06 - Closed Loop Response

Closed loop system with \(G_m=G_f=1\) leads to, \[Y(s) = \frac{G_pG_c}{1+G_pG_c}Y_{sp}(s) + \frac{G_d}{1+G_pG_c}d(s)\] Let us study the effect of setpoint change, i.e., servo problem. Here, \(d(s)=0\).

Let us consider a first order system, controlled by integral only controller. For the integral control action, \[C(t) = K_c\frac{1}{\tau_I}\int_0^t\epsilon(t) dt\] Taking Laplace transform, \[C(s) = K_c\frac{1}{\tau_Is}\epsilon(s) =\qquad \Longrightarrow \quad \frac{C(s)}{\epsilon(s)} = G_c\] \[G_c = K_c\frac{1}{\tau_Is}\] where \(K_c\) is controller gain; and, \(\tau_I\) is integral time.

The closed loop response for the setpoint changes is given by \[\begin{aligned} Y(s) &= \frac{G_pG_c}{1+G_pG_c}Y_{sp}(s) = \frac{\left(\dfrac{K_p}{\tau_p s+1}\right)\left(\dfrac{K_c}{\tau_I s}\right)}{1+\left(\dfrac{K_p}{\tau_p s+1}\right)\left(\dfrac{K_c}{\tau_I s}\right)}Y_{sp}(s) \\ &= \frac{K_pK_c}{\tau_p\tau_I s^2+\tau_Is+K_pK_c}Y_{sp}(s) \\ \Longrightarrow \quad Y(s) &= \frac{1}{\tau^2s^2+2\zeta\tau s+1}Y_{sp}(s) \end{aligned}\] where \[\tau^2 = \frac{\tau_p\tau_I}{K_pK_c} \qquad \Longrightarrow \quad \tau = \sqrt{\frac{\tau_p\tau_I}{K_pK_c}}\]

\[\begin{aligned} 2\zeta\tau = \frac{\tau_I}{K_pK_c} \qquad \Longrightarrow \quad \zeta \sqrt{\frac{\tau_p\tau_I}{K_pK_c}} &= \frac{1}{2}\frac{\tau_I}{K_pK_c} \\ \Longrightarrow \quad \zeta &= \frac{1}{2}\sqrt{\frac{\tau_I}{\tau_pK_pK_c}} \end{aligned}\] With integral controller, order of the system is increased.
Ultimate value for unit step change in point: \(Y_{sp}(s) = 1/s\). \[\lim_{t\rightarrow \infty}Y(t) = \lim{s\rightarrow 0} sY(s) = \lim_{s\rightarrow0}\frac{1}{\tau^2s^2+2\zeta\tau s+1} = 1\] Therefore, \[\text{offset} = Y_{sp} - Y_{t\rightarrow\infty} = 1 - 1 =0\] Integral action eliminates offset.