SOLUTION:
There are two inequalities to satisfy. The first is given by the settling time requirement\begin{equation} t_{s}=\frac{3.2}{\zeta \omega _{n}}<0.25 \tag{1} \end{equation} The second is given by the overshoot requirement\begin{equation} e^{-\frac{\pi \zeta }{\sqrt{1-\zeta ^{2}}}}<0.3 \tag{2} \end{equation} From (2), taking the log of both sides gives\[ -\frac{\pi \zeta }{\sqrt{1-\zeta ^{2}}}<\ln \left ( 0.3\right ) \] Multiplying both sides by \(-1\) changes the inequality from \(<\) to \(>\) \[ \frac{\pi \zeta }{\sqrt{1-\zeta ^{2}}}>-\ln \left ( 0.3\right ) \] Simplifying gives\begin{align*} \frac{\pi \zeta }{\sqrt{1-\zeta ^{2}}} & >\ln \left ( \frac{1}{0.3}\right ) \\ & >1.204 \end{align*}
Squaring both sides and solving for \(\zeta \) \begin{align*} \frac{\zeta ^{2}}{1-\zeta ^{2}} & >\left ( \frac{1.204}{\pi }\right ) ^{2}\\ \zeta ^{2} & >0.14688\left ( 1-\zeta ^{2}\right ) \\ 1.146\,88\zeta ^{2} & >0.14688\\ \zeta ^{2} & >0.12807 \end{align*}
Since \(\zeta \) has to be positive then the positive root is used giving\[ \fbox{$\zeta >0.35787$}\] Back to the (1) specifications, which says\begin{align*} \frac{3.2}{\zeta \omega _{n}} & <0.25\\ \zeta \omega _{n} & >12.8 \end{align*}
For each \(\zeta _{i}>0.3578\), we solve for \(\omega _{n}\) from \(\zeta \omega _{n}>12.8\). This will give full description of where the poles are located.
Here is a plot of the \(\left ( \zeta ,\omega _{n}\right ) \) space showing allowed values of \(\zeta ,\omega _{n}\).
By taking each point \(\left ( \zeta ,\omega _{n}\right ) \) from the above plot, then the pole location with coordinates \(-\zeta \omega _{n}\pm \omega _{n}\sqrt{1-\zeta ^{2}}\) is generated. The following shows the final result, showing the region where the poles have to be located in order to meet the performance requirements.
The above diagram shows the location of each pair of poles as a small dot. Complex poles come in pair of conjugates. One pole will be above the real axis and its pair below the real axis.
SOLUTION:
The closed loop transfer function, in terms of \(K\) and \(K_{t}\) can be found using either Mason rule or simple block reduction. For this problem block reduction seems easier.
Using \(G_{p}=\frac{100}{1+0.2s}\) the above becomes\begin{align*} \frac{C\left ( s\right ) }{R\left ( s\right ) } & =\frac{K\left ( \frac{100}{1+0.2s}\right ) }{20s\left ( 1+K_{t}\frac{100}{1+0.2s}\right ) +K\frac{100}{1+0.2s}}\\ & =\frac{100K}{20s\left ( 1+0.2s+100K_{t}\right ) +100K}\\ & =\frac{100K}{4s^{2}+\left ( 20+2000K_{t}\right ) s+100K}\\ & =\frac{25K}{s^{2}+\left ( 5+500K_{t}\right ) s+25K} \end{align*}
The standard form is \(\frac{\omega _{n}^{2}}{s^{2}+2\zeta \omega _{n}s+\omega _{n}^{2}}\) therefore by comparing to the above we find \begin{align} \omega _{n}^{2} & =25K\nonumber \\ \omega _{n} & =5\sqrt{K} \tag{1} \end{align}
