Step 1: Open-loop transfer function.
\[
L(s) = C(s)G(s) = \frac{3s+1}{s} \cdot \frac{1}{s-1} = \frac{3s+1}{s(s-1)}.
\]
Step 2: Closed-loop transfer function.
\[
T(s) = \frac{L(s)}{1+L(s)}.
\]
Since system has a pole at origin (due to PI controller), system type = 1.
Step 3: Steady-state error for step input.
For type-1 system, steady-state error for unit step is zero. So plant output \(\to 1\).
Step 4: Controller output.
But plant \(G(s) = \frac{1}{s-1}\) is unstable (pole at \(+1\)).
So PI control drives input unbounded \(\to\) controller output tends to \(\infty\).
Final Answer: \[ \boxed{\infty, \infty} \]
Given an open-loop transfer function \(GH = \frac{100}{s}(s+100)\) for a unity feedback system with a unit step input \(r(t)=u(t)\), determine the rise time \(t_r\).
Consider a linear time-invariant system represented by the state-space equation: \[ \dot{x} = \begin{bmatrix} a & b -a & 0 \end{bmatrix} x + \begin{bmatrix} 1 0 \end{bmatrix} u \] The closed-loop poles of the system are located at \(-2 \pm j3\). The value of the parameter \(b\) is: