-1\), this will change the inequality sign from \(\leq \) to \(\geq \) and the above becomes\begin{align*} \frac{\pi \bar{\zeta }}{\sqrt{1-\bar{\zeta }^{2}}} & \geq -\ln \left ( \frac{\gamma \left ( 1+K\right ) -K}{K}\right ) \\ \frac{\pi \bar{\zeta }}{\sqrt{1-\bar{\zeta }^{2}}} & \geq \ln \left ( \frac{K}{\gamma \left ( 1+K\right ) -K}\right ) \end{align*}
Moving all terms to one sides gives\[ 1\geq \ln \left ( \frac{K}{\gamma \left ( 1+K\right ) -K}\right ) \frac{\sqrt{1-\bar{\zeta }^{2}}}{\pi \bar{\zeta }}\]
Or, same as above\begin{equation} \ln \left ( \frac{K}{\gamma \left ( 1+K\right ) -K}\right ) \frac{\sqrt{1-\bar{\zeta }^{2}}}{\pi \bar{\zeta }}\leq 1 \tag{5} \end{equation} The above complete the specification for \(y_{\max }^{new}\leq \gamma \). We now work on the second specification \(\left \vert y_{ss}^{new}-1\right \vert \leq \varepsilon \). Since the new system is scaled by \(\frac{K}{1+K}\), then \(y_{ss}^{new}=\frac{K}{1+K}y_{ss}\) but \(y_{ss}=1\), then \(y_{ss}^{new}=\frac{K}{1+K}\). Therefore, this requirement says\[ \left \vert \frac{K}{1+K}-1\right \vert \leq \varepsilon \] Since \(\frac{K}{1+K}<1\) then \(\left \vert \frac{K}{1+K}-1\right \vert =1-\frac{K}{1+K}\) and the above becomes\begin{align*} 1-\frac{K}{1+K} & \leq \varepsilon \\ \frac{\left ( 1+K\right ) -K}{1+K} & \leq \varepsilon \\ \frac{1}{1+K} & \leq \varepsilon \end{align*}
Therefore we now have two specifications ready for design. They are
We now start the design for finding \(K\). Suppose that user specification is that \(\varepsilon =0.1\) and \(\gamma =1.2\). Assume also that \(\zeta =0.1\). We start with \((B)\) above1 .\begin{align*} \frac{1}{1+K} & \leq \varepsilon \\ \frac{1}{1+K} & \leq 0.1\\ 1+K & \geq 10\\ K & \geq 9 \end{align*}
We now work on \(\left ( A\right ) \). Recall that \(\bar{\zeta }=\frac{\xi }{\sqrt{\left ( 1+K\right ) }}=\frac{0.1}{\sqrt{\left ( 1+K\right ) }}\), hence\begin{align*} \ln \left ( \frac{K}{\gamma \left ( 1+K\right ) -K}\right ) \frac{\sqrt{1-\bar{\zeta }^{2}}}{\pi \bar{\zeta }} & \leq 1\\ \ln \left ( \frac{K}{1.2\left ( 1+K\right ) -K}\right ) \frac{\sqrt{1-\frac{\left ( 0.1\right ) ^{2}}{\left ( 1+K\right ) }}}{\pi \frac{0.1}{\sqrt{\left ( 1+K\right ) }}} & \leq 1\\ \ln \left ( \frac{K}{0.2K+1.2}\right ) \frac{1}{0.1\pi }\sqrt{\left ( 1+K\right ) -\left ( 0.1\right ) ^{2}} & \leq 1\\ \ln \left ( \frac{K}{0.2K+1.2}\right ) \frac{10}{\pi }\sqrt{K+0.99} & \leq 1 \end{align*}
Reader: Plot \(F\left ( K\right ) =\frac{10}{\pi }\sqrt{K+0.99}\ln \left ( \frac{K}{0.2K+1.2}\right ) \)
Here is a plot of \(F\left ( K\right ) \) above
We see from the above, that for \(F\left ( K\right ) \leq 1\) the largest \(K\) is around \(K=1.9\).
But our requirement for \(\left \vert y_{ss}^{new}-1\right \vert \leq \varepsilon \) said we needed \(K\geq 9\). This means we are not able to meet user specifications to find \(K\) which satisfies both A and B at the same time.
Reader: How much does the \(\gamma \) specs has to be relaxed so that we can find \(K\) with \(\varepsilon =0.1\) kept the same as above?
Reader: How much does the \(\varepsilon \) specs has to be relaxed so that we can find \(K\) with \(\gamma =1.2\) kept the same as above?
Reader: With \(\zeta =0.1\), find the region in the \(\left ( \varepsilon ,\gamma \right ) \) space for which a spec meeting \(K\) exist.