Ch.07 Steady-State Error

10/03/2016 System Dynamics and Control 7.01 Steady-State Error 07 Steady-State Error HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.03 Nguyen Tan Tien Steady-State Error System Dynamics and Control 7.02 Steady-State Error Learning Outcome After completing this chapter, the student will be able to • Find the steady-state error for a unity feedback system • Specify a system’s steady-state error performance • Design the gain of a closed-loop system to meet a steady-state error specification • Find the steady-state error for disturbance inputs • Find the steady-state error for nonunity feedback systems • Find the steady-state error sensitivity to parameter changes • Find the steady-state error for systems represented in state space HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.04 Nguyen Tan Tien Steady-State Error §1.Introduction - Control systems analysis and design focus on three specifications • transient response • stability • steady-state errors taking into account the robustness of the design along with economic and social considerations - Control system design entails trade-offs between desired transient response, steady-state error, and the requirement that the system be stable §1.Introduction Definition and Test Inputs - Steady-state error is the difference between the input and the output for a prescribed test input as 𝑡 → ∞ Test inputs used for steady-state error analysis and design are summarized HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.05 Nguyen Tan Tien Steady-State Error System Dynamics and Control 7.06 Nguyen Tan Tien Steady-State Error §1.Introduction - In order to explain how these test signals are used, let us assume a position control system, where the output position follows the input commanded position • Step inputs represent constant position and thus are useful in determining the ability of the control system to position itself with respect to a stationary target, such as a satellite in geostationary orbit An antenna position control is an example of a system that can be tested for accuracy using step inputs Đ1.Introduction ã Ramp inputs represent constant-velocity inputs to a position control system, and can be used to test a system’s ability to follow a linearly increasing input or, equivalently, to track a constant velocity target For example, a position control system that tracks a satellite that moves across the sky at a constant angular velocity, would be tested with a ramp input to evaluate the steady-state error between the satellite’s angular position and that of the control system HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 10/03/2016 System Dynamics and Control 7.07 Steady-State Error Đ1.Introduction ã Parabola inputs represent constant acceleration inputs to position control systems and can be used to represent accelerating targets, such as the missile, to determine the steady-state error performance Application to Stable Systems Since we are concerned with the difference between the input and the output of a feedback control system after the steady state has been reached, our discussion is limited to stable systems, where the natural response approaches zero as 𝑡 → ∞ HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.09 Nguyen Tan Tien Steady-State Error §1.Introduction - A similar example for a ramp input System Dynamics and Control 7.11 7.08 Steady-State Error §1.Introduction Evaluating Steady-State Errors - Let us examine the concept of steady-state errors, two possible outputs for step input • output has zero steady-state error • output has a finite steady-state error, 𝑒2 (∞) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.10 Nguyen Tan Tien Steady-State Error §1.Introduction - Let us now look at the error from the perspective of the most general block diagram • output has zero steady-state error • output has a finite steady-state error, 𝑒2 (∞), as measured vertically after the transients have died down • output has infinite steady-state error, as measured vertically after the transients have died down, if the output’s slope is different from that of the input HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien Steady-State Error §1.Introduction Sources of Steady-State Error - Many steady-state errors in control systems arise from nonlinear sources, such as backlash in gears or a motor that will not move unless the input voltage