Reglerteknik Allmän Kurs. Del 2. Lösningar till Exempelsamling. Läsår 2015/16

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1 Reglerteknik Allmän Kurs Del Lösningar till Exempelsamling Läsår 5/6 Avdelningen för Reglerteknik, KTH, SE 44 Stockholm, SWEDEN

2 AUTOMATIC CONTROL COMMUNICATION SYSTEMS LINKÖPINGS UNIVERSITET Reglerteknik AK Hints Answers Solutions Dictionary

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4 Hints This version: August 3

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6 Dynamic Systems. Start with J θ = f θ + M and try to write M as a function of θ and u using Kirchhoff s voltage law.. What is the relationship between the response of the system and the pole locations?.3 Separate the pure delay and the dynamic response. Use the final value theorem to find the steady state gain and calculate the time constant by estimating the time to reach 63% of the final value (neglecting the time delay)..4 Identify the coefficients ω and ζ in the system description.9 See Glad&Ljung. ω G(s) = s + ζω s + ω. a) Consider what you can control, what is uncontrollable and what is desired. b) Consider the relationship between the signals.. a) Use mass balance and assume that the densities are equal. b) Consider the change in mass and change in component A. c) Assume that all the other independent variables (q, q, c A, ) are constant..3 a) Use mass and component balance.

7 3 Feedback Systems 3. a) Consider the three blocks; tank, valve, and PID. What is the input and the output from each block? Connect the blocks and consider v as a disturbance. For the tank model use the fact that the flow into the tank is x v and the amount of liquid changes as ḣ A. b) Consider the final value and the time constant. d) Put F (s) = K and express the closed loop poles as a function of K. e) Use the final value theorem. f) Put F (s) = K Ps+K I s in the expression for the error, and use the final value theorem. 3. a) Use the expression for the poles from Problem??. b) Put F (s) = K P + K D s in the expression for the closed loop system from Problem??. The relative damping is defined in Glad&Ljung. 3.3 Use Newton s force equation F = ma to derive the transfer function for the astronaut. 3.4 Start with deriving an expression for the transfer function from the disturbance f c to the error e. a) Use F (s) = K and the final value theorem. b) Use F (s) = K + K /s and the final value theorem. 3.6 a) The characteristic equation is s(s + )(s + 3) + K(s + ) = which gives P (s) = s(s + )(s + ) and Q(s) = s +. b) Characteristic equation: P (s) = s(s + s + ), Q(s) =. s(s + s + ) + K = c) Characteristic equation: s(s )(s + 6) + K(s + ) = P (s) = s(s )(s + 6), Q(s) = s Derive the general closed loop transfer function by first deriving the transfer function for the inner loop. a) Let α =. The characteristic equation is then s(s + ) + 4K = Compute the poles explicitly as a function of K. b) The characteristic equation is c) Characteristic equation: d) Characteristic equation: s(s + ) + 4K( + s) = s(s + ) + 4K( + s/3) = s + s αs = 3.8 a) Derive the transfer function from ω ref to ω. b) The characteristic equation is (s + )(s + 4)(s 3) + K(s + ) = 3.9 a) Derive the closed loop transfer function by first deriving the transfer function for the inner loop. The characteristic function is s((s + )(s + ) + K ) + K = We get two principally different root loci when there are complex starting points, and when all starting points are equal. Treat the cases separately.

8 3. a) The characteristic equation is b) Characteristic equation: 3. a) Characteristic equation: (s + )(s )(s + 5) + K = (s + )(s )(s + 5) + K( +.5s) = s 3 + s + a(s + s + 6) = b) First check for which a the system is stable and the steady state requirement is fulfilled. Then use that sinusoid in gives sinusoid out after transients. 3. Check the root locus to find which K-values gives a stable/unstable system, more/less oscillative system. 3.3 Investigate the starting points and end points of the root locus. 3.4 Find the open loop transfer function and use the Nyquist criterion. 3.5 Since G(s) has no poles in the RHP, the closed loop system is stable if the Nyquist path of KG o does not encircle. Note that the K will only modify the distance to the origin, not the shape of the curve. 3.6 Study the amplitude and phase of G(iω). 3.7 a) Draw the complete Nyquist path and use the Nyquist criterion. (Note that G o ( iω) is the mirror image of G o (iω), mirrored in the real axis.) b) Use the final value theorem, and that G o () is known from the Nyquist path. c) Apply the Nyquist criterion to K s G(s) 3.9 Try to find the ω that gives arg F (iω)g(iω) = The Nyquist curve for small ω determines if the system may have an integrator or not. Also check if the system is unstable for some K (from the Nyquist diagram). 3.5 Check the steady state error, the relative damping, etc. 3.6 a) To compute the closed loop transfer function combine and θ(s) = G(s)U(s) U(s) = F (s)(θ ref (s) θ(s)) b) The control error can be computed using E(s) = + F (s)g(s) θ ref(s) To find the steady-state error, use the final value theorem. c) See b). 3.3 a) Consider what you can control, what is uncontrollable and what is desired. b) Use mass and energy balance for both the tank and the heating system. 3.3 a) Compute the poles of the system. b) Note that the system has negative sign in the numerator Check the pole for the closed loop system. 3.8 The system oscillates when the open-loop gain is equal to (check Ke iωt G(iω)). 3

9 4 Frequency Description 4. Determine the angular frequency ω of the signals using the figure. Use the relationship saying that when u(t) = A sin ωt the output becomes to determine G(iω) and arg G(iω). y(t) = G(iω) A sin(ωt + arg G(iω)) 4. a) For K =.5 the open loop system is given by G o (s) = F (s)g r (s)g s (s) =.5( + s/.) s( + s/.)( + s/.5)( + s/.) 4.7 Recall that for stable, linear systems a sinusoid in gives a sinusoid out after initial transients. 4. a) Use the rules in Glad&Ljung to make the Bode plot. b) What can be said of the phase and gain margin when the output of the closed loop system oscillates with constant amplitude? 4. a) What is the stability criterion in the Bode plot? b) What is the current phase margin? Is a lead really necessary? Use the rules in Glad&Ljung to make the Bode plot. b) What can be said about the phase and gain margin when the output of the closed loop system oscillates with constant amplitude? c) When the reference signal is A sin αt the output signal becomes y(t) = G c (iω) A sin(αt + arg G c (iω)) The Bode plot of the open loop system can be used to compute G c (iω). 4.3 a) Check the behavior of G(iω) when ω and ω respectively. See also if the absolute value and the argument decrease monotoneoulsy or not. b) Translate the behavior of the amplitude and phase curves to a pole-zero diagram. 4.4 Check the final values of y(t) against the static gain G(). Check also the overshoots of y(t) against the height of the resonance peaks in G(iω). Check the frequency of the oscillation in y(t) against the resonance frequency in G(iω). 4.5 Use Matlab, in particular the command bode. 4.6 Recall that for stable, linear systems a sinusoid in gives a sinusoid out after initial transients. 4

