Homework 4 - Solutions - University Of California, Berkeley

EE C128 / ME C134 Spring 2014 HW4 - SolutionsUC BerkeleyHomework 4 - SolutionsNote: Each part of each problem is worth 3 points and the homework is worth a total of 42 points.1. State Space Representation To Transfer FunctionFind the transfer function and poles of the system represented in state space below. 8 4 1 4 ẋ 3 20 x 3 u(t)57 94 0i y 2 8 3 x; x(0) 0 0hSolution: G(s) C(sI A) 1 B(sI A) 1 (s 2)(s 9) (4s 29)(s 2)1 3(s2 s 77) 3 3s 27 s s2 91s 675s 317s 76(s2 10s 4) (s 2)(s 9) (4s 29)(s 2) 4hi1 12C(sI A) B 32 8 3 3s 27(s s 77) 3 3 s s2 91s 6725s 317s 76(s 10s 4)4G(s) 44s2 291s 1814s3 s2 91s 672. Transfer Function Analysis For Mechanical SystemsFor the system shown below, do the following:(a) Find the transfer function G(s) X(s)/F (s).(b) Find ζ, ωn , % OS, Ts , Tp and Tr .Solution:(a) Writing the equation of motion yields, (5s2 5s 28)X(s) F (s). Solving the transfer function,1X(s)1 2 2 5F (s)5s 5s 28s s Rev. 1.0, 02/23/20142851 of 9

EE C128 / ME C134 Spring 2014 HW4 - SolutionsUC Berkeley(b) Clearly, ωn2 28/5 rad/s and 2ζωn 1. Therefore, ωn 2.37, ζ 0.211.4 8.01 secTs ζωnπp 1.36 secTp ωn 1 ζ 22%OS e ζπ/ 1 ζ 100 50.7%Tr ω1n (1.76ζ 3 0.417ζ 2 1.093ζ 1) 0.514 sec3. Transfer Function From Unit Step ResponseFor each of the unit step responses shown below, find the transfer function of the system.Solution:K. Using the graph, we can estimate thes a1Ktime constant as T 0.0244 sec. But, a 40.984, and DC gain is 2. Thus 2. Hence,TaK 81.967.Thus,81.967G(s) s 40.984(a) This is a first-order system of the form: G(s) (b) This is a second-order system of the form: G(s) s2percent overshoot and the settling time from the graph.%OS (13.82 11.03) 100 25.3%11.03Ts 2.62secRev. 1.0, 02/23/2014K. We can estimate the 2ζωn s ωn22 of 9

EE C128 / ME C134 Spring 2014 HW4 - SolutionsUC BerkeleyWe can now calculate ζ and ωn from the given information. ln(%OS/100) 0.4ζ q22π ln (%OS/100)4ωn 3.82ζTsKDC Gain 11.03. Therefore, 2 11.03. Hence, K 160.95. Substituting all values, we getωnG(s) s2160.95 3.056s 14.59(c) This is a second-order system of the form: G(s) s2K. We can estimate the 2ζωn s ωn2percent overshoot and the peak time from the graph.%OS (1.4 1.0) 100 40%1.0Tp 4We can now calculate ζ and ωn from the given information. ln(%OS/100)ζ q 0.2822π ln (%OS/100)πωn p 0.818Tp 1 ζ 2KDC Gain 1.0. Therefore, 2 1.0. Hence, K 0.669. Substituting all values, we getωnG(s) s20.669 0.458s 0.6694. State Space AnalysisA system is represented by the state and output equations that follow. Without solving the stateequation, find the characteristic equation and the poles of the system. 0 2 30 ẋ 0 6 5 x 1 u(t)1 4 21hiy 1 2 0 xSolution: 1 0 00 2 3s 2 3 (sI A) s 0 1 0 0 6 5 0 s 6 5 0 0 11 4 2 1 4 s 2Characteristic Equation: det (sI A) s3 8s2 11s 8Factoring yields poles: 9.111, 0.5338 and 1.6448Rev. 1.0, 02/23/20143 of 9

EE C128 / ME C134 Spring 2014 HW4 - SolutionsUC Berkeley5. Block Diagram To Transfer FunctionReduce the system shown below to a single transfer function, T (s) C(s)/R(s).Solution: Push G2 (s) to the left past the summing junction.Collapse the summing junctions and add the parallel transfer functions.Rev. 1.0, 02/23/20144 of 9

