BASIC PROPERTIES OF FEEDBACK

ECE450/550: Feedback Control Systems. 4 BASIC PROPERTIES OF FEEDBACK 4.: Setting up an example to benchmark controllers There are two basic types/categories of control systems: OPEN LOOP: Disturbance r(t) Ctrlr Plant y(t) CLOSED LOOP: Disturbance r(t) Ctrlr Plant y(t) Sensor This chapter of notes is concerned with comparing open-loop and closed-loop control, and showing the potential benefits (and some pitfalls) of closed-loop (i.e., feedback)control. We evaluate these two categories of controller in a number of ways: disturbance rejection, sensitivity, dynamic tracking, steady-state error, and stability. DC motor speed control In order to compare open- and closed-loop control, we will use an extended example.

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 2 Recall equations of motion for a dc motor (pg. 2 9) but add a load torque. e a (t) i a (t) R a L a b e b (t) } {{ } Armature θ(t), τ(t) J }{{} Load Assume that we are trying to control motor speed: J θ + b θ = di a k e θ + L a + R a i a = dt J ẏ + by = di a k e y + L a + R a i a = dt k τ i a + τ l e a k τ i a + w e a let output y = θ, disturbance w = τ l. Solving the mechanical equation for Ia (s) gives sjy(s) + by(s) = k τ I a (s) + W (s) τ l,load sjy(s) + by(s) = k τ I a (s) + W (s) k e Y (s) + sl a I a (s) + R a I a (s) = E a (s) k τ I a (s) = sjy(s) + by(s) W (s) I a (s) = (sj + b) Y (s) W (s) k τ. Substituting into the electrical equation gives k e Y (s) + (sl a + R a ) Some algebra then yields k e Y (s) + sl a I a (s) + R a I a (s) = E a (s) (sj + b) Y (s) W (s) k τ = E a (s).

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 3 (sj + b) Y (s) W (s) k e Y (s) + (sl a + R a ) = E a (s) k τ k τ k e Y (s) + (sl a + R a )(sj + b) Y (s) = k τ E a (s) + (R a + L a s) W (s) ( JLa s 2 + (bl a + JR a ) s + (br a + k τ k e ) ) Y (s) = k τ E a (s) + (R a + L a s) W (s). Dividing both sides by bra + k τ k e gives ( JLa s 2 + bl ) a+jr a k τ s + Y (s) = E a (s) + R a+l a s W (s). br a +k τ k e br a +k τ k e br a +k τ k e br a +k τ k e The left-hand-side can be factored into two parts: ( JL a s 2 + bl ) a + JR a s + = (τ s + )(τ 2 s + ). br a + k τ k e br a + k τ k e Roughly, one of these time constants is mechanical; the other is electrical. If we assume that the mechanical time constant is much larger than the electrical, the right-hand-side can be approximated by where k τ E a (s) + R a + L a s W (s) AE a (s) + BW(s), br a + k τ k e br a + k τ k e Then, we have overall relationship A = k τ /(br a + k τ k e ) B R a /(br a + k τ k e ). (τ s + )(τ 2 s + )Y (s) = AE a (s) + BW(s). So, Y (s) = A (τ s + )(τ 2 s + ) E a(s) + B (τ s + )(τ 2 s + ) W (s).

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 4 4.2: Advantage of feedback: Disturbance rejection We look at how the open-loop and feedback systems respond to a step-like disturbance. If ea (t) = e a (t) (constant) and w(t) = w (t) (constant), What is steady state output? Recall Laplace-transform final value theorem: If a signal has a constant final value, it may be found as y ss = lim sy(s). Note: A signal will have a constant final value iff all of the poles of Y (s) are strictly in the left-half s-plane, except possibly for a single pole at s = 0. For the input signals ea (t) and w(t), wehave So, ( y ss = lim s = Ae a + Bw. E a (s) = e a s, W (s) = w s. A (τ s + )(τ 2 s + ) e a s + B (τ s + )(τ 2 s + ) ) w s This is the response of the open-loop system (without a controller). Let s make a simple controller for the open-loop system. The block diagram looks like: Dist. w(t) B A Motor r(t) Ctrlr e a (t) A (τ s + )(τ 2 s + ) y(t)