10 5 Compensation 5. Try a lead-lag compensator. A table of phase advance versus the N- parameter is found in Glad&Ljung. 5. a) Glad&Ljung gives a good description of how Bode plots can be drawn by hand. b) A proportional controller does not affect the phase curve. c) Try lead compensator. 5.3 a) Draw asymptotic Bode plot (see 5.) by hand or use Matlab. b) Start with calculating the controller and then use the final value theorem. c) Try a lag compensator. c) Use the final value theorem. d) A time delay is described by the transfer function e st. 5. Think of all possible phase curves, for example originating from time delays, and think about the corresponding Nyquist curves or Bode plots. 5.3 See Introduktion till CSTB and previous exercises in this section. 5.4 Try a lead-lag compensator. 5.4 Start with drawing a Bode plot for the open loop transfer function. The final value theorem is a good tool in this exercise. 5.5 Check for signs of dominating poles, pure integrations, resonance frequencies Draw asymptotic Bode plot using the guidelines in Glad&Ljung. See the discussion on lead-lag compensators in Glad&Ljung. 5.7 Use values of G(iω) and arg G(iω) to plot the Nyquist curve G(s). 5.8 The time delay alters the phase curve but not the amplitude curve. Use the Nyquist stability criterion. 5.9 Check steady state level and rise time. Modify Figure?? using G A (s) and adapt a lead-lag compensator. You can use two lead compensators to acchieve a big phase advance. 5. Start by adjusting Figure?? to obtain the Bode plot of G. 5. b) It is possible to derive limits on K using either the Bode plot or the Nyquist curve. 5

11 6 Sensitivity and Robustness 6. The sensitivity function is the transfer function from v to y. 6. Derive the relative model error G (s) = G (s) G(s) G(s) Make a simple plot of G c (iω) using the information in the problem formulation. Compare with the inverse of the relative model error. 6.3 Convert the condition that the amplitude of y is larger than the amplitude of v to the condition + G o (iω) < What does this inequality say about the distance between the Nyquist curve and the origin? 6.4 Compute the transfer function of the closed loop system. Apply the robustness criterion using the given upper bound of the relative model error. 6.5 a) Derive the relative model order and plot / G (iω). G (s) = G (s) G(s) G(s) b) Determine the level that G c (iω) cannot exceed. 6.6 a) Use the robustness criterion and check the condition for G(iω) when ω. b) The characteristic equation of the closed loop system is s ( + 5α) + 5s( + 5α) + 5 = 6.7 a) The characteristic equation becomes b) Derive the relative model error G (s) = G (s) G(s) G(s) Check where the absolute value of the inverse of the relative model error intersects G c (iω) given in the figure. It is sufficient to check the low frequency asymptote. c) What can be said about the necessity and sufficiency of the stability conditions in a) and b)? 6.8 Check where the absolute value of the relative model error intersects G c (iω) given in the figure. 6.9 Derive the closed loop equation relating y(t), r(t), v(t), and n(t) using Y (s) = V (s) + G o (s)(r(s) N(s) Y (s)). Then use the fact that the sensitivity function S(s) and the complementary sensitivity function T (s) are related as S(s) + T (s) =. (Here T (s) coincides with the closed loop system.) 6. a) The relative model error is given by G (s) = G (s) G(s) G(s) b) Use Matlab and results from previous exercises. 6. Recall that for stable, linear systems a sinusoid in gives a sinusoid out after initial transients. 6. Create the loop gain transfer function and use its Bode plot to check stability. 6.3 The sensitivity function is the transfer function from v to y. s (s + ) + α(s + s + 4) = 6

12 7 Special Controller Structures 7. a) Derive the transfer function from w to θ m which then implies the open loop transfer function Θ(s) =.9 ( + s/.33)( + s/.33)( + s) W (s) Draw the Bode plot using the rules from Glad&Ljung. b) Draw the Bode plot using the rules from Glad&Ljung. 7. a) Use the relationship H(s) = As ( ) U(s) V (s) + s/ b) Derive the transfer function from V to H when both feedforward and feedback are used. 7.3 a) Use Y (s) = (G u (s)f f (s) + G v (s)) V (s). b) Recall that for stable, linear systems a sinusoid in gives a sinusoid out after initial transients. 7

13 8 State Space Description 8. Define x = θ, x = θ, and utilize the differential equation for the motor. 8. For the nonlinear equation ẋ = f (x, x, u), the linearized equation is given by ẋ = f (x,, x,, u ) + f x (x,, x,, u ) (x x, ) + f x (x,, x,, u ) (x x, ) + f u (x,, x,, u ) (u u ) 8. b) Compare what happens to the states as t, to the transfer function poles. c) Check if det S and det O are nonzero. 8. The system is minimal, compute the transfer function. 8.3 a) For small deviations around, sin(φ) φ, cos(φ). Take z as input. b) det S = α ( α ) 8.5 a) Combine mass balance with the given equation for reaction speed. 8.3 Define x = y, x = θ, and x 3 = z. Use block diagram algebra to find expressions for s X i (s), then use the inverse Laplace transform. 8.4 Use canonical forms. 8.5 Take the Laplace transform of g(t). 8.6 G(s) = C(sI A) B. 8.7 x(t) = e A(t t) x(t ) + t t e A(t τ) Bu(τ) dτ 8.8 a) Insert the control signals and take Laplace transforms. Use the final value theorem. b) Examine the difference u (t) u (t) for arbitrarily small constant ɛ. 8.9 Check controllability. 8. The controllable subspace is spanned by the linearly independent columns of S. The unobservable subspace is spanned by the null space of O. 8

14 9 State Feedback 9. a) The closed loop system ẋ = Ax + Bu, y = Cx, u = Lx + y ref has characteristic polynomial det(si A + BL) =. b) The observer poles are given by det(si A + KC) = and should be placed to the left of the closed loop poles. 9. a) Write the system in state space form by introducing three state variables corresponding to the outputs of the three left-most integrators in the figure (ż = output). Design a state feedback controller u = Lx + y ref and place the poles in.5. c) Design an observer with poles to the left of the closed loop poles. 9.3 The closed loop poles are given by det(si A + BL) = 9.7 a) The closed loop system ẋ = Ax + Bu, u = Lx has the characteristic polynomial det(si A + BL) =. b) Check oberservability. c) Introduce a new state. 9.3 a) The constant l can be found by using that θ = ω = at steady state. b) Introduce the integrated control error as an auxillary state. 9.4 Decompose the system into two subsystems, one controlled by u and one by u, and check the controllability. 9.5 Is the system observable? 9.6 Is the system observable? 9.7 Is the system controllable? x(t) converges to zero as p(t)e αt if (A KC) has a double eigenvalue in α. Check the observability of the system. 9.9 Study the phase margin of the open loop system. 9. Use the initial value theorem. 9. Compute the transfer function from u(t) to z(t) = Lˆx(t), that is, the loop gain, and check the stability margin given a certain time delay T. 9

