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Industrial Control Systems - Special Structures
 

Industrial Control Systems - Special Structures

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    Industrial Control Systems - Special Structures Industrial Control Systems - Special Structures Presentation Transcript

    • Industrial Control Behzad Samadi Department of Electrical Engineering Amirkabir University of Technology Winter 2009 Tehran, Iran Behzad Samadi (Amirkabir University) Industrial Control 1 / 59
    • Special Control Structures Outline: Set-point Weighting [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 2 / 59
    • Special Control Structures Outline: Set-point Weighting Feedforward Action [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 2 / 59
    • Special Control Structures Outline: Set-point Weighting Feedforward Action Ratio Control [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 2 / 59
    • Special Control Structures Outline: Set-point Weighting Feedforward Action Ratio Control Cascade Control [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 2 / 59
    • Special Control Structures Outline: Set-point Weighting Feedforward Action Ratio Control Cascade Control Override Control [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 2 / 59
    • Special Control Structures Outline: Set-point Weighting Feedforward Action Ratio Control Cascade Control Override Control Selective Control [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 2 / 59
    • Special Control Structures Outline: Set-point Weighting Feedforward Action Ratio Control Cascade Control Override Control Selective Control Split Range Control [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 2 / 59
    • Special Control Structures Set-point Weighting : High load disturbance rejection → too oscillatory set-point step response. [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 3 / 59
    • Special Control Structures Set-point Weighting : High load disturbance rejection → too oscillatory set-point step response. This is actually true especially when the apparent dead time of the process is small (with respect to the dominant time constant). [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 3 / 59
    • Special Control Structures Set-point Weighting : High load disturbance rejection → too oscillatory set-point step response. This is actually true especially when the apparent dead time of the process is small (with respect to the dominant time constant). Set-point weighting for PID controllers: u(t) = Kp(βr(t) − y(t) + 1 Ti t 0 e(τ)dτ − Td dy(t) dt ) u(t): control signal, r(t): reference input, y(t): process output, e(t): tracking error [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 3 / 59
    • Special Control Structures Set-point Weighting: u(t) = Kp(βr(t) − y(t) + 1 Ti t 0 e(τ)dτ − Td dy(t) dt ) Intuitively, given a set of parameters Kp, Ti and Td , the adoption of a set-point weight β < 1 allows the reduction of the overshoot in the set-point response, since the effect of the proportional action is reduced. [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 4 / 59
    • Special Control Structures Set-point Weighting: u(t) = Kp(βr(t) − y(t) + 1 Ti t 0 e(τ)dτ − Td dy(t) dt ) Intuitively, given a set of parameters Kp, Ti and Td , the adoption of a set-point weight β < 1 allows the reduction of the overshoot in the set-point response, since the effect of the proportional action is reduced. Note that this is achieved without affecting the load disturbance rejection performance. [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 4 / 59
    • Special Control Structures Set-point Weighting: U = Kp(βR − Y + 1 Ti s E − Td sY ) Behzad Samadi (Amirkabir University) Industrial Control 5 / 59
    • Special Control Structures Set-point Weighting: U = Kp(βR − Y + 1 Ti s E − Td sY ) U = Kp(β + 1 Ti s )R − Kp(1 + 1 Ti s + Td s)Y Behzad Samadi (Amirkabir University) Industrial Control 5 / 59
    • Special Control Structures Set-point Weighting: U = Kp(βR − Y + 1 Ti s E − Td sY ) U = Kp(β + 1 Ti s )R − Kp(1 + 1 Ti s + Td s)Y U = Kp(1 + 1 Ti s + Td s) βs + 1 1 + Ti s + Ti Td s2 R − Kp(1 + 1 Ti s + Td s)Y Behzad Samadi (Amirkabir University) Industrial Control 5 / 59
