Time t is in seconds; the output y is a shaft angle in radians and the reference r is…

Time \(t\) is in seconds; the output \(y\) is a shaft angle in radians and the reference \(r\) is in radians. The unit step satisfies \(u(t)=1\) for \(t\ge 0\) and \(u(t)=0\) for \(t<0\). Use the unilateral Laplace transform with lower limit \(0^-\). A unity-gain negative-feedback loop with actuating error \(e=r-y\) and forward first-order plant \[ G(s)=\frac{Y(s)}{E(s)}=\frac{4.50}{s+4.50}\qquad\bigl(\mathrm{rad/rad}\bigr) \] is at rest for \(t<0\), so \(y(0^-)=0\). The reference is the ramp \[ r(t)=0.80\,t\,u(t)\qquad(\mathrm{rad}). \] The closed-loop system is known to be asymptotically stable. (a) Derive the closed-loop transfer function \(T(s)=Y(s)/R(s)\) and the error transfer function \(S(s)=E(s)/R(s)\). State the SI units of each. (b) Using the final-value theorem, after first confirming that every pole of \(sE(s)\) lies in the open left half-plane, compute the steady-state tracking error \(e_{\mathrm{ss}}=\lim_{t\to\infty}e(t)\) in radians. Interpret \(e_{\mathrm{ss}}\) as a constant lag: write the large-\(t\) asymptotic form \(y(t)=r(t)-e_{\mathrm{ss}}+o(1)\). (c) Invert \(Y(s)\) to obtain a closed-form expression for \(y(t)\) valid for all \(t\ge 0\). Evaluate \(y(0.50)\) and \(e(0.50)\). Determine \(\lim_{t\to\infty}\dot y(t)\) and compare it with \(\dot r(t)\) for \(t>0\).

** Unity-gain *feedback* means \(H=1\), not that the closed-loop DC gain equals one; compute \(T=G/(1+G)\) first.