Week Content Notes 1 Introduction 2 Frequency Domain Modelling 3 Transient Performance and the s-plane 4 Block Diagrams 5 Feedback System Characteristics Assign 1 Due 6 Root Locus 7 Root Locus 2 Assign 2 Due 8 Bode Plots 9 Bode Plots 2 10 State Space Modeling Assign 3 Due 11 State Space Design Techniques 12 Advanced Control Topics 13 Review Assign 4 Due Dr. Ian R. Manchester Amme 3500 : Introduction Slide 2
Review mathematical justification for feedback over open loop control Introduce the PID controller Performance specification: steady-state error Introduction to parameter sensitivity aka robustness Slide 3
r(t) Desired Value + - error Controller Control Signal Plant Output Feedback Signal Error: e(t) = r(t) y(t) (Assuming perfect measurement) Slide 4
Perfect feed-forward controller: G n Output and error: Only zero when: (Perfect Knowledge) (No load or disturbance) Slide 5
reference r disturbance d output controller plant A sensor unity feedback y = GK 1+ KG r G 1+ GK d Reference to Output Transfer Function Disturbance to Output Transfer Function Slide 6
Equivalent Open Loop Block Diagram r d y = GK 1+ KG r G 1+ GK d Reference to Output Transfer Function Disturbance to Output Transfer Function Slide 7
Simplest possible feedback control useful for thermostats, similarly simple problems Often results in oscillations Rapid changes in control input are often infeasible Nonlinear: can be very difficult to analyse Slide 8
A very large number (at least 95%) of feedback controllers used in practice have the following simple form: u(t) = K P e + K I t e(τ)dτ + K 0 D e This is called a PID Proportional (P) term feeds back on the current error Integral (I) feedback can eliminate steady state error derivative (D) term can improve dynamic response Slide 9
Suppose we just use the P term, i.e. proportional control R(s) E(s) C(s) + - K G(s) We have seen how this form of feedback is able to minimize the effect of disturbances by increasing K This may adversely affect other performance, in particular overshoot Slide 10
k = 5 N m s/rad b = 6 N m s/rad * N.S. Nise (2004) Control Systems Engineering Wiley & Sons Let s compute an example of a torsional massspring-damper system with proportional control J = 1 N m, k = 5 N m/rad and b = 6 N m s/rad Slide 11
Recall the differential equation The transfer function is J θ + b θ + kθ = τ The closed loop response is R(s) E(s) C(s) + - K G(s) Slide 12
K=5 K=100 K=10 K=1000 Slide 13
Proportional (P) control acts like a spring pulling the output towards the reference. A very stiff spring (large gain K) can result in highly oscillatory response. Why not add damping? This is PD control R(s) E(s) U(s) C(s) K p +K d s G(s) + - u(t) = K P e + K D e Slide 14
D term adds damping and differentiates the reference Slide 15
Same torsional setup as before P control: K=100 PD control: K=100+20s Note the anticipatory kick at the beginning, and the damping of oscillations Slide 16
We have looked at transient response for second order systems This can be defined by Rise time, T r Settling time, T s Peak time, T p Percentage overshoot. These can all be specified by PD control c final =steady state response * N.S. Nise (2004) Control Systems Engineering Wiley & Sons Slide 17
Why use integral control? (The I part in PID) Another important characteristic of systems in general is their steady state performance input Steady state error Steady-state error is the difference between the reference input and output for a particular input as t time output Slide 18
Boeing and BAE systems Slide 19
Real systems (with non-linearities): friction/ stiction, backlash, dead-zones, etc. Sensor does not see error Actuator produces no output for command Linear systems: Caused by intrinsic dynamics or disturbance inputs SS error can be examined analytically We can eliminate it through integral action Slide 20
