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Lecture 16: Introduction to Control (Part II) 2. Control requirements 3. Design by pole placement ME 431, Lecture 16 1. Illustrative cruise control example 1 Cruise Control Example F v = output (velocity of car) f(v) = friction/drag force Equation of Motion ME 431, Lecture 16 f(v) F = control input (road force, generated by engine indirectly) F F bv mv mv bv F 2 Cruise Control Example Transfer function first-order response msV ( s ) bV ( s ) F ( s ) F 1 ms b ME 431, Lecture 16 V ( s) 1 1/ b F ( s) ms b (m / b) s 1 DC gain 1/ b m/b 3 v • • • • Response is stable (which is good) Not robust to disturbances, uncertainties May not be fast enough How to make better? Add control! controller plant • Control law is f=Ke, applied force is proportional to the error ME 431, Lecture 16 Cruise Control Example 4 Cruise Control Example K V ( s) forward ms b K R( s) 1 loop 1 ms b K K bK m ms b K s 1 bK K DC gain bK m bK ME 431, Lecture 16 Closed-loop transfer function 5 Cruise Control Example • Making K larger makes system faster and DC gain closer to 1 • Any drawbacks? • Eventually actuator will saturate • Higher-order dynamics have been neglected (sensors, actuators), large K often will increase oscillation • Large K can amplify unwanted inputs (like noise) K V ( s) forward K ms b K N ( s) 1 loop ms b K 1 ms b • Can we do better by trying a more complicated controller? ME 431, Lecture 16 • both are generally desirable 6 • In the preceding, the control was proportional to the error … called a P (proportional) controller • Can add terms for the integral of the error (I) and the derivative of the error (D) 1 C (s) K P K I K D s s • Each term of the PID controller changes system performance in a different way ME 431, Lecture 16 Cruise Control Example 7 • Approach to control design: Translate engineering specifications into control requirements, then design a controller to meet those specifications • Example transient specifications • τ, Mp, tp, ts, tr • Example steady-state specifications • Typically it is required that steady-state error be less than some amount, ess < 0.02 • Can use F.V.T. for different types of reference inputs e lim sE ( s) ss s 0 ME 431, Lecture 16 Control Requirements 8 Pole Placement • Choose controller gains (for ex. KP, KI, and KD) so that poles of the closed-loop system meet given requirements • If system is simple (first order, or second order) can employ algebra ME 431, Lecture 16 • Poles of a system affect the time response • Process is generally iterative 9 Example • Find the closed-loop transfer function for a cruise control system with PI controller Example (continued) 𝑉(𝑠) 𝐾𝑃 𝑠 + 𝐾𝐼 = 𝑉𝑑𝑒𝑠 (𝑠) 𝑚𝑠 2 + 𝑏 + 𝐾𝑃 𝑠 + 𝐾𝐼 • With two parameters (KP and KI) can place poles of this second-order system anywhere we like (2 d.o.f.) • What is the effect of changing the control gains? Example (continued) 𝑉(𝑠) 1 𝐾𝑃 𝑠 + 𝐾𝐼 = ∙ 2 𝑉𝑑𝑒𝑠 (𝑠) 𝑚 𝑠 + 𝑠 𝑏 + 𝐾𝑃 /𝑚 + 𝐾𝐼 /𝑚 • Let m=5 and b=1. Choose KP and KI to achieve a settling time of 2 seconds and a peak time of 1 second. Example (continued) 𝑉(𝑠) 1 𝐾𝑃 𝑠 + 𝐾𝐼 = ∙ 2 𝑉𝑑𝑒𝑠 (𝑠) 5 𝑠 + 𝑠 1 + 𝐾𝑃 /5 + 𝐾𝐼 /5 • Will the closed-loop system with this controller actually have a settling time of 2 seconds and a peak time of 1 second?