exceeds a threshold ⟹ study the steady state errors that arise from the configuration of the system itself and the type of applied input • steady-state error for unity feedback systems • static error constants and system type 𝐺(𝑠) : system transfer function 𝑇(𝑠) : closed-loop transfer function 𝐸(𝑠) : error, the difference between the input and the output - Here we are interested in the steady-state, or final, value of 𝑒(𝑡) • First, study and derive expressions for the steady-state error for unity feedback systems • Then, expand to nonunity feedback systems HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.12 Nguyen Tan Tien Steady-State Error §2.Steady-State Error for Unity Feedback Systems Steady-State Error in Terms of 𝑇(𝑠) The error between the input, 𝑅(𝑠), and the output, 𝐶(𝑠) 𝐸 𝑠 = 𝑅 𝑠 − 𝐶(𝑠) 𝐶 𝑠 = 𝑅 𝑠 𝑇(𝑠) ⟹ 𝐸 𝑠 = 𝑅 𝑠 [1 − 𝑇(𝑠)] Applying the final value theorem 𝑒 ∞ = lim 𝑒(𝑡) (7.2) (7.3) (7.4) 𝑡→∞ = lim 𝑠𝐸(𝑠) (7.5) = lim 𝑠𝑅 𝑠 [1 − 𝑇(𝑠)] (7.6) 𝑠→0 𝑠→0 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 10/03/2016 System Dynamics and Control 7.13 Steady-State Error System Dynamics and Control 7.14 Steady-State Error §2.Steady-State Error for Unity Feedback Systems - Ex.7.1 Steady-State Error in Terms of 𝑇(𝑠) Find the steady-state error for the system 𝑇 𝑠 = 5/(𝑠2 + 7𝑠 + 10) if the input is a unit step Solution From the problem statement 𝑇 𝑠 = , 𝑅 𝑠 = 𝑠 + 7𝑠 + 10 𝑠 𝑠 + 7𝑠 + The error 𝐸 𝑠 = 𝑅 𝑠 − 𝑇 𝑠 = 𝑠(𝑠 + 7𝑠 + 10) Since 𝑇(𝑠) is stable and, subsequently, 𝐸(𝑠) does not have RHP poles or 𝑗𝜔 poles other than at the origin, apply the final value theorem 𝑠 + 7𝑠 + 𝑒 ∞ = lim 𝑠𝐸(𝑠) = lim = 𝑠→0 𝑠→0 𝑠 + 7𝑠 + 10 §2.Steady-State Error for Unity Feedback Systems Steady-State Error in Terms of 𝐺(𝑠) HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 7.15 Steady-State Error §2.Steady-State Error for Unity Feedback Systems Step Input 𝑠(1/𝑠) 𝑒 ∞ = 𝑒step ∞ = lim = 𝑠→0 + 𝐺(𝑠) + lim 𝐺(𝑠) (7.12) 𝑠→0 𝑠→0 (7.13) 𝑠→0 To satisfy Eq (7.13), 𝐺(𝑠) must take on the following form 𝑠 + 𝑧1 ⋯ 𝐺 𝑠 = 𝑛 ⟹ 𝑠 𝑛 𝑠 + 𝑝1 𝑠 + 𝑝2 ⋯ → (7.14) 𝑠 𝑠 + 𝑝1 ⋯ 𝑛 ≥ 1: at least one pure integration in the forward path 𝑧1 𝑧2 ⋯ 𝑛 = 0: lim 𝐺(𝑠) = ≠ ∞ ⟹ finite steady-state error 𝑠→0 𝑝1 𝑝2 ⋯ HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 7.17 Steady-State Error §2.Steady-State Error for Unity Feedback Systems Parabolic Input 𝑠(1/𝑠 3) (7.20) 𝑒 ∞ = 𝑒parabola ∞ = lim = 𝑠→0 + 𝐺(𝑠) lim 𝑠 𝐺(𝑠) 𝑠→0 To have zero steady-state error for a parabolic input lim 𝑠 𝐺(𝑠) = ∞ 𝑠→0 System Dynamics and Control (7.8) (7.9) (7.10) (7.11) Nguyen Tan Tien 7.16 Steady-State Error §2.Steady-State Error for Unity Feedback Systems Ramp Input 𝑠(1/𝑠 2) 𝑒 ∞ = 𝑒ramp ∞ = lim = 𝑠→0 + 𝐺(𝑠) lim 𝑠𝐺(𝑠) (7.16) 𝑠→0 In order to have zero steady-state error, the dc gain, lim 𝐺(𝑠), of the forward transfer function lim 𝐺(𝑠) = ∞ Consider the unity feedback control system 𝐸 𝑠 = 𝑅 𝑠 − 𝐶(𝑠) 𝐶 𝑠 = 𝐸 𝑠 𝐺(𝑠) 𝑅(𝑠) ⟹𝐸 𝑠 = + 𝐺(𝑠) Apply the final value theorem 𝑠𝑅(𝑠) 𝑒 ∞ = lim 𝑠→0 + 𝐺(𝑠) (7.21) To satisfy Eq (7.21), 𝐺(𝑠) must take on the following form 𝑠 + 𝑧1 ⋯ 𝐺 𝑠 = 𝑛 ⟹ 𝑠 𝑛 𝑠 + 𝑝1 𝑠 + 𝑝2 ⋯ → (7.14) 𝑠 𝑠 + 𝑝1 ⋯ 𝑛 ≥ 3: at least three integrations in the forward path 𝑧1 𝑧2 ⋯ 𝑛 = 2: lim 𝑠 𝐺(𝑠) = ≠ ∞ ⟹ finite steady-state error 𝑠→0 𝑝1 𝑝2 ⋯ 𝑛 ≤ 1: lim 𝑠 𝐺 𝑠 = ⟹ infinite steady-state error 𝑠→0 To have zero steady-state error for a ramp input lim 𝑠𝐺(𝑠) = ∞ 𝑠→0 (7.17) To satisfy Eq (7.17), 𝐺(𝑠) must take on the following form 𝑠 + 𝑧1 ⋯ 𝐺 𝑠 = 𝑛 ⟹ 𝑠 𝑛 𝑠 + 𝑝1 𝑠 + 𝑝2 ⋯ → (7.14) 𝑠 𝑠 + 𝑝1 ⋯ 𝑛 ≥ 2: at least two integrations in the forward path 𝑧1 𝑧2 ⋯ 𝑛 = 1: lim 𝑠𝐺(𝑠) = ≠ ∞ ⟹ finite steady-state error 𝑠→0 𝑝1 𝑝2 ⋯ ⟹ diverging ramps 𝑛 = 0: lim 𝑠𝐺(𝑠) = 𝑠→0 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 7.18 Steady-State Error §2.Steady-State Error for Unity Feedback Systems - Ex.7.2 Steady-State Errors for Systems with No Integrations Find the steady-state errors for inputs of 5𝑢(𝑡), 5𝑡𝑢(𝑡), 5𝑡 (𝑡) to the system The function 𝑢(𝑡) is the unit step Solution The closed-loop system is stable 5 = = + lim 𝐺(𝑠) + 20 21 5𝑢(𝑡) or 5/𝑠: 𝑒step ∞ = × 5𝑡𝑢(𝑡) or 5/𝑠 2: 𝑒ramp ∞ = × 𝑠→0 = =∞ lim 𝑠𝐺(𝑠) 𝑠→0 5𝑡 𝑢(𝑡) or 10/𝑠 : 𝑒parabola ∞ = 10 × 10 = =∞ lim 𝑠 𝐺(𝑠) 𝑠→0 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 10/03/2016 System Dynamics and Control 7.19 Steady-State Error §2.Steady-State Error for Unity Feedback Systems - Ex.7.3 Steady-State Errors for Systems with One Integration Find the steady-state errors for inputs of 5𝑢(𝑡), 5𝑡𝑢(𝑡), 5𝑡 (𝑡) to the system The