    • Special Control Structures Set-point Weighting: U = Kp(βR − Y + 1 Ti s E − Td sY ) U = Kp(β + 1 Ti s )R − Kp(1 + 1 Ti s + Td s)Y U = Kp(1 + 1 Ti s + Td s) βs + 1 1 + Ti s + Ti Td s2 R − Kp(1 + 1 Ti s + Td s)Y U = Kp(1 + 1 Ti s + Td s)(FR − Y ) F = βs + 1 1 + Ti s + Ti Td s2 Behzad Samadi (Amirkabir University) Industrial Control 5 / 59
    • Special Control Structures Set-point Weighting: u(t) = Kp(βr(t) − y(t) + 1 Ti t 0 e(τ)dτ − Td dy(t) dt ) C(s) = Kp(1 + 1 Ti s + Td s) F(s) = 1 + βTi s 1 + Ti s + Ti Td s2 [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 6 / 59
    • Special Control Structures Set-point Weighting: F(s) = 1 + βTi s 1 + Ti s + Ti Td s2 From another point of view, the overshoot is reduced by smoothing the set-point signal by means of the filter F. [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 7 / 59
    • Special Control Structures Set-point Weighting: F(s) = 1 + βTi s 1 + Ti s + Ti Td s2 From another point of view, the overshoot is reduced by smoothing the set-point signal by means of the filter F. Example: P(s) = 1 10s + 1 e−4s Ziegler-Nichols: Kp = 3, Ti = 8 and Td = 2 [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 7 / 59
    • Special Control Structures Set-point Weighting: Behzad Samadi (Amirkabir University) Industrial Control 8 / 59
    • Special Control Structures Feedforward Action: Set-point tracking: [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 9 / 59
    • Special Control Structures Feedforward Action: Set-point tracking: M(s) represents the desired performance. [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 9 / 59
    • Special Control Structures Feedforward Action: Set-point tracking: M(s) represents the desired performance. G(s) = M(s) ˜P(s) where tildeP(s) is the minimum phase part of P(s). [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 9 / 59
    • Special Control Structures Feedforward Action: Set-point tracking: Y (s) R(s) = (MC + G)P 1 + PC Therefore if ˜P(s) = P(s) and C(s) = 0 then Behzad Samadi (Amirkabir University) Industrial Control 10 / 59
    • Special Control Structures Feedforward Action: Set-point tracking: Y (s) R(s) = (MC + G)P 1 + PC Therefore if ˜P(s) = P(s) and C(s) = 0 then Behzad Samadi (Amirkabir University) Industrial Control 10 / 59
    • Special Control Structures Feedforward Action: Set-point tracking: Y (s) R(s) = (MC + G)P 1 + PC Therefore if ˜P(s) = P(s) and C(s) = 0 then Y (s) R(s) = G(s)P(s) = M(s) [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 10 / 59
    • Special Control Structures Feedforward Action: Set-point tracking: Example: P(s) = 1 10s + 1 e−5s M(s) = 1 2s + 1 e−5s G(s) = 10s + 1 2s + 1 [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 11 / 59
    • Special Control Structures Feedforward Action: Disturbance rejection: Y = PCE(s) + (H − PG)D(s) [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 12 / 59
    • Special Control Structures Feedforward Action (disturbance rejection): H − PG = 0 ⇒ G = H P [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 13 / 59
    • Special Control Structures Feedforward Action (disturbance rejection): H − PG = 0 ⇒ G = H P Example: P(s) = K Ts + 1 e−Ls H(s) = KH THs + 1 e−LH s [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 13 / 59
    • Special Control Structures Feedforward Action (disturbance rejection): H − PG = 0 ⇒ G = H P Example: P(s) = K Ts + 1 e−Ls H(s) = KH THs + 1 e−LH s If LH = L: G(s) = H P = KH K Ts + 1 THs + 1 [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 13 / 59
    • Special Control Structures Feedforward Action (disturbance rejection): Example: P(s) = K Ts + 1 e−Ls H(s) = KH THs + 1 e−LH s If LH = L: G(s) = H P ≈ KH K (T − LH + L)s + 1 THs + 1 [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 14 / 59
    • Special Control Structures Feedforward Action (disturbance rejection): Example: P(s) = K (T1s + 1)(T2s + 1) H(s) = KH THs + 1 Behzad Samadi (Amirkabir University) Industrial Control 15 / 59
    • Special Control Structures Feedforward Action (disturbance rejection): Example: P(s) = K (T1s + 1)(T2s + 1) H(s) = KH THs + 1 Behzad Samadi (Amirkabir University) Industrial Control 15 / 59
    • Special Control Structures Feedforward Action (disturbance rejection): Example: P(s) = K (T1s + 1)(T2s + 1) H(s) = KH THs + 1 G(s) = H P ≈ KH K (T1 + T2)s + 1 THs + 1 [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 15 / 59
