Lecture 24: Closed-Loop Systems, Steady-State Errors

Massachusetts Institute of Technology Department of Mechanical Engineering 2.004 Dynamics and Control II Spring Term 2008 Lecture 241 Reading: • Nise: Chapter 7 1 The Poles and Zeros of Closed-loop systems: Consider the unity feedback system shown below with a controller Gc (s) and plant Gp (s): R (s ) c o n tr o lle r + G c p la n t (s ) G - p (s ) C (s ) Combine the two cascaded blocks to form a single forward transfer function Gf (s) = Gc (s)Gp (s) R (s ) c o n tr o lle r & p la n t + G c (s ) G p (s ) C (s ) - and write Gf (s) = Nf (s) Df (s) in terms of the numerator polynomial Nf (s) and denominator polynomial Df (s). The closedloop transfer function is Gcl (s) = Gf (s) Nf (s) = 1 + Gf (s) Df (s) + Nf (s) from which we see that 1 c D.Rowell 2008 copyright � 24–1 • The closed-loop poles are the roots of the characteristic equation Nf (s) + Df (s) = 0. • The closed-loop zeros are the same as the zeros of the forward transfer function. Example 1 Find the closed-loop transfer function of the plant Gp (s) = 3/(s + 3) under P-D control where Gc = 10 + 2s. The forward transfer function is Gf (s) = Gc (s)Gp = 6(5 + s) s+3 The closed-loop transfer function is: Nf (s) 6(5 + s) 6(s + 5) Gcl (s) = = = = Df (s) + Nf (s) (s + 3) + 6(5 + s) (7s + 33) � � 6 s+5 7 s + 33/7 so that the closed -loop pole is at s = −33/7 = −4.7143 and the closed-loop zero is at s = −5 (the same as the open loop zero defined by the P-D controller). Aside: The system can be analyzed using the following MATLAB commands: forward_system = zpk(-5, -3, 6) closed_loop =feedback(forward_system,1) pole(closed_loop) %Find the closed-loop system poles zero(closed_loop) %Find the closed-loop system system zeros pzmap(closed_loop) %Make a pole-zero plot step(closed_loop) %Plot the closed-loop step response. Step Response 1 0.9 0.8 Amplitude 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.2 0.4 0.6 Time (sec) 24–2 0.8 1 1.2 1.4 The step response is shown below – note the initial transient caused by the direct feed-through. Now consider a closed-loop system with sensor dynamics H(s) R (s ) + c o n tr o lle r & p la n t G c (s ) G p (s ) C (s ) H (s ) s e n s o r The closed-loop transfer function is Gf (s) Nf (s)DH (s) Ccl (s) = = 1 + Gf (s)H(s)) Df (s)DH (s) + Nf (s)NH (s) where NH (s) and DH (s) are the numerator and denominator polynomials of the sensor transfer function H(s). In this case: • The closed-loop poles are the roots of the characteristic equation Df (s)DH (s) + Nf (s)NH (s) = 0. • The closed-loop zeros are the zeros of the forward transfer function, and the poles of the sensor transfer function. Example 2 Repeat the previous example with a sensor that has a transfer function H(s) = 10/(s + 10). The forward transfer function is Gf (s) = Gc (s)Gp = and H(s) = 6(5 + s) s+3 10 s + 10 The closed-loop transfer function is: Gcl (s) = 6(5 + s)(s + 10) 6(s2 + 15s + 50) Nf (s)DH (s) = = 2 Df (s)DH (s) + Nf (s)NH (s) (s + 10)(s + 3) + 60(5 + s) (s + 73s + 330) so that the closed-loop poles are at s = −4.84, −68.16 and the closed-loop zeros are at s = −5, −10 (the same as the open loop zero defined by the P-D controller, and the pole associated with the sensor). 24–3 2 Steady-State Errors In the lab we have considered steady-state errors for both velocity and position control of the rotary inertia, and we have noted: • There was a finite s.s. error with a constant input under velocity control. • That the s.s. error was eliminated when we used PI (proportional + integral) control. • There was no s.s. error with a constant input for position control. We will look at the steady state error for two basic inputs 1. The step input. The step response measures the ability of a feedback control system to regulate the output to a constant input. Step Response ess = 0 desired yss = 1 Unit step response 1 0.8 ess = 0.25 0.6 0.4 0.2 0 0 1 2 3 Time (sec) 4 5 6 The above figure shows the response to a unit step. One response exhibits ess = 0, while the other shows a constant steady-state error. 2. The ramp input. The steady-state ramp response error is a measure of a feedback control system’s ability to follow a simple time-varying trajectory. 