function 𝑢(𝑡) is the unit step Solution The closed-loop system is stable = =0 + lim 𝐺(𝑠) ∞ 5𝑢(𝑡) or 5/𝑠: 𝑒step ∞ = × 5𝑡𝑢(𝑡) or 5/𝑠 2: 𝑒ramp ∞ = × System Dynamics and Control 7.20 Steady-State Error §2.Steady-State Error for Unity Feedback Systems Skill-Assessment Ex.7.1 Problem A unity feedback system has the following forward TF 10(𝑠 + 20)(𝑠 + 30) 𝐺 𝑠 = 𝑠(𝑠 + 25)(𝑠 + 35) a.Find the steady-state error for the following inputs 15𝑢(𝑡), 15𝑡𝑢(𝑡), and 15𝑡 (𝑡) b.Repeat for 10(𝑠 + 20)(𝑠 + 30) 𝐺 𝑠 = 𝑠 (𝑠 + 25)(𝑠 + 35)(𝑠 + 50) 𝑠→0 = = lim 𝑠𝐺(𝑠) 100 20 𝑠→0 5𝑡 𝑢(𝑡) or 10/𝑠 : 𝑒parabola ∞ = 10 × 10 = =∞ lim 𝑠 𝐺(𝑠) 𝑠→0 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 7.21 Steady-State Error §2.Steady-State Error for Unity Feedback Systems Solution a First check stability 𝐺(𝑠) 10𝑠 + 500𝑠 + 6000 𝑇 𝑠 = = + 𝐺(𝑠) 𝑠 + 70𝑠 + 1375𝑠 + 6000 10(𝑠 + 30)(𝑠 + 20) = (𝑠 + 26.03)(𝑠 + 37.89)(𝑠 + 6.085) ⟹ the closed-loop system is stable 15 5𝑢(𝑡): 𝑒step ∞ = = =0 + lim 𝐺(𝑠) + ∞ 𝑠→0 5𝑡𝑢(𝑡): 15 15 = = 2.1875 lim 𝑠𝐺(𝑠) 10 × 20 × 30 𝑠→0 25 × 35 30 30 ∞ = = =∞ lim 𝑠 𝐺(𝑠) 𝑒ramp ∞ = 5𝑡 𝑢(𝑡): 𝑒parabola HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.22 Nguyen Tan Tien Steady-State Error §2.Steady-State Error for Unity Feedback Systems b First check stability 𝐺(𝑠) 𝑇 𝑠 = + 𝐺(𝑠) 10𝑠2 + 500𝑠 + 6000 𝑠5 + 110𝑠4 + 3875𝑠3 + 4.37 × 104𝑠2 + 500𝑠 + 6000 10(𝑠 + 30)(𝑠 + 20) = (𝑠 +50.01)(𝑠 +35)(𝑠 +25)(𝑠2 −7.189×10−4𝑠 +0.1372) ⟹ From the second-order term in the denominator, the system is unstable Instability could also be determined using the Routh-Hurwitz criteria on the denominator of 𝑇(𝑠) Since the system is unstable, calculations about steady-state error cannot be made = 𝑠→0 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 7.23 Steady-State Error §3.Static Error Constants and System Type Static Error Constants - The relationships for steady-state error 1 + lim 𝐺(𝑠) • For a step input, 𝑢(𝑡) 𝑒step ∞ = • For a ramp input, 𝑡𝑢(𝑡) 𝑒ramp ∞ = HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.24 Nguyen Tan Tien Steady-State Error §3.Static Error Constants and System Type - Ex.7.4 Steady-State Error via Static Error Constants Evaluate the static error constants and find the expected error for the standard step, ramp, and parabolic inputs 𝑠→0 lim 𝑠𝐺(𝑠) 𝑠→0 • For a parabolic input, 𝑡 𝑢(𝑡) 𝑒parabola ∞ = lim 𝑠 𝐺(𝑠) 𝑠→0 - The limits static error constants 𝐾𝑝 = lim 𝐺(𝑠) • Position constant, 𝐾𝑝 𝑠→0 • Velocity constant, 𝐾𝑣 𝐾𝑣 = lim 𝑠𝐺(𝑠) • Acceleration constant, 𝐾𝑎 𝐾𝑝 = lim 𝑠 𝐺(𝑠) HCM City Univ of Technology, Faculty of Mechanical Engineering Solution All closed-loop systems are indeed stable 𝑠→0 𝑠→0 Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 10/03/2016 System Dynamics and Control 7.25 Steady-State Error §3.Static Error Constants and System Type a Steady-State Error - The limits static error constants 𝐾𝑝 = lim 𝐺(𝑠) = ∞ 𝑠→0 500 × × × 𝐾𝑣 = lim 𝑠𝐺(𝑠) = = 31.25 𝑠→0 × 10 × 12 𝐾𝑎 = lim 𝑠 𝐺(𝑠) = 𝑠→0 𝐾𝑎 = lim 𝑠 𝐺(𝑠) = 𝑠→0 𝑠→0 - The steady-state error 1 𝑒step ∞ = = = 0.161 + 𝐾𝑝 + 5.208 1 𝑒ramp ∞ = = =∞ 𝐾𝑣 1 𝑒parabola ∞ = = =∞ 𝐾𝑎 HCM City Univ of Technology, Faculty of Mechanical Engineering 7.27 - The steady-state error 1 𝑒step ∞ = = =0 + 𝐾𝑝 + ∞ 1 𝑒ramp ∞ = = = 0.032 𝐾𝑣 31.25 1 𝑒parabola ∞ = = =∞ 𝐾𝑎 Nguyen Tan Tien Steady-State Error §3.Static Error Constants and System Type c - The limits static error constants 𝐾𝑝 = lim 𝐺(𝑠) = ∞ 𝑠→0 𝐾𝑣 = lim 𝑠𝐺(𝑠) = ∞ 𝑠→0 500 × × × × × 𝐾𝑎 = lim 𝑠 𝐺(𝑠) = = 875 𝑠→0 × 10 × 12 - The steady-state error 1 𝑒step ∞ = = =0 + 𝐾𝑝 + ∞ 1 𝑒ramp ∞ = = =0 𝐾𝑣 ∞ 1 𝑒parabola ∞ = = = 1.14 × 10−3 𝐾𝑎 875 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.26 §3.Static Error Constants and System Type b - The limits static error constants 500 × × 𝐾𝑝 = lim 𝐺(𝑠) = = 5.208 𝑠→0 × 10 × 12 𝐾𝑣 = lim 𝑠𝐺(𝑠) = System Dynamics and Control System Dynamics and Control 7.29 Nguyen Tan Tien Steady-State Error §3.Static Error Constants and System Type System Type - The values of the static error constants, again, depend upon the form of 𝐺(𝑠), especially the number of pure integrations in the forward path - Given the system HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 7.28 Steady-State Error §3.Static Error Constants and System Type Run ch7p1 in Appendix B Learn how to use MATLAB to • test the system for stability • evaluate static error constants • calculate steady-state error • solve Ex.7.4 with System (b) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 7.30 §3.Static Error Constants