Introduction:

The Hold Operation
In the computer-controlled systems, it is necessary to convert the control actions calculated by the computer as a sequence of numbers, to a continuous-time signal that can be applied to the process.

The problem of hold operation may be posed as follows:
Given a sequence $$ {y(0), y(1),..., y(k),...} $$ We have to construct $$ y_a(t), t \ge 0 $$ A commonly used solution to the problem of hold operation is polynomial extrapolation. Using Taylor’s series expansion about , we can express $$ y_a(t) = y_a(kT) + \dot y_a (kT)(t-kT)+ \frac {\ddot y_a(kT)}{2!} (t-kT)^2 +...; kT \le t \lt (k+1)T \tag{1} $$ where, $$ \dot y_a (kT) \cong \frac{1}{T} [y_a(kT)-y_a((k-1)T)] $$ $$ \ddot y_a(kT) \cong \frac{1}{T^2} [y_a(kT)-2y_a((k-1)T)+y_a((k-2)T)] $$ If only the first term in expansion (1) is used, the data hold is called a zero-order hold (ZOH). Here we assume that the function is approximately constant within the sampling interval, at a value equal to that of the function at the preceding sampling instant. Therefore, for a given input sequence , the output of ZOH is given by
$$ y_a(t) = y(k) ; kT \le t \lt (k+1)T \tag{2} $$ The first two terms in expansion (1) are used to realize the first-order hold. For a given input sequence , the output of the first-order hold is given by
$$ y_a(t) = y(k) + \frac {t-kT}{T} [y(k) - y(k-1)] \tag{3} $$ It is obvious from expansion (1) that the higher the order of the derivative to be approximated, the larger will be the number of delay pulses required. The time-delay adversely affects the stability of feedback control systems. Furthermore, a high-order extrapolation requires complex circuitry and results in high costs of construction. The ZOH is the simplest, and most commonly used, data hold device. The standard D/A converters are often designed in such a way that the old value is held constant until a new conversion is ordered.