Homework 11 Solution  AME 30315, Spring 2015


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1 1 Homework 11 Solution  AME 30315, Spring 2015 Problem 1 [10/10 pts] R +  K G(s) Y Gpsq Θpsq{Ipsq and we are interested in the closedloop pole locations as the parameter k is varied. Θpsq Ipsq k ωn 2 s 2 ` 2ζω n s ωn 2 We will use k, ζ, and ω n values for our system found from the hanging configuration so that the transfer function is: Our pole locations are: Θpsq Ipsq p6.23q 2 s 2 ` 2p0.039qp6.23qs p6.23q 2 s 1,2 5.99, 6.48 Rule The root locus starts at the openloop poles pk 0q. By Rule 9.2 we know for our case (no zeros), as k Ñ 8, the root locus will grow unbounded for both poles.
2 2 Rule 9.3 gives the asymptotes along which the root locus will grow unbounded. θ n 180 ` n 360 n z n p θ ` p0q θ ` p1q Next, Rule 9.4 gives where the asymptotes intersect the real axis. s int ř nz i 1 z i řn p i 1 p i n z n p s int 0 p q By Rule 9.5 we recognize that on the real axis the root locus exists to the left of an odd number of poles and zeros. Lastly, Rule 9.6 gives the angle at which a branch of the root locus leaves a pole p j ÿn z =ps p j q =pp j z i q i 1 n ÿ p i 1,i j p r180 ` 0s =pp j p i q 180 p r0 s
3 3 We can make a complete root locus sketch [3 pts, must show steps] Thus, we can see that a proportional controller will be able to attain a stable closedloop system. We will use the magnitude criterion to find the minimum k which will give us a stable closed loop system: k 1 Gpsq Therefore: Gpsq s 0 ˆk Gpsq s 0 p0.248qp q ś nz ś nz i 1 s z i i 1 s p i k 1 Gpsq For any k ą we will have a stable closed loop system [3 pts]. To determine if pcontrol is sufficient to meet our design criteria of %OS ă 25% and t r ă 1.5 sec, we want to map this region in the complex plane. We use the rise time approximation formula:
4 4 t r «1.8 ω n Ñ ω n ą 1.2 And using Fig on p. 361 of Goodwine, we find that the damping ratio must be: Plotting, in Matlab [2 pts] ζ ą 0.4. We can see from the trajectories of the root locus branches that there does not exist a value of k such that all the closedloop poles fall within the design criteria region simultaneously (unshaded region) [2 pts]. 1 % AME HW 11 4/19/ % Problem G = zpk([],[5.99,6.48],0.248*6.23ˆ2) 6 rltool
5 5 Problem 2 [25/25 pts] We choose a point within the specified region which satisfies the design criteria, knowing that: s desired ζω n iω n a 1 ζ2 ζω n iω d We choose ζ 0.7 and ω n 2, so that our desired point is: s «1.4 ` 1.43i which meets our design criteria [2 pts]. We want to design a lead controller to make the root locus go through the desired point in the complex plane. A lead controller has the form: Then for the root locus analysis, we define: And design k such that k P p0`, 8q. K lead psq k s ` z s ` p. Gpsq s ` z 0.248p6.23q 2 s ` p s 2 ` 2p0.039qp6.23qs p6.23q 2 We choose a fixed zvalue, and we need to adjust p to make the root locus go through our desired point. Some rules of thumb include that z can be placed at z ω n ; however, this places the zero in between the poles of our plant. We chose z 8 to the left of the stable plant pole. n z n ÿ ÿ p =Gpsq =ps z i q =ps p i q 180 ` 360 pn 1q i 1 i 1 We find the angles referring to Fig. 1.
6 6 Im sdesired θ X p 1 O z θ θ θ4 2 3 X s2 X s1 Re Fig. 1. Vector representation of G(s). Then, θ 2 pθ 1 ` θ 3 ` θ 4 q 180 ` 360 pn 1q where θ 1 is an unknown function of p, so we solve for θ 1 : θ 1 θ 2 θ 3 θ pn 1q ˆ ˆ ˆ tan 1 ωd tan 1 ω d ˆ180 ` tan 1 ω d pn 1q z ζω n 6.48 ζω n pζω n ` 5.99q So From trig, Our controller has the form where k is still an unknown parameter. θ 1 «7.45. tan θ 1 ω d p ζω n p K lead psq k s ` 8 s ` Before finding k, we can plot the root locus. [10 pts]. By Rule 9.1, the root locus starts at the openloop poles (k=0).
