Control Systems Lect 2 Time Response Basil Hamed

Control Systems Lect. 2 Time Response Basil Hamed

Roadmap (Time Responses) Basil Hamed 2

Why Study Time Responses • Modeling – Some parameters in the system can be estimated or identified by time responses. • Analysis – Evaluate transient and steady-state responses to see if they meets performance requirement (Satisfactory or not? ) • Design – Given design specs in terms of transient and steady state responses, design controllers satisfying all the design specs. Basil Hamed 3

4. 1 Introduction Transient steady state Steady state response is the part of the total response that remains after the transient has died out. Basil Hamed 4

Time Responses – Input and Output • We would like to analyze a system property by applying a test input r(t) and observing a time response y(t). • Time response can be divided as Transient response Steady-state response Basil Hamed 5

4. 2 Poles, Zeros, and System Response • The output response of a system is the sum of two responses: the forced response and the natural response. • Although many techniques, such as solving a differential equation or taking the inverse Laplace transform, enable us to evaluate this output response, these techniques are laborious and time-consuming. • The use of poles and zeros and their relationship to the time response of a system is such a technique. • The concept of poles and zeros, fundamental to the analysis and design of control systems, simplifies the evaluation of a system's response. Basil Hamed 6

Poles and Zeros of a First-Order System: An Example Basil Hamed 7

Poles and Zeros of a First-Order System: An Example Basil Hamed 8

Poles and Zeros of a First-Order System: An Example From the development summarized in pervious Figure, we draw the following conclusions: 1. A pole of the input function generates the form of the forced response 2. A pole of the transfer function generates the form of the natural response. 3. The farther to the left a pole is on the negative real axis, the faster the exponential transient response will decay to zero. 4. The zeros and poles generate the amplitudes for both the forced and natural responses. Basil Hamed 9

Example 4. 1 P 165 PROBLEM: Given the system of Figure below, write the output, c(t), in general terms. Specify the forced and natural parts of the solution. SOLUTION: By inspection, each system pole generates an exponential as part of the natural response. The input's pole generates the forced response. Thus, Taking the inverse Laplace transform, we get Basil Hamed 10

4. 3 First-Order Systems We now discuss first-order systems without zeros to define a performance specification for such a system. A first-order system without zeros can be described by the transfer function shown in Figure below. Basil Hamed 11

4. 3 First-Order Systems If the input is a unit step, where R(s) = 1/s, the Laplace transform of the step response is C(s), where Taking the inverse transform, the step response is given by t C(t) 0 0 1/a 0. 63 2/a 0. 86 3/a 0. 95 4/a 0. 98 Basil Hamed 12

4. 3 First-Order Systems Basil Hamed 13

4. 3 First-Order Systems Basil Hamed 14

4. 4 Second-Order Systems Basil Hamed 15

4. 4 Second-Order Systems There are 4 cases for 2 nd order system; 1. Overdamped Response 2. Underdamped Response 3. Undamped Response 4. Critically Damped Response Basil Hamed 16

Overdamped Response This function has a pole at the origin that comes from the unit step input and two real poles that come from the system. The input pole at the origin generates the constant forced response; each of the two system poles on the real axis generates an exponential natural response whose exponential frequency is equal to the pole location. Hence, the output initially could have been written as Basil Hamed 17

Underdamped Response Basil Hamed 18

Undamped Response This function has a pole at the origin that comes from the unit step input and two imaginary poles that come from the system. The input pole at the origin generates the constant forced response, and the two system poles on the imaginary axis at ±j 3 generate a sinusoidal natural response whose frequency is equal to the location of the imaginary poles. Hence, the output can be estimated as c(t) = K 1 + K 4 cos(3 t - ¢). Basil Hamed 19

Critically Damped Response This function has a pole at the origin that comes from the unit step input and two multiple real poles that come from the system. The input pole at the origin generates the constant forced response, and the two poles on the real axis at — 3 generate a natural response consisting of an exponential and an exponential multiplied by time. Hence, the output can be estimated as Basil Hamed 20