And\begin{align} 5+500K_{t} & =2\zeta \omega _{n}\nonumber \\ & =2\zeta \left ( 5\sqrt{K}\right ) \nonumber \\ \zeta & =\frac{5+500K_{t}}{10\sqrt{K}} \tag{2} \end{align}
Hence the transfer function is \[ \frac{C\left ( s\right ) }{R\left ( s\right ) }=\frac{\omega _{n}^{2}}{s^{2}+2\zeta \omega _{n}s+\omega _{n}^{2}}\] Where \(\omega _{n}=5\sqrt{K}\) and \(\zeta =\frac{5+500K_{t}}{10\sqrt{K}}\). We now apply the user specifications in order to determine \(K\) and \(K_{t}\). From the overshoot requirement, we write\begin{equation} e^{-\frac{\pi \zeta }{\sqrt{1-\zeta ^{2}}}}\leq 0.2 \tag{3} \end{equation} And from the rise time requirements we have\begin{equation} \frac{1-0.4167\zeta +2.917\zeta ^{2}}{\omega _{n}}=0.05 \tag{4} \end{equation} From (3) and (4) we can now solve for \(\omega _{n}\) and \(\zeta \) and this allow us to find \(K\) and \(K_{t}\) by using (1,2). From (3), taking logs and solving for \(\zeta \) gives\begin{align*} -\frac{\pi \zeta }{\sqrt{1-\zeta ^{2}}} & \leq \ln \left ( 0.2\right ) \\ \frac{\pi \zeta }{\sqrt{1-\zeta ^{2}}} & \geq \ln \left ( 5\right ) \\ \frac{\pi \zeta }{\sqrt{1-\zeta ^{2}}} & \geq 1.\,\allowbreak 6094\\ \pi ^{2}\zeta ^{2} & \geq \left ( 1.\,\allowbreak 6094\right ) ^{2}\left ( 1-\zeta ^{2}\right ) \\ \pi ^{2}\zeta ^{2} & \geq 2.5902-2.5902\zeta ^{2}\\ \left ( \pi ^{2}+2.5902\right ) \zeta ^{2} & \geq 2.5902\\ \zeta & \geq \sqrt{\frac{2.5902}{\left ( \pi ^{2}+2.5902\right ) }} \end{align*}
Hence\[ \zeta \geq 0.456 \] Any \(0.456\leq \zeta <1\) can be used. In order to find \(\omega _{n}\), let us choose \[ \fbox{$\zeta =0.46$}\] For the rest of the calculations . From (4) we find\begin{align*} \frac{1-0.4167\left ( 0.46\right ) +2.917\left ( 0.46\right ) ^{2}}{\omega _{n}} & =0.05\\ \omega _{n} & =\frac{1.425\,6}{0.05} \end{align*}
Therefore\[ \fbox{$\omega _n=28.512$ rad/sec}\] Now that we found \(\zeta \) and \(\omega _{n}\,\), we use (1,2) to find the gains. From (1)\begin{align*} \omega _{n} & =5\sqrt{K}\\ K & =\frac{\omega _{n}^{2}}{25}=\frac{28.512^{2}}{25} \end{align*}
Therefore\[ \fbox{$K=32.517$}\] And from (2)\begin{align*} \zeta & =\frac{5+500K_{t}}{10\sqrt{K}}\\ 0.46 & =\frac{5+500K_{t}}{10\sqrt{32.517}} \end{align*}
Hence\[ \fbox{$K_t=0.04246$}\] The final transfer function is \begin{align*} \frac{C\left ( s\right ) }{R\left ( s\right ) } & =\frac{25K}{s^{2}+\left ( 5+500K_{t}\right ) s+25K}\\ & =\frac{25\left ( 32.517\right ) }{s^{2}+\left ( 5+500\left ( 0.04246\right ) \right ) s+25\left ( 32.517\right ) } \end{align*}
Or\[ \fbox{$\frac{C\left ( s\right ) }{R\left ( s\right ) }=\frac{812.93}{s^2+26.23s+812.93}$}\] Matlab is used to simulate the step response, and to also verify the user requirements are met.