7 7 Rule 9.2, n z 1 and n p 3, therefore, the root locus will have one bounded branch and 2 unbounded branches. Rule 9.3, calculate the asymptotes to find that θ 0 90 and θ 1 `90 θ n 180 ` n360 n z n p Rule 9.4, Asymptote intersection with real axis: s int ř nz i 1 z i řn p i 1 p i n z n p z p p 6.48 ` 5.99q s int Rule 9.5, the root locus exists on the real axis to the left of an odd number of poles and zeros. Rule 9.6, we do not need because all of the open loop poles and zeros are on the real axis. Rule 9.7, Break away points from the real axis can be calculated where we let Gpsq Npsq{Dpsq ddpsq dnpsq Npsq Dpsq 0 (1) ds ds The components of this equation are: Npsq 0.248p qps ` 8q dnpsq 0.248p q ds Dpsq ps ` 12.33qps ` 6.48qps 5.99q ddpsq 3s 2 ` 12.82p2qs ds So that we find the roots of Eq. 1 using Mathematica: s i, Then, s is the only breakaway point that makes sense. We can plot our root locus by hand [5 pts, must show steps]:
8 8 Solve for k such that the root locus passes through a given location. We will use the magnitude criterion to find the minimum k which will give us a stable closed loop system: where Gpsq s 0 ˆk k 1 Gpsq ś nz ś nz i 1 s z i i 1 s p i 0.248p q a pz ζω n q 2 ` pω d q a 2 pp ζωn q 2a a p6.48 ζω n q 2 ` ωd 2 p5.99 ` ζωn q 2 ` ωd 2 Gpsq « Ñ k «6.73 [5 pts] Then, the controller by design that will achieve the desired design specifications is: K lead psq 6.73 s ` 8 s ` We verify our sketch using Matlab rltool [3 pts],
9 9 30 Root Locus Editor for Open Loop 1(OL1) Imag Axis Real Axis 1 %Construct the system transfer function 2 G=tf([0.248*6.23ˆ2],[1 2*0.039* ˆ2]) 3 4 % %Assign value for the compensator pole and zero 5 z=8; %Zero 6 p=12.33; %Pole 7 8 %Construct the lead compensator transfer function 9 K_lead=tf([1 z],[1 p]); %Plot the root locus 12 rltool(k_lead*g) We want to evaluate whether the design criteria is actually satisfied, first by evaluating our linear system. When we plot the step response of our system with the controller we designed, we see that:
10 10 14 Step Response Amplitude Time (seconds) Fig. 2. k=6.73, z=8, p=12.33 as designed following the root locus specifications. The overshoot is 4% and the rise time is 1.5 seconds which meets our design criteria. The steadystate value is off; for a step reference input, the output should fall to 1, but we did not set a design criteria for that parameter.
11 11 Problem 3 [10/10 pts] We now want to study the full nonlinear system for the inverted pendulum. Based on homework 9, we can determine the constant parameters for the plant in the provided simulink model: g l ω2 n b ml 2 2ζω n 2p0.039qp6.23q k τ ml 2 kω 2 n These are what we plug into the nonlinear plant diagram. Using our design parameters we get: [5 pts] Reference System Response 80 Theta [deg] Time [sec] This has an overshoot of: OS Ñ %OS 38% and a rise time of 0.25 seconds. This does not meet our overshoot design criteria [5 pts]. To bring the response closer to our desired gain values, we increase the gain and shift the compensator pole and zero. We tried gain k 15, z 7, and p 20.
12 Reference System Response Theta [deg] Time [sec] This decreases the overshoot so that %OS 17% and did not effect the rise time. Our design criteria is satisfied. Time Clock To Workspace2 Step K Gain s+z s+p Transfer Fcn Plant input: i(t) i(t) theta(t) Plant output: theta(t) Plant (doubleclick to edit) System Response Step1 Reference To Workspace1 System_Response To Workspace System Response1 1 close all 2 clear all 3 4 %Construct the system transfer function 5 G=tf([0.248*6.23ˆ2],[1 2*0.039* ˆ2]) 6 7 %Assign value for the compensator pole and zero 8 K = 6.73; 9 z = 8; %zero 10 p = 12.33; %pole 11
13 13 12 %Construct the lead compensator transfer function 13 K_lead=tf([1 z],[1 p]); rltool sim('problem3sim') %Plot system response from simulink data 20 figure(2) 21 plot(time,reference,'b',time,system_response,'r','linewidth',1.5) 22 legend('reference','system Response') 23 xlabel('time [sec]') 24 ylabel('theta [deg]')
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