A linearised liquid-level plant occupies a tank of constant horizontal area A=1.25 m^2

A linearised liquid-level plant occupies a tank of constant horizontal area \(A=1.25\,\mathrm{m^2}\). The outflow through a linear hydraulic resistance \(R_h=4.00\,\mathrm{s/m^2}\) is \(q_{\mathrm{out}}(t)=h(t)/R_h\), with level \(h\) in metres and volumetric flow in cubic metres per second. A valve supplies \[ q_{\mathrm{in}}(t)=K_v u(t)+d(t),\qquad K_v=0.250\,\mathrm{m^3/(s\cdot V)}, \] where \(u(t)\) is the valve command in volts and \(d(t)\) is an unmeasured inflow disturbance in \(\mathrm{m^3/s}\). Mass conservation gives \[ A\dot h=q_{\mathrm{in}}-q_{\mathrm{out}}. \] All initial conditions are zero. Unilateral Laplace transforms are used, and time is in seconds. Unity negative feedback around the level is implemented by the PI law \[ U(s)=C(s)\bigl(R(s)-H(s)\bigr),\qquad C(s)=K_p+\frac{K_i}{s},\qquad K_p=6.00\,\mathrm{V/m},\qquad K_i=2.00\,\mathrm{V/(m\cdot s)}, \] where \(R\) is the commanded level in metres and \(H=\mathcal{L}\{h\}\). (a) Derive the two open-loop maps \(H(s)/U(s)\) and \(H(s)/D(s)\) (each with the other input set to zero) as ratios of coprime polynomials, and state their units. (b) Close the loop and derive \(H/R\) and \(H/D\) as ratios of coprime polynomials with a common monic characteristic polynomial. (c) Using the final-value theorem, and assuming every closed-loop pole lies in the open left half-plane, evaluate \(h(\infty)\) for \(r(t)=0.400\cdot 1(t)\,\mathrm{m}\), \(d\equiv 0\), and for \(r\equiv 0\), \(d(t)=0.0500\cdot 1(t)\,\mathrm{m^3/s}\). (d) If the integral gain is set to \(K_i=0\) with \(K_p\) unchanged, recompute the two steady-state levels of (c). Identify which experiment is affected, and name the loop type (the number of pure integrators in \(C(s)P(s)\), where \(P=H/U\) open-loop) in each of the two designs.

** The tank is self-regulating: the open-loop map \(H/U\) has no free integrator, so loop type is the number of integrators in \(C\).