The step information was also verified using the command stepinfo() which showed the specifications was indeed met.
SOLUTION:
\[ G\left ( s\right ) =\frac{Ks\left ( 20s^{2}+1\right ) }{s^{4}+5s^{2}+10s+15}\] In classical unity feedback, the closed loop transfer function \(T\left ( s\right ) \) is\begin{align*} T\left ( s\right ) & =\frac{G\left ( s\right ) }{1+G\left ( s\right ) }\\ & =\frac{Ks\left ( 20s^{2}+1\right ) }{\left ( s^{4}+5s^{2}+10s+15\right ) +Ks\left ( 20s^{2}+1\right ) }\\ & =\frac{Ks\left ( 20s^{2}+1\right ) }{s^{4}+20Ks^{3}+5s^{2}+\left ( 10+K\right ) s+15} \end{align*}
Applying Routh-Hurwitz to the denominator \(D\left ( s\right ) =\) \(s^{4}+20Ks^{3}+5s^{2}+\left ( 10+K\right ) s+15\) gives
| \(s^{4}\) | \(1\) | \(5\) | \(15\) |
| \(s^{3}\) | \(20K\) | \(\left ( 10+K\right ) \) | \(0\) |
| \(s^{2}\) | \(\frac{20K\left ( 5\right ) -\left ( 10+K\right ) }{20K}\) | \(15\) | \(0\) |
| \(s^{1}\) | \(\frac{\frac{\left ( 20K\left ( 5\right ) -\left ( 10+K\right ) \right ) }{20K}\left ( 10+K\right ) -20K\left ( 15\right ) }{\frac{20K\left ( 5\right ) -\left ( 10+K\right ) }{20K}}\) | \(0\) | \(0\) |
| \(s^{0}\) | \(15\) | ||
Simplifying gives
| \(s^{4}\) | \(1\) | \(5\) | \(15\) |
| \(s^{3}\) | \(20K\) | \(\left ( 10+K\right ) \) | \(0\) |
| \(s^{2}\) | \(\frac{1}{20K}\left ( 99K-10\right ) \) | \(15\) | \(0\) |
| \(s^{1}\) | \(\frac{1}{10-99K}\left ( 5901K^{2}-980K+100\right ) \) | \(0\) | \(0\) |
| \(s^{0}\) | \(15\) | ||
For stability, we need the first column to be positive. Hence the conditions are\begin{align*} 20K & >0\\ \frac{1}{20K}\left ( 99K-10\right ) & >0\\ \frac{1}{10-99K}\left ( 5901K^{2}-980K+100\right ) & >0 \end{align*}
The first just says that \(K>0\). The second says \(99K-10>0\) or \(K>\frac{10}{99}\). Now for the third condition\[ \frac{1}{10-99K}\left ( 5901K^{2}-980K+100\right ) >0 \] Since \(K>\frac{10}{99}\) is required, then \(\frac{1}{10-99K}\) is negative quantity since \(10-99K\) is negative for \(K>\frac{10}{99}\). This means the above becomes\[ 5901K^{2}-980K+100<0 \] Notice the change of inequality from \(>\) to \(<\) since we multiplied both sides by a negative quantity \(\left ( 10-99K\right ) \) to cancel it out. But \(5901K^{2}-980K+100<0\) can not be satisfied with a positive \(K>\frac{10}{99}\). For example, using the minimum allowed \(K\) which is \(\frac{10}{99}\), then the value of \(5901K^{2}-980K+100\) becomes\[ 5901\left ( \frac{10}{99}\right ) ^{2}-980\left ( \frac{10}{99}\right ) +100=61.218 \] But it needs to be negative. So there does not exist \(K\) which makes the closed loop stable.