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 5 We will design the controller to be a gain of Kol such that e a (t) = K ol r(t), Choose Kol so that there is no steady-state error when w = 0. y ss = AK ol r ss + Bw ss K ol = /A. Is closed-loop any better? The block diagram looks like: Dist. w(t) B A r(t) Ctrlr e a (t) A (τ s + )(τ 2 s + ) y(t) Tachometer Let s make a similar controller for the closed-loop system (with the possibility of a different value of K.) e a (t) = K cl (r(t) y(t)). The transfer function for the closed-loop system is: Y (s) = = AK cl (τ s + )(τ 2 s + ) (R(s) Y (s)) + B (τ s + )(τ 2 s + ) W (s) AK cl (τ s + )(τ 2 s + ) R(s) AK cl (τ s + )(τ 2 s + ) Y (s) + B (τ s + )(τ 2 s + ) W (s). Combining Y (s) terms

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 6 ( ) AK cl Y (s) + = (τ s+)(τ 2 s+) ( ) (τ s+)(τ 2 s+)+ak cl Y (s) = (τ s+)(τ 2 s+) This gives Y (s) = AK cl (τ s + )(τ 2 s + ) + AK cl R(s) + AK cl (τ s+)(τ 2 s+) R(s)+ B (τ s+)(τ 2 s+) W (s) AK cl (τ s+)(τ 2 s+) R(s)+ B (τ s+)(τ 2 s+) W (s). Employing the final-value theorem for w = 0 gives B (τ s + )(τ 2 s + ) + AK cl W (s). If AKcl, y ss r ss. Open-loop with load: y ss = = AK cl r ss + AK cl + r ss. AK cl Closed-loop with load: y ss = AK ol r ss + Bw ss = r ss + Bw ss δy = Bw ss. y ss = δy AK cl + AK cl r ss + B + AK cl w ss. B + AK cl w ss which is much better than open-loop since AK cl. ADVANTAGE OF FEEDBACK: Betterdisturbancerejection(byfactorof + AK cl ).

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 7 4.3: Advantage of feedback: Sensitivity and dynamic tracking The steady-state gain of the open-loop system is:.0 How does this change if the motor constant A changes? A A + δ A G ol + δg ol = K ol (A + δ A) δg ol In relative terms: = δ A G ol A = }{{}.0 δ A A. sensitivity = (A + δ A) A = + δ A. }{{} A gain error Therefore, a 0 % change in A results in a 0 % change in gain. Sensitivity=.0. Steady-state gain of closed-loop system is: AK cl + AK cl. G cl + δg cl = From calculus (law of total differential) (A + δ A)K cl + (A + δ A)K cl. or δg cl G cl = δg cl = dg cl da δ A ( ) A dg cl G cl da }{{} sensitivity S G cl A δ A A.

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 8 To calculate this, we first compute Then, dg cl da = d da ( AKcl + AK cl ) = ( + AK cl)k cl K cl (AK cl ) ( + AK cl ) 2 = δg cl G cl = S G cl A = = K cl ( + AK cl ) 2. ( ) A dg cl δ A G cl da A K cl A AK cl /( + AK cl ) ( + AK cl ) 2 + AK cl. ADVANTAGE OF FEEDBACK: Lower sensitivity to modeling error (by a factor of + AK cl ) Dynamic Tracking Steady-state response of closed-loop better than open-loop: Better disturbance rejection, better (lower) sensitivity. What about transient response? Open-loop system: Poles at roots of (τ s + )(τ 2 s + ) s = /τ, s = /τ 2. Closed-loop system: Poles at roots of (τ s + )(τ 2 s + ) + AK cl. s = (τ + τ 2 ) ± (τ + τ 2 ) 2 4τ τ 2 ( + AK cl ) 2τ τ 2.

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 9 FEEDBACK MOVES POLES System may have faster/slower response System may be more/less damped System may become unstable!!! Often a high gain K cl results in instability. For this dc motor example, we can get step responses of the following form:.5 High K cl y(t) 0.5 Low K cl 0 0 2 4 6 8 0 Time (sec) So, we see the first potential downside of feedback if the controller is not well designed, it may make the plant s response worse than it was to begin with. Designing controllers is a main focus of the rest of this course (and of follow-on courses). It s not a trivial task. But, we can get a really good start toward improving the dynamic response of the closed-loop system with a very simple controller We look next at the PID controller, then return to exploring (potential) advantages of feedback.