and System Type - Relationships between input, system constants, and steady-state errors Steady-State Error type, static error define system type to be the value of 𝑛 in the denominator or, equivalently, the number of pure integrations in the forward path • 𝑛 = : type system • 𝑛 = : type system • 𝑛 = : type system HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 10/03/2016 System Dynamics and Control 7.31 Steady-State Error §3.Static Error Constants and System Type Skill-Assessment Ex.7.2 Problem A unity feedback system has the following forward TF 1000(𝑠 + 8) 𝐺 𝑠 = (𝑠 + 7)(𝑠 + 9) a.Evaluate system type, 𝐾𝑝 , 𝐾𝑣 , and 𝐾𝑎 b.Use your answers to (a.) to find the steady-state errors for the standard step, ramp, and parabolic inputs Solution The system is stable System Dynamics and Control 7.32 Steady-State Error §3.Static Error Constants and System Type a The closed-loop transfer function 𝐺(𝑠) 𝑇 𝑠 = + 𝐺(𝑠) 1000(𝑠 + 8) = 𝑠 + 𝑠 + + 1000(𝑠 + 8) 1000(𝑠 + 8) = 𝑠 + 1016𝑠 + 8063 The system is Type Therefore 1000 × 𝐾𝑝 = lim 𝐺(𝑠) = = 127 𝑠→0 7×9 𝐾𝑣 = lim 𝑠𝐺(𝑠) = 𝑠→0 𝐾𝑎 = lim 𝑠 𝐺(𝑠) = 𝑠→0 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.33 Nguyen Tan Tien Steady-State Error HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.34 §3.Static Error Constants and System Type b The steady-state error 1 𝑒step ∞ = = = 7.8 × 10−3 + 𝐾𝑝 + 127 1 𝑒ramp ∞ = = =∞ 𝐾𝑣 1 𝑒parabola ∞ = = =∞ 𝐾𝑎 §3.Static Error Constants and System Type TryIt 7.1 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.35 Nguyen Tan Tien Steady-State Error §4.Steady-State Error Specifications A robot used in the manufacturing of semiconductor randomaccess memories (RAMs) similar to those in personal computers Steady-state error is an important design consideration for assembly-line robots Use MATLAB, the Control System Toolbox, and the following statements to find 𝐾𝑝 ,𝑒step (∞), and the closedloop poles to check for stability for the system of Skill-Assessment Ex.7.2 𝐺 𝑠 = Nguyen Tan Tien Steady-State Error 1000(𝑠 + 8) (𝑠 + 7)(𝑠 + 9) numg=1000*[1 8]; deng=poly([-7 -9]); G=tf(numg,deng); Kp=dcgain(G) estep=1/(1+Kp) T=feedback(G,1); poles=pole(T) System Dynamics and Control 7.36 Nguyen Tan Tien Steady-State Error §4.Steady-State Error Specifications - The specifications for a control system’s transient response • damping ratio, 𝜁 • settling time, 𝑇𝑠 • peak time, 𝑇𝑝 • percent overshoot, %𝑂𝑆 - The specification of a static error constant • the position constant, 𝐾𝑝 • velocity constant, 𝐾𝑣 • acceleration constant, 𝐾𝑎 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 10/03/2016 System Dynamics and Control 7.37 Steady-State Error System Dynamics and Control 7.38 Steady-State Error §4.Steady-State Error Specifications - For example, if a control system has the specification 𝐾𝑣 = 1000, we can draw several conclusions • The system is stable • The system is of Type • A ramp input is the test signal • The steady-state error between the input ramp and the output ramp is 1/𝐾𝑣 per unit of input slope §4.Steady-State Error Specifications - Ex.7.5 Interpreting the Steady-State Error Specification What information is contained in the specification 𝐾𝑝 = 1000? HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 7.39 Steady-State Error §4.Steady-State Error Specifications - Ex.7.6 Gain Design to Meet a Steady-State Error Specification Find the value of 𝐾 so that there is 10% error in the steady state Solution Since the system is Type 1, the error stated in the problem must apply to a ramp input; only a ramp yields a finite error in a Type system Solution The system is stable The system is Type The input test signal is a step The error per unit step is 𝑒step ∞ = System Dynamics and Control 1 = = + 𝐾𝑝 + 1000 1001 Nguyen Tan Tien 7.40 Steady-State Error §4.Steady-State Error Specifications 𝑒 ∞ = = 0.1 𝐾𝑣 ⟹ 𝐾𝑣 = 10 = lim 𝑠𝐺 𝑠 = 𝑠→0 𝐾×5 6×7×8 ⟹ 𝐾 = 672 Applying the Routh-Hurwitz criterion, we see that the system is stable at this gain Although this gain meets the criteria for steady-state error and stability, it may not yield a desirable transient response HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 7.41 §4.Steady-State Error Specifications Run ch7p2 in Appendix B Learn how to use MATLAB to • find the gain to meet specification • solves Ex.7.6 HCM City Univ of Technology, Faculty of Mechanical Engineering Steady-State Error a steady-state error Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.42 