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Q, solution: Given,closed loop transfer function of a system,G(s)=\frac{s^{2}-2 s+21}{s^{3}+s^{2}+2 s+8}For stability of system all poles should lie strictly in the Left half of S-plane.Now. Characteristics equation of CLTF.s^{3}+s^{2}+2 s+8=0-(1)By hit and trial method, we get one solution of Shis equation (1) is (-2). put, s=-2, in equation (I)\begin{array}{c}(-2)^{3}+(-2)^{2}+2(-2)+8=0 \\\Rightarrow-8+4-9+8=0 \\3 \text { O.H.S }=\text { R.H.S } \\\text { S.P. L.H.S }\end{array}50, (S+2) will be a factor of equation (I) Now, divide \left(s^{3}+s^{2}+2 s+8\right) by (s+2).\begin{array}{l}\therefore s+2 \sqrt{s^{3}+s^{2}+2 s+8}\left(s^{2}-s+4\right. \\\frac{s^{3}+2 s^{2}}{-s^{2}+2 s+8} \\-s^{2}-2 s \\\frac{+t+}{+4 s+8} \\\frac{4 s+8}{0} \\\therefore\left(s^{3}+s^{2}+2 s+8\right)=0 \quad[\operatorname{trm}(1)] \\\Rightarrow(s+2)\left(s^{2}-s+4\right)=0 \\\text { eisher. }(s+2)=0 \quad \text { or, } s^{2}-s+4=0 \\\end{array}\begin{array}{l|l}(s, 2)(s+2)=0 \\\therefore s=-2 & \text { or, } s^{2}-s+4=0 \\\text { by quadratic formula } ; \times=\frac{-6 \pm \sqrt{b^{2}+00}}{2 \times 9} \\& s=\frac{-(-1) \pm \sqrt{(-1)^{2}-4 \times 1 \times 4}}{2 \times 1}\end{array}\begin{array}{l}=\frac{1 \pm \sqrt{1-16}}{2} \\=\frac{1}{2} \pm j \sqrt{\frac{15}{4}}\end{array}Q, solution: Given,closed loop transfer function of a system,G(s)=\frac{s^{2}-2 s+21}{s^{3}+s^{2}+2 s+8}For stability of system all poles should lie strictly in the Left half of S-plane.Now. Characteristics equation of CLTF.s^{3}+s^{2}+2 s+8=0-(1)By hit and trial method, we get one solution of Shis equation (1) is (-2). put, s=-2, in equation (I)\begin{array}{c}(-2)^{3}+(-2)^{2}+2(-2)+8=0 \\\Rightarrow-8+4-9+8=0 \\3 \text { O.H.S }=\text { R.H.S } \\\text { S.P. L.H.S }\end{array}50, (S+2) will be a factor of equation (I) Now, divide \left(s^{3}+s^{2}+2 s+8\right) by (s+2).\begin{array}{l}\therefore s+2 \sqrt{s^{3}+s^{2}+2 s+8}\left(s^{2}-s+4\right. \\\frac{s^{3}+2 s^{2}}{-s^{2}+2 s+8} \\-s^{2}-2 s \\\frac{+t+}{+4 s+8} \\\frac{4 s+8}{0} \\\therefore\left(s^{3}+s^{2}+2 s+8\right)=0 \quad[\operatorname{trm}(1)] \\\Rightarrow(s+2)\left(s^{2}-s+4\right)=0 \\\text { eisher. }(s+2)=0 \quad \text { or, } s^{2}-s+4=0 \\\end{array}\begin{array}{l|l}(s, 2)(s+2)=0 \\\therefore s=-2 & \text { or, } s^{2}-s+4=0 \\\text { by quadratic formula } ; \times=\frac{-6 \pm \sqrt{b^{2}+00}}{2 \times 9} \\& s=\frac{-(-1) \pm \sqrt{(-1)^{2}-4 \times 1 \times 4}}{2 \times 1}\end{array}\begin{array}{l}=\frac{1 \pm \sqrt{1-16}}{2} \\=\frac{1}{2} \pm j \sqrt{\frac{15}{4}}\end{array}Nowwe plot poles of CLTF [clused wop Franster finction)Whe con see shat complex poles of CLTF lie in right side of s-plane.Herce, system becomes cmstable.option (c) is correct.Option  A) is incorrect  because all its poles don't lie in left half of s - plane Option  B) is incorrect  because given poles are not correct. Option D) is incorrect because  stability  of closed loop system  doesn't  depend on  zeroes of system.  Hence only Option  c) Is correct.  ...