Question

A unity negative feedback closed loop system has a plant with the transfer function \(G(s) = \dfrac{1}{s^2 + 2s + 2}\) and a controller G_e(s) in the feedforward path. For a unit step input, the transfer function of the controller that gives minimum steady slate error is
1. \(G_e(s) = \dfrac{s + 1}{s + 2}\)
2. \(G_e(s) = \dfrac{s + 2}{s + 1}\)
3. \(G_e(s) = \dfrac{(s + 1)(s+4)}{(s + 2)(s + 3)}\)
4. \(G_e(s) = 1+ \dfrac{2}{s} + 3s\)

Answer 1

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

Concept:

The steady-state error for the second-order system when the input is unit step is defined as:

\(e_{ss}=\dfrac{1}{1+k_p}\)

Where the positional error constant is:

\(\mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right) = {k_p}\)

G(s): Forward path gain

H(s): Feedback gain

To have the less steady-state error the positional error constant must be high.

Calculation:

Check the options and whichever gives the high K_p that is the answer for less error.

Forward gain = G(s) × G_C(s)

Feedback gain = H(s) = 1

Option 1:

\(\mathop {\lim }\limits_{s \to 0} \dfrac{1}{s^2+2s+2}\dfrac{s+1}{s+2} \\= {k_p}=\frac{1}{4}\)

Option 2:

\(\mathop {\lim }\limits_{s \to 0} \dfrac{1}{s^2+2s+2}\dfrac{s+2}{s+1} \\= {k_p}=1\)

Option 3:

\(\mathop {\lim }\limits_{s \to 0} \dfrac{1}{s^2+2s+2}\dfrac{(s+1)(s+4)}{(s+2)(s+3)} \\= {k_p}=\dfrac{2}{6}\)

Option 4:

\(\mathop {\lim }\limits_{s \to 0} \dfrac{1}{s^2+2s+2}{\left(1+\dfrac{2}{s}+3s\right)} \\= {k_p}=\infty\)

If we select controller 4 then the steady-state error will be less.

e_ss = 0