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16 Answers This version: August 3

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18 3 Mathematics. a) A step has Laplace transform A s. b) A ramp has Laplace transform A s. c) s+ d) s s +5 e) su(s) u() f) su(s). (u() = is a common assumption in the course.) g) s U(s) su() u() h) s U(s). (u() = u() = is a common assumption in the course.) i) A time delayed signal has Laplace transform, e st U(s).. a) lim t y(t) = 5/ b) Y (s) = s+ U(s).3 The general solution is given by y(t) = C e t + (C + C 3 t)e t 3 (cos(t) + 7 sin(t)).4 a) y(t) = e t + e t, t b) y(t) =.5e t +.5 sin t.5 cos t.7 Multiplication of the two matrices gives the unit matrix..8.9 λ = 3 v = λ = v = 3 λ 3 = 4 v 3 = T =. A basis for the null space is for example.5 a) e i π 4 b) 5 e i 8 π c) + 3i d) 5 A basis for the range space is.6 decibel (db ) Definition Amplification F log F = F = = 3 log F = 3 F = 3/.78 log F = F = = log F = F =.5 = 3.6 log F = F =.5 =.36 3 The rank of the matrix is hence 3.. a) f(t) = e t ;.

19 b) f(t) =.5e t +.5e t ;. c) f(t) = e t t;.. y (3) + ÿ + ẏ + y = u

20 outflow ph Dynamic Systems. a) Differential equation where b) Transfer function c) Step response. () K =. () K =.5 (3) K = 3 (4) K =.5.3 G(s) =.4 a) a < b) b = e (L/V )s +3s θ + τ θ = k u τ = R af + k a k v JR a k = k a JR a G(s) = θ(s) U(s) = k s(s + /τ) θ(t) = k τt k τ ( e t/τ ).7 α > gives overshoot, α < gives undershoot..8 y(t) = L (G(s) s ).9 a).5 b) M 6% c) T r.5 d) T s 7.8. G C, G 3 B, G 4 A, G 5 D.. a) The signals can be classified as Disturbances signal: Acid process flow (unknown ph and flow) Control signal: NaOH solution Measured and controlled signal: The ph of the outflow b) A block diagram where the control strategy is based on feedback could look like Figure.a Acid flow.5 A B, B F, C A, D C, E E, F D..6 System T r T s M poles G A %, G B %. ± i.98 G C % 4.8,. G D %.5 ± i.87 G E.8.6 6% ± i.73 ref NaOH + Σ F Tank Figure.a. a) q =.5 m 3 /min and c A =. kmol/m3. 3

21 b) The model is nonlinear since the model is described by the following nonlinear equations d(ρv ) = ρ(q in q out ) dt d(v c A ) = q c A, + q c A, qc A dt c) k =. kmol/m 3, k =.3 kmol/m 3, and τ =.5 =.67 min..3 a) The model is given by dm i dt αx i y i = + (α )x i = L i + V i+ L i V i M i dx i dt = L i (x i x i ) + V i+ (y i+ x i ) + V i (x i y i ) b) The linearized model is given by M i dx i dt = L i x i, + V i+y i+, L i x i V i y i + x i L i, + yi+v i+, x L i yi V i α y i = ( + (α )x x i ) i 4

22 3 Feedback Systems v h ref e u x h + Σ F (s) G v (s) + Σ G t (s) c) Using the (approximate) derivative of the error in the regulator increases the damping of the closed loop system. Increasing K D too much, however, gives that an oscillation with higher frequency appears in the step response and finally (approximately when K D > 65) the closed loop system becomes unstable. Figure 3.a 5 3. a) Transfer function of the tank G t (s) = s. Block diagram see Figure 3.a. b) k v =, T = 5 c) H(s) Gt(s)Gv(s)F (s) H ref (s) = +G t(s)g v(s)f (s), H(s) V (s) = G t(s) +G t(s)g v(s)f (s) d) K <.5 e) K f) 3. a). ±.39i b) K D > K < and K = /K. 3.4 a) a/k b) 3.5 a) For small values of K P the step response is slow, well damped and the steady state error is large. For increasing K P the step response becomes faster but more oscillatory, while the error is reduced. For large K P the amplitude of the oscillations increases, that is, the closed loop system becomes unstable. b) The integrator in the regulator eliminates the steady state error. A too small value of K I gives a large settling time while a too large value gives an oscillatory (finally unstable) closed loop system. Imag Axis Real Axis 3.6 Root loci are shown in Figure 3.6a. Imag Axis Imag Axis 8 Real Axis Figure 3.6a Real Axis b) Intersection with the imaginary axis for K = 4, ω = ±. c) Intersection with the imaginary axis for K = 7.5, ω = ±.5. Conclusions about the step response of the corresponding systems: 5

23 Real Axis.5 a) Asymptotically stable all K >. Small K: No oscillations, larger K gives faster system. Larger K: Oscillations. Larger K gives more oscillations. b) Asymptotically stable for < K < 4. Oscillating all K >. Small K: larger K gives faster system. Larger K: larger K gives more oscillating system. Unstable for large K (> 4). c) Asymptotically stable for K > 7.5. Unstable for K < 7.5. Stable and oscillating for K > 7.5. Larger K gives faster system, until the real pole becomes dominating, then larger K gives a slower system..8 The root loci are shown in Figure 3.7a. a) Asymptotically stable for all K >, oscillatory for large K. b) Asymptotically stable for all K >, not oscillatory for any K. c) Asymptotically stable for all K >, no oscillations for small and large K, faster for large K. d) Asymptotically stable for all α >. Oscillatory for small α. Larger α gives more damped system. With the tachometer feedback we can make the system both fast and well damped. The tachometer feedback is equivalent to the D-part in a PID controller. Imag Axis.5.5 Imag Axis a) The system is unstable so ω will grow to infinity. b) The root locus is shown in Figure 3.8a. The system is asymptotically stable for K > Real Axis Real Axis Imag Axis Imag Axis Imag Axis Figure 3.8a Real Axis Real Axis c) No. When K =, s = is one pole but the other two are complex. 3.7 General characteristic equation: Figure 3.7a s(s + ) + 4K( + αs) = 3.9 a) Starting points: s = 5.5 ± 5.5 K. The starting points are all real for K.5, while we have complex starting points for K >.5. The two principal root loci are shown in Figure 3.9a. The system is asymptotically stable for < K < K +. 6

