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ECE-320: Linear Control Systems Homework 8 Due: Tuesday May 10 at 10 AM 1) For the plant 22()121pnnKGsssζωω=++ a) If the plant input is and the output is()ut ()xt, show that we can represent this system with the differential equation 212() () () ()nnxtxtxtKutζωω++=&& & b) Assuming we use states 1() ()qt xt=and 2() ()qt xt=&, and the output is ()xt, show that we can write the state variable description of the system as [] []1122221201 0() ()()2() ()()() 1 0 0 ()()nn nqt qtdutKqt qtdtqtyt utqtωζω ω⎡⎤⎡⎡⎤ ⎡⎤=+⎢⎥⎢⎢⎥ ⎢⎥−−⎣⎦ ⎣⎦⎣⎦⎣⎡⎤=+⎢⎥⎣⎦⎤⎥⎦ or () () ()() () ()qt Aqt Butyt Cqt Dut=+=+& Determine the A, B, C and D matrices. c) Assume we use state variable feedback of the form () () ()pfut G rt kqt=−, where is the new input to the system, is a prefilter (for controlling position error), and is the state variable feedback gain vector. Show that the state variable model for the closed loop system is ()rtpfGk () ( ) () ( )()() ( ) () ( )()pfpfqt A Bkqt BG rtyt C Dkqt DG rt=−+=− +& or () () ()() () ()qt Aqt Brtyt Cqt Drt=+=+%%&%%d) Show that the transfer function (matrix) for the closed loop system between input and output is given by 1()() ( )( ( ))()pf pfYsG s C Dk sI A Bk BG DGRs−==− −− + and if is zero this simplifies to D 1()() ( ( ))()pfYsG s C sI A Bk BGRs−==−− e) Assume and . Show that, in order for () 1rt =0D =lim ( ) 1tyt→∞=, we must have 11()pfGCA Bk B−−=− Note that the prefilter gain is a function of the state variable feedback gain! If matrix is given as PabPcd⎡⎤=⎢⎥⎣⎦ then 11dbPcaad bc−−⎡⎤=⎢⎥−−⎣⎦ and the determinant of is given by Pad bc−. This determinant will also give us the characteristic polynomial of the system. 2) For each of the systems below: • determine the transfer function when there is state variable feedback • determine if and exist (1k2k[]12kkk=) to allow us to place the closed loop poles anywhere. That is, can we make the denominator look like for any and any . If this is true, the system is said to be conrollable. 21sasa++0u1a0a a) Show that for 10 011 1qq⎡⎤⎡⎤=+⎢⎥⎢⎥⎣⎦⎣⎦& []01 [0]yq=+u the closed loop transfer function with state variable feedback is 2(1)()(1)(1 )pfsGGsss k−=−−+b) Show that for 01 001 1qqu⎡⎤⎡⎤=+⎢⎥⎢⎥⎣⎦⎣⎦& []01 [0]yq=+u the closed loop transfer function with state variable feedback is 221()(1)pfsGGssk sk=+−+ c) Show that for 01 011 1qqu⎡⎤⎡⎤=+⎢⎥⎢⎥⎣⎦⎣⎦& []10 [0]yq=+u the closed loop transfer function with state variable feedback is 221()(1)(1pfGGssk sk=)+−+−3) Consider the state variable model of a plant 01 110 1qqu⎡⎤⎡⎤=+⎢⎥⎢⎥⎣⎦⎣⎦& []01 [0]yq=+u a) Show that the plant transfer function is p2(1) 1G()11ssss+==−− b) The eigenvalues of theA matrix are determined by setting the determinant of the matrix AIλ−0 equal to 0, i.e. they are solutions to||AIλ−=. Find the eigenvalues of the plant analytically. The equation ||AI0λ−= is the same as the equation ||. However, the determinant of gives us the characteristic equation of the plant transfer function. Hence the eivenvalues of 0sI A−=sI A−()pGsA are the poles of the plant transfer function. c) Show that the closed loop transfer function with state variable feedback is 1212(1)()()()(1)(pfsGGssksk k k 1)+=++−− − d) Show that, for and 1pfG =120kk== , the closed loop transfer function is the same as that of the original system (plant). This is important! State variable feedback only changes the gain of the system and the location of the closed loop poles. It does not increase the order of the system or add zeros to the transfer function. e) Show that for 12k= and , the closed loop system has poles at -1 and -3. 22k = f) Find the eigenvalues of analytically, and show they are equal to the poles of the closed loop transfer function. The equation |(ABk−) | 0ABk Iλ−−= is the same as the equation|(. However, the determinant of )|sI A Bk−− =0 )(sI A Bk−−gives us the characteristic equation of the closed loop transfer function. Hence the eivenvalues of are the poles of the closed loop transfer function. ABk− g) Show that this system is not controllable.4) For one of the rectilinear systems in lab, I found the following state variable representations: 01 0174.8205 2.6469 5050qq⎡⎤=+⎢⎥−−⎣⎦&u⎡⎤⎢⎥⎣⎦ []10 [0]yq=+u a) Show that the closed loop transfer function with state variable feedback is 2215050()(2.6469 5050 ) (174.8205 5050 )pfoGGssks=++ + +k To get the transfer function from the state space model in Matlab, use the ss2tf command. See closedloop_driver.m b) Design I • Determine feedback gains and so the resulting system has real poles and a bandwidth of 2 Hz. Set the second pole twice as far from the origin as the first pole. 1k2k• Use the final value Theorem to determine the prefilter gain so the steady state error for a step input is zero. • Estimate the settling time for the step response based on the system bandwidth. • Use Matlab to plot the Bode plot of the closed loop system (to verify the bandwidth) and plot the step response (to verify the estimated settling time). You should get numbers like 0.028, 0.0069, 0.06, and 0.32. When you look at the Bode plot, the bandwidth will not be 2 Hz, primarily because the second pole is too close to the first pole. d) Design II • Determine feedback gains and so the resulting (ideal second order) system has a percent overshoot of 15% and a settling time of 0.5 seconds. 1k2k• Use the final value Theorem to determine the prefilter gain so the steady state error for a step input is zero. • Use Matlab to plot the step response (to verify the percent overshoot and settling time). • Determine (analytically) the steady state output if the input to the closed loop system is cm and determine the time delay between the input and the output. r(t) = 1 cos(4t) You should get numbers like 0.0128, 0.0026, 0.047, and a time delay of 0.07 seconds. Your step response should be fairly close to the requirements.5) For one of the rotational systems in lab, I found the following state variable representations: B%01 0329.8427 1.2501 938.3581qq⎡⎤⎡=+⎢⎥⎢−−⎣⎦⎣&u⎤⎥⎦ []10 [0]yq=+u a) Show that the closed loop transfer function with state variable feedback is


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Rose-Hulman ECE 320 - ECE 320 Homework 8

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