Course Outline. Block Diagrams. Block Diagrams. Amme 3500 : System Dynamics and Control. Block Diagrams. Dr. Stefan B. Williams

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1 Amme 3500 : System Dynamics and Control Block Diagrams Dr Stefan Williams Course Outline Week Date Content Assignment Notes Mar Introduction 2 8 Mar Frequency Domain Modelling 3 5 Mar Transient Performance and the splane 4 22 Mar Block Diagrams Assign Due 5 29 Mar Feedback System Characteristics 6 5 Apr Root Locus Assign 2 Due 7 2 Apr Root Locus Apr Bode Plots No Tutorials 26 Apr BREAK 9 3 May Bode Plots 2 Assign 3 Due 0 0 May State Space Modeling 7 May State Space Design Techniques 2 24 May Advanced Control Topics 3 3 May Review Assign 4 Due 4 Spare Dr. Stefan B. Williams Amme 3500 : Introduction Slide 2 Block Diagrams As we saw in the introductory lecture, a subsystem can be represented with an input, an output and a transfer function U(s) H(s) Y(s) Block Diagrams Many systems are composed of multiple subsystems In this lecture we will examine methods for combining subsystems and simplifying block diagrams Input: control surfaces (flap, aileron), wind gust Aircraft output: pitch, yaw, roll Slide 3 Slide 4

2 Examples of subsystems Automobile control: Examples of Subsystems Antenna control Desired course of travel Control Steering Mechanism Automobile Actual course of travel Sensing Measurements Slide 5 Slide 6 Mathematical Modelling Mathematical Modelling u(t) h(t) y(t) U(s) H(s) Y(s) In the time domain, the inputoutput relationship is usually expressed in terms of a differential equation n n m d y() t d y() t d u() t an a 0y() t bm bu 0 () t n L = L n m dt dt dt (a n,, a 0, b m, b 0 ) are the system s parameters, n m The system is LTI (Linear Time Invariant) iff the parameters are timeinvariant n is the order of the system Slide 7 In the Laplace domain, the inputoutput relationship is usually expressed in terms of an algebraic equation in terms of s sy() s a s Y() s L ay() s = bsu() s L bu() s n n m n 0 m 0 Slide 8 2

3 Cascaded systems In time, a cascaded system requires a convolution In the Laplace domain, this is simply a product u u(t) U(s) ( t) h( t) u( τ ) h( t τ ) dτ h (t) H(s) y (t) Y (s) h (t) H (s) Ys () = UsHsH () () () s y(t) yt () = ( ut ()* ht ())* h() t Y(s) Slide 9 A eneral Control System Many control systems can be characterised by these components/subsystems Reference D(s) Error E(s) Feedback H(s) Control Control Signal U(s) Actuator Sensor Plant Sensor Noise Disturbance Process Output Y(s) Slide 0 Simplifying Block Diagrams The objectives of a block diagram are To provide an understanding of the signal flows in the system To enable modelling of complex systems from simple building blocks To allow us to generate the overall system transfer function A few rules allow us to simplify complex block diagrams into familiar forms Components of a Block Diagram A block diagram is made up of signals, systems, summing junctions and pickoff points Slide Slide 2 3

4 Typical Block Diagram Elements Cascaded Systems Typical Block Diagram Elements Parallel Systems Slide 3 Slide 4 Typical Block Diagram Elements Feedback Form Es () = Rs () HsCs () () Cs () = ses () () Cs () = Rs () HsCs () () s () ( ) srs () () = Hss () () Cs () Cs () s () = Rs () Hss () () Moving Past A Summation a) To the left past a summing junction b) To the right past a summing junction Slide 5 Slide 6 4

5 Moving Past PickOff Points a) To the left past a summing junction b) To the right past a summing junction Example I : Simplifying Block Diagrams Consider this block diagram We wish to find the transfer function between the input R(s) and C(s) Slide 7 Slide 8 Example I: Simplifying Block Diagrams a) Collapse the summing junctions b) Form equivalent cascaded system in the forward path and equivalent parallel system in the feedback path c) Form equivalent feedback system and multiply by cascaded (s) Slide 9 Example II : Simplifying Block Diagrams Form the equivalent system in the forward path Move /s to the left of the pickoff point Combine the parallel feedback paths Compute C(s)/R(s) using feedback formula Slide 20 5

