A. The equivalent transfer function of the negative feedback system with G(s) = (s(s+2))/2K and H(s) = 1 is given by H(s)/(1 + G(s)H(s)).
B. To yield closed-loop, overdamped, second-order poles, we need the discriminant of the characteristic equation to be positive. The characteristic equation is 1 + G(s)H(s) = 0, which simplifies to 2Ks² + 2s + 4 = 0. For overdamped poles, we want two real and distinct roots, so the discriminant (Δ) must be greater than zero. Δ = b² - 4ac, where a = 2K, b = 2, and c = 4. Solving Δ > 0 gives K > 1.
C. To yield closed-loop, underdamped, second-order poles, we want the discriminant to be negative. Δ < 0 yields complex conjugate roots. Solving Δ < 0 gives K < 1.
D. To make the system critically damped, the discriminant must be zero. Δ = 0 implies two equal real roots. Solving Δ = 0 gives K = 1.
E. To make the system marginally stable, the gain must be such that the poles lie on the imaginary axis. This occurs when the discriminant is zero (Δ = 0). Solving Δ = 0 gives K = 2. At this value of K, the frequency of oscillation (ω) can be found using the formula ω = √(c/a), where a = 2K and c = 4. Thus, ω = √(4/(2*2)) = √1 = 1.
F. The pole locations and corresponding values of gain, K, are as follows:
- Overdamped poles: Real and distinct roots for K > 1.
- Underdamped poles: Complex conjugate roots for K < 1.
- Critically damped poles: Two equal real roots for K = 1.
- Marginally stable poles: Imaginary axis roots for K = 2.
G. To plot the step response using MATLAB, we need the specific transfer function and the desired values of gain, K. Please provide the transfer function and the desired values of K so that I can assist you with the MATLAB code.