x d x m b F(s) 1 ------ ms 2 +bs X(s) f x d x m b X d (s) + - E(s) kc F(s) 1 ------ ms 2 +bs X(s) k c Slide 21
x d x k m b F(s) 1 ------ ms 2 +bs+k X(s) f x d x k m b X d (s) + - E(s) k c F(s) 1 ------ ms 2 +bs+k X(s) k c Slide 22
m x x d g m x x d b f b k c G(s) = mg/s G(s) F(s) + + 1 ------ ms 2 +bs X(s) X d (s) + - k c + + 1 ------ ms 2 +bs X(s) Slide 23
Consider a unity feedback system. Transfer function T(s) R(s) E(s) C(s) + - G(s) What if we rearrange this to find E(s)? R(s) T(s) C(s) - + E(s) Slide 24
We can now find E(s) in terms of the input R (s) and the closed loop transfer function T(s) Or Although we can use E(s) to find e(t) and compute the error in time, we are often interested in the steady state error Slide 25
Start with LT for the derivative: Take the limit as s 0 Slide 26
The Final Value Theorem can be used to find the steady state value without taking inverse Laplace transforms This only applies if the system is stable Slide 27
* N.S. Nise (2004) Control Systems Engineering Wiley & Sons We ll now look at the steady state errors to some common input functions (polynomials) Slide 28
Position Velocity Acceleration http://www.oceancontrols.com.au Slide 29
For a step input (R(s)=1/s), we find: If lim s 0 G(s) is finite, we define to be the position error constant. If there is a pole at the origin then steady-state error is zero sr(s) 1+ G(s) s(1/s) 1+ G(s) e( ) = lim s 0 = lim s 0 = 1 1+ lim s 0 G(s) Slide 30
For a general G(s) The denominator must be zero for zero steady state error. Therefore n 1 Implies at least one pole must be at the origin In order to have zero steady state error, G(s) must approach as s approaches 0 Slide 31
For a system with no poles at the origin there will be some constant error to a step input For one or more poles at the origin, the error will be integrated away and will reach zero in the steady-state Input e( ) Slide 32
For a ramp input (R(s)=1/s 2 ), we find We can define the velocity error constant, K v, such that Slide 33
Once again, in order to have zero steady state error, sg(s) must approach as s approaches 0 Implies at least two poles must be at the origin Slide 34
Not surprisingly, for a parabolic input (R(s)=1/s 3 ), we find We can define the acceleration error constant, K a, such that Slide 35
Only one is useful for a given system: K step = lim s 0 G(s) K v = lim s 0 sg(s) K a = lim s 0 s 2 G(s) (book calls this K p, don t confuse with proportional control gain!!) They can be determined from the OL transfer function using the formulae presented Only for unity feedback, stable systems. Slide 36
For unity negative feedback, the steady-state errors are dependent on the number of pure integrations in the forward path We define system type to be the number of poles n in the forward path (either from controller or plant) A system with n = 0 is Type 0. If n = 1 or n = 2, the corresponding system type is Type 1 or Type 2 respectively * N.S. Nise (2004) Control Systems Engineering Wiley & Sons Slide 37
This table shows the relationship between input, system type, error constants and steady-state error. Note: only for unity feedback cases! * N.S. Nise (2004) Control Systems Engineering Wiley & Sons Slide 38
This just scratches the surface: a very important result in control theory is the internal model principle: For any feedback system to give zero error for a particular class of reference or disturbance inputs, the system itself must be capable of internally reproducing all signals in this class. This can be interpreted like so: it is necessary for the system to internally estimate the state of the reference or disturbance If you do advanced control, you will learn about state observers and Kalman filters, which explicitly perform this task Slide 39
