Controller Design using Root Locus

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1 Chapter 4 Controller Design using Root Locus 4. PD Control Root locus is a useful tool to design different types of controllers. Below, we will illustrate the design of proportional derivative controllers using root locus technique. PD control shapes the closed-loop transient response of a system by adding a zero to the system dynamics. K(s) =k p + k d s By carefully placing the PD controller zero, the root locus can be shaped to improve damping, rise time, and stability, without any excessively large gains. Let us consider an example. An Example: Consider the following control system. Figure 4.: An example Suppose we use P-control, that is, K(s) = k. The system is in Evan s form. G c (s) = k +k 43

2 Lecture Notes on Control Systems/D. Ghose/ The open-loop poles are p,2 = 0. Since there are no zeros, both poles would go to. Figure 4.2: Open loop poles There are no real axis segments. The asymptotes parameters are, φ α = 80o + α360 o =90 o, 270 o 2 α=0, σ =0 To compute the departure angles, assume the poles to be shifted by an infinitesimal distance along the imaginary axis (see the figure). Then the departure angles can be shown to be θ d =90 o, θ d2 = 270 o. (Work this out yourself.) The root locus is as shown below. Figure 4.3: Root locus Observations:. The closed loop poles are imaginary for all k. 2. There is no damping for any k.

3 Lecture Notes on Control Systems/D. Ghose/ Pure critically damped oscillations. 4. We need to add compensator dynamics to move the pole to the left half plane and get damping. What happens if we use a PD (proportional derivative) controller of the type K(s) =k p + k d s This can be written as, K(s) =k p ( + k ) d s k p = k ( + Ts) Where, k = k p and T = k d k p. Let us assume T = and then find the root locus with respect to k. which is already in Evan s form. G c (s) = k +Ts +k +Ts The open-loop poles are p,2 = 0 and the open-loop zeros are z = T =. So, one pole goes to the zero and the other goes to. Find the real axis segments: The root locus lies on the left of the zero at. Figure 4.4: Open loop pole and zero locations and real axis segment of root locus Asymptote angle is φ α = 80o + α360 o = 80 o α=0 Since there is only one asymptote, it is irrelevant to compute the centroid. Angle of departure from the poles = 90 o and 270 o.

4 Lecture Notes on Control Systems/D. Ghose/ Figure 4.5: Root locus Angle of arrival at the zero = 80 o. Breakaway point is not known, but can be computed. The example shows that when we use a dynamic compensator of the type given above, increasing the value of k makes the closed-loop system move from being unstable to stable but oscillatory and then to stable and overdamped characteristics. A physical interpretation of why a PD controller performs under this condition would be something like this: The PD control responds to changes in command error (because of its derivative component). In a layman s language, in a way it knows what is coming next. This is the reason why a PD controller is also described as an anticipatory controller. Another Example: Let K(s) =k and G(s) =. s(ts+) The closed loop system is which is in Evan s form. G c (s) = k s(ts+) +k s(ts+) Use the steps and you will get the following root locus (try this out yourself). This shows that increasing k takes the system from being neutrally stable to stable and overdamped and then to stable and underdamped. The oscillations increase and so does the overshoot with increasing k. Now consider the same system with PD control K(s) =k(t 2 s +). Let T 2 <T, and both values are given.

5 Lecture Notes on Control Systems/D. Ghose/ Figure 4.6: Root locus with P-control By using the rules in the root locus technique we obtain the following (use the rules to get this locus yourself): Figure 4.7: Root locus with PD-control Here, increasing k moves the system from being neutrally stable to stable overdamped to stable underdamped to stable overdamped again. What are the benefits of PD control?. For the same k, PD control provides faster rise time and peak time with smaller overshoot. 2. One can increase k to decrease steady state error without affecting the transient response.

6 Lecture Notes on Control Systems/D. Ghose/ Lead Control Figure 4.8: Unit step response PD control has a problem that though it improves transient response, the derivative term amplifies noise (why? think about this). So, if there is a small but high frequency disturbance, the PD-control will generate a large control signal. Lead compensation helps to reduce this effect by balancing the PD control zero with an added pole. k(s) =k s + z s + p, 0 z<p Figure 4.9: Lead compensator poles and zeros What does the word lead imply? This has something to do with the phase lead or increase that such a control provides in the forward path. We will understand this better when we cover the frequency response of a LTI system.