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 0 4.4: Proportional-integral-derivative (PID) control (a) General control setup: Disturbance R(s) D(s) U(s) G(s) Y (s) Need to design controller D(s). One option is PID (Proportional Integral Derivative) control design. Extremely popular. 90 + % of all fielded controllers are PID. Doesn t mean that they are great, just popular. We just saw proportional control where u(t) = K p e(t), ord(s) = K p. Proportional control tends to increase speed of response, but: Can allow non-zero steady-state error. Can result in larger transient overshoot. May not eliminate a constant disturbance. Integral control, where D(s) = K i s,caneliminatesteady-stateerror, But, transient response can get worse, and Stability margins can get worse. Derivative control, where D(s) = Kd s,canreduceoscillationsin dynamic response, but Steady-state error can get worse. In the next sections, we look at each of these controllers separately, then consider how to use them together.

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 Proportional control Proportional controllers compute the control effort such that u(t) = K p (r(t) y(t)) = K p e(t)... D(s) = K p. IDEA: For plants with positive gain, if e(t) = r(t) y(t) >0, theni mnot trying hard enough. Multiply error by (positive) K p to try harder. Also, if e(t) <0, theni vetriedtoohardalready.multiply(negative) error by (positive) gain K p to try to pull response back. EXAMPLE: Determine behavior of closed-loop poles for the dc motor. Y (s) R(s) = AK p (τ s + )(τ 2 s + ) + AK p. Poles are roots of (τ s + )(τ 2 s + ) + AK p. Without feedback, K p 0. s = /τ, s 2 = /τ 2. With feedback, Y (s) R(s) = = AK p (τ s + )(τ 2 s + ) + AK p AK p τ τ 2 s 2 + (τ + τ 2 ) s + ( + AK p ). Solving for root locations gives s, s 2 = (τ + τ 2 ) ± (τ + τ 2 ) 2 4τ τ 2 ( + AK p ) 2τ τ 2. We can plot the locations of the poles (a root locus plot) parametrically as K p changes

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 2 K p = (τ τ 2 ) 2 4τ τ 2 A I(s) K p = 0 τ 2 K p = 0 τ R(s) τ + τ 2 2τ τ 2 For 0 < K p < (τ τ 2 ) 2 4τ τ 2 A,polesmovehorizontallytowardeachother along the real axis. Rise time gets faster since dominant pole moves farther from origin, natural frequency increases. Settling time gets faster since real part of dominant pole moves farther from origin. Damping remains same (no overshoot). For K p > (τ τ 2 ) 2 4τ τ 2 A,thepolesgainimaginarypart. Settling time remains same since real part of pole locations is unchanged. Rise time decreases since natural frequency increases. Overshoot increases since damping ratio decreases. For systems having more poles than this example, increasing K p often leads to instability. How do we improve accuracy, but keep stability?

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 3 4.5: Proportional-integral-derivative (PID) control (b) Integral and proportional-integral control Pure integral controllers compute the control effort such that: u(t) = K p T i t 0 e(τ) dτ, D(s) = K p T i s. T i = Integral time = time for output = K p with input e(t) = (t). An alternate formulation has t u(t) = K i e(τ) dτ, 0 D(s) = K i s. Integral feedback can give nonzero control even at points of time when e = 0 because of memory. In many cases this can eliminate steady-state error to step-like reference inputs and step-like disturbances. IDEA: To avoid instability or oscillations with proportional control, the proportional gain K p must be kept small. But, then when error gets small, we no longer try very hard to correct it leads to finite steady-state error. Also, some nonlinearities (e.g., coulombicfriction)cancauseoutput to get stuck even if control effort is nonzero. Integral control can help: If we integrate the error signal, the integrated value will grow over time if the error is stuck. This increases the control signal u(t) until the error starts decreasing making the error converge to zero. EXAMPLE: Substitute: u(t) = K p T i t 0 (r(τ) y(τ)) dτ into dc-motor eqs.

