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Chapter Dynamic Models Problems and Solutions for Section 2.1 Write the differential equations for the mechanical systems shown in Fig 2.38 Solution: The key is to draw the Free Body Diagram (FBD) in order to keep the signs right For (a), to identify the direction of the spring forces on the object, let x2 = and Þxed and increase x1 from Then the k1 spring will be stretched producing its spring force to the left and the k2 spring will be compressed producing its spring force to the left also You can use the same technique on the damper forces and the other mass (a) m1 x ă1 m2 x ă2 = k1 x1 b1 xú − k2 (x1 − x2 ) = −k2 (x2 − x1 ) − k3 (x2 − y) − b2 xú 11 CuuDuongThanCong.com https://fb.com/tailieudientucntt 12 CHAPTER DYNAMIC MODELS Figure 2.38: Mechanical systems CuuDuongThanCong.com https://fb.com/tailieudientucntt 13 m1 x ¨1 m2 x ¨2 m1 x ¨1 m2 x ¨2 = −k1 x1 − k2 (x1 − x2 ) − b1 xú = −k2 (x2 − x1 ) − k3 x2 = −k1 x1 − k2 (x1 − x2 ) − b1 (xú − xú ) = F − k2 (x2 − x1 ) − b1 (xú − xú ) Write the equations of motion of a pendulum consisting of a thin, 2-kg stick of length l suspended from a pivot How long should the rod be in order for the period to be exactly secs? (The inertia I of a thin stick about an endpoint is 13 ml2 Assume θ is small enough that sin θ ∼ = θ.) Solution: Let’s use Eq (2.14) M = Iα, CuuDuongThanCong.com https://fb.com/tailieudientucntt 14 CHAPTER DYNAMIC MODELS l O G mg Moment about point O MO = −mg × = l sin = IO ă 2ă ml ¨θ + 3g sin θ = 2l As we assumed is small, ă + 3g = 2l The frequency only depends on the length of the rod ω2 = 3g 2l s 2l =2 3g T = 2π = 2π ω l = 3g = 1.49 m 2π2 CuuDuongThanCong.com https://fb.com/tailieudientucntt 15 Figure 2.39: Double pendulum q 2l (a) Compare the formula for the period, T = 2π 3g with the well known formula for the period of a point mass hanging with a string with q length l T = 2π gl (b) Important! In general, Eq (2.14) is valid only when the reference point for the moment and the moment of inertia is the mass center of the body However, we also can use the formular with a reference point other than mass center when the point of reference is Þxed or not accelerating, as was the case here for point O Write the equations of motion for the double-pendulum system shown in Fig 2.39 Assume the displacement angles of the pendulums are small enough to ensure that the spring is always horizontal The pendulum rods are taken to be massless, of length l, and the springs are attached 3/4 of the way down Solution: CuuDuongThanCong.com https://fb.com/tailieudientucntt 16 CHAPTER DYNAMIC MODELS G1 l G2 k m l sin G1 m l sin G If we write the moment equilibrium about the pivot point of the left pendulem from the free body diagram, 3 M = −mgl sin θ1 − k l (sin θ1 sin ) cos l = ml2 ă1 4 ml2 ă1 + mgl sin + kl cos θ1 (sin θ1 − sin θ2 ) = 16 Similary we can write the equation of motion for the right pendulem 3 −mgl sin θ2 + k l (sin θ1 − sin θ2 ) cos l = ml2 ă2 4 As we assumed the angles are small, we can approximate using sin θ1 ≈ θ1 , sin θ2 ≈ θ2 , cos θ1 ≈ 1, and cos θ2 ≈ Finally the linearized equations of motion becomes, kl (θ1 − θ2 ) = 16 mlă2 + mg2 + kl (2 ) = 16 mlă1 + mg1 + Or CuuDuongThanCong.com https://fb.com/tailieudientucntt 17 ă1 + g + l g ă2 + + l k (1 θ2 ) = 16 m k (θ2 − θ1 ) = 16 m Write the equations of motion for a body of mass M suspended from a Þxed point by a spring with a constant k Carefully deÞne where the body’s displacement is zero Solution: Some care needs to be taken when the spring is suspended vertically in the presence of the gravity We deÞne x = to be when the spring is unstretched with no mass attached as in (a) The static situation in (b) results from a balance between the gravity force and the spring From the free body diagram in (b), the dynamic equation results mă x = kx mg We can manipulate the equation m mă x = −k x + g , k so if we replace x using y = x + m k g, mă y = ky mă y + ky = CuuDuongThanCong.com https://fb.com/tailieudientucntt 18 CHAPTER DYNAMIC MODELS The equilibrium value of x including the effect of gravity is at x = − m kg and y represents the motion of the mass about that equilibrium point An alternate solution method, which is applicable for any problem involving vertical spring motion, is to deÞne the motion to be with respect to the static equilibrium point of the springs including the effect of gravity, and then to proceed as if no gravity was present In this problem, we would deÞne y to be the motion with respect to the equilibrium point, then the FBD in (c) would result directly in mă y = −ky For the car suspension discussed in Example 2.2, (a) write the equations of motion (Eqs (2.10) and (2.11)) in state-variable form Use the state vector x = [ x xú y yú ]T (b) Plot the position of the car and the wheel after the car hits a “unit bump” (i.e., r is a unit step) using MATLAB Assume that m1 = 10 kg, m2 = 350 kg, kw = 500, 000 N/m, ks = 10, 000 N/m Find the value of b that you would prefer if you were a passenger in the car Solution: (a) We can arrange the equations of motion to be used in the statevariable form b ks b kw kw ks x− xú + y+ yú − x+ r m1 m1 m1 m1 m1 m1 ả kw ks b kw b ks + xú + y+ yú + r x− = − m1 m1 m1 m1 m1 m1 b ks b ks x+ xỳ y yỳ yă = m2 m2 m2 m2 x ă = So, for the given sate vector of x = [ x xú y form will be, xú ³ − ks + x m1 ă = yỳ ks yă m2 kw m1 ´ 0 − mb1 ks m1 b m1 ks −m − mb2 b m2 yú ]T , the state-space x k w xú + m1 r y yú (b) Note that b is not the damping ratio, but damping We need to Þnd the proper order of magnitude for b, which can be done by trial and CuuDuongThanCong.com https://fb.com/tailieudientucntt 19 error What passengers feel is the position of the car Some general requirements for the smooth ride will be, slow response with small overshoot and oscillation Respons e with b = 1000.0 Response with b = 2000.0 1.5 1.5 W heel Car W heel Car 1 0.5 0.5 0 0.5 1.5 Respons e with b = 3000.0 0.5 1.5 Response with b = 4000.0 1.5 1.5 W heel Car W heel Car 1 0.5 0.5 0 0.5 1.5 0.5 1.5 From the Þgures, b ≈ 3000 would be acceptable There is too much overshoot for lower values, and the system gets too fast (and harsh) for larger values CuuDuongThanCong.com https://fb.com/tailieudientucntt 20 CHAPTER DYNAMIC MODELS % Problem 2.5 b clear all, close all m1 = 10; m2 = 350; kw = 500000; ks = 10000; B = [ 1000 2000 3000 4000 ]; t = 0:0.01:2; for i = 1:4 b = B(i); F = [ 0; -( ks/m1 + kw/m1 ) -b/m1 ks/m1 b/m1; 0 1; ks/m2 b/m2 -ks/m2 -b/m2 ]; G = [ 0; kw/m1; 0; ]; H = [ 0 0; 0 ]; J = 0; y = step( F, G, H, J, 1, t ); subplot(2,2,i); plot( t, y(:,1), ’:’, t, y(:,2), ’-’ ); legend(’Wheel’,’Car’); ttl = sprintf(’Response with b = %4.1f ’,b ); title(ttl); end Automobile manufacturers are contemplating building active suspension systems The simplest change is to make shock absorbers with a changeable damping, b(u1 ) It is also possible to make a device to be placed in parallel with the springs that has the ability to supply an equal force, u2, in opposite directions on the wheel axle and the car body (a) Modify the equations of motion in Example 2.2 to include such control inputs (b) Is the resulting system linear? (c) Is it possible to use the forcer, u2, to completely replace the springs and shock absorber? Is this a good idea? Solution: (a) The FBD shows the addition of the variable force, u2 , and shows b as in the FBD of Fig 2.5, however, here b is a function of the control variable, u1 The forces below are drawn in the direction that would result from a positive displacement of x CuuDuongThanCong.com https://fb.com/tailieudientucntt 673 Using Eq (3), (4), and (7), and the maximum radiant power is related to the Þlament length by, Pmax = β2 L α (8) Using Eq (8) together with Eqs (5) and (6), we can then solve for the Þlament length , L, as, # 13 " Pmax α 13 Pmax πVmax L=( ) = 4 )2 4ρe0 ( TTmax )1.2 π ²2 σ (Tmax − T∞ β2 e0 (9) This can be simpliÞed further by assuming that the maximum Þlament temperature is much higher 4 than the ambient temperature, Tmax À T∞ , so that, " Pmax πVmax L∼ = 4ρe0 ( TTmax )1.2 π ²2 σ Tmax e0 # 13 (10) We can solve for the Þlament length and diameter in terms of the given quantities Pmax , Vmax , and Tmax as well, −3.07 3 Vmax Tmax , L ∝ Pmax d ∝ (11) − 23 −0.93 Tmax Pmax Vmax Figures (a), (b), and (c) on top of the next page show plots of the functional relationship between Þlament diameter and length as a function of the maximum power, Pmax , maximum temperature, Tmax , and maximum voltage, Vmax , respectively as well as the nominal operating point corresponding to Pmax = 200W , Tmax = 3000K, and , Vmax = 120V Figure (a) shows that increasing Pmax requires increases in both the Þlament diameter and length As seen from Figure (b), increasing the maximum temperature, Tmax , requires decreasing the Þlament length but is relatively insensitive to the Þlament diameter Figure (c) shows that increasing the maximum voltage, Vmax , requires increasing the Þlament length but decreasing the Þlament diameter Figures (a), (b), and (c) on the bottom of the next page show the same relationships as in the top Þgure but in terms of the normalized diameter, d L dno al , and normalized Þlament length, Lno al CuuDuongThanCong.com https://fb.com/tailieudientucntt 674 CHAPTER CONTROL-SYSTEM DESIGN: PRINCIPLES AND CASE STUDIES d/d nominal L/L 2.5 nominal 1.5 1.5 1 0.5 0 200 400 600 (a) Pmax 800 1000 50 100 150 (b) Vmax(V) 200 250 0.5 2000 2200 2400 2600 (c) Tmax(K) 2800 3000 Lamp design parametrs 1.4 d [mm] L [m] 1.2 1.5 0.8 0.6 0.4 0.5 0.2 0 200 400 600 (a) Pmax(W) 800 1000 50 100 150 (b) Vmax(V) 200 250 2000 2200 2400 2600 (c) Tmax(K) 2800 3000 1.4 1.2 0.8 0.6 0.4 0.2 Lamp design parameters: normalized Assume the nominal operating temperature is denoted by T0 Re-writing Equation (2) in terms CuuDuongThanCong.com https://fb.com/tailieudientucntt 675 of the normalized temperature, we have ( ¸ · V2 4²σT03 T∞ Vmax Tú T )=− ) + ( )4 − ( T T0 ρcd T0 T0 4ρe0 ρcL T0 ( T )1.2 (12) e0 Let us deÞne the normalized temperature, x = system, T T0 ,and re-write Eq (12) as the nonlinear Þrst-order xú = −A(x4 − x4∞ ) + B V2 , x1.2 (13) where, A = B = 4²σT03 , ρcd Vmax 4ρe0 ρcL2 T0 Linearizing Equation (13) about the nominal (normalized) temperature, we Þnd the Þrst-order linear dynamic model for the lamp as, 1.2 V , x2.2 (14) ∼ ρcd = 4Ax0 16²σT03 x30 (15) πd2 Vmax V 4ρe0 L( TTe0 )1.2 (16) xú = −4Ax30 x − B which means that the lamp time constant is, τ= In terms of lamp current, we have, I= Equation (15) implies that fast lamp response requires high Þlament temperature and low Þlament diameter Typical values for the lamp Þlament time constant range from 0.5 to seconds The output of the lamp may