    • Special Control Structures Heat Exchanger Process: heatexdemo - Mathworks.com Behzad Samadi (Amirkabir University) Industrial Control 16 / 59
    • Special Control Structures Heat Exchanger Process: t1 = 21.8 (for 28.3% of the final value) heatexdemo - Mathworks.com Behzad Samadi (Amirkabir University) Industrial Control 17 / 59
    • Special Control Structures Heat Exchanger Process: t1 = 21.8 (for 28.3% of the final value) t2 = 36.0 (for 63.2% of the final value) heatexdemo - Mathworks.com Behzad Samadi (Amirkabir University) Industrial Control 17 / 59
    • Special Control Structures Heat Exchanger Process: t1 = 21.8 (for 28.3% of the final value) t2 = 36.0 (for 63.2% of the final value) τ = 3 2(t2 − t1) heatexdemo - Mathworks.com Behzad Samadi (Amirkabir University) Industrial Control 17 / 59
    • Special Control Structures Heat Exchanger Process: t1 = 21.8 (for 28.3% of the final value) t2 = 36.0 (for 63.2% of the final value) τ = 3 2(t2 − t1) θ = t2 − τ heatexdemo - Mathworks.com Behzad Samadi (Amirkabir University) Industrial Control 17 / 59
    • Special Control Structures Heat Exchanger Process: Gp(s) = 1 23.3s + 1 e−14.7s heatexdemo - Mathworks.com Behzad Samadi (Amirkabir University) Industrial Control 18 / 59
    • Special Control Structures Heat Exchanger Process: Gd (s) = 1 25s + 1 e−35s heatexdemo - Mathworks.com Behzad Samadi (Amirkabir University) Industrial Control 19 / 59
    • Special Control Structures Heat Exchanger Process: PI controller setting using ITAE Kc = 0.859( θ τ )−0.997 , Ti = ( θ τ )0.680 heatexdemo - Mathworks.com Behzad Samadi (Amirkabir University) Industrial Control 20 / 59
    • Special Control Structures Heat Exchanger Process: Kc = 1.23 (blue) and Kc = 0.9 (red) Ti = 24.56 heatexdemo - Mathworks.com Behzad Samadi (Amirkabir University) Industrial Control 21 / 59
    • Special Control Structures Heat Exchanger Process: F(s) = − Gd Gp = − 21.3s + 1 25s + 1 e−20.3s heatexdemo - Mathworks.com Behzad Samadi (Amirkabir University) Industrial Control 22 / 59
    • Special Control Structures Heat Exchanger Process: heatexdemo - Mathworks.com Behzad Samadi (Amirkabir University) Industrial Control 23 / 59
    • Special Control Structures Heat Exchanger Process: heatexdemo - Mathworks.com Behzad Samadi (Amirkabir University) Industrial Control 24 / 59
    • Special Control Structures Ratio Control: Keep a constant ratio between two (or more) process variables, irrespective of possible set-point changes and load disturbances. [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 25 / 59
    • Special Control Structures Ratio Control: Keep a constant ratio between two (or more) process variables, irrespective of possible set-point changes and load disturbances. Chemical dosing, water treatment, chlorination, mixing vessels and waste incinerators [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 25 / 59
    • Special Control Structures Ratio Control: Keep a constant ratio between two (or more) process variables, irrespective of possible set-point changes and load disturbances. Chemical dosing, water treatment, chlorination, mixing vessels and waste incinerators Air-to-fuel ratio for combustion systems [Chau, 2002] [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 25 / 59
    • Special Control Structures Ratio Control: Series metered control: The output y2 is necessarily delayed with respect to y1, due to the closed-loop dynamics of the second loop. [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 26 / 59
    • Special Control Structures Ratio Control: Parallel metered control: A disturbance acting on the first process can cause a large error in the ratio value. [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 27 / 59
    • Special Control Structures Ratio Control: Cross limited control scheme: The two loops are interlocked by using a low and a high selectors that force the fuel to follow the air flow when the set-point increases and that force the air to follow the fuel when the set-point decreases. [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 28 / 59
    • Special Control Structures Ratio Control: Blend Station: r2(t) = a(γr1(t) + (1 − γ)y1(t)) Its use is suggested when no disturbances are likely to occur in the processes and when the two processes exhibit a different dynamics. [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 29 / 59
