Homework 4 - Solutions

1 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 Space Representation To Transfer FunctionFind the transfer function and poles of the system represented in state space below. x= 8 41 32057 9 x+ 4 34 u(t)y=[2 8 3]x;x(0) = 000 Solution:G(s) =C(sI A) 1B(sI A) 1=1s3 s2 91s+ 67 (s 2)(s+ 9) (4s+ 29)(s 2) 3s 27(s2+s 77) 35s 317s 76(s2 10s+ 4) C(sI A) 1B=1s3 s2 91s+ 67[2 8 3] (s 2)(s+ 9) (4s+ 29)(s 2) 3s 27(s2+s 77) 35s 317s 76(s2 10s+ 4) 4 34 G(s) = 44s2+ 291s+ 1814s3 s2 91s+ Function Analysis For Mechanical SystemsFor the system shown below, do the following:(a) Find the transfer functionG(s) =X(s)/F(s).(b) Find , n, % OS,Ts, :(a) Writing the equation of motion yields, (5s2+5s+28)X(s) =F(s). Solving the transfer function,X(s)F(s)=15s2+ 5s+ 28=15s2+s+285 Rev. , 02/23/20141 of 9EE C128 / ME C134 Spring 2014 HW4 - SolutionsUC Berkeley(b) Clearly, 2n= 28/5 rad/s and 2 n= 1.

2 Therefore, n= , = n= secTp= n 1 2= sec%OS =e / 1 2 100 = n( 3 2+ + 1) = Function From Unit Step ResponseFor each of the unit step responses shown below, find the transfer function of the :(a) This is a first-order system of the form:G(s) =Ks+a. Using the graph, we can estimate thetime constant asT= sec. But,a=1T= ,and DC gain is 2. ThusKa= 2. Hence,K= ,G(s) = + (b) This is a second-order system of the form:G(s) =Ks2+ 2 ns+ 2n. We can estimate thepercent overshoot and the settling time from the =( ) 100 = , 02/23/20142 of 9EE C128 / ME C134 Spring 2014 HW4 - SolutionsUC BerkeleyWe can now calculate and nfrom the given information. = ln(%OS/100) 2+ ln2(%OS/100)= n=4 Ts= Gain = Therefore,K 2n= Hence,K= Substituting all values, we getG(s) = + + (c) This is a second-order system of the form:G(s) =Ks2+ 2 ns+ 2n. We can estimate thepercent overshoot and the peak time from the =( ) 100 = 40%Tp= 4We can now calculate and nfrom the given information.

3 = ln(%OS/100) 2+ ln2(%OS/100)= n= Tp 1 2= Gain = Therefore,K 2n= Hence,K= Substituting all values, we getG(s) = + + 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. x= 0 2 30 6 51 4 2 x+ 011 u(t)y=[1 2 0]xSolution:(sI A) =s 1 0 00 1 00 0 1 0 2 30 6 51 4 2 = s 2 30s 6 5 1 4s 2 Characteristic Equation: det (sI A) =s3 8s2 11s+ 8 Factoring yields poles: , and , 02/23/20143 of 9EE C128 / ME C134 Spring 2014 HW4 - SolutionsUC Diagram To Transfer FunctionReduce the system shown below to a single transfer function,T(s) =C(s)/R(s).Solution:PushG2(s) to the left past the summing the summing junctions and add the parallel transfer , 02/23/20144 of 9EE C128 / ME C134 Spring 2014 HW4 - SolutionsUC BerkeleyPushG1(s)G2(s) +G3(s) to the right past the summing the summing junctions and add feedback the feedback formula,T(s) =G3(s) +G1(s)G2(s)1 +H(s)[G3(s) +G1(s)G2(s)] +G2(s)G4(s) Diagram To Transfer FunctionFor the system shown below, find the poles of the closed-loop transfer function,T(s) =C(s)/R(s).

4 Rev. , 02/23/20145 of 9EE C128 / ME C134 Spring 2014 HW4 - SolutionsUC BerkeleySolution:Push 2sto the left past the pickoff point and combine the parallel combination of2 and 1 (2s+ 1)/sto the right past the summing junction and combine summing ,T(s) =2(2s+ 1)1 + 2(2s+ 1)Heq(s)whereHeq= 1 +s2s+ 1+ (s) =4s2+ 2s6s2+ 13s+ Constants To Meet Design SpecificationsFor the system shown below, find the values ofK1andK2to yield a peak time of seconds anda settling time of seconds for the closed loop system s step , 02/23/20146 of 9EE C128 / ME C134 Spring 2014 HW4 - SolutionsUC BerkeleySolution:The closed-loop transfer function is given by:T(s) =10K1s2+ (10K2+ 2)s+ 10K1We know, n=4Ts= And n 1 2= Tp= Therefore, poles are at n n 1 2= Hence, n= + 10K1and (10K2+ 2)/2 = ,K1= andK2= GraphsDraw a signal-flow graph for the following equation. x= 010001 2 4 6 x+ 001 r(t)y=[1 1 0]xSolution: x1=x2 x2=x3 x3= 2x1 4x2 6x3+ry=x1+ GraphsUsing Mason s rule, find the transfer function,T(s) =C(s)/R(s), for the system represented by thefollowing , 02/23/20147 of 9EE C128 / ME C134 Spring 2014 HW4 - SolutionsUC BerkeleySolution:Closed-loop gains:G2G4G6G7H3;G2G5G6G7H3;G3G4G6G7H3;G 6H1;G7H2 Forward-path gains:T1=G1G2G4G6G7;T2=G1G2G5G6G7;T3=G1G 3G4G6G7;T4=G1G3G5G6G7 Nontouching loops 2 at a time:G6H1G7H2 = 1 [H3G6G7(G2G4+G2G5+G3G5) +G6H1+G7H2] + [G6H1G7H2] 1= 2= 3= 4= 1T(s) =T1 1+T2 2+T3 3+T4 4 T(s) =G1G2G4G6G7+G1G2G5G6G7+G1G3G4G6G7+G1G3G5 G6G71 H3G6G7(G2G4+G2G5+G3G4+G3G5) G6H1 G7H2+ Space Representation and Signal-Flow GraphsRepresent the system shown below in state space form and draw its signal-flow (s) =s+ 3s2+ 2s+ 7 Solution:Writing the state equations, x1=x2 x2= 7x1 2x2+ry= 3x1+x2 x=[01 7 2]x+[01]ry=[3 1]xRev.

5 , 02/23/20148 of 9EE C128 / ME C134 Spring 2014 HW4 - SolutionsUC FormsRepresent the system given in Problem 10 in controller canonical form and observer canonical :Controllable canonical form: x=[ 2 710]x+[10]ry=[1 3]xObservable canonical form: x=[ 2 1 7 0]x+[13]ry=[1 0]xRev. , 02/23/20149 of 9