Introduction to Control Theory Including Optimal Control Nguyen Tan Tien - 2002.3 _________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ C.4 Transient and Steady State Response Analysis 4.1 Introduction Notice that when t = a , then y(t) = y(a) = 1− e−1 = 0.63 . The Many applications of control theory are to servomechanisms response is in two-parts, the transient part e−t / a , which which are systems using the feedback principle designed so approaches zero as t → ∞ and the steady-state part 1, which is that the output will follow the input. Hence there is a need for the output when t → ∞ . studying the time response of the system. If the derivative of the input are involved in the differential The time response of a system may be considered in two parts: equation of the system, that is, if a y& + y = b u& + u , then its • Transient response: this part reduces to zero as t → ∞ transfer function is • Steady-state response: response of the system as t → ∞ bs + 1 s + z 4.2 Response of the first order systems Y(s) = U (s) = K U (s) (4.4) as + 1 s + p Consider the output of a linear system in the form where K = b / a Y (s) = G(s)U (s) (4.1) z =1/ b : the zero of the system where p =1/ a : the pole of the system Y(s) : Laplace transform of the output G(s) : transfer function of the system When U (s) =1/ s , Eq. (4.4) can be written as U (s) : Laplace transform of the input K K z z − p Y(s) = 1 − 2 , where K = K and K = K s s + p 1 p 2 p Consider the first order system of the form a y& + y = u , its transfer function is Hence, − pt 1 y(t) = K1 − K 2e (4.5) Y (s) = U (s) { 14243 as +1 steady−state part transient part For a transient response analysis it is customary to use a With the assumption that z > p > 0 , this response is shown in reference unit step function u(t) for which Fig. 4.2. 1 U (s) = K1 s − pt y = K1 − K2e K2 It then follows that 1 1 1 Y (s) = = − (4.2) K − K (as +1)s s s +1/ a 1 2 K e− pt 2 On taking the inverse Laplace of equation (4.2), we obtain t Fig. 4.2 y(t) = 1 − e−t / a (t ≥ 0) (4.3) { 123 steady−state part transient part We note that the responses to the systems (Fig. 4.1 and Fig. 4.2) have the same form, except for the constant terms K1 and Both of the input and the output to the system are shown in K . It appears that the role of the numerator of the transfer Fig. 4.1. The response has an exponential form. The constant 2 a is called the time constant of the system. function is to determine these constants, that is, the size of y(t) , but its form is determined by the denominator. u(t) 1.00 4.3 Response of second order systems An example of a second order system is a spring-dashpot 0.63 arrangement shown in Fig. 4.3. Applying Newton’s law, we find M &y& = −µ y& − k y + u(t) where k is spring constant, µ is damping coefficient, y is the a t distance of the system from its position of equilibrium point, Fig. 4.1 and it is assumed that y(0) = y&(0) = 0 . ___________________________________________________________________________________________________________ Chapter 4 Transient and Steady State Response Analysis 16 Introduction to Control Theory Including Optimal Control Nguyen Tan Tien - 2002.3 _________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ y K − pt − pt k y(t) = (1 − e − p t e ) (4.10) p 2 M µ 2 u(t) Case 3: a1 < 4a2 → under-damped system Fig. 4.3 In this case, the poles p and p are complex conjugate having 1 2 Hence u(t) = M y + µ y + k y the form p = α ± i β where α = a / 2 and β = 1 4a − a 2 . && & 1,2 1 2 2 1 Hence On taking Laplace transforms, we obtain 1 K K1 K 2 K3 Y (s) = U (s) = U (s) Y (s) = + + , 2 2 s s + p1 s + p2 M s + µ s + k s + a1 s + a2 where K K(−β − iα) K(−β + iα) where K =1/ M , a1 = µ / M , a2 = k / M . Applying a unit K = , K = , K = 1 2 2 2 2 2 3 2 2 step input, we obtain α + β 2β (α + β ) 2β (α + β ) K (Notice that K and K are complex conjugates) Y (s) = (4.6) 2 3 s (s + p1)(s + p2 ) It follows that −(α +iβ )t −(α −iβ )t 2 y(t) = K1 + K 2e + K3e