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 4 t ] [ K p τ τ 2 ÿ(t) + (τ + τ 2 )ẏ(t) + y(t) = A Differentiate, T i 0 (r(τ) y(τ)) dτ + Bw(t).... τ τ 2 y(t) + (τ + τ 2 )ÿ(t) +ẏ(t) = AK p (r(t) y(t)) + Bẇ(t) τ τ 2... y(t) + (τ + τ 2 )ÿ(t) +ẏ(t) + AK p T i If r(t) = cst, w(t) = cst, ẇ(t) = 0, T i y(t) = AK p r(t) + Bẇ(t). T i AK p T i y ss = AK p T i r ss no error. Steady-state tracking improves, but dynamic response degrades, especially after poles leave real axis. Very oscillatory; possibly unstable. K i =0 K i =0 I(s) R(s) Can be improved by adding proportional term to integral term. u(t) = K p e(t) + K p T I Poles are at the roots of t 0 τ 2 τ ( e(τ) dτ, D(s) = K p + ). T i s τ τ 2 s 3 + (τ + τ 2 )s 2 + ( + AK p )s + AK p T i = 0. Two degrees of freedom. Derivative and proportional-derivative control Pure derivative controllers compute the control effort such that: where T d = derivativetime. u(t) = K p T d ė(t), D(s) = K p T d s,

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 5 An alternate formulation has u(t) = K d ė(t), D(s) = K d s. IDEA: Would like to anticipate momentum, which mechanically is proportional to velocity, and subtract out its predicted contribution. Contribution to control effort acts as braking force when approaching reference value quickly. WARNING: PURE DERIVATIVE CONTROL IMPRACTICAL SINCE DERIVATIVE MAGNIFIES SENSOR NOISE! Practical version = lead control, which we will study later. Derivative control tends to stabilize a system. Does nothing to reduce constant error! If ė(t) = 0, thenu(t) = 0, even if e(t) = 0. Motor control: Poles at roots of τ τ 2 s 2 + (τ + τ 2 + AK p T d )s + = 0. T d enters ζ term, can make damping better. PD = Proportional plus derivative control where D(s) = K p ( + T d s) or D(s) = K p + K d s. Root locus for dc motor, PD control. I(s) R(s) τ 2 T d τ Great damping, possibly poor steady-state error.

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 6 4.6: Proportional-integral-derivative (PID) control (c) Proportional Integral Derivative Control ( D(s) = K p + ) T i s + T ds or D(s) = K p + K i s + K ds. Need ways to design parameters {K p, T i, T d } or {K p, K i, K d }. In general (i.e., not always), K p, T i T d error, stability stability For speed control problem, u(t) = K p [ (r(t) y(t)) + T i t (math happens). Solve for poles 0 ] (r(τ) y(τ)) dτ + T d (ṙ(t) ẏ(t)). τ τ 2 T i s 3 + T i ((τ + τ 2 ) + AK p T d )s 2 + T i ( + AK p )s + AK p = 0 [ ] [ ] s 3 τ + τ 2 + AK p T d + + s 2 AKp + s + AK p = 0. τ τ 2 τ τ 2 τ τ 2 T i Three coefficients, three parameters. We can put poles anywhere! Complete control of dynamics in this case. Entire transfer functions are: Y (s) W (s) = T i Bs T i τ τ 2 s 3 + T i (τ + τ 2 )s 2 + T i ( + AK p )s + AK p. Y (s) R(s) = AK p (T i s + ) T i τ τ 2 s 3 + T i (τ + τ 2 )s 2 + T i ( + AK p )s + AK p. We can plot responses in MATLAB:

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 7 y(t) (rad/sec) num = [TI*B 0]; den = [TI*TAU*TAU2 TI*(TAU+TAU2) TI*(+A*KP) A*KP]; step(num, den).6.4.2 0.8 0.6 0.4 Step Reference Response PI P PID y(t) (rad/sec).6.4.2 0.8 0.6 0.4 Step Disturbance Response P PI PID 0.2 0.2 0 0 0.0 0.02 0.03 0.04 0.05 Time (sec) Ziegler Nichols tuning of PID controllers Rules-of-thumb for selecting K p, T i, T d. 0 0 0.0 0.02 0.03 0.04 0.05 Time (sec) Not optimal in any sense but often provide good performance. METHOD I: If system has step response like this, A Slope, A/τ Y (s) U(s) = Ae τds τs +, (first-order system plus delay) τ d τ We can easily identify A, τd, τ from this step response. Don t need complex model! Tuning criteria: Ripple in impulse response decays to 25 % of its value in one period of ripple