Nguyen Tan Tien Steady-State Error §4.Steady-State Error Specifications Skill-Assessment Ex.7.3 Problem A unity feedback system has the following forward TF 𝐾(𝑠 + 2) 𝐺 𝑠 = (𝑠 + 14)(𝑠 + 18) Find the value of 𝐾 to yield a 10% error in the steady state Solution The system is stable for positive 𝐾 For a step input 𝑒step ∞ = = 0.1 + 𝐾𝑝 12 × 𝐾 ⟹ 𝐾𝑝 = = lim 𝐺 𝑠 = 𝑠→0 14 × 18 ⟹ 𝐾 = 189 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 10/03/2016 System Dynamics and Control 7.43 Steady-State Error §4.Steady-State Error Specifications TryIt 7.2 Use MATLAB, the Control 𝐾(𝑠 + 2) System Toolbox, and the 𝐺 𝑠 = (𝑠 + 14)(𝑠 + 18) following statements to solve numg=[1 12]; deng=poly([-14 -18]); G=tf(numg,deng); Kpdk=dcgain(G); estep=0.1; K=(1/estep-1)/Kpdk T=feedback(G,1); poles=pole(T) System Dynamics and Control 7.45 Nguyen Tan Tien Steady-State Error 𝑠→0 𝑠 𝑠𝐺2 𝑠 𝑅 𝑠 − lim 𝐷 𝑠 𝑠→0 + 𝐺1 𝑠 𝐺2 𝑠 𝑠 𝐺2 𝑠 = 𝑒𝑅 ∞ + 𝑒𝐷 (∞) = lim 𝑠→0 + 𝐺1 where, • 𝑒𝑅 ∞ : the steady-state error due to 𝑅(𝑠) 𝑠 𝑒𝑅 ∞ = lim 𝑅 𝑠 𝑠→0 + 𝐺1 𝑠 𝐺2 𝑠 • 𝑒𝐷 (∞) : the steady-state error due to the disturbance 𝐷(𝑠) 𝑠𝐺2 𝑠 𝑒𝐷 ∞ = − lim 𝐷 𝑠 𝑠→0 + 𝐺1 𝑠 𝐺2 𝑠 1 𝑠 𝐺2 𝑠 𝐺2 𝑠 𝑠 𝐺2 𝑠 𝑅 𝑠 − 1+𝐺 𝐷 𝑠 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.46 Nguyen Tan Tien Steady-State Error §5.Steady-State Error for Disturbances - Assume a step disturbance, 𝐷 𝑠 = 1/𝑠 Substituting this value into the second term of Eq (7.61), 𝑒𝐷 (∞) , the steady-state error component due to a step disturbance is found to be 𝑠𝐺2 𝑠 1 (7.62) 𝑒𝐷 ∞ = − lim =− 𝑠→0 + 𝐺1 𝑠 𝐺2 𝑠 𝑠 lim + lim 𝐺1 𝑠 𝑠→0 𝐺2 𝑠 𝑠→0 This equation shows that the steady-state error produced by a step disturbance can be reduced by increasing the dc gain of 𝐺1 𝑠 or decreasing the dc gain of 𝐺2 𝑠 (7.90) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Steady-State Error - Consider a feedback control system with a disturbance, 𝐷(𝑠), injected between the controller and the plant 𝐶 𝑠 = 𝐸 𝑠 𝐺1 𝑠 𝐺2 𝑠 + 𝐷(𝑠) 𝐺2 (𝑠) (7.58) 𝐶 𝑠 = 𝑅 𝑠 − 𝐸(𝑠) (7.59) 𝐺2 𝑠 (7.60) 𝐸 𝑠 = 𝑅 𝑠 − 𝐷 𝑠 + 𝐺1 𝑠 𝐺2 𝑠 + 𝐺1 𝑠 𝐺2 𝑠 §5.Steady-State Error for Disturbances - To find the steady-state value of the error, apply the final value theorem to Eq (7.60) and obtain 𝑒 ∞ = lim 𝑠𝐸(𝑠) 𝐸 𝑠 = 1+𝐺 7.44 §5.Steady-State Error for Disturbances - Feedback control systems are used to compensate for disturbances or unwanted inputs that enter a system, result that regardless of these disturbances, the system can be designed to follow the input with small or zero error Skill-Assessment Ex.7.3 and check the resulting system for stability HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.47 Nguyen Tan Tien Steady-State Error §5.Steady-State Error for Disturbances - Rearrange the system so that the disturbance, 𝐷(𝑠), is depicted as the input and the error, 𝐸(𝑠), as the output, with 𝑅 𝑠 = To minimize the steady-state value of 𝐸(𝑠), we must either • increase the dc gain of 𝐺1 (𝑠) so that a lower value of 𝐸(𝑠) will be fed back to match the steady-state value of 𝐷(𝑠), or • decrease the dc value of 𝐺2 (𝑠), which then yields a smaller value of 𝑒(∞) as predicted by the feedback formula HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.48 Nguyen Tan Tien Steady-State Error §5.Steady-State Error for Disturbances - Ex.7.7 Steady-State Error Due to Step Disturbance Find the steady-state error component due to a step disturbance for the system Solution The system is stable 1 𝑒𝐷 ∞ = − =− =− + 1000 1000 lim + lim 𝐺1 𝑠 𝑠→0 𝐺2 𝑠 𝑠→0 The result shows that the steadystate error produced by the step disturbance is inversely proportional to the dc gain of 𝐺1 (𝑠) The dc gain of 𝐺2 (𝑠) is infinite in this example HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 10/03/2016 System Dynamics and Control 7.49 Steady-State Error §5.Steady-State Error for Disturbances Skill-Assessment Ex.7.4 Problem Evaluate the steady-state error component due to a step disturbance for the system lim 𝑠→0 𝐺2 + lim 𝐺1 𝑠 𝑠 𝑠→0 = + 1000 = −9.98 × 10−4 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.51 7.50 Steady-State Error §6.Steady-State Error for Nonunity Feedback Systems - A general feedback system, showing the input transducer, 𝐺1 (𝑠), controller and plant, 𝐺2 (𝑠), and feedback, 𝐻1 (𝑠) - Pushing the input transducer to the right past the summing junction yields the general nonunity feedback system, where 𝐺 𝑠 = 𝐺1 (𝑠)𝐺2 (𝑠) 𝐻 𝑠 = 𝐻1 (𝑠)/𝐺1 (𝑠) 𝐸𝑎 (𝑠) : actuating signal, 𝐸𝑎 𝑠 ≠ 𝐸 𝑠 = 𝐶 𝑠 − 𝑅(𝑠) If 𝑟(𝑡) and 𝑐(𝑡) have the same units, the