a1 ± a1 − 4a2 where p = , p and p are the poles of the −αt 1,2 2 1 2 = K1 + e [(K 2 + K3 ) cos β t + (K3 − K 2 )i sin β t] K (using the relation eiβ t = cos β t + i sin β t ) transfer function G(s) = , that is, the zeros of the s 2 + a s + a K K α 1 2 = + e−αt (− cos β t − sin β t (4.11) denominator of G(s). α 2 + β 2 α 2 + β 2 β There are there cases to be considered: K K α 2 = − 1 + e−αt sin(β t + ε) (4.12) α 2 + β 2 α 2 + β 2 β 2 2 Case 1: a1 > 4a2 → over-damped system where tan ε = β /α In this case p1 and p2 are both real and unequal. Eq. (4.6) can be written as Notice that whent = 0 ,0y(t) = . The there cases discussed above are plotted in Fig. 4.4. K K K Y (s) = 1 + 2 + 3 (4.7) y(t) under damped system s s + p1 s + p2 u(t) where 1 K K K K K1 = = , K 2 = , K3 = p1 p2 a2 p1( p1 − p2 ) p2 ( p2 − p1) critically damped system over damped system (notice that K1 + K 2 + K3 = 0 ). On taking Laplace transform of Eq.(4.7), we obtain t Fig. 4.4 − p1t − p2t y(t) = K1 + K 2e + K3e (4.8) From Fig. 4.4, we see that the importance of damping (note The transient part of the solution is seen to be that a1 = µ / M ,µ being the damping factor). We would − p1t − p2t expect that when the damping is 0 (that is, a = 0 ) the system K 2e + K3e . 1 should oscillate indefinitely. Indeed when a1 = 0 , then 2 Case 2: a1 = 4a2 → critically damped system α = 0 , and β = a2 In this case, the poles are equal: p1 = p2 = a1 / 2 = p , and and since sinε = 1 and cosε = 0 , then ε = π / 2 , Eq. (4.12) K K1 K 2 K3 becomes Y (s) = = + + (4.9) s (s + p)2 s s + p (s + p)2 K ⎡ ⎛ π ⎞⎤ K y(t) = ⎢1− sin⎜ a2 t + ⎟⎥ = []1− cos a2 t − pt − pt 2 a2 ⎣ ⎝ 2 ⎠⎦ a2 Hence y(t) = K1 + K 2e + K3 t e , where K1 = K / p , 2 This response of the undamped system is shown in Fig.4.5. K 2 = −K / p and K3 = −K / p so that ___________________________________________________________________________________________________________ Chapter 4 Transient and Steady State Response Analysis 17 Introduction to Control Theory Including Optimal Control Nguyen Tan Tien - 2002.3 _________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ y(t) (1) Overshoot defined as maximum overshoot ×100% final desired value (2) Time delay td , the time required for a system response to reach 50% of its final value. 2π 4π t (3) Rise time, the time required for the system response to a a rise from 10% to 90% of its final value. 0 2 2 Fig. 4.5 (4) Settling time, the time required for the eventual settling down of the system response to be within (normally) 5% There are two important constants associated with each second of its final value. order system. (5) Steady-state error ess , the difference between the steady • The undamped natural frequency ωn of the system is the state response and the input. frequency of the response shown in Fig. 4.5: ωn = a2 In fact, one can often improve one of the parameters but at the expense of the other. For example, the overshoot can be • The damping ratio ξ of the system is the ratio of the decreased at the expense of the time delay. actual damping µ (= a M ) to the value of the damping 1 In general, the quality of a system may be judged by various µc , which results in the system being critically damped standards. Since the purpose of a servomechanism is to make µ a1 the output follow the input signal, we may define expressions (that is, when a1 = 2 a2 ). Hence ξ = = . which will describe dynamics accuracy in the transient state. µ 2 a c 2 Such expression are based on the system error, which in its simplest form is the difference between the input and the We can write equation (4.12) in terms of these constants. We output and is denoted bye(t ) , that is, e(t) = y(t) − u(t) , where 2 note that a1 = 2ωnξ and a2 = ωn . Hence y(t) is the actual output and u(t) is the desired output (u(t ) is the input). a2 2 a 1 1+α 2 / β 2 = 1+ 1 = 2 = The expression called the performance index can take on 4a − a2 2 2 various forms, typically among them are: 2 1 4a2 − a1 1−ξ ∞ Eq.