Question 13.10: Consider a continuous time plant with nominal transfer funct...
Consider a continuous time plant with nominal transfer function G_{o}(s) given by
G_{o}(s) = \frac{2}{(s + 1)(s + 2)} (13.7.5)
Assume that this plant has to be digitally controlled with sampling period ∆ = 0.2[s] in such a way that the plant output tracks a periodic reference, r[k], given by
r\left[k\right] = \sum\limits_{i=0}^{\infty }{r_{T}\left[k – 10i\right] } \Longleftrightarrow R_{q}(z) = R_{Tq}(z)\frac{z^{10}}{z^{10} – 1} (13.7.6)
where {r_{T} [k]} = {0.0; 0.1; 0.25; 0.6; 0.3; 0.2; −0.1; −0.3; −0.4; 0.0} and³ R_{Tq}(z) = Z [r_{T} [k]].
Synthesize the digital control which achieves zero steady state at-sample errors.
From (13.7.6) we observe that the reference generating polynomial, Γ_{qr}(z), is given by z^{10} − 1. Thus the IMP leads to the following controller structure
C_{q}(z)=\frac{P_{q}(z)}{L_{q}(z)}=\frac{P_{q}(z)}{\bar{L}_{q}(z)\Gamma _{qr}(z) } (13.7.7)
We then apply pole assignment with the closed loop polynomial chosen as A_{clq}(z) = z^{12}(z − 0.2). The solution of the Diophantine equation yields
P_{q}(z)=13.0z^{11}+11.8z^{10}-24.0z^{9}+19.7z^{8}-16.1z^{7}+13.2z^{6}-10.8z^{5}+8.8z^{4}-7.3z^{3}+36.4z^{2}-48.8z+17.6 (13.7.8)
L_{q}(z)=(z^{10}-1)(z+0.86) (13.7.9)
Figure 13.9 shows the performance of the resulting digital control loop.
In Figure 13.9 we verify that,after a transient period,the plant output, y(t) follows exactly the periodic reference, at the sampling instants. Notice the dangers of a purely at-sample analysis and design. The intersample behavior of this example can be predicted with the techniques for hybrid systems which will be presented in Chapter 14.