    • Special Control Structures Ratio Control: Modified Blend Station: P1(s) = K1 T1s + 1 e−L1s , P2(s) = K2 T2s + 1 e−L2s Behzad Samadi (Amirkabir University) Industrial Control 30 / 59
    • Special Control Structures Ratio Control: Modified Blend Station: P1(s) = K1 T1s + 1 e−L1s , P2(s) = K2 T2s + 1 e−L2s u1(t) = Kp1 βr1(t) − y1(t) + 1 Ti1 t 0 (r1(τ) − y1(τ))dτ u2(t) = Kp2 βr2(t) − y2(t) + 1 Ti2 t 0 (r2(τ) − y2(τ))dτ [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 30 / 59
    • Special Control Structures Ratio Control: Modified Blend Station: γ = 0 if L1 > L2 and t < t0 + L1 − L2 γ + Kp(er (t) + 1 Ti t 0 er (τ)dτ) otherwise Behzad Samadi (Amirkabir University) Industrial Control 31 / 59
    • Special Control Structures Ratio Control: Modified Blend Station: γ = 0 if L1 > L2 and t < t0 + L1 − L2 γ + Kp(er (t) + 1 Ti t 0 er (τ)dτ) otherwise er (t) = ay1(t) − y2(t) Behzad Samadi (Amirkabir University) Industrial Control 31 / 59
    • Special Control Structures Ratio Control: Modified Blend Station: γ = 0 if L1 > L2 and t < t0 + L1 − L2 γ + Kp(er (t) + 1 Ti t 0 er (τ)dτ) otherwise er (t) = ay1(t) − y2(t) Kp1 Ti1 β1 Kp2 Ti2 β2 γ Kp Tp 0.9T1 K1L1 3L1 0 0.9T2 K2L2 3L2 0 Ti2 Ti1 0.5 L2 T2 T1 L1 T1 L1 [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 31 / 59
    • Special Control Structures Ratio Control: Adaptive Blend Station (Hagglund, 2001): dγ dt = S Ta (ay1 − y2) Behzad Samadi (Amirkabir University) Industrial Control 32 / 59
    • Special Control Structures Ratio Control: Adaptive Blend Station (Hagglund, 2001): dγ dt = S Ta (ay1 − y2) Ta = 10 max{T1, T2} Behzad Samadi (Amirkabir University) Industrial Control 32 / 59
    • Special Control Structures Ratio Control: Adaptive Blend Station (Hagglund, 2001): dγ dt = S Ta (ay1 − y2) Ta = 10 max{T1, T2} If r1 > max{y1, y2/a} + eps then S = 1 Else if r1 < min{y1, y2/a} − eps then S = −1 Else S = 0 [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 32 / 59
    • Special Control Structures Cascade Control: Behzad Samadi (Amirkabir University) Industrial Control 33 / 59
    • Special Control Structures Cascade Control: [Chau, 2002] Behzad Samadi (Amirkabir University) Industrial Control 33 / 59
    • Special Control Structures Cascade Control: Primary (master) loop: outer loop [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 34 / 59
    • Special Control Structures Cascade Control: Primary (master) loop: outer loop Secondary (slave) loop: inner loop [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 34 / 59
    • Special Control Structures Cascade Control: Primary (master) loop: outer loop Secondary (slave) loop: inner loop P1: slow dynamics [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 34 / 59
    • Special Control Structures Cascade Control: Primary (master) loop: outer loop Secondary (slave) loop: inner loop P1: slow dynamics P2: fast dynamics [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 34 / 59
    • Special Control Structures Cascade Control: Primary (master) loop: outer loop Secondary (slave) loop: inner loop P1: slow dynamics P2: fast dynamics The approach can be generalized to more than two loops. [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 34 / 59
    • Special Control Structures Cascade Control: It appears that the improvement in the cascade control performance is more significant when disturbances act in the inner loop and when the secondary sensor is placed in order to separate as far as possible the fast dynamics of the process from the slow dynamics (Krishnaswami et al., 1990). [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 35 / 59
    • Special Control Structures Cascade Control: It appears that the improvement in the cascade control performance is more significant when disturbances act in the inner loop and when the secondary sensor is placed in order to separate as far as possible the fast dynamics of the process from the slow dynamics (Krishnaswami et al., 1990). Additional advantage: The nonlinearities of the process in the inner loop are handled by that loop and therefore they are removed from the more important outer loop. [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 35 / 59
