Dr. J. Tani, Prof. Dr. E. Frazzoli 5-059-00 Control Systems I (Autumn 208) Exercise Set 0 Topic: Specifications for Feedback Systems Discussion: 30.. 208 Learning objectives: The student can grizzi@ethz.ch, 30th November 208 Classify a transfer function to see which order or ramp it can follow and with which expected error. Understand time domain specifications: steady-state-value, overshoot, rise time and settling time. Calculate these values for a second order system. Apply the dominant pole approximation to simplify a system as second-order and understand the utility of being able to use the dominant pole approximation. Remember conditions on S(s), T (s) to ensure noise rejection and allow for accurate reference tracking. Apply frequency domain specifications to assess the robustness of a control system. Background So far we have looked at analysis issues (how to determine closed loop stability). We will now concentrate on control synthesis, i.e. how to design a feedback control system with a desired behavior. The closed loop behavior can be given using Time Domain Specifications Frequency Domain specifications Steady State Error We derive the conditions that a unit ramp of order m can be tracked, i.e., a reference of the form r(t) := m! tm. Assuming that S(s) is stable we can use the final value theorem. Furthermore we write L(s) = s ˆL(s) with ˆL(s) having no integrators in the origin and γ being the Type of γ L(s). We can summarize the steady state error for all possible cases in table. γ = 0 + ˆL(0) m = 0 m = m = 2 m = 3 γ = 0 ˆL(0) γ = 2 0 0 ˆL(0) γ = 3 0 0 0 ˆL(0) Table : Steady-state tracking error depending on the Type γ and the order m of the m-th order unit ramp given as reference.
Time Domain Specifications Many closed loop systems have a dominant pole pair such that their step responses are approximated well by a second-order system with the same pole configuration: T (s) = ω n s 2 + 2ζω n s + ω 2 n y(t) = e σt cos(ω n ζ 2 ) While in class you have seen how to compute many important features of the closed loop time response looking at the property of the closed loop system, it is possible to directly consider the open loop transfer function L(s). (derivation in Analysis and Synthesis of Single-Input Single Output Control Systems, L. Guzzella Chapter 0.3. We can in fact relate the rise time or T 90 (time to reach the 90 % of the steady state output) and the percentage of overshoot M p (with respect to the steady state value) using the following approximations T 90 =.7 ω c M p = 7 ϕ 7 where ϕ and ω c are respectively the open loop phase margin and the crossover frequency. Frequency Domain specifications Frequency domain specifications relate requirements such as disturbance/noise rejection, reference tracking and stability to the frequency response of the open loop transfer function L(s). Noise and Disturbance Rejection Recall that S(s) is the transfer function from disturbance/reference to error T (s) is the transfer function from noise to output. Reference tracking/disturbance rejection and noise rejection can be met simultaneously if S(s) and T (s) are both small but S(s) + T (s) = + L(s) + L(s) + L(s) = Fortunately it generally happens that disturbances and reference have a low frequency content while noise is most concentrated at high frequencies. A practical requirements is to make S(jω) at low frequencies and T (jω) at high frequencies. Furthermore, the crossover frequency is close to the bandwidth ω b of T (s). Although this is not a fixed dependency, in most designs ω b will be very close to ω c. So the previous requirements can be summarized by the following rules of thumb: ω b ω c < 0. ω noise ω b ω c > 0 ω d where the noise frequency spectrum N(jω) > 0 db only for ω > ω noise. the disturbance frequency spectrum D(jω) > 0 db only for ω < ω d.
