Question 21.30: Many engineering systems can be modelled by a first-order di......
Many engineering systems can be modelled by a first-order differential equation. The time constant is a measure of the rapidity with which these systems respond to a change in input. Suppose an electronic thermometer is used to measure the temperature of an oven. The sensing element does not respond instantly to changes in the oven temperature because it takes time for the element to heat up or cool down. Provided the electronic circuitry does not introduce further time delays then the differential equation that models the thermometer is given by
\tau \frac{\mathrm{d} v_{\mathrm{m}}}{\mathrm{d} t}+v_{\mathrm{m}}=v_{\mathrm{o}}where v_m = measured temperature, v_0 = oven temperature, π = time constant of the sensor. For convenience the temperature is measured relative to the ambient room temperature, which forms a ‘base line’ for temperature measurement. Suppose the sensing element of an electronic thermometer has a time constant of 2 seconds. Ifthe temperature ofthe oven increases linearly atthe rate of 3^{\circ} \mathrm{C} \mathrm{s}^{-1} starting from an ambient room temperature of 20^{\circ} \mathrm{C} \text { at } t=0, calculate the response of the thermometer to the changing oven temperature. State the maximum temperature error.
Taking Laplace transforms of the equation gives
\tau\left(s V_{\mathrm{m}}(s)-v_{\mathrm{m}}(0)\right)+V_{\mathrm{m}}(s)=V_{\mathrm{o}}(s).
v_{\mathrm{m}}(0)=0 as the oven temperature and sensor temperature are identical at t = 0. Therefore,
\begin{aligned} & \tau s V_{\mathrm{m}}(s)+V_{\mathrm{m}}(s)=V_{\mathrm{o}}(s) \\ & V_{\mathrm{m}}(s)=\frac{V_{\mathrm{o}}(s)}{1+\tau s}&&(21.1) \end{aligned}For this example, the input to the thermometer is a temperature ramp with a slope of 3^{\circ} \mathrm{C} \mathrm{s}^{-1} \text {. Therefore, } v_{\mathrm{o}}=3 t \text { for } t \geqslant 0:
V_{\mathrm{o}}(s)=\mathcal{L}\left\{v_{\mathrm{o}}(t)\right\}=\frac{3}{s^2} (21.2)
Combining Equations (21.1) and (21.2) yields
V_{\mathrm{m}}(s)=\frac{3}{s^2(1+\tau s)}=\frac{3}{s^2(1+2 s)} \quad \text { since } \tau=2Then using partial fractions, we have
V_{\mathrm{m}}(s)=\frac{3}{s^2}-\frac{6}{s}+\frac{12}{1+2 s}Taking the inverse Laplace transform yields
v_{\mathrm{m}}=3 t-6+6 \mathrm{e}^{-0.5 t} \quad t \geqslant 0This response consists of three parts:
(1) a decaying transient, 6 \mathrm{e}^{-0.5 t}, which disappears with time;
(2) a ramp, 3t, with the same slope as the oven temperature;
(3) a fixed negative temperature error, —6.
Therefore, after the transient has decayed the measured temperature follows the oven temperature with a fixed negative error. It is instructive to obtain the temperature error by an alternative method. Given that the temperature error is v_e, then
v_{\mathrm{e}}=v_{\mathrm{m}}-v_{\mathrm{o}}and
V_{\mathrm{e}}(s)=V_{\mathrm{m}}(s)-V_{\mathrm{o}}(s) (21.3)
Combining Equations (21.1) and (21.3) yields
V_{\mathrm{e}}(s)=\frac{V_{\mathrm{o}}(s)}{1+\tau s}-V_{\mathrm{o}}(s)=\frac{-\tau s V_{\mathrm{o}}(s)}{1+\tau s} (21.4)
Combining Equations (21.2) and (21.4) yields
V_{\mathrm{e}}(s)=\frac{-\tau s}{1+\tau s} \frac{3}{s^2}=\frac{-3 \tau}{s(1+\tau s)}The final value theorem can be used to find the steady-state error:
\lim _{t \rightarrow \infty} v_{\mathrm{e}}(t)=\lim _{s \rightarrow 0} s V_{\mathrm{e}}(s)=\lim _{s \rightarrow 0}\left[\frac{-3 \tau s}{s(1+\tau s)}\right]=-3 \tau=-6that is, the steady-state temperature error is -6^{\circ} \mathrm{C}. It is important to note that the final value theorem can only be used if it is known that the time function tends to a limit as t \rightarrow \infty. In many cases engineers know this is the case from experience.