we really need to use SRL to guide the proper selection of poles. The bottom line is that SRL (or LQR) is the method of choice! Consider the second-order servomechanism system described by G(s) = 1 / s(s + 1) and with state description x1 = x2, x2 = −x2 + u. Design a controller using pole placement

Question

Servomechanism: Increasing the Velocity Constant through Zero Assignment
Consider the second-order servomechanism system described by

[latex] G(s) = \frac{1}{s(s + 1)} [/latex]

and with state description

[latex] \dot{x}_1 = x_2, \ \dot{x}_2 = −x_2 + u. [/latex]

Design a controller using pole placement so that both poles are at [latex] s = −2 [/latex] and the system has a velocity constant [latex] K_v = 10.[/latex] Derive an equivalent discrete controller with a sampling period of [latex] T_s = 0.1 ext{ sec } (20 × ω_n = 20 × 0.05 = 0.1 ext{ sec}),[/latex] and compare the continuous and digital control outputs, as well as the control efforts.

Accepted Answer

For this problem, the state feedback gain

[latex] \pmb K = \begin{bmatrix} 8 & 3 \end{bmatrix} [/latex]

results in the desired control poles. However, with this gain, [latex] K_v = 2, [/latex] and we need [latex] K_v = 10. [/latex] What effect will using estimators designed according to the three methods for [latex] \pmb M [/latex] and [latex] \overset{-}{N} [/latex] selection have on our design? Using the first strategy (the autonomous estimator), we find that the value of [latex] K_v [/latex] does not change. If we use the second method (error control), we introduce a zero at a location unknown beforehand, and the effect on [latex] K_v [/latex] will not be under direct design control. However, if we use the third option (zero placement) along with Truxal’s formula [Eq. (7.197)], we can satisfy both the dynamic response and the steady-state requirements.

[latex] \frac{1}{K_v} = \sum{\frac{1}{z_i}} − \sum{\frac{1}{p_i}}. [/latex] (7.197)

First we must select the estimator pole [latex] p_3 [/latex] and the zero [latex] z_3 [/latex] to satisfy Eq. (7.197) for [latex] K_v = 10. [/latex] We want to keep [latex] z_3−p_3 [/latex] small, so that there is little effect on the dynamic response, and yet have [latex] 1/z_3 − 1/p_3 [/latex] be large enough to increase the value of [latex] K_v.[/latex] To do this, we arbitrarily set [latex] p_3 [/latex] small compared with the control dynamics. For example, we let

[latex] p_3 = −0.1. [/latex]

Notice that this approach is opposite to the usual philosophy of estimation design, where fast response is the requirement. Now, using Eq. (7.197) to get

[latex] \frac{1}{K_v} = \frac{1}{z_3} − \frac{1}{p_1} − \frac{1}{p_2} −\frac{1}{p_3}, [/latex]

where [latex] p_1 = −2+2j, p_2 = −2−2j, [/latex] and [latex] p_3 = −0.1, [/latex] we solve for [latex] z_3 [/latex] such that [latex] K_v = 10 [/latex] we obtain

[latex] \frac{1}{K_v} = \frac{4}{8} + \frac{1}{0.1} + \frac{1}{z_3} = \frac{1}{10} , [/latex]

or

[latex] z_3 = − \frac{1}{10.4} = −0.096. [/latex]

We thus design a reduced-order estimator to have a pole at [latex] −0.1 [/latex] and choose [latex] \pmb M/ \overset{-}{N} [/latex] such that [latex] γ (s) [/latex] has a zero at [latex] −0.096. [/latex] A block diagram of the resulting system is shown in Fig. 7.50(a).You can readily verify that this system has the overall transfer function

[latex] \frac{Y(s)}{R(s)} = \frac{8.32(s + 0.096)}{(s^2 + 4s + 8)(s + 0.1)}, [/latex] (7.198)

for which [latex] K_v = 10, [/latex] as specified.

The compensation shown in Fig. 7.50(a) is nonclassical in the sense that it has two inputs ([latex] e ext{ and } y [/latex] ) and one output. If we resolve the equations to provide pure error compensation by finding the transfer function from [latex] e ext{ and } u,[/latex] which would give Eq. (7.198), we obtain the system shown in Fig. 7.50(b). This can be seen as follows: The relevant controller equations are

[latex] \dot{x}_c = 0.8 e − 3.1 u, [/latex]

[latex] u = 8.32 e + 3.02 y + x_c, [/latex]

where [latex] x_c [/latex] is the controller state. Taking the Laplace transform of these equations, eliminating [latex] X_c(s), [/latex] and substituting for the output [latex] [Y(s) = G(s)U(s)],[/latex] we find that the compensator is described by

[latex] \frac{U(s)}{E(s)} = D_c(s) = \frac{(s + 1)(8.32s + 0.8)}{(s + 4.08)(s + 0.0196)}. [/latex]

This compensation is a classical lag–lead network. The root locus of the system in Fig. 7.50(b) is shown in Fig. 7.51. Note the pole–zero pattern near the origin that is characteristic of a lag network. The Bode plot in Fig. 7.52 shows the phase lag at low frequencies and the phase lead at high frequencies. The step response of the system is shown in Fig. 7.53 and shows the presence of a “tail” on the response due to the slow pole at [latex] −0.1. [/latex] Of course, the system is Type [latex] 1 [/latex] and the system will have zero tracking error eventually.

The discrete equivalent for the controller is obtained from MATLAB by using the [latex] ext{ c2d } [/latex] command, as in the following code:

Question 7.EX.34: we really need to use SRL to guide the proper selection of p...

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