Unstable Poles Consider the plant P (s) = a s, a > 0 with one unstable pole π+ = a. The closed loop L(s) = k p P (s) can be stabilized only if the crossover frequency is larger than π + since for stability, we need to encircle the point. Using Nyquist, this is possible with a gain k p smaller than a. Informally, we can say that the controller must be faster than the fastest unstable pole, leading to the rule of thumb ω c > 2 π + being π + the fastest unstable pole. Non Minimum Phase Zeros Using Root Locus we can see that a simple proportional gain will move the closed loop poles to the open loop zeros. In case of non minimum phase zeros this is equivalent to eventually reach closed loop instability for k = k critic. This shows that the loop gain and hence the bandwidth is limited and particularly upper bounded by the non minimum phase zeros. This consideration applies to general plants but for a detailed derivation, the interested reader is reminded to Analysis and Synthesis of Single-Input Single Output Control Systems, L. Guzzella (Chapter 9.5). As a rule of thumb we choose ω c < 0.5 ζ + with ζ + being the slowest non minimum phase zero. Plant Delay Delay can be approximated as a first order system using Taylor approximation e st st Thus its effect can be approximated as a non minimum phase zero at the frequency ω delay = /T. We apply the same rule of thumb and we get ω c < 0.5 ω delay = (4) 2T Plant Uncertainty Given that the nominal closed loop plant is stable we would like to account for modeling errors as well. This means that the actual loop shape is in a range of possible plants: L real (s) = L(s) + L(s)W 2 (s) with < (5) being W 2 (s) a multiplicative uncertainty weight which quantifies the uncertainty about the plant at each frequency. E.g. if W 2 (s) = 0 at the specific complex frequency s, we restore L real (s) = L(s). Then we would like that a stable design based on L(s) is still stable for all possible plants. Alternatively, using Nyquist, we might ask that the number of encirclements of the point does not change for all L real (s). Since L(s)W 2 (s) draws a circle of possible plants around L(s), this is equivalent to ask that the distance of L(s) from the point is larger than this radius or L(jω) ( ) = L(jω) + > L(jω)W 2 (jω) T (jω) < W 2 (jω) (6) This requirement can be fulfilled if ω b ω c < ω 2 being ω 2 the first frequency where W 2 (jω) > 0 db (percentage uncertainty is larger than ). The applied rule of thumb is to choose ω c < 0.5 ω 2 () (2) (3) (7) (8)
Exercise (Noise Amplification and Bandwidth) Consider the following first order system G(s) = σ s + (9) a) Design a controller such that the closed loop step response is as in figure Desired step response.0 0.8 y(t) 0.6 0.4 0.2 0.0 0.0 0. 0.2 0.3 0.4 0.5 0.6 t[s] Figure : Desired step response for the closed loop system. constant is approximately equal to 0.s. As shown in the plot, the time b) What would you do to get a faster response? c) Derive the transfer function from the input to the noise. d) According to the two previous answers, which limitation does the noise impose in terms of a faster system response? Assume that the noise has the following frequency content: N(jω) = 0.00jω + (0) Can you think to a practical solution and as a consequence, what would be a requirement for the feasibility of a good control system? Exercise 2 (Constraints on controller design) Please use the following constraints on the crossover frequency ω c (outlined at the end of chapter 0 of the accompanying book) to answer the following questions max{0 ω d, 2 ω π } < ω c < min{0.5 ω 2, 0.5 ω delay, 0.5 ω ζ, 0. ω n },
where ω d refers to the relevant disturbance frequency, ω π refers to the maximal frequency of unstable poles, ω 2 is the earliest point at which the uncertainty reaches 00%, i.e. W 2 (s) =, ω delay = T delay is the frequency of the delay, ω ζ is the minimal frequency of all non-minimum phase zeros and ω n is the frequency of the noise. All these frequencies impose design constraints on a potential controller. The task in this exercise is then to determine whether such a system can or cannot fulfill the design constraints summarized above. You are given a system plant defined by the following equations: ẋ (t) = x 2 (t) ẋ 2 (t) = x (t).5 x 2 (t) + u(t 0.0) y(t) = 4 x (t) + x 2 (t). Disturbances occur mainly in the frequency range 0 rad/s to 0. rad/s. The present noise has high frequency of ω n = 6000 rad/s. Furthermore, the transfer function of uncertainty W 2 (s) is: W 2 (s) = (s + ). 0 The goal now is to design a controller (not part of this exercise) which always reduces all disturbances d(t) by at least a factor of 0. a) What conditions must the crossover frequency satisfy in order to achieve disturbance and noise rejection? b) Derive the plant transfer function P (s). c) Derive the frequency constraints eventually introduced by non-minimum phase zeros and unstable poles. d) Which are the constraints on the bandwidth imposed by the delay and model uncertainty? e) Find the range of admissible frequencies according to the previously identified constraints. Exercise 3 (Homework: Implementation Considerations) You are in charge of designing a controller for a small electronic motor whose plant P (s) has the usual second order transfer function: P (s) = ω 2 n s 2 + 2ζω n s + ω 2 n () with the parameters being: ω n = 2 rad/s ζ = 0.9 After presenting your system model and controller to the board, some concerns are raised. Address them individually using theoretical and considerations and graphical illustrations and if necessary propose a way to solve the issue. a) An engineer form the electrical engineering department is worried that you did not consider the dynamics of the sensor in your system model. The sensor has the transfer function P Sensor = 0 s+0.
