Question 13.4: Consider a continuous time plant with transfer function Go(s...
Consider a continuous time plant with transfer function
G_{o}(s) =\frac{50}{(s+2)(s+5)} (13.6.11)
Synthesize a minimum prototype controller with sampling period Δ = 0.1[s].
The sampled transfer function is given by
C_{oq}(z) =\frac{0.0398(z+0.7919)}{(z-0.8187)(z-0.6065)} (13.6.12)
Notice that C_{oq}(z) is stable and minimum phase, with m = 2 and n = 3. Then, on applying (13.6.10), we have that
C_{q}(z) = [G_{oq}(z)]^{-1} \frac{1}{z^{n-m} -1} and T_{o}(z) =\frac{1}{z^{n-m}} (13.6.10)
C_{q}(z) =\frac{25.124 (z-0.8187)(z-0.6065)}{(z-1)(z+0.7919)} and T_{oq}(z) =\frac{1}{z} (13.6.13)
The performance of the resultant control loop is evaluated for a unit step reference at t = 0.1[s]. The plant output is shown in Figure 13.5. We see that the sampled response settles in exactly one sample period. This is as expected, since T_{oq}(z)= z^{-1} . However, Figure 13.5 illustrates one of the weaknesses of minimal prototype control: perfect tracking is only guaranteed at the sampling instants. Indeed, we see a very substantial intersample ripple! We will analyze the cause of this problem in the next chapter. Another drawback of this approach is the large control effort required: since the controller is biproper, it reacts instantaneously to the step reference with an initial value equal to 48.73 times the magnitude of the step.