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 8 0.25 Period RESULTING TUNING RULES: P PI PID K p = τ 0.9τ K K p = p =.2τ Aτ Aτ d d Aτ d T i = τ d T i = 2τ d 0.3 T d = 0.5τ d METHOD II: Configure system as r(t) K u Plant y(t) Turn up gain Ku until system produces oscillations (on stability boundary) K u = ultimategain. Period, P u RESULTING TUNING RULES: P PI PID K p = 0.5K u K p = 0.45K u T i =.2 P u K p T i T d = 0.6K u = 0.5P u = P u 8 Practical Problem: Integrator Overload Integrator in PI or PID control can cause problems. For example, suppose there is saturation in the actuator. Error will not decrease. Integrator will integrate a constant error and its value will blow up.

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 9 Solution = integrator anti-windup. Turn off integration when actuator saturates. K e(t) u(t) u min K T I s u max K a y(t) Doing this is NECESSARY in any practical implementation. Omission leads to bad response, instability..6.4.2 0.8 0.6 0.4 0.2 Step Response Without antiwindup With antiwindup u(t) 0.8 0.6 0.4 0.2 0 0.2 0.4 Control Effort Without antiwindup With antiwindup 0 0 2 4 6 8 0 Time (sec) 0.6 0 2 4 6 8 0 Time (sec)

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 20 4.7: Steady-state error (a) System error is any difference between r(t) and y(t). Twosources:. Imprecise tracking of r(t). 2. Disturbance w(t) affecting the system output. Steady-state error (w.r.t. reference input) We have already seen examples of CL systems that have some tracking error (proportional ctrl) or not (integral ctrl) to a step input. We will formalize this concept here. Start with very general control structure with closed-loop transfer function T (s) that computes Y (s) from R(s): r(t) T (s) y(t) We don t care what is inside the box T (s). Itcouldbeanyblock diagram, and we may need to compute T (s) from the block diagram. The error is E(s) = R(s) Y (s) = R(s) T (s)r(s) = [ T (s)] R(s). Assume conditions of final value theorem are satisfied (i.e., [ T (s)]r(s) has poles only in LHP except perhaps for a single pole at the origin) e ss = lim t e(t) = lim s[ T (s)]r(s).

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 2 When considering steady-state error to different reference inputs, we restrict ourselves to inputs of the type r(t) = t k k! (t); R(s) = s k+. The first few of this family are drawn below: (t) t (t) t 2 2 (t) t As k increases, reference tracking is progressively harder. (It is easier to track a constant reference than it is to track a moving reference.) We define a concept called system type to describe the ability of the closed-loop system to track inputs of different kinds. If system type = 0, constant steady-state error for step input, infinite s.s. error for ramp or parabolic input. If system type =, no steady-state error for step input, constant s.s. error for ramp input, infinite s.s. error for parabolic input. If system type = 2, no steady-state error for step or ramp inputs, constant s.s. error for parabolic inputs. And so forth, for higher-order system types. To find the system type in general, we must compute the following equation for values of k = 0,,...until we calculate a finite nonzero value for e ss : e ss = lim T (s) s k = t 0, type > k; constant, type = k;, type < k. t

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 22 As an example, consider ramp responses of different system types: t Type 0 System t Type System t Type 2 System t DANGER: Higher order sounds better but they are harder to stabilize and design. Transient response may be poor. Summary table for computing steady-state error for different system types (where the limits evaluate to finite nonzero values) EXAMPLES: Steady-state tracking errors e ss for generic closed-loop T (s) Sys. Type Step Input Ramp Input Parabola Input Type 0 lim ( T (s)) T (s) Type 0 lim s T (s) Type 2 0 0 lim s 2 () Consider T (s) = s + (s + 2)(s + 3).Whatisthesystemtype? Try k = 0. Evaluate (s + 2)(s + 3) (s + ) lim [ T (s)] = lim = 5 = 0. (s + 2)(s + 3) 6 Therefore, the system type is zero, ess = 5/6 to unit step input. (2) Consider T (s) = s + 6 (s + 2)(s + 3).Whatisthesystemtype? t t

