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: \[ \boxed{\infty, \infty} \]
In the Wheatstone bridge shown below, the sensitivity of the bridge in terms of change in balancing voltage \( E \) for unit change in the resistance \( R \), in V/Ω, is __________ (round off to two decimal places).
The relationship between two variables \( x \) and \( y \) is given by \( x + py + q = 0 \) and is shown in the figure. Find the values of \( p \) and \( q \). Note: The figure shown is representative.