Use the one-sided Laplace transform with rest initial conditions and s in s^(-1)

Use the one-sided Laplace transform with rest initial conditions and \(s\) in \(\mathrm{s}^{-1}\). Negative unity feedback is \(U=C(R-Y)\). A type-0 plant is \[ G(s)=\frac{11.2}{s^2+5.60 s+6.40}\quad(\mathrm{m/V}), \] with \(y\) in metres and \(u\) in volts. The series controller is the ideal parallel PID law \[ C(s)=K_p+\frac{K_i}{s}+K_d s, \] with \(K_p\) in \(\mathrm{V/m}\), \(K_i\) in \(\mathrm{V/(m\cdot s)}\), and \(K_d\) in \(\mathrm{V\cdot s/m}\). The closed-loop characteristic polynomial of \(1+C(s)G(s)=0\) is required to equal exactly \[ (s+10.0)\bigl(s^2+2\zeta\omega_n s+\omega_n^2\bigr),\qquad\zeta=0.500,\qquad\omega_n=8.00\,\mathrm{rad/s}. \] The position-error constant of \(L=CG\) is \(K_p^{\mathrm{err}}=\lim_{s\to 0}L(s)\) (dimensionless). Percent overshoot and \(2\%\) settling time of the quadratic factor alone, if it were an uncancelled no-zero second-order system, are the standard formulae \[ \mathrm{PO}=100\exp\bigl(-\zeta\pi/\sqrt{1-\zeta^2}\bigr),\qquad t_s=4/(\zeta\omega_n). \] Determine (a) the unique real triple \((K_p,K_i,K_d)\), (b) \(L(s)\), \(K_p^{\mathrm{err}}\), and the steady-state error to a unit-step reference of \(1.00\,\mathrm{m}\), (c) the closed-loop \(T(s)=Y(s)/R(s)\) in monic-denominator form, listing every finite zero, and (d) \(\mathrm{PO}\) and \(t_s\) of the specified quadratic factor together with a statement of whether those two numbers are exact for the unit-step response of this \(T(s)\). Decide whether the closed-loop system is asymptotically stable and whether \(|e_{\mathrm{ss}}|\le 0.0100\,\mathrm{m}\) on the unit step.

** Equate the cubic \(s\cdot\text{plant den}+11.2\cdot\text{PID num}\) to the expanded target \((s+10)(s^2+8s+64)\). Integral action makes the loop type 1.