Försättsblad till skriftlig tentamen vid Linköpings universitet

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1 Försättsblad till skriftlig tentamen vid Linköpings universitet Datum för tentamen Sal (1) TER3 (Om tentan går i flera salar ska du bifoga ett försättsblad till varje sal och ringa in vilken sal som avses) Tid 8:00 13:00 Kurskod TSRT15 Provkod TEN1 Kursnamn/benämning Reglerteknik M (Engelsk) Institution ISY Antal uppgifter som ingår 5 i tentamen Jour/kursansvarig Johan Löfberg (Ange vem som besöker salen) Telefon under skrivtiden , Besöker salen cirka kl. 9:30 och 12:00 Kursadministratör/ Ninna Stensgård, , kontaktperson ninna.stensgard@liu.se (Namn, telefonnummer, mejladress) Tillåtna hjälpmedel 1. Bok i reglerteknik (anteckningar tillåtna) 2. Tabeller, t.ex.: L. Råde & B. Westergren: Mathematics handbook C. Nordling & J. Österman: Physics handbook S. Söderkvist: Formler & tabeller 3. Miniräknare Övrigt Vilken typ av papper Rutigt ska användas, rutigt eller linjerat Antal exemplar i påsen

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3 SAL: TER3 TENTAMEN I TSRT15 REGLERTEKNIK M (ENGELSK) TID: kl. 8:00 13:00 KURS: TSRT15 Reglerteknik M (Engelsk) PROVKOD: TEN1 INSTITUTION: ISY ANTAL UPPGIFTER: 5 ANTAL BLAD: 4 ANSVARIG LÄRARE: Johan Löfberg, tel , BESÖKER SALEN: cirka kl. 9:30 och 12:00 KURSADMINISTRATÖR: Ninna Stensgård, , ninna.stensgard@liu.se TILLÅTNA HJÄLPMEDEL: 1. Bok i reglerteknik (anteckningar tillåtna) 2. Tabeller, t.ex.: L. Råde & B. Westergren: Mathematics handbook C. Nordling & J. Österman: Physics handbook S. Söderkvist: Formler & tabeller 3. Miniräknare LÖSNINGSFÖRSLAG: Finns på kursens websida efter skrivningens slut. VISNING av tentan äger rum kl i Ljungeln, B-huset, ingång 27, A-korridoren till höger. PRELIMINÄRA BETYGSGRÄNSER: betyg 3 23 poäng betyg 4 33 poäng betyg 5 43 poäng Solutions to all problems must be presented in such detail that all steps (except trivial calculations) can be followed. Missing motivations will reduce the points given. Lycka till! Good luck! Viel Glück! Bonne chance!

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5 1. (a) Give examples of situations where an open-loop controller would work. (2p) (b) Make a rough, well-motivated, sketch of the step-response of the system 2000 G(s) = (s + 1)(s + 10)(s + 100) (2p) (c) Three systems are driven by the input u(t) = sin(t). Match the systems below with the outputs A, B and C in Figure 1 G 1 (s) = G 2 (s) = G 3 (s) = 1 s s s + 1 s s s + 1 (3p) (d) Name three things which limit us from making controllers with arbitrarily good performance. (3p) 1

6 0.8 A 1 B 6 C y 1 0 y 2 0 y t t t Figur 1: Outputs from the three systems in 1c when driven with the input u(t) = sin(t). 2

7 2. (a) A closed-loop system is described by where G c = F yg i G F y G i G 1 G i = F ig F i G 2 Draw a block-diagram of this, using the blocks F i, G 1, G 2 and F y. (3p) (b) Draw a block-diagram representation of ẏ = y + x, ẋ = y + u (2p) (c) Write the equations in 2b in standard state-space form using matrices (A, B, C, D). The input is u and the output is y. (2p) (d) A system G(s) is controlled in standard error feedback format with a controller F(s). It is known that there is massive measurement noise with most of its energy in the frequency 10 rad/s. Two controllers are available, F 1 (s) and F 2 (s) with F 1 (10i)G(10i) = 12 18i F 2 (10i)G(10i) = i Which controller would you use if you want to have minimal impact from the measurement noise on the output.? (3p) 3