24 Root locus Root locus Imag Axis Imag Axis Imag Axis Imag Axis 4 6 Figure 3.a Real Axis Real Axis 5 5 Figure 3.a Real Axis Real Axis.5.5 Figure 3.9a Imag Axis.5.5 b) A larger K gives stability for larger K Root loci in Figure 3.a..5 Real Axis a) The system is unstable for all K. b) Asymptotically stable for K > a) The root locus is shown in Figure 3.a. The system is asymptotically stable for a >. b) The smallest amplitude is.. 3. K Step 4 C D 8 B 5 A 3.3 The poles of the system all tends to points in the LHP or to for large K. 3.4 The system is stable for K/A < π. 3.5 a) The closed loop system is stable in (i), (ii), and (iv). b) Stable when: (i) K <.5, (ii) K >, (iii) K < /, and (iv) K < /4 or K > /. 3.6 The Nyquist curves are shown in Figure 3.6a. 3.7 a) K < /3 b) +K c) K < /3 when K < /3 7

25 a) Im b) Im.5.5 ω.5 ω Re Re Imag Axis τ =.69 T = π arctan τ =.53 T = π arctan τ = K < 3. Root locus. 3. P b = b = I b = b = D b = b = Figure 3.6a 3. a) The root locus with respect to K P is shown in Figure??. When K P increases the two complex poles move towards the imaginary axis, that is, the closed loop system becomes more oscillatory. Finally, for K P 6., the poles cross the imaginary axis and the closed loop system becomes unstable. This result is in accordance with Problem??. For small values of K P the properties of the step response are mainly determined by the real pole close to the origin. For larger values the complex poles start to dominate and when the complex poles cross the imaginary axis the amplitude of the oscillations in the step response increases and the system becomes unstable. Note, however, that the root locus alone does not give sufficient information to tell how the steady state error changes with the parameter..5 Real Axis Figure 3.a b) The root locus with respect to K I is shown in Figure??. For small K I the response of the closed loop system is dominated by the poles on the real axis close to the origin. When K I increases the poles become complex and move towards the imaginary axis, that is, the closed loop system becomes more oscillatory. Finally, for K I.5, the poles cross the imaginary axis, that is, the closed loop system becomes unstable. As can be seen in Problem?? a small value of K I, that is, a pole close to the origin, gives a slow step response. When K I increases the dominating poles become complex and the step response becomes oscillatory. A large settling time will typically follow if the system is slow or has poor damping. Here, the large settling time for small K I is due to the system being slow. That the steady state error is eliminated cannot easily be seen in the root locus. c) The root locus with respect to K D is shown in Figure??. When K D increases the complex poles closest to the origin move towards the origin and and at the same time the damping of the poles is increased. When K D increases even more the second pair of complex poles moves towards the imaginary axis giving a high frequency oscillation which finally gives instability. 8

26 Imag Axis Imag Axis Real Axis 4 Real Axis Figure 3.c Figure 3.b 3.3 a) The Nyquist curve is far away from the point for all frequencies and the step response of the closed loop system is well damped. As K P increases the Nyquist curve grows in size and for K P = 6. the Nyquist curve reaches and thus is the limit of stability. b) For low frequencies the Nyquist curve is now far away from the origin since the integrating part makes G(iω) large for low frequencies. The Nyquist curve now passes closer to which results in a more oscillatory closed loop system. The system becomes unstable around K I =.44. c) The Nyquist curve is now further away from which corresponds to an improved damping of the closed loop system. The system becomes unstable around K D = a) K P small Both poles on the real axis, but one pole very close to the origin Slow but not oscillatory system. K P = /(4τ k ) Both poles in /(τ), that is, faster than in () but still no oscillations. K P large Complex poles with large imaginary part relative to the real part, that is, oscillatory system. b) If the reference is a step, If the reference is a ramp, lim e(t) = t lim e(t) = A t K P k τ 3.4 a) ω c =.38, ω p =., ϕ m = 94 and A m = 3.. b) The closed loop system is now much more oscillatory due to the reduced phase and gain margins. c) K P = A iii, B i, C iv, D ii. c) lim t e(t) = 3.7 G c = Go +G o 3.8 a) G o = F G b) G c = F G +F G 9

27 Figure 3.3b K =.95 - Im Re c) G ny = F G +F G d) G re = +F G 3.9 a) 3A 3+K b) F (s) = s (for example) c) Poles in,. No zeros. 3.3 A 4, B, C 3, D F t,in, T t,in T ref + Σ F F c,in, T c,in G T t Figure 3.3a 3.3 a) See the block diagram in Figure 3.3a. There, the signals are classified as: Input F c,in and T c,in Output T t Disturbance F t,in and T t,in 3.3 a) The system has a pole in 3. b) The system is stable for K K > µ 3.34 b) The model is given by c) T t (s) = 3 (s+3.675)(s+.85) F c (s) dt t V t dt = F t(t t,in T t ) + U c p t,in ρ (T c T t ) t dt c V c dt = F c(t c,in T c ) U c p (T c T t ) c ρ c d,e) The root locus for a P controller is shown in Figure 3.3b.

28 Bode Diagram Frequency (rad/sec) Figure 4.a 4 Frequency Description 4. G(s) =.6 s a) See figure in the solution. ω c =.5, ϕ m = 3, A m =.5. b) The period time will be 8 seconds, K =.5. c) B = 8, β =. rad/s and ϕ = a) Figure 4.3a in Solutions. b) Figure 4.3b in Solutions. 4.8 a,b) ω G(iω) arg G(iω) = db. rad = 5.8 =.9 db.9 rad = 5.5 = 6 db.6 rad = 9. = 4 db. rad = 6 c) See Figure 4.8a in Solutions. 4.9 G B, G D, G 3 A, G 4 C, G 5 E. 4. Bode gain step response pairs: A D, B C, C A, D B. 4.4 A B, B C, C D, D A. 4.5 a) System G() ω B ω r M p G A.64 G B.5.5 G C. G D G E b) The bandwidth of a system is (approximately) inversely proportional to the rise time. The damping is inversely proportional to the height of the resonance peak. A large peak implies low damping and large overshoot. Phase (deg) Magnitude (abs) y(t) = 5 sin(t / 4 π arctan ). 4.7 a).45 sin(t.) b) Unstable system. c). sin(t.4) d).45 sin(t.) 4. a) The bode digram of the system is shown in Figure 4.a. b) K =.946 = a) K 5.4 b) F (s) =.58 s+. s

