Performance of FCS2018 mk lecture notes

CONTROL SYSTEM DESIGN Lecture Notes The Performance of Feedback Control Systems Page Notes - To analyze & design a control system, we must define & measure its performance - Generally, for a given p, M1 is large and M2 is small, or vice versa - A compromise value of p should be selected to obtain M1 and M2 large (or small) at the same time - R( s) = : the Laplace transform of the unit impulse A s A R( s) = : the Laplace transform of the ramp s 2A R( s) = : the Laplace transform of the parabolic s - R( s ) = : the Laplace transform of the step - R(s) is a unit step input - 10 11 13 14 15 16 ωn2 : the standard form of a second – order transfer function s + 2ζωn s + ωn2 - y(t): the time response of the system - The steady – state of the output y(t) is - The response in the time domain of a second – order system to a unit step input - For a fixed ωn , the less the value of ζ , the more oscillatory the time response; and the time response crosses the steady – state value (1) several times - R(s) is a unit impulse input - The steady – state of the output y(t) is - For a fixed ωn , the less the value of ζ , the more oscillatory the time response; and the time response crosses the steady – state value (0) several times - The swiftness of the response is measured by the rise time Tr & the peak time Tp - The settling time is defined as the time required for the system to settle within a certain percentage delta of the input amplitude - Percent overshoot & peak time should be small, but they can not be small at the same time - When ζ varies from to 1, percent overshoot decreases and peak time increases; these two criteria can not get minimum at the same time - The less the value of ωn , the faster the time response goes to steady state - s1,2 = −ζωn ± jωn − ζ : roots of the denominator of the transfer function T(s) Page Notes - Ts = 23 25 ζωn : settling time, see page #13 - s1,2 = −1 ± j1 : a pair of conjugate roots is selected, their real part satisfies the condition of settling time (less than seconds) - T1(s) has a negative real pole - This real pole dampens the overshoot and increases the settling time M N - y (t ) = +  Ai e −σ t +  Dk e −α t sin(ωk t + θ k ) : time response to a unit step i i =1 27 A 1+ Kp : the steady – state error is inversely proportional to the position error constant - Kv: the velocity error constant - ess = 29 k =1 - The position of a pole (in the s – domain) determines properties of the time response - E(s): the tracking error - Kp: the position error constant - ess = 28 k A Kv : the steady – state error is inversely proportional to the velocity error constant - Ka: the acceleration error constant - ess = A Ka : the steady – state error is inversely proportional to the acceleration error constant 30 - Number of Integrations: the order of s in 31 - ess = : for a step input, the steady – state error is zero - ess = 33 35 A K2 K s : for a ramp input, the steady – state error is nonzero - The larger the value of K2, the less the value of error - 1 + 2lim 40 s →0 s+5 : suppose that K1 = 39 - ISE is the accumulation of the square of the error - In steady state, e(t) and e2(t) are zero, but ISE is nonzero 41 - T ( s) = : s + 2ζ s + transfer function of the system, obtained directly from the block diagram model - ISE, ITAE, and ITSE obtain their own minima at different values of ζ - For instance, if we want to use ITAE to evaluate the performance of the system, we should select ζ = 0.75 Page 42 Notes - If ITSE is applied, then ζ should be 0.60 - Y (s) = Td ( s )  P∆ k k k ∆ : + The transfer function, see Mathematical Models of Systems + This is the transfer function of the system when the input is the disturbance Td(s) - The effect of the disturbance is minimized if the output Y(s) in Y ( s) = Td ( s)  P∆ k k k ∆ is minimized N - ∆ = −  Ln + n =1 43 44 Ln Lm −  n ,m nontouching  Ln Lm Lp + : the determinant, see Mathematical n ,m , p nontouching Models of Systems - P1 = 1: the unique path connecting the input Td(s) to the ouput Y(s) - K1 = 0.5; K1K2 K p = 2.5 : for instance s - Td ( s) = : suppose that the disturbance is a step 45 51 52 54 57 dI = − K3−2 + 0.1 = : dK3 to minimize I with respect to K3, the derivative of I with respect to K3 is set to - Figure of two plots: applying the obtained K3, the steady state of the output Y(s) (the red curve) responsing to the disturbance Td(s) (the blue line) is zero, so the effect of the disturbance is minimized - The step responses of the two transfer function are nearly identical - It can be observed that the term (s + 10) has nearly no effect on the response of T1(s), hence it can be eliminated - A higher – order system GH(s) is simplified to a lower – order system GL(s) - k!: the factorial of the positive integer k - By definition, 0! = - The step responses of the two transfer function are nearly identical ... error is nonzero - The larger the value of K2, the less the value of error - 1 + 2lim 40 s →0 s+5 : suppose that K1 = 39 - ISE is the accumulation of the square of the error - In steady state, e(t)...Page Notes - Ts = 23 25 ζωn : settling time, see page #13 - s1,2 = −1 ± j1 : a pair of conjugate roots is selected, their real part satisfies the condition of settling time (less... function of the system, obtained directly from the block diagram model - ISE, ITAE, and ITSE obtain their own minima at different values of ζ - For instance, if we want to use ITAE to evaluate the performance