System Dynamics Ch2 PDF

C H 2 A P T E R Dynamic Response and the Laplace Transform Method CHAPTER OUTLINE 2.1 2.2 2.3 Differential Equations 31 The Laplace Transform Method 37 Solving Equations with the Laplace Transform 47 2.4 Partial-Fraction Expansion 61 2.5 Response Parameters and Stability 70 2.6 Transfer Fun...

C H 2 A P T E R Dynamic Response and the Laplace Transform Method CHAPTER OUTLINE 2.1 2.2 2.3 Differential Equations 31 The Laplace Transform Method 37 Solving Equations with the Laplace Transform 47 2.4 Partial-Fraction Expansion 61 2.5 Response Parameters and Stability 70 2.6 Transfer Functions 82 2.7 The Impulse and Numerator Dynamics 85 2.8 Additional Examples 91 2.9 Computing Expansion Coefficients with MATLAB 98 2.10 Transfer-Function Analysis in MATLAB 100 2.11 Chapter Review 107 References 108 Problems 108 CHAPTER OBJECTIVES When you have finished this chapter you should be able to do the following: 1. Apply direct integration, separation of variables, or the Laplace transform method, whichever is appropriate, to obtain the solution of a linear differential equation model. 2. When applying the Laplace transform method, be able to perform the appropriate expansion and apply the appropriate transform properties to obtain the inverse transform. 3. Identify the free, forced, transient, and steady-state components of the complete response. 4. Be able to identify and interpret the time constant, oscillation frequency, and damping ratio. 5. Obtain transfer functions from models expressed as single equations or as sets of equations. 6. Evaluate the effects of impulse inputs and input derivatives on the response. 7. Use MATLAB to apply the chapter’s methods. ynamic models are differential equations that describe a dynamic system. In this chapter, we develop methods for obtaining analytical solutions to differential equations commonly found in engineering applications. In Section 2.1, we introduce some important concepts and terminology associated with differential equations, and present methods for quickly solving simple differential equations. D 30 2. 1 Differential Equations An important tool for obtaining solutions in general is the Laplace transform, introduced in Section 2.2 and applied in Section 2.3. Section 2.4 introduces the concepts of free, forced, transient, and steady-state response, and explains the important concept of stability. The transform also forms the basis for the important concept of the transfer function, covered in Section 2.5. The effects of impulse inputs and input derivatives on a system’s dynamic response are treated in Section 2.6. Section 2.7 presents additional, in-depth examples. Sections 2.8 and 2.9 show how to use MATLAB to obtain partial-fraction expansions and to obtain responses to step, impulse, and other input function types. A review of the chapter’s main concepts is given in Section 2.10. ■ 2.1 DIFFERENTIAL EQUATIONS An ordinary differential equation (ODE) is an equation containing ordinary, but not partial, derivatives of the dependent variable. Because the subject of system dynamics is time-dependent behavior, the independent variable in our ODEs will be time t. We will often denote the time derivative with an overdot, as d2 y dx ẍ = 2 dt dt In a standard form for expressing an ODE, all functions of the dependent variable are placed on the left-hand side of the equal (=) sign, and all isolated constants and isolated functions of time are placed on the right-hand side. The quantities in the righthand side are called the input, or forcing function. The dependent variable is called the solution or the response. For example, consider the equation 3 ẍ +7ẋ +2t 2 x = 5 + sin t, where x is the dependent variable. The input is 5 + sin t and the response is x(t). If the right-hand side is zero, the equation is said to be homogeneous; otherwise, it is nonhomogeneous. ẋ = 2.1.1 INITIAL CONDITIONS An example of an ODE is 2ẋ + 6x = 3, where x is the dependent variable. “Solving the equation” means to obtain the function x(t) that satisfies the equation. For this example, the function is x(t) = Ce−3t + 0.5, where C is a constant. We cannot determine a numerical value for C unless we are given a specified value for x at some time t. Most commonly x is specified at some starting time, usually denoted t0 . The specified value of x at t0 is denoted x0 and is called the initial condition. Often the starting time t0 is taken to be at t = 0. When solving differential equations, you need never wonder if your answer is correct, because you can always check your answer by substituting it into the differential equation and by evaluating the solution at t = t0 . 