6 Example III: Shuttle Pitch Control Example III: Shuttle Pitch Control Manipulating block diagrams is important but how do they relate to the real world? Here is an example of a real system that incorporates feedback to control the pitch of the vehicle (amongst other things) The control mechanisms available include the body flap, elevons and engines Measurements are made by the vehicle s inertial unit, gyros and accelerometers Slide 2 Slide 22 Example III : Shuttle Pitch Control Example III : Shuttle Pitch Control A simplified model of a pitch controller is shown for the space shuttle Assuming other inputs are zero, can we find the transfer function from commanded to actual pitch? Pitch Reference K K 2 (s) 2 (s) _s_ K Combine and 2. Push K to the right past the summing junction s 2 Output Slide 23 Slide 24 6

7 Example III: Shuttle Pitch Control Example III: Shuttle Pitch Control Pitch Reference K K 2 (s) 2 (s) _s 2 _ K K 2 _s_ K Output Matlab provides us with a tool called Simulink that allows us to represent systems by their block diagram We ll take a bit of time to see how we construct a Simulink model of this system Push K K 2 to the right past the summing junction Hence KK 2 () s2() s T() s = 2 s s KK 2 ( s ) 2( s) K KK2 Slide 25 Slide 26 Closed Loop Feedback We have suggested that many control systems take on a familiar feedback form Intuition tells us that feedback is useful imagine driving your car with your eyes closed Let us examine why this is the case from a mathematical perspective SingleLoop Feedback System Error Signal et () = dt () f() t = dt () Kyt () The goal of the Controller K(s) is: Ø To produce a control signal u(t) that drives the error e(t) to zero dt () et () ut () yt () Ks () s () error Desired Value f() t Feedback Signal Controller H Control Signal Transducer Plant Output Slide 27 Slide 28 7

8 Controller Objectives Controller cannot drive error to zero instantaneously as the plant (s) has dynamics Clearly a large control signal will move the plant more quickly The gain of the controller should be large so that even small values of e(t) will produce large values of u(t) However, large values of gain will cause instability Feedforward & Feedback Driving a Formula car Feedforward control: preemptive/predictable action: prediction of car s direction of travel (assuming the model is accurate enough) Feeback control: corrective action since the model is not perfect noise Slide 29 Slide 30 Why Use Feedback? Perfect feedforward controller: n dt () lt () yt () / n Output and error: Only zero when: y = d l n e= d y= d l n n = l = 0 (Perfect Knowledge) (No load or disturbance) Why Use Feedback? load l d e u demand K controller plant No dynamics Here! unity feedback So: e= d y, y= ( u l), u = Ke e= d y= d ( u l) = d ( Ke l) y output A sensor Slide 3 Slide 32 8

9 Collect terms: Or: Closed Loop Equations e= d Ke l ( K) e = d l e= d l K K Demand to Error Transfer Function Load to Error Transfer Function Similarly d Closed Loop Equations K y = d l K K Demand to Output Transfer Function Equivalent Open Loop Block Diagram K l Load to Output Transfer Function K y Slide 33 Slide 34 Rejecting Loads and Disturbances Load to Output y = l K Transfer Function Tracking Performance K Demand to Output y= d Transfer Function K If K is big: K >> If K is big: K >> If K is big: If K is big: y l l 0 K = K Perfect Disturbance Rejection Independent of knowing Regardless of l K y d = d K Perfect Tracking of Demand Independent of knowing Regardless of l Slide 35 Slide 36 9

10 Two Key Problems Two Key Problems d e u K l y d e u K l y n Power: u = K( d y) Large K requires large actuator power u Noise: u = K( d y n) u Kd Large K amplifies sensor noise In practise a Compromise K is required Slide 37 Slide 38 Conclusions Further Reading We have examined methods for computing the transfer function by reducing block diagrams to simple form We have also presented arguments for using feedback in control systems Next week, we will look at more closely on feedback control Nise Sections Franklin & Powell Section 3.2 Slide 39 Slide 40 0