29 B A Figure 5.3a Am =.5 ϕm = Compensation arg G(iω) G(iω) Am = 5. ϕm = 88.3 arg G(iω) G(iω) ω c = ω p =.8 Figure 5.a ω [rad/s] -7 = 5 5 e + 3 ω ca =.4 ω p 5 ω [rad/s] ω cb = For example, the following controller fulfills the requirements: 5. a) See Figure 5.a. F (s) = 3.33 s +.85 s +.3 s s +.36 b) Largest crossover frequency:.4 rad/s. c) One controller that fulfills the requirements is the lead compensator (with gain adjustment) F (s) =.9 7 s +.6 s a) See Figure 5.3a. b) Smallest value of ramp error.67 and crossover frequency 5 rad/s. c) One controller that fulfills the constraints is F (s) =.75 s +.4 s One controller which fulfills the requirements is F (s) =. 4 s s + 5 s +.6 s A E C, B C E, C A B, D D D, E B A. 5.6 One controller which satisfies the demands is F (s) =. 5 s + 8. s s +.8 s +.8/84

30 .9 Figure 5.a Im 33 Re 5.7 The system is stable when < K <. or.67 < K < a) T <.698 s b). s < T <.79 s ω p = 3. G(iω) - - G o G m.473 ω c = arg G(iω) -8-7 G o G m - - ω c =.78 ω c,d =.4 ω [rad/s] ϕ = c) lim t e(t) = 5 d) T <.4 5. a) Impossible to determine. Figure 5.9a b) It is stable. 5.9 a) k A =.5 and a =.5. The Bode plot is given in Figure 5.9a. b) One controller that does the job is ( (s +.) F (s) =.6 4 (s +. 4) 5. The following compensator fulfills the requirements: ( F (s) = s +.53 ) s +.5 s s +.5/95 5. a) See Figure 5.a. b) Asymptotically stable for < K <. ) 5.3 a) ω c = 5 rad/s, ω p = 9.5 rad/s, A m = 3.5 and ϕ m = 7. b)-d) See solution. 5.4 The following controller will do: F (s) = s +. s +. s + 3. s 5.5 a) e =, e = 4 K, provided K < 4. Larger K results in an unstable system. b) The following controller will do: F (s) = s s s + s +.9 3

31 6 Sensitivity and Robustness 6. The gain of the sensitivity is: S(i) = and the requirement on K becomes K >. 6. The maximum bandwidth is ω B =. 6.3 See the solution, Figure 6.3a. (K ) + b) Stability cannot be guaranteed for F (s) =, while it can be guaranteed for the regulator from Problem??. 6. The amplitude of the steady state error will be.. 6. The controller also stabilizes the system for the stirring speed 4 r/min. 6.3 K Yes. 6.5 a) See the solution, Figure 6.5a. b) F (iω)g(iω) + F (iω)g(iω) < 6.6 a) No, stability cannot be guaranteed when G(s) =. b) α > /5. This is not contradictory since the robustness criterion is a sufficient but not necessary condition. 6.7 a) Asymptotically stable for α > 3. See the solution, Figure 6.7a. b) α > 4 c) The robustness criterion gives a sufficient but not necessary condition. 6.8 γ < y(t) = sin(t π 4 ) sin(t) 6. a) G (s) = s+ s 4

32 7 Special Controller Structures 7. a) See Figure 7.a in the solution. ω c and ϕ m are undefined and A m = The stability requirement gives K =.75 which implies where a is the size of the step. lim e(t) =.487 a t b) See Figure 7.b in the solution. ω c and ϕ m are undefined, and A m = 6. The requirement gives K = 8 which implies lim e(t) =. a t 7. a) F f (s) =, and h(t) =. A ( e t ). b) Zero steady state error. 7.3 a) F f = G v 3(s + 3) = G u (s + 4) b) The amplitude of the control signal is 3. c) lim t y(t) = 9( b/) +4Kb 7.4 a) F f (s) = 4(s+) 3(s+)(s+5) b) lim t y(t) =. c) lim t y(t) =. 3K+ d) y(t) doesn t have a final value. 5

33 and and 8 State Space Description 8. ẋ = ( ) x + /τ y = ( ) x ( ) u K 8.5 ẋ (t) = x (t) + u(t) ẋ (t) = 4x (t) + 3u(t) y(t) = x (t) + x (t) 8. ẋ = x 8.6 G(s) = s (s+)(s+3) a) b) c) ẋ = ωx + u y = x ẋ (t) = K x (t) + M l (t) ẋ (t) = x (t) + x 3 (t) ẋ 3 (t) = K x (t) + K i(t) ẋ (t) = x (t) ẋ (t) = x 3 (t) ẋ 3 (t) = 6x (t) x (t) 6x 3 (t) + 6u(t) y(t) = x (t) ẋ (t) = x (t) + x 3 (t) + 4u(t) ẋ (t) = 3x (t) + u(t) ẋ 3 (t) = 5x (t) + x (t) + u(t) y(t) = x (t) ẋ (t) = x (t) u(t) ẋ (t) = 3x (t) + 3u(t) y(t) = x (t) + x (t) 8.7 x(t + T ) = e AT x(t ) + ( t+t t e A(t+T s) ds) 8.8 a) The state space description of the closed loop system ẋ(t) = 3 x(t) + h ref (t) h(t) = ( ) x(t) ) Bu b,c) The state space description of the closed loop system with noise 3 ẋ(t) = x(t) + n(t) h(t) = ( ) x(t) 8.9 Yes, since the system is controllable. { ( ) T 8. a) Dimensions: and. Subspaces:, ( 3 6 ) } T { ( ) } T. b) Dimensions: and. Subspaces: { ( ) } T. { ( 4 ) T, ( 8 8 ) T } 6

34 ) ( ) ( q ) ca, + V u c B, 8. a) x = e t, x =.5(e t ) b) No. Yes. c) Controllable, not observable. d) Unobservable growing state simulation collapses. b) The linearized model is given by ( ) ( d q 3kc ca, A V = dt c B, k c A V q V y = ( ) ( ) c A, c B, 8. Poles: ± i. Zeros:. 8.3 a) ẋ = x ẋ = α x u α ẋ 3 = x 4 ẋ 4 = x 3 u b) det S = α ( α ). Thus, the system is controllable except for the case α =, that is, when the two pendulums have the same lengths. 8.4 a) ẋ = ( ) x + 3 y = ( ) x ( ) u b) u = 5x + x + 3.r c) Y (s) = 3.(s+5) (s+4) R(s) 8.5 a) The model is given by V dc A dt V dc B dt = V k c 3 A + qc A,in qc A = V k c 3 A 3 qc B 7

