Dr. Ian R. Manchester


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1 Dr Ian R. Manchester
2 Week Content Notes 1 Introduction 2 Frequency Domain Modelling 3 Transient Performance and the splane 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 Assign 3 Due 10 State Space Modeling 11 State Space Design Techniques 12 Advanced Control Topics 13 Review Assign 4 Due Slide 2
3 The concept of a linear time invariant (LTI) Response of LTI systems to basic inputs Impulse Response Step Response Frequency Response The Laplace Transform Transfer Functions of linear systems Assignment 1 Slide 3
4 M f(t) y(t) In the last lecture, we considered this mechanical system We derived the following differential equation: m y = " F m y = f (t) # Ky(t) # K d y (t) m y + K d y (t) + Ky(t) = f (t) Slide 4
5 u(t) Plant y(t) The inputoutput relationship of physical systems can usually be modelled as a differential equation: Can we solve it? If we know u(t) can we find y(t)? Slide 5
6 This is called a Linear Time Invariant (LTI) System u 1 u 2 Plant y u 3 Superposition y(t) = f (u 1 (t) + u 2 (t) +...) = f (u 1 (t)) + f (u 2 (t)) +... Homogeneity y(t) = f (au(t)) = af (u(t)) Slide 6
7 vid Slide 7
8 Slide 8
9 Response of a car suspension system to a pothole Flexing modes of an aircraft wing Or a highprecision industrial robot arm The response of blood glucose concentration, insulin production, etc after eating a meal Slide 9
10 An impulse is an infinitely short pulse at t =0 Any signal can be thought of as the summation (integral) of many impulses at different points in time By the principal of superposition, if we can find the response of the system to one impulse, we will be able to find the response to an arbitrary input f(t) t Slide 10
11 Suppose the input consisted of just three impulses at times t = 0, 1, 2 of size 7, 8, 9. What is the value of y(t) at time t = 5? y = f (u 1 + u 2 + u 3 ) = f (u 1 ) + f (u 2 ) + f (u 3 ) = 7h(5) + 8h(4) + 9h(3) Slide 11
12 The output is the sum (integral) of each impulse response of the system to each individual impulse of the input, delayed by the appropriate time y(t) = % $ 0 h(")u(t #")d" We call this convolution, and it is written as y(t) = h(t) "u(t) Slide 12
13 Assume we have a system described by the following differential equation y (t) + ky(t) = u(t) The impulse response for this system is Slide 13
14 If we want to find the response of the system to a sinusoidal input, sin(!t) Slide 14
15 Now what happens if we have multiple components in the system? u(t) h (t) y 1 (t) h 1 (t) y(t) Slide 15
16 This provides us with a rich basis function for describing functions Through Euler s formula, we find Fourier analysis tells us that this is sufficient for representing any signal Slide 16
17 We can also consider the response of a system to sinusoidal inputs Fourier theory tells us that all signals can be decomposed into a sums of sinusoids Our basis signals are complex exponentials: u(t) = e st = e (" + j# )t Slide 17
18 In general Where Slide 18
19 Step input f (t) =1 Laplace Tr lim R " # % 0 R & = lim ( R " #' e $st dt e $st $s ) + * R 0 e $sr $1 = lim R " # $s If s > 0 = 1 s Slide 19
20 What about that nasty integral in the Laplace operation? We normally use tables of Laplace transforms rather than solving the preceding equations directly This greatly simplifies the transformation process Slide 20
21 Slide 21
22 Convolution in time is equivalent to multiplication in the Laplace domain U(s) H(s) Y 1 (s) H 1 (s) Y(s) Slide 22
23 Starting with an impulse response, h(t), and an input, u(t), find y(t) u(t), h(t) convolution y(t) L L 1 U(s), H(s) Multiplication, algebraic manipulation Y(s) Slide 23
24 The transfer function H(s) of a system is defined as the ratio of the Laplace transforms output and input with zero initial conditions H(s) = Y(s) U(s) It is also Laplace transform of the impulse response Slide 24
25 Recall the system we examined earlier y (t) + ky(t) = u(t) To find the transfer function for this system, we perform the following steps sy(s) " y(0) + ky(s) = U(s) Y(s)(s + k) = U(s) H(s) = Y(s) U(s) = 1 s + k or Slide 25