Natural Responses and Found Their Characteristics Basil Hamed 21

Step Responses for Second-Order System Damping Cases Basil Hamed 22

4. 5 The General Second-Order System • In this section, we define two physically meaningful specifications for second-order systems. • These quantities can be used to describe the characteristics of the second-order transient response just as time constants describe the first-order system response. • The two quantities are called natural frequency and damping ratio. Basil Hamed 23

4. 5 The General Second-Order System Basil Hamed 24

4. 5 The General Second-Order System Basil Hamed 25

4. 5 The General Second-Order System Basil Hamed 26

Second-Order Response as a Function of Damping Ratio Basil Hamed 27

Example 4. 4 P. 176 PROBLEM: For each of the systems shown below, find the value of ξ and report the kind of response expected. Basil Hamed 28

Example 4. 4 P. 176 SOLUTION: First match the form of these systems to the forms shown in Eqs. we find ξ = 1. 155 for system (a), which is thus overdamped, since ξ > 1; ξ = 1 for system (b), which is thus critically damped; and ξ = 0. 894 for system (c), which is thus underdamped, since ξ < 1. Basil Hamed 29

4. 6 Underdamped Second-Order Systems Basil Hamed 30

4. 6 Underdamped Second-Order Systems Let us begin by finding the step response for the general secondorder system of Eq. The transform of the response, C(s), is the transform of the input times the transfer function, or Taking the inverse Laplace transform, Basil Hamed 31

4. 6 Underdamped Second-Order Systems • A plot of this response appears in Figure below for various values of ξ , plotted along a time axis normalized to the natural frequency. • The lower the value of ξ, the more oscillatory the response. Basil Hamed 32

4. 6 Underdamped Second-Order Systems Basil Hamed 33

Second-Order Underdamped Response Specifications Basil Hamed 34

Typical Unit Step Response Basil Hamed 35

Typical Unit Step Response Basil Hamed 36

Second-Order Underdamped Response Specifications Peak time, Tp Percent overshoot, %OS: Basil Hamed 37

Second-Order Underdamped Response Specifications Basil Hamed 38

Peak Value, Peak Time, and Overshoot (%) Basil Hamed 39

Delay, Rise, and Settling Times Basil Hamed 40

2 nd Order System Properties Basil Hamed 41

2 nd Order System Remarks • Percent overshoot depends on z, but NOT wn. • From 2 nd-order transfer function, analytic expressions of delay & rise time are hard to obtain. • Time constant is 1/(zwn), indicating convergence speed. • For z >1, we cannot define peak time, peak value, percent overshoot (no overshoot). Basil Hamed 42

Example 4. 5 P. 182 Basil Hamed 43

Pole plot for an underdamped second-order system The pole plot for a general, underdamped second-order system, is shown in Figure below. Poles (0<z<1) cos θ=ξ Basil Hamed 44

Example 4. 6 P. 184 Basil Hamed 45

Example 4. 6 P. 184 Basil Hamed 46

Example 4. 7 P 185 PROBLEM: Given the system shown below, find J and D to yield 20% overshoot and a settling time of 2 seconds for a step input of torque T(t). SOLUTION: First, the transfer function for the system is Basil Hamed 47

Example 4. 7 P 185 From the transfer function, from the problem statement, a 20% overshoot implies ξ = 0. 456. Therefore From the problem statement, K = 5 N-m/rad. , D = 1. 04 N-ms/rad, and J = 0. 26 kg-m 2. Basil Hamed 48

4. 7 System Response with Additional Poles • In the last section, we analyzed systems with one or two poles. It must be emphasized that the formulas describing percent overshoot, settling time, and peak time were derived only for a system with two complex poles and no zeros. • If a system that has more than two poles or has zeros, we cannot use the formulas to calculate the performance specifications that we derived. • However, under certain conditions, a system with more than two poles or with zeros can be approximated as a second-order system that has just two complex dominant poles. • Once we justify this approximation, the formulas for percent overshoot, settling time, and peak time can be applied to these higher -order systems by using the location of the dominant poles. Basil Hamed 49

4. 7 System Response with Additional Poles or, in the time domain, Basil Hamed 50

