Storage Elements in the Control Loop

Let us take a system with P behaviour as a first example:

\[ \begin{gathered} A = k_P = 1 \\[6pt] y = \frac{A}{1 + A} \cdot w = \frac{w}{2} \end{gathered} \]

First, let us look at an A that only shows P behaviour with k_P = 1. With k_P = 1, the system has a control error of e = 0.5 ∙ w. If there were no control error, e would be 0 and therefore always y = 0. But we want y = w.

How, then, can the control error disappear completely as soon as we use storage elements in the control loop? Let us look at a second example with I behaviour:

Control loop with k_I = 1 and integrator 1/s

\[ \begin{gathered} A = k_I \cdot \frac{1}{s} \rightarrow \frac{1}{A} = \frac{s}{k_I} \\[6pt] y = e \cdot A \\[6pt] e = w - y \\[6pt] y = \frac{1}{1 + \frac{1}{A}} \cdot w = \frac{1}{1 + \frac{1}{k_I} \cdot s} \cdot w \end{gathered} \]

The answer lies in the equation of storage elements in the time domain. If we insert an A with pure I behaviour and k_I = 1, we obtain:

\[ y(t) = y_0 + \int e(t)\,dt \]

So the value of y(t) depends not only on the current value of e(t), but on its entire past history. The parameter y₀ contains the complete past of the control error over time. So it is not necessary for the control error e always to be non-zero; it is enough for it to have been non-zero at some point in the past. With e = 0, the equation above gives

\[ \begin{gathered} y(t) = y_0 + \int e(t)\,dt \xrightarrow[e = 0]{} y_0 \\[6pt] y(t) = y_0 \end{gathered} \]

A storage element keeps its output y constant when the input e = 0. The fill level of a bucket stays constant if there is no inflow. So a bucket has a non-zero fill level at its output y when the input e = 0. If the bucket is to be filled to the target level y_Soll = “half full”, an inflow e runs in until y = y_Soll. After that nothing more flows in (e = 0), and the bucket keeps its state at the output constant.