24–4 10 zero e 9 ss 8 finite ess Unit ramp response 7 6 infinite ess 5 4 3 2 1 0 0 1 2 3 4 5 Time (sec) 6 7 8 9 10 The above figure shows three responses to a unit ramp input r(t) = t. In one case there is no steady-state error - as t becomes large, the response follows the input exactly. In the second case there is a finite steady state error, the response has unit slope but exhibits a constant offset from the input. The third case shows a response in which the error is growing without bound, and the steady-state error is infinite. We now look at the whole question of steady-state errors under closed-loop control, and methods to eliminate them. Consider the unity feedback system: R (s ) c o n tr o lle r & p la n t + E (s ) G (s ) C (s ) - The error signal E(s) is defined to be E(s) = R(s) − C(s), and the transfer function relating the error to the input command is found from: C(s) = G)s)E(s) E(s) = R(s) − C(s) giving E(s) 1 D(s) = = R(s) 1 + G(s) D(s) + N (s) where N (s) and D(s) are the numerator and denominator polynomials of G(s) respectively. We recall the final value theorem: lim f (t) = lim sF (s) t→∞ s→0 24–5 so that � D(s) ess = lim e(t) = lim sR(s) t→∞ s→0 D(s) + N (s) � Now consider the two cases: Step input: In this case, when r(t) = us (t), R(s) = 1/s so that � � � � 1 D(s) D(s) ess = lim s = lim s→0 s→0 s D(s) + N (s) D(s) + N (s) The condition to ensure that ess = 0 therefore must be that lim D(s) = 0 s→0 If �m (s − zi ) G(s) = K �ni=1 i=1 (s − pi ) � or D(s) = ni=1 (s − pi ), to ensure that lims→0 D(s) = 0 we require that at least one of the pi = 0 (one or more poles of the system be at the origin). This is equivalent to saying that the forward transfer function must be of the form �m (s − zi ) G(s) = K k �i=1 k≥1 n s i=k+1 (s − pi ) jw F o r z e ro s te a d y -s ta te e rro r fo r a s te p in p u t, th e r e m u s t b e a t le a s t o n e p o le a t th e o r ig in . x x o x jw x Ramp input: In this case, when the input is a unit ramp r(t) = t, R(s) = 1/s2 � � � � 1 D(s) 1 D(s) ess = lim s 2 = lim s→0 s→0 s D(s) + N (s) s D(s) + N (s) The condition to ensure that ess = 0 therefore must be that D(s) =0 s→0 s lim 24–6 If as before �m (s − zi ) G(s) = K �ni=1 i=1 (s − pi ) � or D(s) = ni=1 (s − pi ), to ensure that lims→0 D(s)/s = 0 we require that at least two of the pi = 0 (one or more poles of the system be at the origin). This is equivalent to saying that the forward transfer function must be of the form �m (s − zi ) G(s) = K k �i=1 k ≥ 2. n s i=k+1 (s − pi ) jw x F o r z e ro s te a d y -s ta te e rro r fo r a r a m p in p u t, th e r e m u s t b e a t le a s t tw o p o le s a t th e o r ig in . x o xx jw x The above argument can be extended as follows: For zero steady-state error to a waveform with a Laplace transform 1/sk , the forward transfer function must have at least k poles at the origin. 2.1 System Type Poles at the origin s = 0 are known as free integrators. The System Type is defined as the number of free integrators in the system. Type 0: Type 1: Type 2: - (s − z1 ) . . . (s − zm ) (s − p1 ) . . . (s − pn ) (s − z1 ) . . . (s − zm ) G(s) = K s(s − p2 ) . . . (s − pn ) (s − z1 ) . . . (s − zm ) G(s) = K 2 s (s − p1 ) . . . (s − pn ) G(s) = K and we can say • For a system under proportional control, to ensure ess = 0 for a step input, the system must be at least Type 1. 24–7 • For a system under proportional control, to ensure ess = 0 for a ramp input, the system must be at least Type 2. and we can make the following table showing the steady-state error conditions: step ramp parabola Type 0 Type 1 ess = constant ess = 0 ess = ∞ ess = constant ess = ∞ ess = ∞ Type 2 ess = 0 ess = 0 ess = constant Type ess = ess = ess = 3 0 0 0 Example 3 In the lab you (should have) observed that with proportional control (1) that the velocity control gave a finite steady state error for a constant input, whereas (2) the position control had zero steady-state error. For velocity control: Ω(s) Kp = Ωd Js + B which is a Type 0 system, which will have a finite steady-state error. For position control: θ(s) Kp G(s) = = θd (s) s(Js + B) G(s) = which is a Type 1 system, and from the above argument will have a zero steadystate error. Example 4 Show why PI control reduces the steady-state error to zero for a step input with a Type 0 system. R (s ) + P I c o n tr o lle r K p + K i s 1 G p la n t - 24–8 p (s ) C (s ) For a PI controller, the transfer function is 1 Kp s + Ki = s s so that the controller introduces (1) a pole at the origin, and(2) a zero at s = −Ki /Kp so that the forward transfer function is Gc (s) = Kp + Ki (s + Ki /Kp ) Gp (s) s which is Type 1, and will have zero steady-state error for a constant input. Gf (s) = Gc (s)Gp (s) = Kp 2.2 Static Error Constants Recall that the transfer function relating the error to the input is 1 E(s) . 1 + G(s) For a step input � � 1 1 1 ess = lim s = s→0 s 1 + G(s) 1 + lims→0 G(s) If we define a static position constant Kp (not to be confused with a controller gain) as Kp = lim G(s) then ess = s→0 1 1 + Kp Similarly, for a ramp input (constant velocity) � � 1 1 1 ess = lim s 2 = s→0 s 1 + G(s) lims→0 sG(s) and, if we define a static velocity constant Kv as Kv = lim sG(s) then ess = s→0 1 . Kv We can also define an acceleration constant Ka for parabolic inputs, since � � 1 1 1 ess = lim s 3 = s→0 s 1 + G(s) lims→0 s2 G(s) so that if Ka = lim s2 G(s) then ess = s→0 Input step ramp parabola 1 . Ka Error Type 0 Type 1 1/(1 + Kp ) ess = 1/(1 + Kp ) Kp = ∞, ess = 0 1/Kv Kv = 0, ess = ∞ ess = 1/Kv Kv = 0, ess = ∞ Kv = 0, ess = ∞ 1/Ka 24–9 Type 2 Kp = ∞, ess = 0 Kv = ∞, ess = 0 1/Ka