35 ẏ() = β α 9 State Feedback 9. a) State feedback. Poles in { 3, 5 } gives the state feedback 9. a) u = 6x 4x + y ref Poles in {, 5 } gives the state feedback u = 3x 49x + y ref b) Observer poles in gives the observer gain ( ) 38 K = 399 b) u = 8K K x 3 4K x 3 K x 3 K ẋ = x + u K c) Observer gain K T = ( 6 /K 8/K ) 9.3 a) u = c τ θ τc ω + c τ θ ref b) u = c τ θ τc ω + c τ θ ref c c ˆx State feedback gain L = ( 6 ). Observer gain K T = ( 6 9 ). 9.5 The system is observable and the poles of the observer may be placed arbitrarily. c) The system is observable with the sensor at x or x 3. The sensor at x and observer poles in 4 give K T = ( ). 9.7 ˆX 3 (s) = K s+k U(s) + K s s+k X (s) 9.8 a) L = ( ) 9.9 T < b) In steady state: h =.. c) F f (s) = gives, in steady state, h =. d) h = (k ) k v e) Introduce the integral of the height as a new state arctan ωc ω c =.65s 9. a) K T = ( 3 38 ) z(t) = t b) The transfer function from v to x is where C = ( ). h(s) ds ż = h C (si A + KC) K = 9. a) The initial value theorem gives 3s s + 5s a) Yes, since the system is controllable. b) Closed loop poles in 3 gives u = 3x 5x 4x 3 + y ref and hence ẏ() decreases as α decreases. b) No, since the zero is not affected by the feedback. 8

36 u = l r l x s = l ± 4 9. A very fast closed loop system: implies that the poles are far into the LHP which implies a need for generating large input signals. easily becomes unstable in case of model uncertainties. becomes sensitive to measurement noise. has a sensitivity function with a large peak. 9.3 a) L = ( ) b) r(t) = r e 5t 9.4 a) x = y (motor angle) and x = ẏ (angular velocity). b) The pole locations give similar rise and settling times. With complex poles the maximum value of the input is lower. c) Larger weight on the motor angle gives faster response. d) Increasing weight on the input makes the system slower. 9.6 Are the specifications 4 fulfilled???. The bandwidth requirement is not fulfilled.??. The system is stable despite the model errors.??. The gain is different from when κ??. Both measurement and process noise are amplified for some frequency. 9.7 a) u = 4x + x b) Yes! It is essential that the input u is known since u is required in the observer design to get an asymptotically vanishing state estimation error. c) Yes, by introducing a third state x 3 = u. This new system is observable hence a observer can be designed to estimate u. 9.8 a) The poles are pure complex and thus the system doesn t have a well defined stationary error or speed of response. b) A linear combination of r and x is given by e) Increasing weight on the velocity makes the system slower. 9.5 a) Yes the system is controllable. with this controller the poles can be placed with l as l 4 b) Poles in. gives the state feedback u =.3x.8x c) It is desirable that the estimation error converges to zero faster than the dynamics of the system. Thus, we should place the eigenvalues of the observer to the left of the poles of the closed loop system. To avoid large amplification of the measurement noise the poles of the observer should not be placed too far into the left hand plane. d) Observer poles in. gives the observer gain ( ).45 K =.33 and by setting l = the stationary error will be zero when w =. If w and l = then there will be stationary error of size l w. c) Designing a observer with the following observer gains k =, k = 6, and k 3 = 8. Let the control law be u = l r l ˆx l 3 ˆx 3. With l 3 = l and l = there will be no error. Place the poles to the closed loop with l. 9

37 Implementation. β =.95, α = 9.4, and α = a) y k+ y k = T u k b) < K < T.3 a) A = b) T = T/π gives A =. +(ωt/π) + (ω T ) ω = π T ω ϕ = π + arctan ω T

38 AUTOMATIC CONTROL COMMUNICATION SYSTEMS LINKÖPINGS UNIVERSITET Reglerteknik: Solutions Solutions This version: August 3

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40 Solutions This version: August 3

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42 ẏ(t) + y(t) = u(t) Mathematics. a) A step has Laplace transform A s. The differential equation b) A ramp has Laplace transform A s. c) s+ d) s s +5 e) su(s) u() f) su(s). (u() = is a common assumption in the course.) g) s U(s) su() u() h) s U(s). (u() = u() = is a common assumption in the course.) i) A time delayed signal has Laplace transform, e st U(s).. a) Insert ẏ(t) = och u(t) = 5 directly into the differential equation y(t) = 5/. It is also possible to solve the differential equation and let t, or to use b) and the final value theorem. b) Use Laplace transform on the differential equation Y (s) = s+u(s). The denominator coincides with the characteristic polynomial of the differential equation. Note that we also have assumed y() =..3 The general solution is given by.4 a) y(t) = C e t + (C + C 3 t)e t 3 (cos(t) + 7 sin(t)) b) The Laplace transform of the input y(t) = e t + e t, t may be represented by the transfer function G(s) = Y (s) U(s) = s + Hence, the Laplace transform of the system output is given by Y (s) = s s + }{{} Y (s) + s + s + }{{} Y (s) Rewriting the first term using partial fractions leads to with inverse transform Y (s) = s s + = s s + y (t) = e t Rewriting the second term using partial fractions leads to with inverse transform Y (s) = s + s + =.5 s +.5s s s + Hence, the system output is y (t) =.5e t.5 cos t +.5 sin t y(t) =.5e t +.5 sin t.5 cos t yields u(t) = + sin t U(s) = s + s +.5 a) The abolute value is + i =, and the argument is arctan = π 4 = 45. Hence, the polar form is e i π 4

43 λ = 3 v = 3 b) The absolute value is and the corresponding eigenvectors (v) are given by the equation (λi A)v =. + i 5 + = (3)i 5.4 The argument is ( ) + i arg 5i( + = arg ( + i) arg 5i arg ( + 3i) 3i) Hence, the polar form is = arctan 9 arctan 3 = = 5 c) e i π 3 = cos π 3 + i sin π 3 = + 3i 5 e i 8 π.9 λ = v = λ 3 = 4 v 3 = T =. A basis for the null space is for example d) 5e iπ = 5 cos( π) + 5i sin( π) = 5.6 The amplification in decibel is computed as log F = log F, where F is the absolute value of the amplification. The amplification F = hence corresponds to log = 4 db. decibel (db ) Definition Amplification F log F = F = = 3 log F = 3 F = 3/.78 log F = F = = log F = F =.5 = 3.6 log F = F =.5 =.36 A basis for the range space is 3 The rank of the matrix is hence a) Writing the function with partial fractions yields.7 Multiplication of the two matrices gives the unit matrix..8 The eigenvalues (λ) of the matrix A are given by the equation det(λi A) =, F (s) = s s + The inverse transform is then computed by use of a Laplace transform table: f(t) = e t