4. 7 System Response with Additional Poles Basil Hamed 51

4. 7 System Response with Additional Poles Basil Hamed 52

Example 4. 8 P 189 PROBLEM: Find the step response of each of the transfer functions shown in below and compare them. SOLUTION: The step response, Ci(s), for the transfer function, Ti(s), can be found by multiplying the transfer function by 1/S, a step input, and using partial-fraction expansion Basil Hamed 53

Example 4. 8 P 189 the results are The three responses are plotted below. Notice that c 2(t), with its third pole at -10 and farthest from the dominant poles, is the better approximation of c 1(t), Basil Hamed 54

4. 10 Laplace Transform Solution of State Equations Taking the Laplace transform of both sides of the state equation yields Taking the Laplace transform of the output equation yields Basil Hamed 55

Eigenvalues and Transfer Function Poles We have The roots of denominator are the poles of the system The system poles = eigenvalues= det(s. I - A) = 0. Basil Hamed 56

Example 4. 11 P. 200 PROBLEM: Given the system represented in state space do the following: a. Solve the preceding state equation and obtain the output for the given exponential input. b. Find the eigenvalues and the system poles. Basil Hamed 57

Example 4. 11 P. 200 SOLUTION: we have then Basil Hamed 58

Example 4. 11 P. 200 The output equation is found b. The denominator of Eq. , which is det(s. I - A), is also the denominator of the system's transfer function. Thus, det(. sl - A) = 0 furnishes both the poles of the system and the eigenvalues - 2 , - 3, and -4. Basil Hamed 59

4. 11 Time Domain Solution of State Equations Basil Hamed 60

4. 11 Time Domain Solution of State Equations Basil Hamed 61

4. 11 Time Domain Solution of State Equations Natural Zero Input Response Basil Hamed Forced Zero State Response 62

Example 4. 12 P 204 PROBLEM: For the state equation and initial state vector shown below, where u(t) is a unit step, find the state-transition matrix and then solve for x(t). SOLUTION: Since the state equation is in the form Basil Hamed 63

Example 4. 12 P 204 Basil Hamed 64

Example 4. 12 P 204 Basil Hamed 65

Example Consider the state equation Solution: The coefficient matrices are identified to be Basil Hamed 66

Example The state-transition matrix of A is found by taking the inverse Laplace transform of above eq. Basil Hamed 67

Example The state-transition equation for t > 0 is obtained; OR Basil Hamed 68

Stability

Stability – A Simple Example We want the mass to stay at x = 0, but wind gave some initial speed (f(t) = 0). What will happen? How to characterize different behaviors with TF? Basil Hamed 70

Stability – Importance • The most basic and important specification in control analysis and synthesis! • Unstable systems have to be stabilized by feedback. • Unstable closed-loop systems are useless. – What happens if a system is unstable? • may hit mechanical/electrical “stops” (saturation) • may break down or burn out Basil Hamed 71

6. 1 Introduction • Three requirements enter into the design of a control system: transient response, stability, and steady-state errors. Thus far we have covered transient response, which we will revisit in Chapter 8. We are now ready to discuss the next requirement, stability. • Stability is the most important system specification. If a system is unstable, transient response and steady-state errors are moot points. • An unstable system cannot be designed for a specific transient response or steady-state error requirement. Basil Hamed 72

6. 1 Introduction What is stability? There are many definitions for stability, depending upon the kind of system or the point of view. In this section, we limit ourselves to linear, time-invariant systems. A system is stable if every bounded input yields a bounded output. We call this statement the bounded-input, bounded -output (BIBO) definition of stability. A system is unstable if any bounded input yields an unbounded output. Basil Hamed 73

Stability – Definition BIBO (Bounded-Input-Bounded-Output) stability : Any bounded input generates a bounded output Asymptotic stability : Any ICs generates y(t) converging to zero. Basil Hamed 74

6. 1 Introduction we present the following definitions of stability, instability, and marginal stability: • A linear, time-invariant system is stable if the natural response approaches zero as time approaches infinity. • A linear, time-invariant system is unstable if the natural response grows without bound as time approaches infinity. • A linear, time-invariant system is marginally stable if the natural response neither decays nor grows but remains constant or oscillates as time approaches infinity. Basil Hamed 75

