Minimizing Steady-State Error in Control Systems

The problem asks us to find the transfer function of a controller, \(G_e(s)\), that will result in the minimum steady-state error for a unit step input in a unity negative feedback closed-loop system. We are given the plant transfer function \(G(s) = \dfrac{1}{s^2 + 2s + 2}\).

Steady-State Error for Unit Step Input

For a unity negative feedback system, the open-loop transfer function is given by \(G_{OL}(s) = G_e(s) G(s)\). For a unit step input, \(R(s) = \dfrac{1}{s}\), the steady-state error (\(e_{ss}\)) is determined using the Final Value Theorem:

\(e_{ss} = \lim_{s \to 0} s E(s)\)

Where \(E(s)\) is the error signal in the s-domain, given by:

\(E(s) = \dfrac{R(s)}{1 + G_e(s) G(s)}\)

Substituting \(R(s) = \dfrac{1}{s}\), we get:

\(e_{ss} = \lim_{s \to 0} s \left( \dfrac{1/s}{1 + G_e(s) G(s)} \right)\)

\(e_{ss} = \lim_{s \to 0} \dfrac{1}{1 + G_e(s) G(s)}\)

To achieve minimum steady-state error, ideally zero steady-state error, the denominator term \(1 + G_e(s) G(s)\) must approach infinity as \(s \to 0\). This implies that the open-loop transfer function \(G_e(s) G(s)\) must approach infinity as \(s \to 0\).

System Type and Steady-State Error

The steady-state error for a step input is inversely related to the "type" of the system. The type of a system is defined by the number of pure integrators (poles at \(s=0\)) in its open-loop transfer function \(G_{OL}(s)\). To have zero steady-state error for a step input, the system must be at least Type 1.

Let's examine the given plant transfer function:

\(G(s) = \dfrac{1}{s^2 + 2s + 2}\)

This plant has no poles at \(s=0\). Therefore, the plant itself is a Type 0 system. For the overall closed-loop system to be Type 1 or higher (and thus achieve zero steady-state error for a step input), the controller \(G_e(s)\) must introduce at least one pole at the origin (\(s=0\)).

Analyzing Controller Options

We will now evaluate each given option for \(G_e(s)\) to determine which one, when combined with \(G(s)\), results in an open-loop transfer function \(G_{OL}(s)\) with a pole at the origin, leading to minimum steady-state error.

Option 1: \(G_e(s) = \dfrac{s + 1}{s + 2}\)

• Open-loop transfer function: \(G_{OL}(s) = G_e(s)G(s) = \left( \dfrac{s + 1}{s + 2} \right) \left( \dfrac{1}{s^2 + 2s + 2} \right) = \dfrac{s + 1}{(s + 2)(s^2 + 2s + 2)}\)
• Limit as \(s \to 0\): \( \lim_{s \to 0} G_{OL}(s) = \dfrac{0 + 1}{(0 + 2)(0^2 + 2(0) + 2)} = \dfrac{1}{2 \cdot 2} = \dfrac{1}{4}\)
• Steady-state error: \(e_{ss} = \dfrac{1}{1 + \lim_{s \to 0} G_{OL}(s)} = \dfrac{1}{1 + 1/4} = \dfrac{1}{5/4} = \dfrac{4}{5}\)
• This option gives a non-zero steady-state error.

Option 2: \(G_e(s) = \dfrac{s + 2}{s + 1}\)

• Open-loop transfer function: \(G_{OL}(s) = G_e(s)G(s) = \left( \dfrac{s + 2}{s + 1} \right) \left( \dfrac{1}{s^2 + 2s + 2} \right) = \dfrac{s + 2}{(s + 1)(s^2 + 2s + 2)}\)
• Limit as \(s \to 0\): \( \lim_{s \to 0} G_{OL}(s) = \dfrac{0 + 2}{(0 + 1)(0^2 + 2(0) + 2)} = \dfrac{2}{1 \cdot 2} = 1\)
• Steady-state error: \(e_{ss} = \dfrac{1}{1 + \lim_{s \to 0} G_{OL}(s)} = \dfrac{1}{1 + 1} = \dfrac{1}{2}\)
• This option also gives a non-zero steady-state error.

Option 3: \(G_e(s) = \dfrac{(s + 1)(s+4)}{(s + 2)(s + 3)}\)

• Open-loop transfer function: \(G_{OL}(s) = G_e(s)G(s) = \left( \dfrac{(s + 1)(s+4)}{(s + 2)(s + 3)} \right) \left( \dfrac{1}{s^2 + 2s + 2} \right) = \dfrac{(s + 1)(s+4)}{(s + 2)(s + 3)(s^2 + 2s + 2)}\)
• Limit as \(s \to 0\): \( \lim_{s \to 0} G_{OL}(s) = \dfrac{(0 + 1)(0 + 4)}{(0 + 2)(0 + 3)(0^2 + 2(0) + 2)} = \dfrac{1 \cdot 4}{2 \cdot 3 \cdot 2} = \dfrac{4}{12} = \dfrac{1}{3}\)
• Steady-state error: \(e_{ss} = \dfrac{1}{1 + \lim_{s \to 0} G_{OL}(s)} = \dfrac{1}{1 + 1/3} = \dfrac{1}{4/3} = \dfrac{3}{4}\)
• This option also results in a non-zero steady-state error.

Option 4: \(G_e(s) = 1+ \dfrac{2}{s} + 3s\)

• First, simplify \(G_e(s)\) by finding a common denominator: \(G_e(s) = \dfrac{s}{s} + \dfrac{2}{s} + \dfrac{3s^2}{s} = \dfrac{s + 2 + 3s^2}{s}\)
• This controller has an integral term (\(\frac{2}{s}\)), which means it introduces a pole at \(s=0\). This is characteristic of a PID (Proportional-Integral-Derivative) controller.
• Open-loop transfer function: \(G_{OL}(s) = G_e(s)G(s) = \left( \dfrac{3s^2 + s + 2}{s} \right) \left( \dfrac{1}{s^2 + 2s + 2} \right) = \dfrac{3s^2 + s + 2}{s(s^2 + 2s + 2)}\)
• Limit as \(s \to 0\): \( \lim_{s \to 0} G_{OL}(s) = \lim_{s \to 0} \dfrac{3s^2 + s + 2}{s(s^2 + 2s + 2)}\) As \(s \to 0\), the numerator approaches \(3(0)^2 + 0 + 2 = 2\). As \(s \to 0\), the denominator approaches \(0 \cdot (0^2 + 2(0) + 2) = 0 \cdot 2 = 0\). Therefore, \( \lim_{s \to 0} G_{OL}(s) = \dfrac{2}{0} = \infty\)
• Steady-state error: \(e_{ss} = \dfrac{1}{1 + \lim_{s \to 0} G_{OL}(s)} = \dfrac{1}{1 + \infty} = \dfrac{1}{\infty} = 0\)
• This option results in zero steady-state error, which is the minimum possible.

Conclusion

By including an integral term (\(\frac{2}{s}\)), the controller \(G_e(s) = 1+ \dfrac{2}{s} + 3s\) introduces a pole at the origin into the open-loop transfer function \(G_{OL}(s)\). This effectively increases the system type to at least Type 1. For a Type 1 system, the steady-state error for a unit step input is zero, which is the minimum possible error.