44 This means that f(t) as t. The same result can also be obtained by use of the final value theorem, that is, by computing lim s sf (s). b) Writing the function with partial fractions yields F (s) =.5 s s The inverse transform is then computed by use of a Laplace transform table: f(t) =.5e t +.5e t This means that f(t) will grow without bound as t. Here, the final value theorem cannot be used since f(t) lacks a final value. c) The inverse transform can be computed by use of the relation L {G(s + a)} = e at g(t) Here, G(s) = s and a =. The inverse transform of G is g(t) = t, so f(t) = L { (s + ) } = e t t which tends to as t. This result can also be obtained by use of the final value theorem.. The relation between inflow and water level is given by the transfer function Y (s) = s + Z(s) and the relation between control signal and inflow may be written as Z(s) = s + s + U(s) This means that the Laplace transforms of the control signal and water level are related by Y (s) = (s + ) (s + s + ) U(s) = s 3 + s + s + U(s) which corresponds to the differential equation y (3) + ÿ + ẏ + y = u 3

45 Dynamic Systems. a) We start from the equations Voltage equilibrium gives where L a =. Equation (.) in (.) gives From (.4) and (.3) we get which in (.5) gives J θ = f θ + M (.) M = k a i (.) v = k v θ (.3) u R a i L a di dt v = (.4) J θ + f θ = k a i (.5) i = (u k v θ)/ra c) Suppose that u is a unit step, that is, {, t < u = t that is This gives θ(s) = G(s)U(s) = U(s) = s Inverse Laplace transformation gives k s(s + /τ) ( s = k τ s k ) τ s + /τ s θ(t) = k τt k τ ( e t/τ ) that is, θ will grow unlimited when t increases. that is Let which gives J θ + f θ = k a (u k v θ)/ra θ + R af + k a k v JR a θ = k a JR a u τ = R af + k a k v JR a k = k a JR a θ + τ θ = k u (.6). () Asymptotically stable system. Monotonic step response, that is, real poles: K =.. () Very oscillative system. Poles close to the imaginary axis: K =.5. (3) Unstable system. Poles in the right half plane: K = 3. (4) Asymptotically stable system. Oscillative step response, that is, complex poles in the left half plane: K =.5. b) Laplace transformation of (.6) gives and this gives the transfer function (s + τ s)θ(s) = k U(s) G(s) = θ(s) U(s) = k s(s + /τ).3 The inverse Laplace transform gives the step response { β d (t) = L + st } = β( e t/t ) s For the final value, we have d (t) β, t 4

46 Amplitude Step Response The figure gives β =. At the time t = T, the system time constant, the step response has reached 63% of the final value, that is, d (T ) =.63 The figure gives T = 3, which gives the total transfer function G(s) = + 3s If we measure the signal d (t) we introduce an additional time delay of L V time units. The total transfer function then becomes Answer:.4 Use the system description G(s) = e L V s + 3s G(s) = e L V s + 3s.5 The pairs of plots that belong to the same system will be written in the form pole-zero-letter step-response-letter. Pole-zero diagram B has a single pole in the origin which gives a ramp as step response, that is, B F. Pole-zero diagram D also has a pole in the origin which gives an infinitely growing step response, D C. Pole-zero diagram F has complex poles which gives an oscillative step response, F D. Pole-zero diagram A has a zero in the origin which gives final value zero, A B. Polezero diagram C cannot be step response E, since two real poles and no zeros give no overshoot. Hence C A, and step response E is the only alternative left for pole-zero diagram E. Answer: A B, B F, C A, D C, E E, F D..6 a) Enter the systems. >> s = tf( s ); >> GA = / ( s^ + *s + ); >> GB = / ( s^ +.4*s + ); >> GC = / ( s^ + 5*s + ); >> GD = / ( s^ + s + ); >> GE = 4 / ( s^ + *s + 4 ); Compute and plot the step response. >> step( GA ); grid In the first figure ω = and ζ =.5. a) For the system ω G(s) = s + ζω s + ω G(s) = s + as + we have ω = and ζ =.5a. The step response is more oscillative than in the case ζ =.5, that is, ζ <.5. This gives a <. b) For the system G(s) = b s + bs + b we have ω = b and ζ =.5. The step response is in this case pure time scaling compared to the case ω =. The figures show that the step response is twice as fast as in the case ω =. This gives b = ω =. 5 5 Time (sec.) The systems G B (s), G C (s), G D (s), and G E (s) can be simulated in a similar way. The values of T r, T s, and M for the different step responses can be found by a right click in the figure and selecting Characteristics and then selecting the desired property. Use the Properties... menu item 5

47 >> s = tf( s ); >> G = ( *s + ) / ( s^ + *s + ); (of the right click menu) to change the interval for the settling time. (The default interval is %, while we use 5% in the course.) b) Compute the poles. >> pole( GA ) ans = - - The other systems are handled in the same way. c) The results from a) and b) can be summarized in the following table. System T r T s M poles G A %, G B %. ± i.98 G C % 4.8,. G D %.5 ± i.87 G E % ± i.73 y Using this table we can draw the following conclusions. (i): The speed of the step response (mainly) depends on the distance between the poles and the origin. Poles further away from the origin give a faster step response and shorter rise time. (ii): The damping of the system depends on the relationship between the imaginary part and the real part of the poles. Poles with large imaginary part relative to the real part give a poorly damped (oscillatory) step response. Remark: We see that even though the distance to the origin is nearly the same in system G A and G B the rise time is almost 3 times faster in system B. Note that speed is not only rise time, also the settling time should be considered! Look at the following system ω t Figure.6a ω G(s) = s + ζω s + ω The poles of this system are given by s = ω ( ζ ± i ζ ) = ω ( cos φ ± i sin φ) where cos φ = ζ. The parameter ζ is called relative damping and ζ. We see that ω is the distance from the origin to the poles and in Figure.6a the step responses for different ζ are shown when ω is constant. We see clearly that the rise time is faster when ζ is small but when ζ is small the settling time is big!.7 Enter the system. Here we consider the case α =, that is the system has a zero in.5. 6

48 Plot the step response. >> step( G, ); grid Step Response.4. c) Find the time points where the output is % (.5) and 9% (.35) of the steady state value. The rise time is the difference between these values, here approximately T r.5 s. d) Find the earliest time such that the output then lies within ±5% of the steady state value. Here, the interval is [.45,.575 ], and the settling time is T s 7.8. Amplitude Time (sec.) A zero located close to the origin on the negative real axis causes an overshoot in the step response. A zero on the positive real axis causes the step response to initially move in the negative direction. This means that in some cases the zeros of the system can have significant influence in the system properties. Systems with zeros in the right half plane normally imply extra difficulties for the design of control systems..8 The Laplace transform of a step is U(s) = s. The step response is hence given by y(t) = L (G(s) s ). If G(s) is a rational function the inverse Laplace transform can be computed by first doing a partial fraction expansion and then using a transform table. When the system is available one can let the input u(t) be a step and measure y(t)..9 a) The steady state value is.5. b) The output signal almost reaches.9, which is slightly less than.4 over the final value. The overshoot is hence.4.5 6%.. G C: G is poorly damped, which gives an oscillatory behavior. G : Can be excluded since it is the only system having static gain, and among the step responses there is always more than one match for each of the present final values. G 3 B: This case has the shortest rise time, and some overshoot due to the pair of complex poles. The static gain is. G 4 A: The pole in dominates, which gives slower step response than systems G 3 and G 5. The static gain is. G 5 D: The dominating pole is in 3, which is slower than for G 3 but faster than for G 4. The static gain is. G 6 : Can be excluded due to instability.. a) The signals can be classified as Disturbances signal: Acid process flow (unknown ph and flow) Control signal: NaOH solution Measured and controlled signal: The ph of the outflow b) A block diagram where the control strategy is based on feedback could look like Figure.a. a) At steady state the inflow is equal to the outflow (constant volume). From mass balance ρ q = ρ q + ρ q Assuming the densities are equal (ρ = ρ = ρ ) gives q = q + q = +.5 =.5 m 3 /min. From component balance for component A q c A = q c A, + q c A, 7