SOLUTION:
Given \(G\left ( s\right ) =\frac{K}{s\left ( 1+Ts\right ) }\) then the closed loop transfer function is\begin{align*} G_{closed}\left ( s\right ) & =\frac{G}{1+G}\\ & =\frac{K}{s\left ( 1+Ts\right ) +K}\\ & =\frac{K}{Ts^{2}+s+K} \end{align*}
Therefore \(D\left ( s\right ) =Ts^{2}+s+K\). For the closed loop poles with real part to be less than \(-a\), let \(s_{1}=s+a\). Then \(s=s_{1}-a\). We apply Routh-Hurwitz to \(D\left ( s\right ) \) but with \(s=s_{1}-a\). The new denominator polynomial becomes\[ D\left ( s_{1}\right ) =T\left ( s_{1}-a\right ) ^{2}+\left ( s_{1}-a\right ) +K \] Expanding gives\begin{align*} D\left ( s_{1}\right ) & =T\left ( s_{1}^{2}+a^{2}-2s_{1}a\right ) +s_{1}-a+K\\ & =Ts_{1}^{2}+s_{1}\left ( 1-2Ta\right ) +\left ( Ta^{2}-a+K\right ) \end{align*}
Routh table applied to the above polynomial is
| \(s_{1}^{2}\) | \(T\) | \(Ta^{2}-a+K\) |
| \(s_{1}^{1}\) | \(1-2Ta\) | \(0\) |
| \(s_{1}^{0}\) | \(Ta^{2}-a+K\) | |
We need all entries in the first column to be same sign (positive in this case, since \(T=1\)) for stability to hold (This is in addition to having the poles be with real part less than \(-a\)). For \(T=1\) the above becomes
| \(s^{2}\) | \(1\) | \(a^{2}+a+K\) |
| \(s^{1}\) | \(1-2a\) | \(0\) |
| \(s^{0}\) | \(a^{2}-a+K\) | |
The conditions for stability are\begin{align*} 1-2a & >0\\ a\left ( 1-a\right ) +K & >0 \end{align*}
The first condition gives \(a>\frac{1}{2}\). The second condition gives \[ \fbox{$K>a^2-a$}\] Here is plot of the region in the \(\left ( a,K\right ) \) plane associated with closed loop stability.
SOLUTION:
The closed loop transfer function is\[ T\left ( s\right ) =\frac{KG\left ( s\right ) \frac{1}{s}}{1+HKG\left ( s\right ) \frac{1}{s}}\] Replacing \(H\left ( s\right ) =1\) and \(G=\frac{\left ( s+2\right ) ^{2}}{s^{2}+0.01}\) the above becomes\begin{align*} T\left ( s\right ) & =\frac{\frac{K\left ( s+2\right ) ^{2}}{s^{2}+0.01}\frac{1}{s}}{1+\frac{K\left ( s+2\right ) ^{2}}{s^{2}+0.01}\frac{1}{s}}\\ & =\frac{K\left ( s+2\right ) ^{2}}{s\left ( s^{2}+0.01\right ) +K\left ( s+2\right ) ^{2}}\\ & =\frac{K\left ( s+2\right ) ^{2}}{s^{3}+Ks^{2}+\left ( 0.01+4K\right ) s+4K} \end{align*}
The Routh table for \(D\left ( s\right ) =s^{3}+Ks^{2}+\left ( 0.01+4K\right ) s+4K\) is
| \(s^{3}\) | \(1\) | \(0.01+4K\) |
| \(s^{2}\) | \(K\) | \(4K\) |
| \(s^{1}\) | \(4K-3.99\) | \(0\) |
| \(s^{0}\) | \(4K\) | |
Therefore for stability we need\begin{align*} K & >0\\ 4K & >3.99\\ 4K & >0 \end{align*}
The first and the third conditions give \(K>0\). From the second condition, \(K>\frac{3.99}{4}=0.9975\). Therefore\[ \fbox{$K>0.9975$}\] To verify, here is the step response for \(k=0.9974\) and \(k=0.9976\), showing one is unstable and the second is stable.
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