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 . ______

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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

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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. (4.12) becomes (1) integral of error squared (IES) e2 (t)dt ∫0 ⎛ ⎞ ∞ K ⎜ 1 −ω ξ t ⎟ (2) integral of absolute error (IAS) e (t)dt y(t) = ⎜1− e n sin(ωt + ε)⎟ (4.13) ∫0 ω 2 ⎜ 2 ⎟ n ⎝ 1−ξ ⎠ (3) integral of time multiplied absolute error criterion (ITAE) where ∞ t e (t)dt 2 ∫0 2 1−ξ ω = ωn 1−ξ and tanε = . ξ Having chosen an appropriate performance index, the system which minimizes the integral is called optimal. The object of It is conventional to choose K / a2 = 1 and then plot graphs of modern control theory is to design a system so that it is the ‘normalised’ responsey(t ) againstωt for various values of optimal with respect to a performance index and will be discussed in the part II of this course. the damping ratioξ . There typical graphs are shown in Fig. 4.6. 4.4 Response of higher order systems

Some definitions We can write the transfer function of an nth - order system in

the form y(t) maximum overshoot steady-state error e ss K (s m + b s m −1 + + b ) G (s) = 1 K m 1.0 n n−1 (4.14) 0.9 s + a1s + K + a n

0.5 Example 4.1______rise time With reference to Fig. 2.11, calculate the close loop transfer 0.1 1 function G(s) given the transfer functions A(s) = and t t d s + 3 Fig. 4.7 B(s) = 2/ s ______

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Introduction to Control Theory Including Optimal Control Nguyen Tan Tien - 2002.3 ______

R(s) E(s) C(s) such as resistanceless circuits. We then divide the numerator until we obtain a proper fraction so that when applying a unit A(s) R(s) C(s) A(s) step input, we can write Y(s) as G(s)= 1+ A(s)B(s) B(s) K (sn + d sn−1 + + d ) Y (s) = K(c + c s + + csn−m−1) + 1 1 K n 1 2 K n n−1 Fig. 2.11 s(s + a1s +K+ an ) (4.20) We obtain where cs ,ds , K and K1 are all constants.

s s G(s) = = The inverse Laplace transform of the first right term of (4.20) s2 + 3s + 2 (s +1)(s + 2) involves the impulse function and various derivatives of it. ______The second term of (4.20) is treated as in Case 1 or Case 2 above. The response of the system having the transfer function (4.14) to a unit step input can be written in the form Example 4.2______

K(s + z1)(s + z2 )L(s + zm ) Find the response of the system having a transfer function Y(s) = (4.15) s(s + p1)(s + p2 )L(s + pn ) 5(s2 + 5s + 6) where G(s) = 3 2 z1, z2 ,L, zm : the zeros of the numerator s + 6s +10s + 8 p , p , , p : the zeros of the denominator 1 2 L n to a unit step input.

We first assume that n ≥ m in equation (4.14); we then have In this case, two cases to consider: 5(s2 + 5s + 6) 5(s + 2)(s + 3) Y (s) = = s3 + 6s2 +10s + 8 s(s + 4)[s + (1+ i)][s + (1− i)] Case 1: p1, p2 ,L, pn are all distinct numbers. The partial fraction expansion of equation (4.15) has the form The partial fraction expansion as

K K K Y(s) = 1 + 2 + + n+1 (4.16) K K K K s s + p L s + p Y (s) = 1 + 2 + 3 + 4 1 n s s + 4 s + (1+ i) s + (1− i)

K , K , , K are called the residues of the expansions. The 1 2 L n+1 15 1 −7 + i −7 − i response has the form where K = , K = − , K = , K = . 1 4 2 4 3 4 4 4

− p1t − pnt y(t) = K1 + K2e +K+ Kn+1e Hence 15 1 1 y(t) = − e−4t + e−t (−14cost + 2sin t) Case 2: p1, p2 ,L, pn are not distinct any more. Here at least 4 4 4 one of the roots, say p1 , is of multiplicity r , that is 15 1 2000 = − e−4t + e−t sin(t + 352) 4 4 4 K(s + z )(s + z ) (s + z ) ______Y(s) = 1 2 L m (4.17) r s(s + p1) (s + p2 )L(s + pn ) 4.5 Steady state error