EE C128 / ME C134 Spring 2014 HW4 - SolutionsUC BerkeleyPush G1 (s)G2 (s) G3 (s) to the right past the summing junction.Collapse the summing junctions and add feedback paths.Applying the feedback formula,T (s) G3 (s) G1 (s)G2 (s)1 H(s)[G3 (s) G1 (s)G2 (s)] G2 (s)G4 (s)6. Block Diagram To Transfer FunctionFor the system shown below, find the poles of the closed-loop transfer function, T (s) C(s)/R(s).Rev. 1.0, 02/23/20145 of 9

EE C128 / ME C134 Spring 2014 HW4 - SolutionsSolution:2 and 1/s.UC BerkeleyPush 2s to the left past the pickoff point and combine the parallel combination ofPush (2s 1)/s to the right past the summing junction and combine summing junctions.Hence,T (s) 2(2s 1)s5where Heq 1 .1 2(2s 1)Heq (s)2s 1 2sT (s) 4s2 2s6s2 13s 57. Finding Constants To Meet Design SpecificationsFor the system shown below, find the values of K1 and K2 to yield a peak time of 1.5 seconds anda settling time of 3.2 seconds for the closed loop system’s step response.Rev. 1.0, 02/23/20146 of 9

EE C128 / ME C134 Spring 2014 HW4 - SolutionsUC BerkeleySolution: The closed-loop transfer function is given by:T (s) s210K1 (10K2 2)s 10K1pπ4 1.25. And ωn 1 ζ 2 2.09. Therefore, poles are at ζωn We know, ζωn TsTpp ωn 1 ζ 2 1.25 j2.09. Hence, ωn 1.252 2.092 10K1 and (10K2 2)/2 1.25.Therefore, K1 0.593 and K2 0.05.8. Signal-Flow GraphsDraw a signal-flow graph for the following equation. 0100 ẋ 001 x 0 r(t) 2 4 61hiy 1 1 0 xSolution:x 1 x2x 2 x3x 3 2x1 4x2 6x3 ry x1 x29. Signal-Flow GraphsUsing Mason’s rule, find the transfer function, T (s) C(s)/R(s), for the system represented by thefollowing figure.Rev. 1.0, 02/23/20147 of 9

EE C128 / ME C134 Spring 2014 HW4 - SolutionsUC BerkeleySolution:Closed-loop gains: G2 G4 G6 G7 H3 ; G2 G5 G6 G7 H3 ; G3 G4 G6 G7 H3 ; G6 H1 ; G7 H2Forward-path gains: T1 G1 G2 G4 G6 G7 ; T2 G1 G2 G5 G6 G7 ; T3 G1 G3 G4 G6 G7 ; T4 G1 G3 G5 G6 G7Nontouching loops 2 at a time: G6 H1 G7 H2 1 [H3 G6 G7 (G2 G4 G2 G5 G3 G5 ) G6 H1 G7 H2 ] [G6 H1 G7 H2 ] 1 2 3 4 1T (s) T1 1 T2 2 T3 3 T4 4 T (s) G1 G 2 G4 G 6 G7 G1 G 2 G5 G6 G7 G1 G3 G4 G6 G7 G1 G3 G5 G6 G 71 H3 G6 G7 (G2 G4 G2 G5 G3 G4 G3 G5 ) G6 H1 G7 H2 G6 H1 G7 H210. State Space Representation and Signal-Flow GraphsRepresent the system shown below in state space form and draw its signal-flow graph.G(s) s 3s2 2s 7Solution: Writing the state equations,x 1 x2x 2 7x1 2x2 ry 3x1 x2"#" #010ẋ x r 7 21hiy 3 1 xRev. 1.0, 02/23/20148 of 9

EE C128 / ME C134 Spring 2014 HW4 - SolutionsUC Berkeley11. Canonical FormsRepresent the system given in Problem 10 in controller canonical form and observer canonical form.Solution:Controllable canonical form:"#" # 2 71ẋ x r100hiy 1 3 xObservable canonical form:Rev. 1.0, 02/23/2014"#" # 2 11ẋ x r 7 03hiy 1 0 x9 of 9

Homework 4 - Solutions Note: Each part of each problem is worth 3 points and the homework is worth a total of 42 points. 1. State Space Representation To Transfer Function Find the transfer function and poles of the system represented in state space below. x_ 2 6 4 8 4 1 3 2 0 5 7 9