Understanding the Control System Components

The problem describes a closed-loop control system with unity-gain negative feedback. It consists of a plant, represented by the transfer function \( \rm G(s)=\frac{1}{s-1} \), and a Proportional-Integral (PI) controller.

The PI controller's transfer function is given by:

$$ C(s) = K_p + \frac{K_i}{s} $$

Given the proportional gain \( K_p = 3 \) and the integral gain \( K_i = 1 \), the controller transfer function becomes:

$$ C(s) = 3 + \frac{1}{s} = \frac{3s+1}{s} $$

Calculating the Open-Loop Transfer Function (OLTF)

The open-loop transfer function, denoted by \( L(s) \), is the product of the controller and plant transfer functions:

$$ L(s) = C(s)G(s) $$

Substituting the expressions for \( C(s) \) and \( G(s) \):

$$ L(s) = \left(\frac{3s+1}{s}\right) \left(\frac{1}{s-1}\right) = \frac{3s+1}{s(s-1)} $$

Determining the Closed-Loop Transfer Function (CLTF)

For a unity-gain negative feedback system, the closed-loop transfer function \( T(s) \) relates the output \( Y(s) \) to the reference input \( R(s) \), where \( Y(s) = T(s)R(s) \). It is calculated as:

$$ T(s) = \frac{L(s)}{1 + L(s)} $$

Substituting the expression for \( L(s) \):

$$ T(s) = \frac{\frac{3s+1}{s(s-1)}}{1 + \frac{3s+1}{s(s-1)}} $$

To simplify, multiply the numerator and denominator by \( s(s-1) \):

$$ T(s) = \frac{3s+1}{s(s-1) + (3s+1)} $$

$$ T(s) = \frac{3s+1}{s^2 - s + 3s + 1} = \frac{3s+1}{s^2 + 2s + 1} $$

The denominator \( s^2 + 2s + 1 \) can be factored as \( (s+1)^2 \):

$$ T(s) = \frac{3s+1}{(s+1)^2} $$

This \( T(s) \) represents the transfer function from the reference input \( R(s) \) to the plant output \( Y(s) \).

Final Value of the Plant Output

We need to find the final value of the plant output, \( y(t) \), for a unit step reference input. The reference input is \( r(t) = 1 \), and its Laplace transform is \( R(s) = \frac{1}{s} \).

The plant output is \( Y(s) = T(s)R(s) \):

$$ Y(s) = \frac{3s+1}{(s+1)^2} \cdot \frac{1}{s} $$

We use the Final Value Theorem, which states that if the system is stable and \( \lim_{s \to 0} sY(s) \) exists, then the final value of \( y(t) \) is \( y_{final} = \lim_{s \to 0} sY(s) \).

$$ y_{final} = \lim_{s \to 0} s \left( \frac{3s+1}{(s+1)^2} \cdot \frac{1}{s} \right) $$

$$ y_{final} = \lim_{s \to 0} \frac{3s+1}{(s+1)^2} $$

Substitute \( s=0 \):

$$ y_{final} = \frac{3(0)+1}{(0+1)^2} = \frac{1}{1^2} = 1 $$

The final value of the plant output is 1.

Final Value of the Controller Output

Let the controller output be \( u(t) \), with its Laplace transform being \( U(s) \). The relationship between the plant output \( Y(s) \) and the controller output \( U(s) \) is:

$$ Y(s) = G(s)U(s) $$

Therefore, the controller output can be expressed as:

$$ U(s) = \frac{Y(s)}{G(s)} $$

Substituting the expressions for \( Y(s) \) and \( G(s) \):

$$ U(s) = \frac{\frac{3s+1}{(s+1)^2} \cdot \frac{1}{s}}{\frac{1}{s-1}} = \frac{3s+1}{s(s+1)^2} \cdot (s-1) $$

Now, apply the Final Value Theorem to find \( u_{final} \):

$$ u_{final} = \lim_{s \to 0} sU(s) $$

$$ u_{final} = \lim_{s \to 0} s \left( \frac{3s+1}{s(s+1)^2} \cdot (s-1) \right) $$

Cancel out the \( s \) terms:

$$ u_{final} = \lim_{s \to 0} \frac{(3s+1)(s-1)}{(s+1)^2} $$

Substitute \( s=0 \):

$$ u_{final} = \frac{(3(0)+1)(0-1)}{(0+1)^2} = \frac{(1)(-1)}{1^2} = -1 $$

The final value of the controller output is -1.

Conclusion

The final value of the controller output is -1, and the final value of the plant output is 1. This matches option 4.