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 23 Try k = 0. Evaluate (s + 2)(s + 3) (s + 6) lim [ T (s)] = lim = 0 (s + 2)(s + 3) 6 = 0. Therefore the system type must be greater than zero. Try k =. Evaluate T (s) lim = lim s s (s + 2)(s + 3) (s + 6) (s + 2)(s + 3) s 2 + 4s = lim s (s + 2)(s + 3) = 4 = 0. 6 Therefore, the system is type, ess = 2/3 to unit ramp input. 5s + 6 (3) Consider T (s) = (s + 2)(s + 3).Whatisthesystemtype? Try k = 0. Evaluate (s + 2)(s + 3) (5s + 6) lim [ T (s)] = lim = 0 (s + 2)(s + 3) 6 = 0. Therefore, the system type must be greater than zero. Try k =. Evaluate T (s) lim = lim s s (s + 2)(s + 3) (5s + 6) (s + 2)(s + 3) = lim s (s + 2)(s + 3) = 0 6 = 0. Therefore, the system type must be greater than one. Try k = 2. Evaluate T (s) lim s 2 s 2 (s + 2)(s + 3) (5s + 6) = lim s 2 (s + 2)(s + 3) s 2 = lim s 2 (s + 2)(s + 3) = = 0. 6 Therefore, the system type is two, e ss = /6 to unit parabola input.

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 24 4.8: Steady-state error w.r.t. reference input, unity feedback WARNING: The following method is a special case of the above general method. Always use the appropriate method for the problem at hand! Unity-feedback is when the control system looks like: r(t) G ol (s) y(t) That is, the feedback loop has a gain of exactly one. If we are fortunate enough to be considering a unity-feedback scenario, the prior rules have a simpler solution. But, if there are any dynamics in the feedback loop, we DO NOT have aunity-feedbacksystem,andmustusethemoregeneralrulesfrom Section 4.7. For unity-feedback systems, there are some important simplifications: T (s) = G ol(s) + G ol (s) [ ] + Gol (s) T (s) = G ol(s) + G ol (s) + G ol (s) So, = E(s) = + G ol (s). For test inputs of the type R(s) = s k+ + G ol (s) R(s). e ss = lim se(s) = lim [ + G ol (s)]s k.

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 25 For a system that is type 0, e ss = lim + G ol (s) = + K p, K p = lim G ol (s). For a system that is type, e ss = lim + G ol (s) s = lim sg ol (s) =, K v For a system that is type 2, e ss = lim + G ol (s) s = lim 2 s 2 G ol (s) =, K a K v = lim sg ol (s). K a = lim s 2 G ol (s). These formulas meaningful only for unity-feedback! K p = lim G ol (s). K v = lim sg ol (s). K a = lim s 2 G ol (s). position error constant velocity error constant acceleration error constant Steady-state tracking errors e ss for unity-feedback case only. Sys. Type Step Input Ramp Input Parabola Input Type 0 + K p Type 0 K v Type 2 0 0 K a EXAMPLES: () Consider G ol (s) = s + (s + 2)(s + 3).Whatisthesystemtype? G ol (0) = 2 3 = 6.

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 26 Therefore, system type= 0, ess to unit step= + /6 = 6 7. (s + )(s + 0)(s 5) (2) Consider G ol (s) = (s 2 + 3s)(s 4 + s 2 + ).Whatisthesystemtype? 0 ( 5) G ol (0) = = Type > 0. 0 (s + )(s + 0)(s 5) sg ol (s) = (s + 3)(s 4 + s 2 + ) sg ol (s) s=0 = 0 ( 5) 3 = 50 3. Therefore, system type=, ess to unit ramp= 3 50. (3) Consider G ol (s) = s2 + 2s + s 4 + 3s 3 + 2s 2.Whatisthesystemtype? G ol (0) = 0 = Type > 0 sg ol (s) = s2 + 2s + s 3 + 3s 2 + 2s sg ol (s) s=0 = 0 = Type > s 2 G ol (s) = s2 + 2s + s 2 + 3s + 2 s 2 G ol (s) s=0 = 2 Type = 2. Therefore, system type= 2, ess to unit parabola= 2. KEY POINT: Open-loop G ol (s) tells us about closed-loop s.s. response. KEY POINT: For unity-feedback systems, number of poles of G ol (s) at s = 0 is equal to the system type.

