Question 21.36: There are many examples of position control systems in engin......
There are many examples of position control systems in engineering, for example control of the position of a plotter pen, and control of the position of a radio telescope. The common term for these systems is servo-systems. Consider the block diagram of Figure 21.16 which represents a simple servo-system.
Consider the block diagram of Figure 21.16 which represents a simple servo-system.
The actual position of the motor is denoted by \Theta_{\mathrm{a}}(s) in the s domain and \Theta_{\mathrm{a}}(t) in the time domain. The desired position is denoted by \Theta_{\mathrm{d}}(s) \text { and } \theta_{\mathrm{d}}(t), respectively. The system is a closed loop with negative feedback. The difference between the desired and the actual position generates an error signal which is fed to a controller with gain K. The output signal from the controller is fed to a servo-motor and its associated drive circuitry. The aim of the control system is to maintain the actual position of the motor at a value corresponding to the desired position. In practice, if a new desired position is requested then the system will take some time to attain this new position. The engineer can choose a value of the controller gain to obtain the best type of response from the control system. We will examine the effect of varying K on the response of the servo-system. We can use Rules 1 and 2 to obtain an overall transfer function for the system. The forward transfer function is
The overall transfer function \frac{\Theta_{\mathrm{a}}(s)}{\Theta_{\mathrm{d}}(s)} is obtained by Rule 2 with H(s) = 1. So,
\frac{\Theta_{\mathrm{a}}(s)}{\Theta_{\mathrm{d}}(s)}=\frac{\frac{0.5 K}{s(s+1)}}{1+\frac{0.5 K(1)}{s(s+1)}}=\frac{0.5 K}{s(s+1)+0.5 K}=\frac{0.5 K}{s^2+s+0.5 K}Let us now examine the effect of varying K. We will consider three values, K = 0.375, K = 0.5, K =5, and examine the response ofthe system to a unit step input in each case.
For K = 0.375
With 1 \Theta_{\mathrm{d}}(s)=\frac{1}{s}, then
\Theta_2(s)=\frac{0.1875}{\left(s^2+s+0.1875\right) s}=\frac{1}{s}+\frac{0.5}{s+0.75}-\frac{1.5}{s+0.25}using partial fractions. So,
\theta_a(t)=1+0.5 \mathrm{e}^{-0.75 t}-1.5 \mathrm{e}^{-0.25 t} \quad t \geqslant 0This is shown in Figure 21.17. Engineers usually refer to this as an overdamped response. The response does not overshoot the final value.
For K = 0.5
This is shown in Figure 21.17 and is termed a critically damped response. It corresponds to the fastest rise time of the system without overshooting.
For K = 5
Rearranging to enable standard forms to be inverted gives
\begin{aligned} \Theta_a(s) & =\frac{1}{s}-\frac{s+0.5}{(s+0.5)^2+1.5^2}-\frac{0.5}{\left(s+0.5^2\right)+1.5^2} \\ \theta_a(t) & =1-\mathrm{e}^{-0.5 t} \cos 1.5 t-\frac{1}{3} \mathrm{e}^{-0.5 t} \sin 1.5 t \end{aligned}The trigonometric terms can be expressed as a single sinusoid using the techniques given in Section 3.7. Thus,
\theta_a(t)=1-1.054 \mathrm{e}^{-0.5 t} \sin (1.5 t+1.249) \quad \text { for } \quad t \geqslant 0This is shown in Figure 21.17 and is termed an underdamped response. The system overshoots its final value.
In a practical system it is common to design for some overshoot, provided it is not excessive, as this enables the desired value to be reached more quickly. It is interesting to compare the system response for the three cases with the nature of their respective transfer function poles. For the overdamped case the poles are real and unequal, for the critically damped case the poles are real and equal, and for the underdamped case the poles are complex. Engineers rely heavily on pole positions when designing a system to have a particular response. By varying the value of K it is possible to obtain a range of system responses and corresponding pole positions.