    • Special Control Structures Cascade Control: It appears that the improvement in the cascade control performance is more significant when disturbances act in the inner loop and when the secondary sensor is placed in order to separate as far as possible the fast dynamics of the process from the slow dynamics (Krishnaswami et al., 1990). Additional advantage: The nonlinearities of the process in the inner loop are handled by that loop and therefore they are removed from the more important outer loop. When the secondary process exhibits a significant dead time or there is an unstable (positive) zero, the use of cascade control is not useful in general. [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 35 / 59
    • Special Control Structures Cascade Control: Design procedure: First: Design the controller for the secondary loop [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 36 / 59
    • Special Control Structures Cascade Control: Design procedure: First: Design the controller for the secondary loop Second: Design the controller for the primary loop [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 36 / 59
    • Special Control Structures Cascade Control: Example: Gp = 0.8 2s + 1 , Gv = 0.5 s + 1 , GL = 0.75 s + 1 [Chau, 2002] Behzad Samadi (Amirkabir University) Industrial Control 37 / 59
    • Special Control Structures Cascade Control: Example: Let us consider a proportional controller Gc2(s) = Kc2 and make the time constant of the secondary loop equal to 0.1 second. [Chau, 2002] Behzad Samadi (Amirkabir University) Industrial Control 38 / 59
    • Special Control Structures Cascade Control: Example: Let us consider a proportional controller Gc2(s) = Kc2 and make the time constant of the secondary loop equal to 0.1 second. 1 + 0.5Kc2 = 10 ⇒ Kc2 = 18 [Chau, 2002] Behzad Samadi (Amirkabir University) Industrial Control 38 / 59
    • Special Control Structures Cascade Control: Example: Let us consider a proportional controller Gc2(s) = Kc2 and make the time constant of the secondary loop equal to 0.1 second. 1 + 0.5Kc2 = 10 ⇒ Kc2 = 18 Gv = 9 s+10 [Chau, 2002] Behzad Samadi (Amirkabir University) Industrial Control 38 / 59
    • Special Control Structures Cascade Control: Example: Let us consider a proportional controller Gc2(s) = Kc2 and make the time constant of the secondary loop equal to 0.1 second. 1 + 0.5Kc2 = 10 ⇒ Kc2 = 18 Gv = 9 s+10 Gv Gp = 0.9 0.1s+1 0.8 2s+1 [Chau, 2002] Behzad Samadi (Amirkabir University) Industrial Control 38 / 59
    • Special Control Structures Cascade Control: Example: Let us consider a proportional controller Gc2(s) = Kc2 and make the time constant of the secondary loop equal to 0.1 second. 1 + 0.5Kc2 = 10 ⇒ Kc2 = 18 Gv = 9 s+10 Gv Gp = 0.9 0.1s+1 0.8 2s+1 Gv Gp ≈ 0.72 2.05s+1 e−0.05s [Chau, 2002] Behzad Samadi (Amirkabir University) Industrial Control 38 / 59
    • Special Control Structures Cascade Control: Example: Let us consider a proportional controller Gc2(s) = Kc2 and make the time constant of the secondary loop equal to 0.1 second. 1 + 0.5Kc2 = 10 ⇒ Kc2 = 18 Gv = 9 s+10 Gv Gp = 0.9 0.1s+1 0.8 2s+1 Gv Gp ≈ 0.72 2.05s+1 e−0.05s SIMC tuning rule: Kc = 1 k τ1 τc + θ = 1 0.72 2.05 0.15 = 18.98 τi = min{τ1, 4(τc + θ)} = min{2.05, 4(0.15)} = 0.6 [Chau, 2002] Behzad Samadi (Amirkabir University) Industrial Control 38 / 59
    • Special Control Structures Cascade Control: Simultaneous tuning of the controllers: Assume: P2(s) = P2m(s)P2a(s) P2a(s) is the all-pass portion of the transfer function containing all the nonminimum phase dynamics (P2a(0) = 1) [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 39 / 59
    • Special Control Structures Cascade Control: Simultaneous tuning of the controllers: Desired inner loop transfer function: ¯Tr2y2 (s) = P2a(s) (λ2s + 1)n2 λ2 and n2 are design parameters. C2(s) = P−1 2m (s) (λ2s + 1)n2 − P2a(s) To approximate the controller with a PID controller: C2(s) = 1 s k(s) ≈ 1 s k(0) + ˙k(0)s + ¨k(0) 2 [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 40 / 59
    • Special Control Structures Cascade Control: Simultaneous tuning of the controllers: Assume: P21(s) = P1(s) P2a(s) (λ2s + 1)n2 = P12m(s)P12a(s) P12a(s) is the nonminimum phase part in all-pass form. Desired outer loop transfer function ¯Tr1y1 (s) = P12a(s) (λ1s + 1)n1 Primary controller transfer function C1(s) = P−1 12m(s)(λ2s + 1)n2 P2a(s)((λ1s + 1)n1 − P12a(s)) [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 41 / 59