b) Another engineer from the marketing department wants to sell the component as a replacement for the well-known Maneki-neko ( ) (Figure 2), so he wants it to wave with a given frequency ω wave (0, 0] depending on the real-time turnover of the shop. He would like to know if the system can fulfill that requirement and if this is not the case which controller would make this possible. Figure 2: Manek-Neki lucky, waiving cat product.
Exercise 4 (Homework: Time domain specifications) The bode plot of the open loop gain L(jω) of a control system is depicted in figure 3. Magnitude (db) 50 40 30 20 0 0 0 20 30 40 50 60 70 80 90 00 0 20 30 40 50 60 70 L(s) 4 L(s) 0 0 0 0 0 2 0 3 Phase (deg) 90 95 00 05 0 5 20 25 30 35 40 45 50 55 60 65 70 75 80 L(s) 4 L(s) 0 0 0 0 0 2 0 3 Frequency (rad/sec) Figure 3: Bode plot - The relationship of the crossover frequency ω c, phase margin ϕ to time domain specifications. In the plot we show the magnitude and the frequency of both the original plant (blue) and the plant obtained halving the controller gain (orange). Please assume the closed loop system is approximately similar to a second-order system. Please also note that this approximation is only valid for simple loop gains L(s) (see chapter 0 of [?]). If non-minimum phase zeros or unstable poles exist in P (s), the second-order system approximation may not yield satisfactory results. a) How large is the static error e ss of the system depicted in figure 3? Please assume a unit step as reference input (r(t) = h(t)). b) Please use the following formulas to answer the next questions. The relationship of the overshoot ˆɛ and phase margin can be approximated as follows M p = 7 ϕ 7 Similarly, the relationship of rise time T 90 and crossover frequency ω c can be approximated as T 90 =.7 ω c
Determine the maximum overshoot M p and the rise time T 90 of the closed loop control system. Make use of the approximation formulas for M p and T 90. c) The control gain k p of the controller is reduced by a factor of 4. Redetermine the values M p and T 90. How do the values of M p and T 90 change compared to the previous control gain? d) Sketch the step response of the control system for both cases. e) Approximately which control gain k p is required to achieve a phase margin of 60? Which control gain k p is needed to obtain a crossover frequency ω c =? Exercise 5 (Homework: Static error e ss ) The transfer functions of plant P (s) and of two controllers C (s) and C 2 (s) are provided. P (s) = s + ( C (s) = 2 C 2 (s) = 2 + ) s a) Calculate the open loop transfer functions (L (s), L 2 (s)), the resulting sensitivities (S (s), S 2 (s)) and the complementary sensitivities (T (s), T 2 (s)) using controllers C (s) and C 2 (s), respectively. b) During the calculation to obtain L 2 (s) a zero-pole cancellation occurs. Does this pose a problem? Explain. c) Calculate the system response to a step input of height. Only calculate the static response (t ). How large is the static error e ss? d) Calculate the system response of both proposed controllers to a step disturbance d of height. Again only calculate the static state response (t ). How large is the static error e ss? e) Simulate the step responses from c) and d) using Python. For this purpose you can use the control library which can be installed using conda. Run the conda terminal as Administrator and type in the following commands: conda install -c python-control -c cyclus slycot control Then you can use the following functions: sys = control.tf(num, den): it returns a transfer function whit numerator coefficients num and denominator coefficients den in decreasing power order. This transfer functions can then be summed and multiplied together. y, t, x = control.matlab.lsim(sys, U, T): it returns the output y, for the internally computed time t and the state evolution x for the specified system sys simulated for the time vector T under the input U.