8 3. The uptake of a substance in the body (e.g., a medicine) can be described by the following model dq(t) = k 1 q(t) + u(t) dt dy(t) = k 1 q(t) k 2 y(t) dt where the input u(t) is the supply of the substance, the output y(t) is the amount of the substance in the blood, and q(t) is the amount of the substance in the intestines. The constants k 1 and k 2 describe metabolism and satisfy k 1 > k 2 > 0. Here, we assume k 1 = 0.05 and k 2 = (a) Is the system controllable? (2p) (b) Derive a state-feedback controller with closed-loop poles in 0.1, such that a constant reference signal can be tracked without any static error (5p) (c) q(t) can not be measured. Extend your controller to account for this. (3p) 4

9 4. A mechanical system consisting of a load and a flexible shaft can be described by G(s) = k s(τs + 1) ω0 2 (s 2 + 2ζω 0 s + ω0 2) where k = 0.3, τ = 0.2, ω 0 = 1 och ζ = 0.4. The Bode plot of the system is given in Figure 2. (a) Assume proportional control is used U(s) = K(R(s) Y (s)) How large can the gain K be if we require a phase margin of at least 30. What will the cross-over frequency be? (3p) (b) When using the proportional controller U(s) = K(R(s) Y (s)) with K = 1, a step-response with acceptable damping is obtained, but it is too slow. Compute a controller of the form such that U(s) = F(s)(R(s) Y (s)) The system is twice as fast as the case when a proportional controller with K = 1 was used. The step-response should have no more overshoot than what is obtained with the proportional controller using K = 1. The closed-loop system should satisfy e 1 < 0.01, i.e., when the reference is a ramp (r(t) = At) the error should in stationarity be less than 1%. (7p) 5

10 5. Let us control an unstable exoterm chemical process. The input is the supplied power to heat or cool the reaction. The transfer function from input u to the output temperature y is given by G(s) = 2 s 1 (a) The system is controlled using a P-controller. Which requirement must the gain in the P-controller satisfy to obtain stability from reference signal r to output y? (1p) (b) The chemical process is not heated/cooled directly, but from heating/cooling up the tank, which has transfer function G v (s) = 1 2s + 4 hence, the full dynamical model is given by G(s)G v (s) = 2 (s 1)(2s + 4) A P-controller is still used. Draw a root-loci w.r.t to the gain in the controller. Mark, compute and comment points of interest. (4p) (c) The measurement sensors have dynamics given by G m (s) = 10 s + 10 Hence, the complete model is actually G(s)G v (s)g m (s) = 20 (s 1)(2s + 4)(s + 10) Once again, draw a root-loci w.r.t gain in a P-controller, and mark, compute and comment points of interest. (5p) 6

11 Figur 2: Bode plot for exercise 4. 7

12 Robustness criteria Given a feedback controller F(s) stabilizing a model G(s). Let the real system be given by G 0 (s) = G(s)(1 + (s)), Now assume that G 0 (s) and G(s) have the same number of poles in the right half plane (including the origin), and F(s)G(s) and F(s)G 0 (s) both tend to 0 when s goes to infinity. Let T (s) denote the complimentary sensitivity function arising when G(s) is controlled using F(s). If (iω) 1 T (iω) ω the closed loop system is stable when G 0 (s) is controlled using F(s) Frequency compensation F lead (s) = K τ Ds + 1 βτ D s + 1, F lag = τ Is + 1 τ I s + γ 1 10 τ D =, τ I = ω c,desired β ω c,desired β Figur 3: Phase advancement as a function of β. Frequency response y(t) = G(iω) sin(ωt + arg(g(iω)) State-feedback u(t) = Lx(t) + l 0 r(t) Observer ˆx(t) = Aˆx(t) + Bu(t) + K(y(t) Cˆx(t)) 8