    • Special Control Structures Cascade Control: Example: P1(s) = K1 T1s + 1 e−L1s P2(s) = K2 T2s + 1 e−L2s ¯Tr2y2 (s) = e−L2s λ2s + 1 ¯Tr1y1 (s) = e−(L1+L2)s λ1s + 1 [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 42 / 59
    • Special Control Structures Cascade Control: Example: Kp2 = T2 + L2 2 2(λ2+L2) K2(λ2 + L2) Ti2 = T2 + L2 2 2(λ2 + L2) Td2 = L2 2 6(λ2 + L2)  3 − L2 T2 + L2 2 2(λ2+L2)   [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 43 / 59
    • Special Control Structures Cascade Control: Example: Kp1 = T1 + λ2 + (L1+L2)2 2(λ1+L1+L2) K1(λ1 + L1 + L2) Ti1 = T1 + λ2 + (L1 + L2)2 2(λ2 + L1 + L2) Td1 = λ2T1 − (L1+L2)3 6(λ1+L1+L2) T1 + λ2 + (L1+L2)2 2(λ1+L1+L2) + (L1 + L2)2 2(λ1 + L1 + L2) The suggestion is to set: λ2 = 0.5L2 λ1 = 0.5(L1 + L2) [Visioli, 2006] Behzad Samadi (Amirkabir University) Industrial Control 44 / 59
    • Special Control Structures Override Control: There are two modes of operation Normal operation: One process variable is the controlling variable. Abnormal operation: Some other process variable becomes the controlling variable to prevent it from exceeding a process or equipment limit. The limiting controller is said to override the normal process controller. http://pse.che.ntu.edu.tw/chencl/Process_Control Behzad Samadi (Amirkabir University) Industrial Control 45 / 59
    • Special Control Structures Override Control: [Smith, 2002] Behzad Samadi (Amirkabir University) Industrial Control 46 / 59
    • Special Control Structures Override Control: The set point to LC50 is somewhat above h2, as shown in the figure. [Smith, 2002] Behzad Samadi (Amirkabir University) Industrial Control 47 / 59
    • Special Control Structures Override Control: The set point to LC50 is somewhat above h2, as shown in the figure. The FC50 is a reverse-acting controller, while the LC50 is a direct-acting controller. [Smith, 2002] Behzad Samadi (Amirkabir University) Industrial Control 47 / 59
    • Special Control Structures Override Control: The set point to LC50 is somewhat above h2, as shown in the figure. The FC50 is a reverse-acting controller, while the LC50 is a direct-acting controller. The low selector (LS50) selects the lower signal to manipulate the pump speed. [Smith, 2002] Behzad Samadi (Amirkabir University) Industrial Control 47 / 59
    • Special Control Structures Override Control: Reset Feedback (RFB) or external reset feedback: This capability allows the controller not selected to override the controller selected at the very moment it is necessary. When FC50 is being selected, its integration is working, but not that of LC50 (its integration is being forced equal to the output of LS50). When LC50 is being selected, its integration is working but not that of FC50 (its integration is being forced equal to the output of LS50). [Smith, 2002] Behzad Samadi (Amirkabir University) Industrial Control 48 / 59
    • Special Control Structures Selective Control: The high selector in this scheme selects the transmitter with the highest output, and in so doing the controlled variable is always the highest, or closest to the highest, temperature. [Smith, 2002] Behzad Samadi (Amirkabir University) Industrial Control 49 / 59
    • Special Control Structures Split range: A very common control scheme is split range control in which the output of a controller is split to two or more control valves. For example: Controller output 0% Valve A is fully open and Valve B fully closed. Controller output 25% Valve A is 75% open and Valve B 25% open. Controller output 50% Both valves are 50% open. Controller output 75% Valve A is 25% open and Valve B 75% open. Controller output 100% Valve A is fully closed and Valve B fully open. www.contek-systems.co.uk Behzad Samadi (Amirkabir University) Industrial Control 50 / 59
    • Special Control Structures Split range: X: Steam Valve, air-to-open Y: Cooling Water Valve, air-to-close TC47: Temperature Controller TY47: I/P converter [Love, 2007] Behzad Samadi (Amirkabir University) Industrial Control 51 / 59
    • Special Control Structures Split range: ZC47A,B: Valve positioners TV47A,B: Valves [Love, 2007] Behzad Samadi (Amirkabir University) Industrial Control 52 / 59