49 ref + Σ which gives c A =. kmol/m3. F NaOH Figure.a Tank Acid flow outflow ph b) The amount of mass in the tank is given by ρv (assuming ρ is constant). The change in mass is given by the mass coming in subtracted by the mass going out of the tank d(ρv ) dt = ρ(q in q out ) (.) where q in = q + q and q out = q. Assuming that the volume is constant gives d(ρv ) dt which means q + q = q (.) The amount of component A contained in the tank is given by V c A. The change is then given by d(v c A ) dt Constant V and (.) gives V dc A dt = q c A, + q c A, qc A (.3) = q (c A, c A ) + q (c A, c A ) (.4) The model (.), (.3) is nonlinear since it contains products between variables. Assuming volumes and flows to be constant gives a linear model. c) Assume that all the other independent variables (q, q, c A, ) are constant. Take their values from a). Equation (.4) then gives V dc A dt = q (c A, (t) c A (t)) + q ( c A, c A (t) ) =.5c A (t) + c A, (t) + The equation can be written dc A dt =.5c A (t) + 3. for t. The corresponding Laplace transform equation is or (Lc A )(s) = s(lc A )(s) (Lc A )() =.5(Lc A )(s) + 3. s ( + 3. ) s +.5 s = s s +.5 which transforms back to c A (t) = e.5t + 3. ( e.5t ).5 Rearranging yields ( 3..5 ) ( e.5t = 3. ) ( ) e t.5.5 where the sought constants can be identified: k =. kmol/m 3, k =.3 kmol/m 3, and τ =.5 =.67 min..3 a) The equilibrium equation is Mass balance gives dm i dt Component balance gives d(m i x i ) dt y i = αx i + (α )x i (.) = L i + V i+ L i V i (.) dx i = M i dt + x dm i i = L i x i + V i+ y i+ L i x i V i y i (.3) dt Combining (.) (.3) gives M i dx i dt = xi (Li + Vi+ Li Vi) + Li xi + Vi+yi+ Lixi Viyi = L i (x i x i) + V i+ (y i+ x i) + V i (x i y i) s (.4) The dynamic model for M i (t) and x i (t) is described by (.), (.), and (.4). 8

50 b) The stationary point for (.) gives L i + V i+ L i V i =. Introduce the difference variables x i = x i x i x i+, = x i+ x i+ y i = y i yi y i, = y i yi L i+, = L i+ L i+ V i, = V i Vi L i = L i L i V i = V i Vi The assumption that the change of mass on the plate is zero gives which means that this will simplify (.4) to Linearization of (.5) gives M i dx i dt Linearization of (.) gives dm i dt = L i + V i+ L i V i = M i dx i dt = L i x i + V i+ y i+ L i x i V i y i (.5) = L i x i, + V i+y i+, L i x i V i y i + x i L i, + y i+v i+, x L i y i V i (.6) y i = The linearized model is described by (.6) (.7). α ( + (α )x i ) x i (.7).4 Från blockdiagramet fås Y (s) = G (s)[f (s)y (s) + G (s)u(s) + F (s)u(s)], vilket ger överföringsfunktionen Y (s) U(s) = G (s)(g (s) + F (s)). F (s)g (s).5 D: En integrator vars stegsvar är en ramp. Ger. B: Nollställe i högra halvplanet vilket ger ett stegsvar som initialt går åt fel håll. Ger 5. A: Polerna till A och B är samma, vilket ger samma relativa dämpning. Ger. C: Polerna har relativ dämpning ζ =.5 vilket är mindre än alla andra. Ger 4. F: Polerna har relativ dämpning ζ = och snabbhet ω = 3. Inget annat system är så snabbt. Ger 3. E: Enda systemet kvar. Ger 6. Svar: A-, B-5, C-4, D-, E-6 and F-3 9

51 3 Feedback Systems 3. a) To begin with, the transfer function for the tank system is derived. The mass balance equation is, assuming that the bottom area of the tank is m ḣ(t) = x(t) v(t) that is (note that all initial conditions are zero when deriving transfer functions) sh(s) = X(s) V (s) Hence where H(s) = G t (s)(x(s) V (s)) G t (s) = s The block diagram becomes like in Figure 3.a. b) The transfer function for the valve is G v (s) = k v + T s With the input taken as a unit step signal, that is, h ref e u x h + Σ F (s) G v (s) + Σ G t (s) Figure 3.a c) By using the controller F (s), the closed loop system shown in Figure 3.a is obtained. From the block diagram, the following equations are obtained: E(s) = H ref (s) H(s) H(s) = G t (s) ( F (s)g v (s)e(s) V (s) ) This leads to ( ) H(s) = G t (s) G v (s)f (s) [H ref (s) H(s)] V (s) ( ) ( ) H(s) + G t (s)g v (s)f (s) = G t (s) G v (s)f (s)h ref (s) V (s) v it follows that The final value theorem gives U(s) = s X(s) = k v + T s s lim x(t) = lim sx(s) = k v t s The time constant T is the time it takes for the step response to reach 63% of its final value. From the plot it follows that T = 5 and k v =, that is G v (s) = + 5s H(s) = G t(s)g v (s)f (s) G t (s) H ref (s) V (s) + G t (s)g v (s)f (s) + G t (s)g v (s)f (s) }{{}}{{} G c(s) G v,h (s) That the expression for the output is a sum over all inputs, with each term given by a rational transfer function multiplied by the input, is no coincidence; this will always be true of any transfer function between points in a block diagram with rational transfer functions and summation points. In particular, the output is a linear (dynamic) function of the inputs. This leads to a conclusion that will be used frequently hereafter: When computing the transfer function from one input to the output, all other inputs may be set to zero. The reader is encouraged to try this by taking H ref (s) = in the first equation above.