The partial fraction expansion of equation (4.17) has the form Consider a unity feedback system as in Fig. 4.8

K1 K21 K2r Kn−r+2 Y(s) = + +L+ +L+ (4.18) s s + p (s + p )r s + p R(s) E(s) C(s) 1 1 n−r+1 A(s)

⎪⎧ K ⎪⎫ K Since −1 = t j−1e− pt , ( j = 1,2, ,r) , the L ⎨ j ⎬ L ⎩⎪(s + p) ⎭⎪ ( j −1)! Fig. 4.8 response has the form where

r(t) : reference input K − p1t − p1t 2r r−1 − p1t y(t) = K1 + K21e + K22e + + t e + c(t) : system output K (r −1)! e(t) : error − p2t − pn−r+1t K3e +L+ Kn−r+2e (4.19) We define the error function as We now consider that n < m in equation (4.14); which is the case when the system is improper; that is, it can happen when e(t) = r(t) − c(t) (4.21) we consider idealized and physically non-realisable systems, ______

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Introduction to Control Theory Including Optimal Control Nguyen Tan Tien - 2002.3 ______

hence, ess = lim e(t) . Since E(s) = R(s) − A(s)E(s) , it c(t) steady-state t→∞ error e R(s) ss follows that E(s) = and by the final value theorem 1+ A(s) sR(s) ess = lim sE(s) = lim (4.22) s→0 s→0 1+ A(s) k u(t)

We now define three error coefficients which indicate the steady state error when the system is subjected to three different standard reference inputs r(s) . t

Fig. 4.11 (1) step input, r(t) = ku(t) ( k is a constant) Example 4.3______s k / s k ess = lim = Find the (static) error coefficients for the system having a s→0 1+ A(s) 1+ lim A(s) 8 s→0 open loop transfer function A(s) = s(4s + 2) lim A(s) = K p , called the position error constant, then s→0 K p = lim A(s) = ∞ s→0 k k − e ss Kv = lim sA(s) = 4 ess = or K p = (4.23) s→0 1+ K p ess 2 Ka = lim s A(s) = 0 s→0 c(t) steady-state ______k u(t) error ess From the definition of the error coefficients, it is seen that ess k depends on the number of poles at s = 0 of the transfer function. This leads to the following classification. A transfer function is said to be of type N if it has N poles at the origin. Thus if

K(s − z ) (s − z ) A(s) = 1 L m (4.24) t j s (s − p1)L(s − pn )

Fig. 4.9 K1 K(−z1)L(−zm ) At s = 0 , A(s) = lim where K1 = (4.25) s→0 j (− p ) (− p ) (2) Ram input, r(t) = ktu(t) ( k is a constant) s 1 L n K is called the gain of the transfer function. Hence the steady k k k 1 In this case, R(s) = , so that e = or K = , where 2 ss v state error ess depends on j and r(t) as summarized in Table s Kv ess 4.1 Kv = lim sA(s) is called the velocity error constant. s→0 Table 4.1

ess c(t) steady-state j System r(t)=ku(t) r(t)=ktu(t) r(t)=½kt2u(t) error ess 0 Type 1 Finite ∞ ∞ 1 Type 2 0 finite ∞ k u(t) 2 Type 3 0 0 finite

4.6 Feedback Control

Consider a negative feedback system in Fig. 4.12 t

Fig. 4.10 R(s) E(s) C(s) A(s)

1 (3) Parabolic input, r(t) = kt 2 u(t) ( k is a constant) 2 B(s) k k k In this case, R(s) = , so that e = or K = , where Fig. 4.12 3 ss a s Ka ess 2 The close loop transfer function is related to the feed-forward Ka = lim s A(s) is called the acceleration error constant. s→0 transfer function A(s) and feedback transfer function B(s) by ______