7 Lecture Notes on Control Systems/D. Ghose/ Lead compensation shifts the root locus to the left. Consider the angle criterion. θz θ p = 80 o Suppose we have two poles as shown in the next figure. Figure 4.0: A system with two poles subjected to lead control What do we achieve by shifting the root locus to the left? For the same damping ratio ζ, we can increase the controller gain k to obtain better steady state response. We can also have higher ω n for faster rise time. Example of Lead Control: G(s) = 2 s(s +2) We have the following design specifications to meet:. Maximum Overshoot (M p ) Rise time T r 0.5 sec 3. Steady state error with ramp input 0.2 (20 percent) Let us try P-control first. Then, G c (s) = 2k +2s +2k

8 Lecture Notes on Control Systems/D. Ghose/ From maximum overshoot specification: M p = e πζ ζ ζ 0.46 Now obtain the root locus of the system (shown in the figure below). Figure 4.: Root locus Suppose the point s lies on the root locus so that the damping ratio ζ =sinφ 0.46, then φ =sin 0.46 = 27.4 o. On every point on the root locus, if we use our polar representation, the following equation should be satisfied: ( ) 2 0 = +kg(s) =+k s(s +2) =+k 2 e j(θp +θp 2 ) R p R p2 2k R p R p2 = For the damping ratio ζ =0.46, the point on the root locus corresponds to R p sin φ = R p2 sin φ = R p = R p2 =2.7 So, the corresponding value of k is obtained from, 2k =(2.7) 2 k =2.36 Note that this is the maximum permissible value of k. since any larger value will produce a damping ratio less than the prescribed value. So, from maximum overshoot specification we get 0 <k 2.36

9 Lecture Notes on Control Systems/D. Ghose/ From rise time specification: We should meet T r 0.5 sec. Now, T r = π/2+sin ζ ω n ζ ω n π/2+sin ζ 0.5 ζ 2 Note that the RHS of the above inequality increases asζ increases and its minimum value is attained when ζ =0.46. So substituting ζ =0.46 in the above inequality, we obtain, But, ω n π/2+sin =4.67 and so, 2k = ω 2 n ω n 4.67 k =0.65 So, the rise time specification implies that k We can immediately see that this contradicts the maximum overshoot specifications and so both these specifications can never be met using any value of k. We can actually stop here and say that P-control cannot meet the design specifications, but just for the sake of completion, let us also look at the steady state error specification. 3. Steady state error specification: We want the steady state error to ramp input to be less than 20 percent. Applying the final value theorem, we get, e( ) = lim se(s) = lim = lim s + k 2 (s+2) s = k +kg(s) = lim s +k 2 s(s+2) Then e( ) < 0.2 k 5.

10 Lecture Notes on Control Systems/D. Ghose/ To summarize: Maximum overshoot specification 0 <k 2.36 Rise time specification k 0.65 Steady state error specification k 5.0 Hence, all the design specifications cannot be met with P-control. Lead Compensation: Let us try lead compensation here. For simplicity let us put the lead compensator zero on top of the system pole. Let, K(s) =k s +2 s +5 (Note that the pole-zero cancelation is not very desirable, but it is used here to make it simpler for the root locus to be plotted.) Thus, K(s)G(s) =k s +2 2 s +5s(s +2) = k 2 s(s +5) So, all we did was to kick the open loop pole further to the left from (-2 to -5). The closed loop system is, G c (s) = k 2 s(s+5) +k 2 s(s+5) Let us obtain the root locus as shown below: Figure 4.2: Root locus

11 Lecture Notes on Control Systems/D. Ghose/ Using the same reasoning, for the point on the root locus that corresponds to ζ =0.46 (overshoot specification), R p = R p2 = 2.5 sin 27.4 =5.44 o So, the maximum permissible value of k is, k = =4.8 So, maximum overshoot specification says that 0 <k 4.8. From the rise time specification (as before): From steady state specification: ω n π/2+sin =4.67 k 0.65 e( ) = lim se(s) = lim = lim s + k 2 (s+5) s +kg(s) = lim s +k 2 = 5 2k s(s+5) Then e( ) < 0.2 k 2.5. To summarize: Maximum overshoot specification 0 <k 4.8 Rise time specification k 0.65 Steady state error specification k 2.5 Hence a range of k such that 2.5 k 4.8 will meet all the design specifications.

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