be considered to be current, normalized Þlament temperature, or radiative power, ylamp = I x Pmax (17) A nonlinear simulation for the lamp model may be implemented in Simulink as shown on the next page The results of the simulation show the temperature and current response of the lamp to a step voltage command input as shown The Þgure shows the fast lamp Þlament temperature response with a time constant of 0.07 seconds, and also shows that the current initially surges but quickly drops to a steady-state within approximately 0.3 seconds CuuDuongThanCong.com https://fb.com/tailieudientucntt CuuDuongThanCong.com Constant3 Step Math Function uv Gain2 b u Math Function4 Product Gain1 a Simulink diagram for lamp model https://fb.com/tailieudientucntt To Workspace1 Icurrent Out2 u Math Function5 Scope1 Out1 To Workspace Tnorm Scope Product1 Constant4 1.2 Constant1 Constant Constant2 Gain -K- Math Function3 v u Integrator s Math Function2 v u Math Function1 v u 0.3 To File1 Icurrent.mat To File ToT.mat 676 CHAPTER CONTROL-SYSTEM DESIGN: PRINCIPLES AND CASE STUDIES 677 Response to step voltage command 14 12 Filament temperature, current 10 0 0.05 0.1 0.15 0.2 0.25 Time (sec) 0.3 0.35 0.4 0.45 0.5 Lamp response to a step voltage command References [1] Reynolds, W C., and H C Perkins, Engineering Thermodynamics, McGraw-Hill, 1977 [2] J L Ebert, A Emami-Naeini, R L Kosut, “Thermal Modeling of Rapid Thermal Processing Systems,” in Proc RTP’95, September 1995 [3] Lide, D R., Ed., Handbook of Chemistry and Physics, CRC Press, 1993-1994 [4] Modest, M F., Radiative Heat Transfer, McGraw-Hill, 1993 21 Develop a nonlinear model for a pyrometer Show how temperature can be deduced from the model Solution: Temperature measurement can be done by a variety of methods including thermocouples, resistive temperature detectors (RTDs), and pyrometers [1] A pyrometer is a non-contact temperature sensor and measures the Infrared (IR) radiation which is directly a function of the temperature It is known that objects emit radiant energy proportional to T where T is the temperature of the object Among the advantages of pyrometers are that they have very fast response time, can be used to measure the temperature of moving objects (e.g., a rotating semiconductor wafer), and in vacuum for semiconductor manufacturing The single-wavelength pyrometer measures the total energy emitted from a surface at a given wavelength To understand the operation of a pyrometer, we need to review some concepts from radiation heat transfer [2-3] The emissivity of an object, ², is deÞned as the ratio of the energy ßux emitted by a surface to that from a black body at the same temperature: R ²λ Ibλ (T )dλ ²= R , Ibλ (T )dλ CuuDuongThanCong.com (18) https://fb.com/tailieudientucntt 678 CHAPTER CONTROL-SYSTEM DESIGN: PRINCIPLES AND CASE STUDIES where, ² ²λ T λ Ibλ = = = = = total emissivity, spectral emissivity, absolute temperature, K, wavelength of radiation, µm, spectral black body intensity The frequency, ν, is given by, ν= c0 , λ0 where, c0 λ0 = speed of light in vacuum = 2.998 × 108 m/s, = wavelength of light in vacuum The Plank’s law of radiation states that the spectral radiance of a blackbody, or spectral intensity, Ibν , in a dielectric medium as a function of the wavelength and temperature is, Ibν (T ) = where, 2hν n2 hν c2o (e kT − 1) , (19) h = Planck0 s constant = 6.626 × 10−34 Js, k = Boltzmann0 s constant = 1.3806 × 10−23 J/K, n = real refractive index (n = for most gases), and the frequency and wavelength are related by, ν= c , λ (20) where c is the speed of