1 (30 pts) Dominant Pole
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1 EECS C8/ME C34 Fall Problem Set 9 Solutions (3 pts) Dominant Pole For the following transfer function: Y (s) U(s) = (s + )(s + ) a) Give state space description of the system in parallel form (ẋ = Ax + Bu, y = Cx). (s + )(s + ) = /9 s + /9 s + ẋ = y = /9 x + /9 x b) Draw the block diagram for the system in parallel form. u c) For a step input, determine which mode of the parallel system dominates. ( /9 Y (s) = s + /9 ) ( ) s + s y(t) = 9 ( e t )u(t) 9 ( e t )u(t) = ( 9 e t + 9 e t )u(t) Hence, we can see that the mode at s = is dominant. d) Consider approximating the system by keeping only the dominant mode. (That is, in the parallel form, keeping only one state variable.) Considering the dominant mode, determine the approximate dynamic system x = ã x + bu and ỹ = c x, where ã, b, c are scalar constants. Remove the pole at s = : Ỹ (s) U(s) = s + Note the DC gain has been adjusted so that the final value is the same. In state space form, x = x + u ỹ = x
2 e) In Matlab, compare the step response for the original system {A, B, C} and the approximate system {ã, b, c}. The original system does not rise as quickly as the approximation near t = this is from when the mode at s = has an effect. Since it is opposite sign as the dominant mode, it will cause the output to settle slower. > figure; hold on; > step(ss(- ; -,;,/9 -/9,)); > step(ss(-,,,)); > legend( Original System, Single-Pole approximation ); (35 pts) Controllability and Observability For the circuit below, input is voltage u(t). u(t) R C x y(t) u(t) R C x y(t) AGND a) Write state and output equations for the circuit. AGND Y (s) U(s) = C s R + C s = (R C ) s + (R C ) Y (s) U(s) = (R C ) s + (R C )
3 These are two systems running in parallel with no coupling besides the input. Writing the state space form: (R C ẋ = ) (R C ) x + (R C = ) y (R C ) x y b) Find conditions for R, R, C, C that make the system controllable. Using the controllability matrix: C = B AB (R C = ) (R C ) For the two columns to be linearly independent, (R C ) (R C ) R C R C Under these conditions, the system is controllable. c) For output y = x, is the system observable? u Using the observability matrix: O = = C CA (R C ) Rank is less than and O =. The system is not observable since only the first state and it s derivative can be observed and there is no coupling from the second state. d) For R C = sec and R C =. sec, find gains k k such that the closed-loop system using state feedback with u = k k (r x) has closed loop poles at, 5. (Suggestion, Matlab place). >> A = - ; -; >> B = ;; >> K = place(a,b,- -5) K = i.e. u = x e) Use Matlab to plot x(t) for initial condition x() =, and reference input r(t) = 5e 5t e t T for < t < 4 sec. (Suggestion Matlab lsimplot.) >> SYS = ss(a-b*k,b*k,eye(),); >> T = :.:4 ; >> R = 5*exp(-5*T) -exp(-*t); >> lsimplot(sys,r,t); 3
4 Note at time t = that the two systems have opposite signs. f) Explain how x and x could have opposite signs for some time t although both are driven by the same input u(t). A few arguments could be made:. Consider u(t) = step(t). At time t =, both capacitors will begin discharging from V to V, however Circuit will discharge faster that Circuit. Therefore, Circuit will cross the axis before Circuit and the two will have opposite signs for some period of time until Circuit also crosses the axis.. Because the system is controllable, we know that any state can be reached with an appropriate input, including those states where x and x have opposite sign. 3. The reference input in effect has two modal responses super-imposed, with opposite sign. The closedloop system has a transient response to the superposition of these modes, thus there can be an opposite sign on the outputs although both have the same inputs. 4. A very loose analogy would be dribbling a tennis ball and a ping-pong ball on a tennis raquet - both could be hit with the same velocity, but the tennis ball could go higher and take longer to recontact the raquet. In this analogy both balls get the same contact velocity input, but have different dynamic response. Here the different time constants of the RC networks provides a difference of output even with the same input. 4