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A(s) where q(t) is output signal from the controller G(s) = (4.26) 1+ A(s)B(s) Q1 , Q2 are some constants

We consider a simple example of a first order system for The on-off controller is obviously a nonlinear device and K it cannot be described by a transfer function. which A(s) = and B(,s) = c so that as +1 (2) Proportional Controller For this control action K K / a G(s) = = q(t) = K e(t) as + Kc +1 Kc +1 p s + a where K p is a constant, called the controller gain. The On taking Laplace inverse transform, we obtain the impulse transfer function of this controller is response of the system, where Q(s) B(s) = = K p (4.28) E(s) K (1) c = 0 (response of open loop system): g(t) = e−t / a a (3) Integral Controller K − K c+1 t K − t a t (2) c ≠ 0 : g(t) = e a = e α , where α = In this case q(t) = K e(t)dt , hence a a K c +1 ∫0 a andα are respectively the time-constants of the open loop and closed loop systems. a is always positive, but α can be B(s) = K / s (4.29) either positive or negative. (4) Derivative Controller Fig. 4.13 shows how the time responses vary with different de In this case q(t) = K , hence values of K c . dt g(t) Kc ≤ −1 Kc = −5 Kc = −3 B(s) = Ks (4.30)

(5) Proportional-Derivative Controller (PD) K Kc = −1 de In this case q(t) = K e(t) + K , hence a Kc = 0 p 1 dt ⎛ K ⎞ Kc = 4 B(s) = K ⎜1+ 1 s⎟ = K (1+ K s) (4.30) stable region p ⎜ ⎟ p ⎝ K p ⎠ t

Fig. 4.13 (6) Proportional-Integral Controller (PI) t If the impulse response does not decay to zero as t increase, In this case q(,t) = K pe(t) + K1 e(t)dt hence ∫0 the system is unstable. From the Fig. 4.13, the instability ⎛ K 1 ⎞ ⎛ K ⎞ region is defined byKc ≤ − 1. B(s) = K ⎜1+ 1 ⎟ = K ⎜1+ ⎟ (4.31) p ⎜ ⎟ p ⎝ K p s ⎠ ⎝ s ⎠ In many applications, the control system consists basically of a plant having the transfer functionA(s ) and a controller (7) Proportional-Derivative-Integral Controller (PID) having a transfer function B(s) , as in Fig. 4.14. de t In this case q(t) = K pe(t) + K1 + K2 e(t)dt , hence dt ∫0 R(s) E(s) Q(s) C(s) B(s) A(s) ⎛ K K 1 ⎞ B(s) = K ⎜1+ 1 s + 2 ⎟ = K (1+ k s + k / s) p ⎜ ⎟ p 1 2 controller plant ⎝ K p K p s ⎠

Example 4.4______Fig. 4.14

Design a controller for a plant having the transfer function With the closed loop transfer function A(s) = 1/(s + 2) so that the resulting closed loop system has a A(s)B(s) zero steady state error to a reference ramp input. G(s) = (4.27) 1+ A(s)B(s) For zero steady state error to a ramp input, the system must be of type 2. Hence if we choose an integral controller with The controllers can be of various types. B(s) = K / s then the transfer function of the closed loop

system including the plant and the controller is (1) The on-off Controller

The action of such a controller is very simple. A(s)B(s) K ⎧Q1 if e(t) > 0 = q(t) = ⎨ 1+ A(s)B(s) s3 + 2s2 + K ⎩Q2 if e(t) < 0 ______

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Introduction to Control Theory Including Optimal Control Nguyen Tan Tien - 2002.3 ______

The response of this control system depends on the roots of the denominator polynomials3 + 2s2 + K = 0 .

If we use PI controller, B(s) = K p (1+ K / s ) the system is of type 2 and response of the system depends on the roots of 3 2 s + 2s + K p s + KK p = 0 ______

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Chapter 4 Transient and Steady State Response Analysis 22