* N.S. Nise (2004) Control Systems Engineering Wiley & Sons Find the steady state error for inputs of r(t)=5u(t), r(t) =5tu(t) and r(t)=5t 2 u(t), where u(t) refers to step input First verify that the system is stable Yes System type? Type 1 Slide 40
For step r(t)=5u(t) For ramp r(t)=5tu(t) For parabola r(t)=5t 2 u(t) Slide 41
Just as damping ratio ζ, settling time T s, peak time T p, and percent overshoot are used as specifications for transient response, The position constant, K step, velocity constant, K v, and acceleration constant, K a, can be used as specifications for steady-state error Slide 42
u(t) = K P e + K I t e(τ)dτ + K 0 D e Intuitively: if the controller continues to see an error signal over a long time, then the integral term builds up, and increases control signal Slowly it builds up to the level required to completely kill the error Slide 43
For a PD controller, if the error and its derivative are both zero, then u = K P e + K D e = 0 So, a zero-steady-state-error response to a step must result in zero control input. This is not possible if there is some force (e.g. a spring, or gravity, or wind) pulling the system away from the reference Slide 44
Given this control system, find K so that there is 10% error in the steady state response to a ramp input System Stable? Yes System Type? Type 1 * N.S. Nise (2004) Control Systems Engineering Wiley & Sons Slide 45
For LTI systems we have assumed that the parameters of the system are perfectly known However, there are instances in which the parameters of a system may change due to, for example: changes in ambient temperature, Altitude wear Manufacturing tolerances It is important to understand the degree to which changes in system parameters affect system performance. This is called sensitivity analysis Slide 46
Sensitivity is defined as the normalised ratio of the change in the transfer function T to the change in the plant G: S(s) = T(s) /T(s) G(s) /G(s) = T(s) G(s) Note that S is itself a transfer function: G(s) T(s) The sensitivity to plant changes is dynamic the feedback system may be more sensitive to plant changes when subjected to some inputs than others Slide 47
T(s) = Y(s) R(s) = G(s) G n (s) S = T(s) G(s) G(s) T(s) T(s) G(s) = 1 G n (s) S = 1 G n 1 G n =1 Slide 48
r d T(s) = Y(s) R(s) = T(s) G(s) = = KG 1+ KG K 1+ KG K 2 G 1+ KG K ( 1+ KG) 2 ( ) 2 S = = T /T G /G = 1 1+ GK With very high gains: K G(1+ KG) (1+ KG) 2 GK lim S = 0 K Slide 49
We can see a pattern in many of the sensitivity relations of feedback systems Y(s) R(s) = G(s)K(s) 1+ G(s)K(s) Response to reference command: (should = 1) Y(s) N(s) = G(s)K(s) 1+ G(s)K(s) Response to measurement noise: (should = 0) Y(s) D(s) = 1 1+ G(s)K(s) Response to disturbance at output: (should = 0) dy(s) /Y(s) dg(s) /G(s) = 1 1+ G(s)K(s) Sensitivity to model uncertainty: (should = 0) Note also that 1 1+ G(s)K(s) + G(s)K(s) 1+ G(s)K(s) =1 Slide 50
Since they show up so much, they have been given names: The Sensitivity Function S(s) = The Complementary Sensitivity Function T(s) = 1 1+ G(s)K(s) G(s)K(s) 1+ G(s)K(s) Complementary because S+T=1 for all s Many advanced control techniques are based on shaping these transfer functions Slide 51
PID: u(t) = K P e + K I t 0 e(τ)dτ + K D e U(s) E(s) = K(s) = K P + K I s + K D s Error e(t) Slope of e(t) ~ derivative control term Area under curve ~ integral control term e(t) ~ proportional control Time t Slide 52
Slide 53
Same torsional setup as before P control: K=100 PD control: K(s)=100+10s PID control: K(s)=80+60/s+6s Note the zero steady-state error for PID Slide 54
PID with Different spring constants: k =5 (nominal), 7.5, and 2.5 Nm/rad Slide 55
r(t) G d G n d(t) Slide 56
Nise Sections 7.1-7.7 (Steady state errors) Franklin & Powell Section 4.1-4.3 Åström & Murray (free online) Chapter 10 Slide 57