Automatic Control (MSc in Mechanical Engineering) Lecturer: Andrea Zanchettin Date: 29..23 Given and family names......................solutions...................... Student ID number.......................... Signature.......................... Duration: 2 minutes Number of exercises/questions: 5 Maximum score: 32 pts. All solutions must be written (either in English or Italian) within the available blank space (and not on additional paper sheets) It is forbidden to use any electronic device, apart from a non programmable and non graphic calculator Books and lecture material are not allowed It is not allowed to leave the class room within the first 3 minutes
This page has been intentionally left blank (to be used only if necessary). 2
Question (for all students) The unit step responses of the output of three closed-loop systems are shown in the following:.2.2.2.8.8.8 Output.6 Output.6 Output.6.4.4.4.2.2.2 5 5 2 Time (s) 5 5 2 Time (s) 5 5 2 Time (s) Knowing that the three loop transfer functions are the following ones: L (s) = s, L 2 (s) = s ( +.s), L 3 (s) = 3 + 3s, match each loop transfer function with the corresponding closed-loop response. Solution: The first two step responses have zero steady state error, whilst the third does not. The sole transfer function with type is the third one. Therefore, the last step response corresponds to L 3. The first two step responses differ from the settling time and from the presence/absence of overshoots. L surely produces a smooth response and hence corresponds to the second response. Finally, the first response corresponds to L 2. 3
Exercise 2 (for all students) Consider the closed-loop system below, where G (s) = s/3 + s/3. w + + - R(s) u G(s) d + y + + n Design the controller R (s) that satisfies the following specifications: a null steady-state error for a step reference, i.e. e = when w (t) = step (t); a phase margin of at least 6 deg, i.e. ψ m 6 deg; a crossover frequency between and rad/s, i.e. ω c ; attenuation of d (t) on the output of at least 3 db in the bandwidth ω 3 rad/s; attenuation of n (t) on the output of at least 4 db in the bandwidth ω rad/s. Finally, propose a suitable sampling time for the digital implementation of the control law. Solution: The static requirement on the steady-state error is satisfied with R (s) = µ R. We can then set s R (s) = /s, obtaining L (s) = s/3 s + s/3 which however does not satisfy all the other requirements. By selecting: which corresponds to L (s) = s R (s) = s one obtains the Bode diagrams reported in the next page. ( + s/) ( s/3) ( + s/3) ( + s/3) 2 ( + s/3) ( + s/) ( + s/3) ( + s/3) 2 4
Magnitude (db) 6 4 2-2 -4-6 -8 - -9 Bode Diagram -35 Phase (deg) -8-225 -27-35 -36 2 3 4 5 Frequency (rad/s) The resulting loop transfer function now satisfies all the requirements with ω c = 32 rad/s, ψ m = 59.9 deg. As for the sampling frequency for the digital implementation we can select ω N = ω c = 32 rad/s which corresponds to T s = π/ω N ms (which corresponds to a phase margin lag of 9 deg). 5
Exercise 3 (for all students) Consider the following transfer function G (s) = s (s + ) (s + 2) (s + 3) Using the root locus method discuss whether it is possible or not to design a proportional controller ρ R which stabilises the closed-loop system. If so, determine the admissible range of ρ such that the closed-loop system is asymptotically stable. Solution: The direct and inverse loci of ρg (s) are sketched in the following: 2 Root Locus 2 Root Locus 5.5 Imaginary Axis (seconds - ) 5-5 - Imaginary Axis (seconds - ).5 -.5 - -5 -.5-2 -4-3 -2-2 -2-2 -5 - -5 5 Real Axis (seconds - ) Real Axis (seconds - ) It is therefore possible to implement such a controller. The range of ρ such that the closed-loop system is asymptotically stable is of the kind ρ min < ρ < ρ max, where ρ max = 2 3 = 6 while for ρ min < we can evaluate the inverse locus in 6 6 + 6 + 2 6 + 3 ρ min = = 8.574 6 6
Exercise 4.A (for students of Automatic Control A, only) Consider a servomechanism with K = Nm/A (torque constant), J m =.2 kgm 2 (motor inertia), n = (reduction ratio), J l = 2 kgm 2 (load inertia). A current impulse exhibits the behaviour shown in the following figure. 5 Impulse Response 45 4 35 3 Amplitude 25 2 5 5.5..5.2.25.3.35.4 Time (seconds) Based on the acquired data, estimate the the elasticity of the transmission K el and propose a proper tuning for the velocity controller that maximises the damping of closed-loop poles. Solution: The oscillation has a period of 2 ms and hence corresponds to ω p = 34.6 rad/s. The inertia ratio is ρ = J lr /J m = and finally the locked frequency results ω z = ω p / + ρ = 222.4 rad/s. Finally the elasticity of the transmission can be computed as K el = ω 2 zj lr = 987 Nm/rad. As for the tuning of the PI velocity controller we can select T iv = /ω z = 4.5 ms and K pv =.7ω z / (J m + J lr ) =.622 A/rad. 7
Exercise 4.B (for students of Automatic Control B, only) The following controller R (s) = +.s + s has been design for a given LTI system in order to obtain a crossover frequency of ω c = rad/s. Design a suitable sampling time T s, compute the corresponding discretisation using the Tustin (bilinear) method, and finally write the corresponding pseudo-code (difference equation, u (k) =... ). Solution: In order to meet the Shannon s criterion we can set ω N = ω c and T s = π/ω N =.34 s. In order to obtain the discrete time realisation of the controller we need to substitute s with 2 T s z z +, obtaining R (z) = +. 2 z T s z + z + +.6369 (z ).6369z.363.222 +.4927z + 2 = = = z z + + 6.3694 (z ) 7.3694z 5.3694.7286z T s z + As for the implementation, we have u (k) (.7286z ) = e (k) (.222 +.4927z ) and therefore u (k) =.7286u (k ) +.222e (k) +.4927e (k ) 8
Exercise 5.A (for students of Automatic Control A, only) A closed-loop system is characterised by the following loop transfer function L (s) = s ( +.s) compute the crossover frequency and the corresponding phase margin. Then design a suitable fist-order analog anti-aliasing filter which contribute in a phase margin lag not higher than 5 deg and draw the corresponding circuit. Solution: The crossover frequency is approximately ω c = rad/s while the corresponding phase margin is of 84.3 deg. A first order anti-aliasing filter is a transfer function of the form: H (s) = + st Therefore, the parameter T > must be tuned in order to have a phase lag not higher than 5 deg at rad/s and therefore: H (j) = atant 5, T <.875 s We can select T =. s, while a possible analog implementation is the following one C V R R R v out which has the following transfer function H (s) = C = µf.. One could select, for example, R = kω and + src 9
Exercise 5.B (for students of Automatic Control B, only) Given the following continuous time signal y (t) = + sin (t) + sin (7t), answer to the following questions: find the maximum sampling time T s for which there is no aliasing effect; assuming T s =.5 s, find the frequency of the aliasing harmonics; Solution: The highest frequency harmonic happen at ω max = 7 rad/s, therefore in order to satisfy the Shannon s criterion, T s < π =.4488 s. ωmax Clearly T s =.5 s does not satisfy the Shannon s criterion. Therefore, an aliasing harmonic will be present on the acquired of signal. The corresponding sampling frequency is ω s = 2π/T s = 4π and therefore the aliasing harmonic is at 7 ω s /2 =.768 rad/s.