Question Construct the root loci for K ≥ 0. Find the value of K that makes the relative damping ratio of the closed-loop system (measured by the dominant complex characteristic equation roots) equal to 0.707, if such a solution exists. please solve (d) 9-16. The forward-path transfer functions of a unity-feedback control system are given in the following (a) G(s) = K(s+3) s(s* + 4s+4)(s+5)(s+6) K (b) G(s)=- s(s+2)(s+4)(s+10) K(s? +2s+8) (c) G(s)= s(s+5)(s+10) K(s?+4) (d) G(s) = (s+2)*(s+5)(+6) (e) G(s)= K(+10) s(s+2.5)(s' +2s+2) K (f) G(s) =- (s+1)(s* +45+5) K(s+2) (g) G(s)=- (s+1)(s* +6s+10) (h) G(s) = K(s+2)(s+3) s(s+1) K (i) G(s)= s(s? +4s+5)

PLEASE FIND THE SOLUTION ATTACHED IN JPEG FORMAT.THANK YOU.PLEEEASE UPVOTE, IF YOU FIND THE SOLUTIONHELPFUL.G H(s))=\frac{K\left(s^{2}+44\right)}{(s+2)^{2}(8+5)(s+6)} \quad 0<k<\inftyNo of poles =4No of zeroes =2No of branelus of root lous : Poez which ever is gerater. Here no of tranches =4Let us plot Polezero diagran and rootlous on real axis. At poles k=0 &amp; at zeroes k=\infty we place ontgoing aroious 1-2 j at poler &amp; in coning aovous at goroel. Root lows on sual axis exists on left side of oold no. of poles &amp; for zerou.Asyniptotes:The ecoot lows moves parallel to asymptote &amp; muts then@ @ no of asymptote: p-z=2Angle of asymptote (2 q+1) 180^{\circ} \quad q=0,1 Hure angles are 90^{\circ} &amp; 270^{\circ}.Centroid \sigma=\frac{\sum p o l e s-\sum z^{2}-\bar{s}}{p-z}\begin{array}{l}\sigma=\frac{-2-2-5-6-0}{2} \\\sigma=-7.5\end{array}Brakawoy point:form 1+G H=0\begin{array}{c}s^{4}+15 s^{3}+78 s^{2}+164 s+120+k\left(s^{2}+4\right)=0 \\\therefore k=-\frac{\left(s^{4}+15 s^{3}+78 s^{2}+164 s+120\right)}{s^{2}+4}\end{array}Find s sueh that \frac{d k}{d s}=0\begin{array}{l}\left(s^{2}+4\right)\left(4 s^{3}+45 s^{2}+156 s+164\right)-\left(s^{4}+15 s^{3}+78 s^{2}\right. \\+1648+120)(28) \\\end{array}2 s^{5}+15 s^{4}+16 s^{3}+16 s^{2}+384 s+65 d=0. =0conputer simulation roots aieUsings=-2 / 1-3.39,-5.53,1.71 \pm{ }^{2} 2.41Rerreated poles have brealcaway point s=-2 adjacent poles hove breakavay point 8=-5.53Hunce we have two ralid breakpoints s=-2, s=-5.53 At thee points K=\varnothing \quad k=0.1167 {minimum values}Angle of Arenival:\begin{array}{l} \phi_{p_{1}}= 45^{\circ}, 45^{\circ} \\\phi_{p_{2}}= 21.8^{\circ} \\\left\{\tan ^{-1} \frac{2}{5}\right\} \\\phi_{p_{3}}= 18.4^{\circ} \\\left\{\tan ^{-1} 1 / 3\right\} \quad \phi_{z_{1}}=90^{\circ} \\\phi_{a}= 180^{\circ}-\sum \phi_{z}+\sum \phi_{p} \phi_{p_{1}}=\phi^{\circ} \phi_{1} \\\phi_{a}=220^{\circ}=-\phi_{a}^{\prime}\end{array}ju intuupt.Root lows teminates at finite zeroes which are locatid on \pm j 2 on the imaginany axis. Hue K=\infty. Hunc we can't say that there is jw intrupt.given \xi=0.107^{\circ} \quad \therefore \theta=\cos ^{-1} 0.707=45^{\circ} Using donninant pole fromgraph the point A is -1.2+j 1.2 \&amp; k=8.25 ...