light in the medium and is given by, c0 n After some manipulation, we can re-write Eq (19) as [2], c= Ibλ (T ) = 2hc20 hc0 n2 λ5 (e nλkT − 1) (21) The total black-body radiation intensity, Ib , is obtained by integrating over all frequencies or wavelength [2] Z Ib (T ) = Iνb (T )dν (22) = CuuDuongThanCong.com n2 σT π https://fb.com/tailieudientucntt 679 The black body emissive ßux is given by, qλb (T ) = C1 C2 n2 λ (e nλT where, C1 C2 , (23) − 1) = 2πhc20 = 3.7419 × 10−16 , W/m2 , hc0 = = 14388 µmK k (24) Integrating over all wavelengths λ, we obtain the total black body emissive ßux [2], qb (T ) = Z ∞ qλb (T )dλ (25) = n σT , where σ is the Stefan-Boltzman constant, σ = 5.67 × 10−8 W/(m2 K ) The temperature may be determined from Eq (19) as, T = C2 λ ln(1 + ²λ C I ) , (26) where, C= C1 λ5 (27) Schematic of temperature measurement using pyrometry CuuDuongThanCong.com https://fb.com/tailieudientucntt 680 CHAPTER CONTROL-SYSTEM DESIGN: PRINCIPLES AND CASE STUDIES The Þgure on the bottom of the previous page shows the schematic of temperature measurement of a semiconductor wafer using pyrometry where, for this particular application, response time and view angle are very important A two color pyrometer is also used for applications where absolute temperature measurement is important The measurement can then be used for feedback control purposes, e.g., pyrometers are now routinely used in control of rapid thermal processing (RTP) systems in semiconductor manufacturing [4] References: [1] Fraden, J., Handbook of Modern sensors: Physics, Designs, and Applications, Springer, 1996 [2] Ozisik, M N., Radiative Transfer and Interactions with Conduction and Convection, WileyInterscience, 1973 [3] Siegel, R and J R Howell, Thermal Radiation Heat Transfer, Second Ed., Hemisphere Publishing Corp., 1981 [4] J L Ebert, A Emami-Naeini, R L Kosut, “Thermal Modeling of Rapid Thermal Processing Systems,” in Proc RTP’95, September 1995 22 Repeat the RTP case study design by summing the three sensors to form a single signal to control the average temperature Demonstrate the performance of the linear design, and validate the performance on the nonlinear Simulink simulation Solution: A linear model for the system was derived in the text as, ˙ = F3 T + G3 u, T y = H T + J3 u, (28) where y = [Ty1 Ty2 Ty3 ]T and, −0.0682 0.0149 0.0000 0.3787 0.1105 0.0229 F3 = 0.0458 −0.1181 0.0218 , G3 = 0.0000 0.4490 0.0735 , 0.0000 0.04683 −0.1008 0.0000 0.0007 0.4177 0 0 H3 = , J3 = 0 0 0 The three open-loop poles are computed from MATLAB and are located at −0.0527, −0.0863, and −0.1482 Since we tied the three lamps into one actuator and are only using the average temperature for feedback, the linear model is then: −0.0682 0.0149 0.0000 0.5122 F = 0.0458 −0.1181 0.0218 , G = 0.5226 , 0.0000 0.04683 −0.1008 0.4185 Havg = CuuDuongThanCong.com £ 3 Ô , J = [0] , https://fb.com/tailieudientucntt 681 resulting in the transfer function, G(s) = Tyavg (s) 0.4844(s + 0.0878)(s + 0.1485) = Vcmd (s) (s + 0.1482)(s + 0.0527)(s + 0.0863) We may try a simple PI controller of the form, Dc (s) = (s + 0.0527) , s so as to cancel the effect of the slower pole The linear closed-loop response is shown as well as the associated control effort The system response follows the commanded trajectory with a time delay of approximately sec and no overshoot The lamp has its normal response until 75 sec and goes negative (shown in dashed) to try to follow the sharp drop in commanded temperature As mentioned in the text, this behavior is not possible in the system as there is no means of