The stability of a system can be determined by the roots of its characteristic equation. If all roots of the characteristic equation have negative real parts, the system is stable. If any root has a positive real part, the system is unstable. If a root has a zero real part, the system is marginally stable. The characteristic equation given is a quartic equation (4th degree polynomial) in s: 2s^4 + 20ks^3 + 3s^2 + 5s + 7 = 0. The Routh-Hurwitz criterion is a mathematical test that is used to tell whether a linear system is stable. It does not require computation of the roots of the characteristic equation. The Routh array for this characteristic equation is: s^4: 2, 3 s^3: 20k, 5 s^2: (60k-5)/20k, 7 s^1: 5 - (140k/(60k-5)) s^0: 7 For the system to be stable, all the first elements in each row must be positive. From the s^2 row, we have (60k-5)/20k 0. Solving this inequality gives k 5/60 = 0.0833. From the s^1 row, we have 5 - (140k/(60k-5)) 0. Solving this inequality gives k

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The stability of a system can be determined by the roots of its characteristic equation. If all roots of the characteristic equation have negative real parts, the system is stable. If any root has a positive real part, the system is unstable. If a root has a zero real part, the system is marginally stable. The characteristic equation given is a quartic equation (4th degree polynomial) in s: 2s^4 + 20ks^3 + 3s^2 + 5s + 7 = 0. The Routh-Hurwitz criterion is a mathematical test that is used to tell whether a linear system is stable. It does not require computation of the roots of the characteristic equation. The Routh array for this characteristic equation is: s^4: 2, 3 s^3: 20k, 5 s^2: (60k-5)/20k, 7 s^1: 5 - (140k/(60k-5)) s^0: 7 For the system to be stable, all the first elements in each row must be positive. From the s^2 row, we have (60k-5)/20k > 0. Solving this inequality gives k > 5/60 = 0.0833. From the s^1 row, we have 5 - (140k/(60k-5)) > 0. Solving this inequality gives k < 5/140 = 0.0357. Therefore, the system is stable for 0.0357 < k < 0.0833. This does not match any of the given options (a) through (d). Please note that this is a simplified explanation and the actual process involves more detailed calculations.

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The stability of a system can be determined by the roots of its characteristic equation. If all roots of the characteristic equation have negative real parts, the system is stable. If any root has a positive real part, the system is unstable. If a root has a zero real part, the system is marginally stable. The characteristic equation given is a quartic equation (4th degree polynomial) in s: 2s^4 + 20ks^3 + 3s^2 + 5s + 7 = 0. The Routh-Hurwitz criterion is a mathematical test that is used to tell whether a linear system is stable. It does not require computation of the roots of the characteristic equation. The Routh array for this characteristic equation is: s^4: 2, 3 s^3: 20k, 5 s^2: (60k-5)/20k, 7 s^1: 5 - (140k/(60k-5)) s^0: 7 For the system to be stable, all the first elements in each row must be positive. From the s^2 row, we have (60k-5)/20k > 0. Solving this inequality gives k > 5/60 = 0.0833. From the s^1 row, we have 5 - (140k/(60k-5)) > 0. Solving this inequality gives k < 5/140 = 0.0357. Therefore, the system is stable for 0.0357 < k < 0.0833. This does not match any of the given options (a) through (d). Please note that this is a simplified explanation and the actual process involves more detailed calculations.

the feedback system with the characteristic equation 2s4 + 20ks3 + 3s2 + 5s + 7 = 0(a)stable for all value of k(b)stable fo rpositive value of k(c)stable for infinity>k>7.0(d)unstable for all values of k

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