6. 1 Introduction How do we determine if a system is stable? Let us focus on the natural response definitions of stability • Stable systems have closed-loop transfer functions with poles only in the left half-plane. • If the closed-loop system poles are in the right half of the splane and hence have a positive real part, the system is unstable. Basil Hamed 76

6. 1 Introduction Basil Hamed 77

6. 1 Introduction Basil Hamed 78

Stability – “s” Domain Stability For a system represented by a transfer function G(s), system is BIBO stable All the poles of G(s) are in the open left half of the complex plane system is asymptotically stable Basil Hamed 79

Stability – Remarks on Stability Definition • For a general system (nonlinear etc. ), BIBO stability condition and asymptotic stability condition are different. • For linear time-invariant (LTI) systems (to which we can use Laplace transform and we can obtain a transfer function), the conditions happen to be the same. • In this course, we are interested in only LTI systems, we use simply “stable” to mean both BIBO and asymptotic stability. Basil Hamed 80

Stability – Examples Basil Hamed 81

Stability – Summary • Stability for LTI systems – (BIBO and asymptotically) stable, marginally stable, unstable – Stability for G (s) is determined by poles of G. • Next – Routh-Hurwitz stability criterion to determine stability without explicitly computing the poles of a system. Basil Hamed 82

6. 2 Routh-Hurwitz Criterion In this section, we learn a method that yields stability information without the need to solve for the closed-loop system poles. Using this method, we can tell how many closed-loop system poles are in the left half-plane, in the right half-plane, and on the jw-axis. (Notice that we say how many, not where. ) The method requires two steps: (1) Generate a data table called a Routh table (2) interpret the Routh table to tell how many closed-loop system poles are in the LHP, the RHP, and on the jw-axis. Basil Hamed 83

Generating a Basic Routh Table Look at the equivalent closed-loop transfer function shown in Fig Since we are interested in the system poles, we focus our attention on the denominator. In order the above characteristic eq. does not have roots in RHP, it is necessary but not sufficient that the following hold; 1. All the coefficient of the polynomial have the same sign. 2. None of the coefficient vanishes. Basil Hamed 84

Generating a Basic Routh Table We first create the Routh table shown in Table below Basil Hamed 85

Generating a Basic Routh Table Only the first 2 rows of the array are obtained from the characteristic eq. the remaining are calculated as follows; Basil Hamed 86

Stability – Routh-Hurwitz Example Find the stability of the characteristic eq; Basil Hamed 87

Example 6. 1 P. 306 PROBLEM: Make the Routh table for the system shown in Figure Basil Hamed 88

Example 6. 1 P. 306 SOLUTION: The first step is to find the equivalent closed-loop system because we want to test the denominator of this function. System is unstable and has 2 poles in RHP. Basil Hamed 89

Example 6. 9 P. 318 PROBLEM: Find the range of gain, K, for the system shown, that will cause the system to be stable, unstable, and marginally stable. Assume K > 0. SOLUTION: First find the closed-loop transfer function as i) For stable System 0<k<1386 ii) For Unstable System k >1386 iii) For marginal stable k= 1386 Basil Hamed 90

6. 5 Stability in State Space • Up to this point we have examined stability from the s-plane viewpoint. Now we look at stability from the perspective of state space. • Stability of State Space depends on eigenvalues of matrix A. Because the values of the system's poles are equal to the eigenvalues of the system matrix, A. • eigenvalues of matrix A is found by det(s. I - A) = 0 Basil Hamed 91

Example 6. 11 P. 321 PROBLEM: Given the system Find out how many poles are in the left half-plane, in the right halfplane, and on the jw-axis. SOLUTION: First form (s. I - A): Basil Hamed 92

Example 6. 11 P. 321 Now find the det(s. I — A): Since there is one sign change in the first column, the system has one right half- plane pole and two left-halfplane poles. It is therefore unstable. Basil Hamed 93

HW # 2 4. 12, 4. 26, 4. 29, 4. 39, 5. 13, 5. 16, 5. 27, 6. 31, 6. 35 Due next class Basil Hamed 94