steady-state error can be found, 𝑒 ∞ = 𝑟 ∞ − 𝑐(∞) Solution The system is stable For a step input 𝑒𝐷 ∞ = − System Dynamics and Control Nguyen Tan Tien Steady-State Error §6.Steady-State Error for Nonunity Feedback Systems - Form a unity feedback system by adding and subtracting unity feedback paths This step requires that input and output units be the same HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.52 Nguyen Tan Tien Steady-State Error §6.Steady-State Error for Nonunity Feedback Systems - Combine 𝐻(𝑠) with the negative unity feedback - Combine the feedback system consisting of 𝐺(𝑠) and [𝐻 𝑠 − 1], leaving an equivalent forward path and a unity feedback Notice that the final figure shows 𝐸 𝑠 = 𝑅 𝑠 − 𝐶(𝑠) explicitly HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.53 Nguyen Tan Tien Steady-State Error §6.Steady-State Error for Nonunity Feedback Systems - Ex.7.8 Steady-State Error for Nonunity Feedback Systems Find the system type, the appropriate error constant associated with the system type, and the steady-state error for a unit step input Assume input and output units are the same Solution Using the equivalent forward TF 100 𝐺(𝑠) 𝑠(𝑠 + 10) 𝐺𝑒 𝑠 = = 100 + 𝐺 𝑠 𝐻 𝑠 − 𝐺(𝑠) + 100 − 𝑠(𝑠 + 10) 𝑠 + 𝑠(𝑠 + 10) 100(𝑠 + 5) ⟹ 𝐺𝑒 𝑠 = 𝑠 + 15𝑠 − 50𝑠 − 400 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.54 Nguyen Tan Tien Steady-State Error §6.Steady-State Error for Nonunity Feedback Systems The equivalent forward TF 100(𝑠 + 5) 𝐺𝑒 𝑠 = 𝑠 + 15𝑠 − 50𝑠 − 400 Thus, the system is Type 0, since there are no pure integrations The appropriate static error constant is then 𝐾𝑝 100 × 5 𝐾𝑝 = lim 𝐺𝑒 𝑠 = =− 𝑠→0 −400 The steady-state error 1 𝑒(∞) = = = −4 + 𝐾𝑝 − 5/4 The negative value for steady-state error implies that the output step is larger than the input step HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 10/03/2016 System Dynamics and Control 7.55 Steady-State Error §6.Steady-State Error for Nonunity Feedback Systems TryIt 7.3 System Dynamics and Control 7.56 Steady-State Error §6.Steady-State Error for Nonunity Feedback Systems - Consider a nonunity feedback control system with disturbance Use MATLAB, the Control System Toolbox, and the following statements to find 𝐺𝑒 (𝑠) in Ex.7.8 G=zpk([],[0 -10],100); H=zpk([],- 5,1); Ge=feedback(G,(H-1)); 'Ge(s)' Ge=tf(Ge) T=feedback(Ge,1); 'Poles of T(s)' pole(T) - The steady-state error for this system, 𝑒 ∞ = 𝑐(∞) − 𝑟(∞) 𝑒 ∞ = lim 𝑠𝐸(𝑠) 𝑠→0 = lim 𝑠 𝑠→0 𝑠→0 Nguyen Tan Tien 7.57 Steady-State Error §6.Steady-State Error for Nonunity Feedback Systems lim(𝐺1 𝐺2 ) lim 𝐺2 𝑠→0 𝑒 ∞ = − 𝑠→0 − lim(1 + 𝐺1 𝐺2 𝐻) lim(1 + 𝐺1 𝐺2 𝐻) 𝑠→0 For zero error lim(𝐺1 𝐺2 ) 𝑠→0 lim(1 + 𝐺1 𝐺2 𝐻) (7.69) 𝑠→0 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.58 Nguyen Tan Tien Steady-State Error §6.Steady-State Error for Nonunity Feedback Systems (7.70) 𝑠→0 lim 𝐺2 𝑠→0 = 1, lim(1 + 𝐺1 𝐺2 𝐻) 𝑠→0 =0 (7.71) 𝑠→0 The above two equations can always be satisfied if • the system is stable • 𝐺1 (𝑠) is a Type system • 𝐺2 (𝑠) is a Type system • 𝐻(𝑠) is a Type system with a dc gain of unity HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 𝐺1 𝐺2 𝐺2 𝑅− 𝐷 + 𝐺1 𝐺2 𝐻 + 𝐺1 𝐺2 𝐻 with step inputs and step disturbances, 𝑅 𝑠 = 𝐷 𝑠 = 1/𝑠 lim(𝐺1 𝐺2 ) lim 𝐺2 𝑠→0 (7.70) 𝑒 ∞ = − 𝑠→0 − lim(1 + 𝐺1 𝐺2 𝐻) lim(1 + 𝐺1 𝐺2 𝐻) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 1− 7.59 Nguyen Tan Tien Steady-State Error - The steady-state value of the actuating signal, 𝐸𝑎1 (𝑠), for a general feedback system 𝑠𝑅(𝑠)𝐺1 (𝑠) (7.72) 𝑒𝑎1 ∞ = lim 𝑠→0 + 𝐺2 (𝑠)𝐻1 (𝑠) Note: There is no restriction that the input and output units be the same, since we are finding the steady-state difference between signals at the summing junction, which have the same units HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.60 §6.Steady-State Error for Nonunity Feedback Systems - Ex.7.9 Steady-State Actuating Signal for Nonunity Feedback Systems Find the steady-state actuating signal for the system with a unit step input Repeat for a unit ramp input Solution 𝑠𝑅(𝑠)𝐺1 (𝑠) 𝑠𝑅(𝑠) × 𝑒𝑎 ∞ = lim = lim 100 𝑠→0 + 𝐺2 (𝑠)𝐻1 (𝑠) 𝑠→0 1+ × 𝑠(𝑠 + 10) 𝑠 + 𝑠(1/𝑠) × For step input 𝑒𝑎 ∞ = lim =0 100 𝑠→0 1+ × 𝑠(𝑠 + 10) 𝑠 + 𝑠(1/𝑠 ) × 1 For ramp input 𝑒𝑎 ∞ = lim = 100 𝑠→0 1+ ì ( + 10) + Đ6.Steady-State Error for Nonunity Feedback Systems Skill-Assessment Ex.7.5 Problem HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien Steady-State Error a.Find the steady-state error, 𝑒 ∞ = 𝑐(∞) − 𝑟(∞), for a unit step