5 3 (35 pts) State Feedback versus Observer Feedback 3. (35 pts) State Feedback versus Observer Feedback Given the following system (e.g. inverted pendulum with position sensor).3 ẋ = Ax + Bu = x + u(t), y = x, x() = 4 a) Design a state feedback controller u = k k x such that the closed loop system has ζ =.5 and ω n = 5. ζ =.5 means θ = 3. The poles should be at.5 ± 4.33j. We will find k, k so that the eigenvalues of A BK have these values. A BK = 4 k k s A BK si = 4 k k s A BK si = s(k + s) 4 + k k =.5 k ± 4(k 4) = ±4.33j k = 5 ± k = ±8.66j = s + k s + k 4 s = k ± k 4(k 4) 4 4k = 75 k = 9 K = 9 5 b) Use Matlab to plot the states x(t) for t > for the system with state feedback. Using lsimplot, and a C matrix of I, both x and x are plotted. Out refers to x and Out refers to x. 5
6 Linear Simulation Results.3.5. To: Out().5..5 Amplitude.5..5 To: Out() Time (seconds) c) Design a critically damped observer ẑ = Aẑ + Bu + L(y ŷ) with both observer poles at s = 6. With ŷ = Cẑ, we can rearrange the design: and so we need to design ẑ = (A LC)ẑ + Bu + Ly L = so that A LC has poles at 6, 6. l A LC = 4 l l s A LC si = 4 l s l l A LC si = s(l + s) 4 + l = s + sl + l 4 s = l ± l 4l + 6 l = 6 l 4l + 6 = l = l = l = 4 L = 4 d) Write the state space equations for the controller with u = Kẑ. (This should have 4 state variables.) From the separation principle, what are the eigenvalues of the combined system with observer and feedback control? 6
7 The system equations with combined state feedback and observer are given by: ẋ = Ax BK ẑ and ẑ = (A LC BK)ẑ + LCx where we note that we are looking for the zero input response (ZIR, r(t) = ). One standard approach is to augment the x system with e = ẑ x: ẋ A BK BK x = ė A LC e ẋ x ẋ ė = x e ė 36 e y = This form makes the separation principle easy to see, because the block in the lower left hand corner means that the eigenvalues of the total A matrix are just the eigenvalues of A BK and A LC; and we know that these eigenvalues are j, j, 6, 6 because that s how we designed them. The disadvantage of this formulation is that it s a little bit harder to get the ẑ signal (which we need for part (e) below). Another approach is to augment the x system with ẑ: ẋ A BK x = ẑ LC A LC BK ẑ ẋ x ẋ ẑ = x ẑ ẑ ẑ x x e e x y = x ẑ ẑ This formulation is just as good as the one above. We could calculate the eigenvalues with MATLAB to verify that they are j, j, 6, 6. However, we also know that by the separation principle, the eigenvalues of this matrix must be the eigenvalues of A BK and A LC. (Why is this true, even if the matrix doesn t have that nice form with the block of zeros in the lower left hand corner? The two formulations are related by a similarity transform; similarity transforms don t affect eigenvalues). e) Use Matlab to plot the states x(t) and ẑ(t) for t > for the closed loop system using u = k k ẑ. (Suggestion, use sys = ss(ao,bo,co,do) and lsimplot(sys), where AO, BO, CO, DO are the matrices for the system with observer). lsimplot again. Out and Out are x and x. Out 3 and Out 4 are ẑ and ẑ. 7
8 Linear Simulation Results.5 To: Out().5 To: Out() Amplitude.5 To: Out(3).5.5 To: Out(4) Time (seconds) f) Compare the responses in part b) and e). What differences are there? The effect of the observer s estimate locking on to the true state is most easily seen in the very first tenth of a second of x. In the case with no observer, the x signal just oscillates starting from. But in the case with the observer, the x signal takes a little time before it starts oscillating; this is the e 6t transient of the observer going away. Otherwise, the x signals are very similar. The observer estimate clearly differs from the true state x at the beginning, but matches it very closely after about.5 seconds. 8
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