26 The output response of a system is the sum of two related responses The natural response describes dissipation, oscillation, or unstable growth from initial conditions The forced response describes how the system reacts to external inputs Together these elements determine the overall response of the system Slide 26
27 M f(t) y(t) Earlier, we considered this mechanical system Now we can attempt to find y(t) M y (t) + K d y (t) + Ky(t) = f (t) M(s 2 Y(s) " sy(0) " y (0)) + K d (sy(s) " y(0)) + KY(s) = F(s) Slide 27
28 The natural response of the system can be found by assuming no input force Taking inverse Laplace transform will give us the system response Slide 28
29 Suppose we wish to find the natural response of the system to an initial displacement of 0.5m Assume the mass of the system is 1kg and the damping constant k d is 4 Nsec/m and the spring constant k is 3 N/m Slide 29
30 By Partial Fraction expansion we find Slide 30
31 What if we change the spring constant to 20 N/m? Slide 31
32 Assuming zero initial conditions, the forced response will be This is the transfer function for this system Slide 32
33 Suppose we wish to find the response of the system to a unit step input force f(t) Assume the mass of the system is 1kg and the systems constant k d is 4 Nsec/m and k is 3 N/m Slide 33
34 We have found We wish to write Y(s) in terms of its partialfraction expansion Slide 34
35 So We now take inverse Laplace transforms of these simple transforms Slide 35
36 This represents the step response for this system This plot shows the response y(t) vs time Slide 36
37 %! % Example 1: Step response for! % Y(s)/F(s) = 1/(s^2 + 4s + 3)! %! sys = tf(1,[1,4,3]);! step(sys);! hold on;! pause;! %! % Solving for time domain response yields! % y(t) = 1/31/2*exp(t) + 1/6*exp(3*t)! %! t = 0:0.1:6;! y_t = 1/31/2*exp(t) + 1/6*exp(3*t);! plot(t, y_t, 'r');! pause;! Slide 37
38 The poles of a transfer function are the values of the LT variables, s, that cause the transfer function to become infinite The zeros of a transfer function are the values of the LT variables, s, that cause the transfer function to go to zero A qualitative understanding of the effect of poles and zeros can help us to quickly estimate performance Slide 38
39 Recall Im{s} x x x Re{s} Slide 39
40 What if we change the spring constant k to 20 N/m? Im{s} x x Re{s} x Slide 40
41 Notice that the step response has changed in response to the change in the modified parameter k The steady state response and transient behaviour are different Slide 41
42 %! % Example 1: Step response for! % Y(s)/F(s) = 1/(s^2 + 4s + 20)! %! sys = tf(1,[1,4,20]);! step(sys);! hold on;! pause;! Slide 42
43 A first order system without zeros can be described by: The resulting impulse response is When! > 0, pole is located at s < 0, exponential decays = stable! < 0, pole is at s > 0, exponential grows = unstable Step Response Slide 43
44 From the preceding we can conclude that in general, if the poles of the system are on the left hand side of the splane, the system is stable Poles in the right hand plane will introduce components that grow without bound stable Im{s} unstable Re{s} Slide 44
45 Consider the modelling of an automobile suspension system. Wheel can be modelled as a stiff spring with the suspension modelled as a spring and damper in parallel Interested in describing the motion of the mass in response to changing conditions of the road K s K w M K d y(t) u(t) Slide 45
46 The Linear Time Invariant abstraction allows us to completely understand system response by looking at certain basic responses (impulse, step, frequency) The Laplace transform (of signals) and transfer functions (of systems) are a very convenient representations for analysis Slide 46
47 Nise Section and Franklin & Powell Section 3.1 Murray and Åström (online) Chapter 8 (somewhat different presentation) Slide 47
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