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 27 EXAMPLE: DC-motor example with proportional control, D(s) = K p. r(t) Ctrlr A (τ s + )(τ 2 s + ) y(t) D(s)G(s) = K p A (τ s + )(τ 2 s + ), lim D(s)G(s) = K p A. So system is type 0, with s.s. error to step input of + K p A. This agrees with prior results. [ EXAMPLE: DC-motor example with PI control, D(s) = K p + ]. T i s D(s)G(s) = lim D(s)G(s) = lim sd(s)g(s) = K p A. T i K p A + K p A T i s (τ s + )(τ 2 s + ). System is type, with s.s. error to ramp input of T i K p A. EXAMPLE: DC-motor with two-integrator controller, [ D(s) = K p + T i s + ]. T i s 2 D(s)G(s) = K p A + K p A T i s lim s2 D(s)G(s) = K p A. T i + K p A T i s 2 (τ s + )(τ 2 s + ) System is type 2, with s.s. error to parabolic input of T i K p A.

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 28 4.9: Steady-state error w.r.t. disturbance Recall, system error is any difference between r(t) and y(t). We have just spent considerable time considering differences between r(t) and y(t) because the system is not capable of tracking r(t) perfectly. This is an issue with T (s) for the general case, or G ol (s) for the unity-feedback case. But, another source of steady-state error can be uncontrolled inputs to the system disturbances. EXAMPLE: Consider a vehicle cruise-control system. We may set the reference speed r(t) = 55 mph, but we find that the steady-state vehicle speed y ss is affected by wind and road grade (in addition to the cruise-control system s ability to track the reference input.) We can think of a system s overall response to both the reference input and the disturbance input as Y (s) = T (s)r(s) + T w (s)w (s). We find the system type with regard to the reference input by examining T (s); similarly,we find the system type with respect to the disturbance input by examining T w (s). Due to linearity, we can consider these two problems separately. When thinking about system type with respect to reference input, we consider W (s) = 0 and follow the procedures outlined earlier. When thinking about system type with respect to disturbance input, we consider R(s) = 0 and follow the procedure below.

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 29 We do not wish the output to have ANY disturbance term in it, so the output error due to the disturbance is equal to whatever output is caused by the disturbance. e ss = r ss y ss = 0 lim st w (s)w (s). We say that the system is type 0 with respect to disturbance if it has nonzero steady-state error when the disturbance is /s. Type 0 system (w.r.t. disturbance) has constant e ss = lim T w (s). We say that a system is type with respect to disturbance if it has nonzero steady-state error when the disturbance is /s 2. T w (s) Type system (w.r.t. disturbance) has constant e ss = lim. s We say that a system is type 2 with respect to disturbance if it has nonzero steady-state error when the disturbance is /s 3. Type 2 system (w.r.t. disturbance) has constant e ss = lim T w (s) s 2. Summary table for computing steady-state error for different system types with respect to disturbance (where the limits evaluate to finite nonzero values) Steady-state errors e ss due to disturbance for generic closed-loop T w (s) Sys. Type Step Input Ramp Input Parabola Input Type 0 lim T w (s) T w (s) Type 0 lim s T w (s) Type 2 0 0 lim s 2

ECE450/ECE550, BASIC PROPERTIES OF FEEDBACK 4 30 NOTE: There are no special cases for unity feedback when computing system type with respect to disturbance. You must always compute the closed-loop transfer function T w (s) = Y (s)/w (s), andthenperformthetests. EXAMPLE: Consider the motor-control problem from before. Dist. w(t) B A r(t) Ctrlr e a (t) A (τ s + )(τ 2 s + ) y(t) Tachometer We ll use a proportional controller with gain Kcl,so e a (t) = K cl (r(t) y(t)). When we looked at the example earlier, we found that AK cl B Y (s) = R(s) + W (s). (τ s + )(τ 2 s + ) + AK cl (τ s + )(τ 2 s + ) + AK cl From this equation, we gather AK cl T (s) = (τ s + )(τ 2 s + ) + AK cl B T w (s) =. (τ s + )(τ 2 s + ) + AK cl Starting with system type 0, test for finite nonzero steady-state error: e ss = lim T w (s) = B = 0. + AK cl Therefore, the system is type 0 with respect to disturbance (it s also type 0 with respect to reference input, but the two system types ARE NOT the same in general).