    • Special Control Structures Split range: Different arrangements are possible. For example, the split can be configured as follows: Controller output 0% Both valves are closed. Controller output 25% Valve A is 50% open and Valve B still closed. Controller output 50% Valve A is fully open and Valve B closed. Controller output 75% Valve A is fully open and Valve B 50% open. Controller output 100% Both valves are fully open. www.contek-systems.co.uk Behzad Samadi (Amirkabir University) Industrial Control 53 / 59
    • Special Control Structures Split range: Example: www.contek-systems.co.uk Behzad Samadi (Amirkabir University) Industrial Control 54 / 59
    • Special Control Structures Split range: Example: www.contek-systems.co.uk Behzad Samadi (Amirkabir University) Industrial Control 55 / 59
    • Special Control Structures Split range: In this application, the flare valve will need to open quickly in response to high pressures, but the compressor suction valve will need to move much more slowly to prevent instability in the compressors. The main problem with split range control is that the controller only has one set of tuning parameters. www.contek-systems.co.uk Behzad Samadi (Amirkabir University) Industrial Control 56 / 59
    • Special Control Structures Split range: In this application, the flare valve will need to open quickly in response to high pressures, but the compressor suction valve will need to move much more slowly to prevent instability in the compressors. The main problem with split range control is that the controller only has one set of tuning parameters. The solution is to replace the split range controller with two independent controllers, both reading the same pressure transmitter, but one controlling the flare valve and the other the suction valve. www.contek-systems.co.uk Behzad Samadi (Amirkabir University) Industrial Control 56 / 59
    • Special Control Structures Two controller implementation: www.contek-systems.co.uk Behzad Samadi (Amirkabir University) Industrial Control 57 / 59
    • Special Control Structures Marriage analogy: Split - range control is like a good marriage. One partner may be doing 90of the work, but both partners are occasionally going to share the work. [Lieberman, 2008] Behzad Samadi (Amirkabir University) Industrial Control 58 / 59
    • Special Control Structures Marriage analogy: Split - range control is like a good marriage. One partner may be doing 90of the work, but both partners are occasionally going to share the work. Override control is like a bad marriage. One partner plays a potentially dominating role, even though the other partner is doing all the work. [Lieberman, 2008] Behzad Samadi (Amirkabir University) Industrial Control 58 / 59
    • Special Control Structures Marriage analogy: Split - range control is like a good marriage. One partner may be doing 90of the work, but both partners are occasionally going to share the work. Override control is like a bad marriage. One partner plays a potentially dominating role, even though the other partner is doing all the work. Cascade control is more like my marriage. I do the best I can, but my wife Liz constantly and lovingly recalibrates my efforts. She dampens down the extremes in my behavior so as to promote a stable relationship and home life. [Lieberman, 2008] Behzad Samadi (Amirkabir University) Industrial Control 58 / 59
    • Special Control Structures Smith predictor: To handle systems with a large dead time Behzad Samadi (Amirkabir University) Industrial Control 59 / 59
    • Special Control Structures Smith predictor: To handle systems with a large dead time Behzad Samadi (Amirkabir University) Industrial Control 59 / 59
    • Special Control Structures Smith predictor: To handle systems with a large dead time [Chau, 2002] Behzad Samadi (Amirkabir University) Industrial Control 59 / 59
    • Chau, P. C. (2002). Process Control: A First Course with MATLAB (Cambridge Series in Chemical Engineering). Cambridge University Press, 1 edition. Lieberman, N. (2008). Troubleshooting Process Plant Control. Wiley. Love, J. (2007). Process Automation Handbook: A Guide to Theory and Practice. Springer, 1 edition. Smith, C. A. (2002). Automated Continuous Process Control. Wiley-Interscience, 1 edition. Visioli, A. (2006). Practical PID Control (Advances in Industrial Control). Springer, 1 edition. Behzad Samadi (Amirkabir University) Industrial Control 59 / 59