input given the nonunity feedback system Repeat for a unit ramp input Assume input and output units are the same b.Find the steady-state actuating signal, 𝑒𝑎 (∞), for a unit step input given the nonunity feedback system Repeat for a unit ramp input Nguyen Tan Tien 10 10/03/2016 System Dynamics and Control 7.61 Steady-State Error §6.Steady-State Error for Nonunity Feedback Systems Solution The system is stable Create a unity-feedback system System Dynamics and Control 7.62 Steady-State Error §6.Steady-State Error for Nonunity Feedback Systems a.The steady-state error, 𝑒 ∞ = 𝑐(∞) − 𝑟(∞) The system is Type 0, 𝐾𝑝 = lim 𝐺 𝑠 = 100/4 = 25 𝑠→0 The steady-state error 𝑒step ∞ = 1/(1 + 𝐾𝑝 ) = 0.0385 𝑒ramp ∞ = ∞ −𝑠 −1= 𝑠+1 𝑠+1 𝐺(𝑠) 100(𝑠 + 1) 𝐺𝑒 𝑠 = = + 𝐺(𝑠)𝐻𝑒 (𝑠) 𝑠 − 95𝑠 + 𝐻𝑒 𝑠 = HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.63 Nguyen Tan Tien Steady-State Error §6.Steady-State Error for Nonunity Feedback Systems b.The steady-state actuating signal, 𝑒𝑎 (∞) 𝑠𝑅(𝑠)𝐺1 (𝑠) 𝑠𝑅(𝑠) × 𝑒𝑎 ∞ = lim = lim 100 𝑠→0 + 𝐺2 (𝑠)𝐻1(𝑠) 𝑠→0 1+ × 𝑠+4 𝑠+1 𝑠(1/𝑠) × 1 step input 𝑒𝑎 ∞ = lim = = 0.0385 100 𝑠→0 104 1+ × 𝑠+4 𝑠+1 𝑠(1/𝑠2) × 1 ramp input 𝑒𝑎 ∞ = lim = =∞ 100 𝑠→0 1+ × 𝑠+4 𝑠+1 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.65 Nguyen Tan Tien Steady-State Error HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.64 Nguyen Tan Tien Steady-State Error §7.Sensitivity Sensitivity: the ratio of the fractional change in the function to the fractional change in the parameter as the fractional change of the parameter approaches zero Fractional change in the function,𝐹 𝑆𝐹:𝑃 = lim ∆𝑃→0 Fractional change in the parameter,𝑃 ∆𝐹/𝐹 = lim ∆𝑃→0 ∆𝑃/𝑃 𝑃∆𝐹 = lim ∆𝑃→0 𝐹∆𝑃 𝑃 𝛿𝐹 ⟹ 𝑆𝐹:𝑃 = (7.75) 𝐹 𝛿𝑃 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.66 Nguyen Tan Tien Steady-State Error §7.Sensitivity - Ex.7.10 Sensitivity of a Closed-Loop Transfer Function Calculate the sensitivity of the closed-loop transfer function to changes in the parameter 𝑎 How would you reduce the sensitivity? Solution The closed-loop transfer function 𝐾 𝑇 𝑠 = 𝑠 + 𝑎𝑠 + 𝐾 The sensitivity 𝑎 𝛿𝑇 𝑎 −𝐾𝑠 −𝑎𝑠 𝑆𝑇:𝑎 = = = 𝑠 + 𝑎𝑠 + 𝐾 𝐾 𝑇 𝛿𝑎 𝐾 𝑠 + 𝑎𝑠 + 𝐾 𝑠 + 𝑎𝑠 + 𝐾 An increase in 𝐾 reduces the sensitivity of the closed-loop transfer function to changes in the parameter 𝑎 §7.Sensitivity - Ex.7.11 Sensitivity of Steady-State Error with Ramp Input Find the sensitivity of the steady-state error to changes in parameter 𝐾 and parameter a with ramp inputs Solution The steady-state error for the system 𝑒 ∞ = 1/𝐾𝑣 = 𝑎/𝐾 The sensitivity of 𝑒(∞) to changes in parameter 𝑎 𝑎 𝛿𝑒 𝑎 𝑆𝑒:𝑎 = = =1 𝑒 𝛿𝑎 𝑎/𝐾 𝐾 The sensitivity of 𝑒(∞) to changes in parameter 𝐾 𝐾 𝛿𝑒 𝐾 −𝑎 𝑆𝑒:𝐾 = = = −1 𝑒 𝛿𝐾 𝑎/𝐾 𝐾 There is no reduction or increase in sensitivity HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 11 10/03/2016 System Dynamics and Control 7.67 Steady-State Error System Dynamics and Control 7.67 Steady-State Error §7.Sensitivity - Ex.7.12 Sensitivity of Steady-State Error with Step Input Find the sensitivity of the steady-state error to changes in parameter 𝐾 and parameter 𝑎 for the system with a step input Solution The steady-state error for this Type system 1 𝑎𝑏 𝑒 ∞ = = = + 𝐾𝑝 + 𝐾 𝑎𝑏 + 𝐾 𝑎𝑏 The sensitivity of 𝑒(∞) to changes in parameter 𝑎 𝑎 𝛿𝑒 𝑎 𝑎𝑏 + 𝐾 𝑏 − 𝑎𝑏 𝐾 𝑆𝑒:𝑎 = = = 𝑎𝑏 𝑒 𝛿𝑎 (𝑎𝑏 + 𝐾)2 𝑎𝑏 + 𝐾 𝑎𝑏 + 𝐾 §7.Sensitivity The steady-state error for this Type system 1 𝑎𝑏 𝑒 ∞ = = = + 𝐾𝑝 + 𝐾 𝑎𝑏 + 𝐾 𝑎𝑏 The sensitivity of 𝑒(∞) to changes in parameter 𝑎 𝑎 𝛿𝑒 𝑎 𝑎𝑏 + 𝐾 𝑏 − 𝑎𝑏 𝐾 𝑆𝑒:𝑎 = = = 𝑎𝑏 𝑒 𝛿𝑎 (𝑎𝑏 + 𝐾)2 𝑎𝑏 + 𝐾 𝑎𝑏 + 𝐾 The sensitivity of 𝑒(∞) to changes in parameter 𝐾 𝐾 𝛿𝑒 𝐾 −𝑎𝑏 −𝐾 𝑆𝑒:𝐾 = = = 𝑎𝑏 (𝑎𝑏 + 𝐾)2 𝑎𝑏 + 𝐾 𝑒 𝛿𝐾 𝑎𝑏 + 𝐾 The sensitivity to changes in parameter 𝐾 and parameter 𝑎 is less than unity for positive 𝑎 and 𝑏 Thus, feedback in this case yields reduced sensitivity to variations in both parameters HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.69 Nguyen Tan Tien Steady-State Error §7.Sensitivity TryIt 7.4 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.71 The system is Type 𝐾𝑝 = lim 𝐺 𝑠 = 7𝐾/10 𝑠→0 The steady-state error for this Type system 1 10 𝑒 ∞ = = = + 𝐾𝑝 + 7𝐾/10 10 + 7𝐾 The sensitivity of 𝑒(∞) to changes in parameter 𝐾 𝐾 𝛿𝑒 𝐾 −10 × 7𝐾 𝑆𝑒:𝐾 = = =− 10 (10 + 7𝐾)2 𝑒 𝛿𝐾 10 + 7𝐾 10 + 7𝐾 Nguyen Tan Tien Steady-State Error §8.Steady-State Error for Systems in State Space Analysis via Final Value Theorem - Consider the closed-loop system represented in state space 𝒙ሶ = 𝑨𝒙 + 𝑩𝑟, 𝑦 = 𝑪𝒙 (7.84) - The Laplace transform of the error 𝐸 𝑠 = 𝑅 𝑠 − 𝑌(𝑠) (7.85) 𝑌 𝑠 = 𝑅 𝑠 𝑇(𝑠) (7.86) 𝑇(𝑠) : the closed-loop transfer function ⟹ 𝐸 𝑠 = 𝑅 𝑠 [1 − 𝑇 𝑠 ] (7.87) - Using Eq.