active cooling and the lamps saturate low There is no explicit means of controlling the temperature nonuniformity using the PI controller PID (average) temperature tracking response 30 25 Temperature (K) 20 15 10 -5 10 20 30 40 50 Time (sec) 60 70 80 90 100 Linear closed-loop RTP response for PI controller CuuDuongThanCong.com https://fb.com/tailieudientucntt 682 CHAPTER CONTROL-SYSTEM DESIGN: PRINCIPLES AND CASE STUDIES PID (average) temperature : control effort Lamp voltage (V) -5 -10 -15 -20 -25 10 20 30 40 50 Time (sec) 60 70 80 90 100 RTP Linear response for PI: control effort Next we design a state-space based controller As in the text, we use the error space approach for inclusion of integral control and employ the linear quadratic gaussian technique of Chapter The error system is, · ¸ · ¸· ¸ · ¸ eú Havg e J = + µ, (29) F ξ G ξ˙ where, A= · Havg F ¸ ,B = Ã J G with = u and e = y − r, ξ = T ú For state feedback design, the LQR formulation of Chapter is used Z ∞ J = {zT Q z+ρµ2 }dt, T where z =[e ξ T ] Note that J has been chosen in such a way as to penalize the tracking error, e, the control, u, as well as the differences in the three temperatures— therefore, the performance index should include a term of the form, © ª 10 (T1 − T2 )2 + (T1 − T3 )2 + (T2 − T3 )2 , and hence minimizes the temperature non-uniformity As in the text, the factor of ten is used as the relative weighting between the error state and the plant state The state and control weighting matrices, Q and R, are then, 0 20 −10 −10 Q= −10 20 −10 , R = ρ = −10 −10 20 CuuDuongThanCong.com https://fb.com/tailieudientucntt 683 The following MATLAB command is used to design the feedback gain, [K]=lqr(A,B,Q,R) The resulting feedback gain matrix computed from MATLAB is, K = [K1 : K0 ], where, K1 = 1, K0 = Ê 0.7344 0.9344 0.3921 Ô , which results in the internal model controller of the form, xú c = Bc e, u = Cc xc − K0 T, (30) with xc denoting the controller state and, Bc = −K1 = −1, Cc = The resulting state-feedback closed-loop poles computed using MATLAB’s eig command are at −0.5395 ± 0.4373j, −0.1490, and −0.0879 The full-order estimator was designed with the same process and sensor noise intensities used in the text as the estimator design knobs, Rw = 1, Rv = 0.001 The following MATLAB command is used to design the estimator, [L]=lqe(F,G,H,Rw,Rv) The resulting estimator gain matrix is, 16.142 L = 16.4667 , 13.1975 with estimator error poles at −15.3197, −0.1485, and −0.0878 The estimator equation is, ˆú = FT ˆ + Gu+L(y−HT) ˆ T (31) With the estimator, the internal model controller equation is modiÞed as in the text xú c = Bc e, ˆ u = Cc xc − K0 T (32) The closed-loop system equations are given in the text, x˙ cl y CuuDuongThanCong.com = Acl xcl + Bcl r, = Ccl xcl + Dcl r, (33) https://fb.com/tailieudientucntt 684 CHAPTER CONTROL-SYSTEM DESIGN: PRINCIPLES AND CASE STUDIES ˆ T ]T and where r is the reference input temperature trajectory, the closed-loop state vector is xcl = [TT xTc T the system matrices are, −GK0 F GCc , Bcl = −Bc , 0 = Bc H LH GCc F − GK0 −LH Ê Ô H 0 , = Dcl = [0], Acl Ccl with closed-loop poles (computed with MATLAB) located at −0.5395±0.4373j, −0.1490, −0.0879, −15.3197, −0.1485 and −0.0879 as expected text Figure 9.85 The closed-loop control structure is as shown in The linear closed-loop response and the associated control effort are shown The commanded temperature trajectory, r, is a ramp from 0◦ C to 25◦ C with a 1◦ C/sec slope followed by 50 sec soak time and drop back to 0◦ C The system tracks the commanded temperature trajectory — albeit with a time delay of approximately seconds for the ramp and a maximum of 0.0216◦ C overshoot As expected the system tracks a constant input asymptotically with zero steady-state error The lamp command increases as expected to allow for tracking the ramp input, reaches a maximum value at 25 sec and then drops to a steady-state value around 35 sec The normal response of the lamp is seen from to 75 sec followed by negative commanded voltage for a few seconds corresponding to fast cooling Again, the negative control effort voltage (shown in dashed lines) is physically impossible as there is no active cooling in the system Hence in the nonlinear simulations, commanded lamp power must be constrained to be strictly non-negative Note that the response from 75-100 sec is that of the (negative) step response of the system temperature tracking response 30 25 temperature (K) 20 15 10 -5 10 20 30 40 50 Time (sec) 60 70 80 90 100 RTP linear (average) temperature tracking response CuuDuongThanCong.com https://fb.com/tailieudientucntt 685 control effort u lamp voltage -2 -4 -6 -8 -10 -12 -14 10 20 30 40 50 Time (sec) 60 70 80 90 100 RTP linear (average) : control eort The nonlinear Êclosed-loop Ôsystem was simulated in Simulink as shown on the next page In the diagram, Gain4 = 13 31 31 As in the text, the model was implemented in temperature units of degrees Kelvin and the ambient temperature is 301K The nonlinear plant model is the implementation of text Eq 9.48 The voltage range for system operation is between to volts as seen from the diagram As in the text, a saturation nonlinearity is included for the lamp as well as integrator anti-windup logic to deal with lamp saturation The nonlinear dynamic response and the control effort are shown Note that the nonlinear response is in general agreement with the linear response CuuDuongThanCong.com https://fb.com/tailieudientucntt CuuDuongThanCong.com Repeating Sequence k1 s Scope3 Scope4 Matrix Gain1 K*u Matrix Gain2 K*u Anti-windup Integrator Saturation Out1 Matrix Gain K*u ko Subsystem2 In1 InvLamp Scope1 To Workspace2 To Workspace1 Scope2 u r Out1 Estimator Matrix Gain3 K*u Lamp Uncertainty State-Space1 x' = Ax+Bu y = Cx+Du Subsystem1 In1 VtoPower Matrix Gain5 K*u Out2 Matrix Gain4 K*u Subsystem In1 Nonlinear Plant Scope5 To Workspace3 yavg Scope To Workspace y 686 CHAPTER CONTROL-SYSTEM DESIGN: PRINCIPLES AND CASE STUDIES Simulink diagram for nonlinear closed-loop RTP system to control average temperature https://fb.com/tailieudientucntt 687 Average temperature tracking response 335 Temperature (K) 330 325 320 315 310 20 40 60 80 100 Time (sec) 120 140 160 180 Nonlinear closed-loop response Average temperature tracking : control effort 3.5 Lamp voltage (V) 2.5 1.5 0.5 20 40 60 80 100 Time (sec) 120 140 160 180 Nonlinear closed-loop response: control effort CuuDuongThanCong.com https://fb.com/tailieudientucntt ... all % data J1 = 5e-5; B1 = 1e-2; r1 = 2e-2; Kt = 3e-2; K = 2e4; B = 20; r2 = 2e-2; J2 = 2e-5; B2 = 2e-2; % state-variable form F = [ r1 0; -K*r1/J1 -( B1+B*r1ˆ2)/J1 K1*r1/J1 -B*r1*r2/J1; 0 r2;... Þnd the response of x1 to a step input in ia Solution: - K ( x1 - x2 ) - B( x1 - x2 ) - K ( x2 - x1 ) - B( x2 - x1 ) r1 r2 - B1M1 - B2M Tm = K t ia F (a) J1 ωú J2 ωú Tm xú xú = = = = = Tm... https://fb.com/tailieudientucntt 10 29 second clear all, close all F = [ -0 .00643 0.0263 -3 2.2 0; -0 .0941 -0 .624 820 0; -0 .000222 -0 .00153 -0 .668 0; 0 0; -1 830 ]; G = [ 0; -3 2.7; -2 .08; 0; ]; J = 0; x0 = [ 0; 0; 0; 0; ];
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