(3.73) for 𝑇(𝑠) 𝐸 𝑠 = 𝑅 𝑠 − 𝑪 𝑠𝑰 − 𝑨 −1 𝑩 (7.88) - Applying the final value theorem lim 𝑠𝐸(𝑠) = lim{𝑠𝑅 𝑠 − 𝑪 𝑠𝑰 − 𝑨 −1 𝑩 } (7.89) 𝑇 𝑠 = = 𝑪 𝑠𝑰 − 𝑨 Steady-State Error Solution syms K a b s G=K/((s+a)*(s+b)); Kp=subs(G,s,0); e=1/(1+Kp); Sea=(a/e)*diff(e,a); Sea=simple(Sea); 'Sea' pretty(Sea) 𝑠→0 7.70 §7.Sensitivity Skill-Assessment Ex.7.6 Problem Find the sensitivity of the steady-state error to changes in 𝐾 Use MATLAB, the Symbolic Math Toolbox, and the following statements to find 𝑆𝑒:𝑎 in Ex.7.12 𝑌(𝑠) 𝑅(𝑠) System Dynamics and Control Nguyen Tan Tien 𝑠→0 −1 𝑩 + 𝑫 HCM City Univ of Technology, Faculty of Mechanical Engineering (3.73) Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 7.72 Nguyen Tan Tien Steady-State Error §8.Steady-State Error for Systems in State Space - Ex.7.13 Steady-State Error Using the Final Value Theorem Evaluate the steady-state error for the system with unit step and unit ramp inputs Use the final value theorem −5 𝑨= −2 , 𝑩 = , 𝑪 = −1 20 −10 1 Solution The steady-state error 𝑠+4 𝑠 + 6𝑠 + 13𝑠 + 20 𝑠 + 6𝑠 + 12𝑠 + 16 = lim 𝑠𝑅(𝑠) 𝑠→0 𝑠 + 6𝑠 + 13𝑠 + 20 𝑒 ∞ = lim 𝑠𝑅(𝑠) − 𝑠→0 For a unit step, 𝑅 𝑠 = 1/𝑠, and 𝑒 ∞ = 4/5 For a unit ramp, 𝑅 𝑠 = 1/𝑠 2, and 𝑒 ∞ = ∞ Notice that the system behaves like a Type system HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 12 10/03/2016 System Dynamics and Control 7.73 Steady-State Error §8.Steady-State Error for Systems in State Space TryIt 7.5 Use MATLAB, the Symbolic −5 Math Toolbox, and the −2 , 𝑩 = , following statements to find 𝑨 = 20 −10 1 the steady-state error for a step input to the system of 𝑪 = −1 syms s A= [-5 0; -2 1; 20 -10 1]; B=[0;0;1]; C=[-1 0]; I=[1 0; 0; 0 1]; E=(1/s)*[1-C*[(s*I-A)^-1]*B]; error=subs(s*E,s,0) HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 7.75 Steady-State Error §8.Steady-State Error for Systems in State Space Ramp Inputs If the input is an unit ramp, 𝑟 = 𝑡, a steady-state solution, 𝒙𝑠𝑠 𝒙𝑠𝑠 = 𝑉1 𝑡 + 𝑊1 𝑉2 𝑡 + 𝑊2 ⋯ 𝑉𝑛 𝑡 + 𝑊𝑛 𝑇 = 𝑽𝑡 + 𝑾, 𝑊𝑖 , 𝑉𝑖 are constants (7.97) 𝒙ሶ 𝑠𝑠 = 𝑉1 𝑉2 ⋯ 𝑉𝑛 𝑇 = 𝑽 (7.98) ⟹ 𝑽 = 𝑨 𝑽𝑡 + 𝑾 + 𝑩𝑡, 𝑦𝑠𝑠 = 𝑪 𝑽𝑡 + 𝑾 (7.99) In order to balance Eq.(7.99) equate the matrix coefficients of 𝑡 𝑨𝑽 = −𝑩 or 𝑽 = −𝑨−1 𝑩 equating constant terms 𝑨𝑾 = 𝑽 or 𝑾 = 𝑨−1 𝑽 Substituting into (7.99) yields 𝑦𝑠𝑠 = 𝑪 −𝑨−1𝑩𝑡 + 𝑨−1(−𝑨−1𝑩) = −𝑪 𝑨−1𝑩𝑡 + (𝑨−1)2𝑩) The steady-state error 𝑒 ∞ = lim (𝑡 − 𝑦𝑠𝑠) = lim [ + 𝑪𝑨−1𝑩 𝑡 + 𝑪(𝑨−1)2𝑩] (7.103) 𝑡→∞ 𝑡→∞ HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 7.77 Steady-State Error §8.Steady-State Error for Systems in State Space Skill-Assessment Ex.7.7 Problem Find the steady-state error for a step input using both the final value theorem and input substitution methods 𝑨= ,𝑩= ,𝑪= 1 −3 −6 Solution Using the final value theorem 𝑒step ∞ = lim 𝑠𝑅 𝑠 − 𝑪 𝑠𝑰 − 𝑨 −1 𝑠→0 𝑠 −1 = lim − 1 𝑠→0 𝑠+6 = lim − 1 𝑠→0 𝑠 + 5𝑠 + + 6𝑠 + = 2/3 7.74 Steady-State Error §8.Steady-State Error for Systems in State Space Analysis via Input Substitution Consider the closed-loop system represented in state space 𝒙ሶ = 𝑨𝒙 + 𝑩𝑟, 𝑦 = 𝑪𝒙 (7.84) Step Inputs If the input is a unit step, 𝑟 = 1, a steady-state solution, 𝒙𝑠𝑠 𝒙𝑠𝑠 = 𝑉1 𝑉2 ⋯ 𝑉𝑛 𝑇 = 𝑽, 𝑉𝑖 is constant (7.92) 𝒙ሶ 𝑠𝑠 = 𝟎 (7.93) Eq.(7.84)⟹𝟎 = 𝑨𝑽 + 𝑩, 𝑦𝑠𝑠 = 𝑪𝑽 (7.94) 𝑽 = −𝑨−1 𝑩 (7.95) The steady-state error 𝑒 ∞ = − 𝑦𝑠𝑠 = − 𝑪𝑽 = + 𝑪𝑨−1 𝑩 (7.96) Ex.7.13 System Dynamics and Control System Dynamics and Control HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 7.76 Steady-State Error §8.Steady-State Error for Systems in State Space - Ex.7.14 Steady-State Error Using Input Substitution Evaluate the steady-state error for the system with unit step and unit ramp inputs Use input substitution −5 𝑨= −2 , 𝑩 = , 𝑪 = −1 Solution 20 −10 1 For a unit step input, the steady-state error 𝑒 ∞ = + 𝑪𝑨−1 𝑩 = − 0.2 = 0.8 For a ramp input, the steady-state error 𝑒 ∞ = lim [ + 𝑪𝑨−1 𝑩 𝑡 + 𝑪(𝑨−1 )2 𝑩] 𝑡→∞ = lim 0.8𝑡 + 0.08 𝑡→∞ =∞ HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 7.78 Steady-State Error §8.Steady-State Error for Systems in State Space Using input subtituition 𝑒step ∞ = + 𝑪𝑨−1 𝑩 −1 −3 −6 −6 −1 0 =1+ 1 = + 1 −3 = =1+ 1 𝑩 −1 𝑠+6 −3𝑠 𝑠 𝑠 + 6𝑠 + = lim 𝑠→0 𝑠 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 13 ... 7.13 Steady-State Error System Dynamics and Control 7.14 Steady-State Error §2 .Steady-State Error for Unity Feedback Systems - Ex.7.1 Steady-State Error in Terms of