2.1.2 CLASSIFICATION OF DIFFERENTIAL EQUATIONS We can categorize differential equations as linear or nonlinear. Linear differential equations are recognized by the fact that they contain only linear functions of the dependent variable and its derivatives. Nonlinear functions of the independent variable do not make a differential equation nonlinear. For example, the following equations are linear. ẋ + 3x = 5 + t 2 ẋ + 3t 2 x = 5 3ẍ + 7ẋ + 2t 2 x = sin t 31 32 CHAPTER 2 Dynamic Response and the Laplace Transform Method whereas the following equations are nonlinear: 2ẍ + 7ẋ + 6x 2 = 5 + t 2 , because of x 2 3ẍ + 5ẋ 2 + 8x = 4, because of ẋ 2 ẍ + 4x ẋ + 3x = 10, because of x ẋ The equation ẋ + 3t 2 x = 5 is a variable-coefficient differential equation, so named because one of its coefficients is a function of the independent variable t. By contrast, the equation ẋ + 2x = 5 is a constant-coefficient differential equation. When solving constant-coefficient equations, the initial time t0 can always be chosen to be 0. This simplifies the solution form. The order of the equation is the order of the highest derivative of the dependent variable in the equation. The equation 3ẍ + 7ẋ + 2x = 5 is thus called a second-order differential equation. A model can consist of more than one equation. For example, the model 3ẋ 1 + 5x1 − 7x2 = 5 ẋ 2 + 4x1 + 6x2 = 0 consists of two equations that must be solved simultaneously to obtain the solution for the two dependent variables x1 (t) and x2 (t). The equations are said to be coupled. Although each equation is first order, the set can be converted into a single differential equation of second order. Thus, solving a set of two coupled first-order equations is equivalent to solving a single second-order equation. In general, a coupled set of differential equations can be reduced to a single differential equation whose order is the sum of the orders of the individual equations in the set. 2.1.3 SOLUTION BY DIRECT INTEGRATION With some first-order equations, we can isolate the derivative on the left, as dx = f (t) dt In this case, we can integrate both sides to obtain  x(t)  t dx = f (t)dt x(t0 ) or t0  x(t) = x(t0 ) + t f (t)dt (2.1.1) t0 For example, the relation between the acceleration a(t) and velocity v(t) of an object is a differential equation, and we can obtain the velocity v(t) with direct integration as follows:  t dv = a(t), v(t) − v(t0 ) = a(t)dt dt t0 The relation between the velocity v(t) and the displacement x(t) is also a differential equation, and we can obtain the displacement x(t) as follows:  t dx = v(t), x(t) − x(t0 ) = v(t)dt dt t0 2. 1 Differential Equations 33 Suppose that a(t) = 6t 2 , t0 = 0, and v(0) = 5. Then  t 6t 2 dt = 2t 3 v(t) − 5 = 0 Thus v(t) = 5 + 2t . Now if the initial displacement is x(0) = 7,  t t4 x(t) − 7 = (5 + 2t 3 )dt = 5t + 2 0 3 2.1.4 SEPARATION OF VARIABLES You can solve the equation ẋ = g(t) f (x) (2.1.2) by separating the variables x and t as follows. First write the equation as dx = g(t) dt f (x) Then integrate both sides to obtain  x(t) x(0) dx dx = f (x)  t g(t) dt (2.1.3) 0 The solution x(t) can be found if the integrals on the left and on the right can be evaluated and if the resulting expression can be solved for x as a function of t. Separation of Variables for a Linear Equation ■ Problem Use separation of variables to solve the following problem for t ≥ 0. ẋ + 2x = 20 x(0) = 3 (1) ■ Solution Comparing this with (2.1.2), we see that f (x) = 20 − 2x and g(t) = 1. First write the equation as dx dx = 20 − 2x or = dt dt 20 − 2x Integrate both sides to obtain  3 x(t) dx dx = 20 − 2x  t dt = t 0 The integral on the left can be evaluated as follows: ln[20 − 2x(t)] − ln[20 − 2(3)] = −2t Solve for x(t) to obtain x(t) = 10 − 7e−2t The plot of this function is shown in Figure 2.1.1. (2) E X A M P L E 2.1.1 Figure 2.1.1 Response for Example 2.1.1. CHAPTER 2 Dynamic Response and the Laplace Transform Method 12 10 8 x(t ) 34 6 4 2 0 0 0.2 0.4 0.6 0.8 1 t 1.2 1.4 1.6 1.8 2 2.1.5 TRIAL-SOLUTION METHOD We can use the results of Example 2.1.1 to gain insight into the solution of the equation ẋ + ax = b, where a = 0. Its solution has the form x(t) = C + Dest (2.1.4) where C, D, and s are constants to be determined. We can verify that this form is the solution by substituting x(t) into the differential equation, as follows:   ẋ + ax = s Dest + a C + Dest = (s + a)Dest + aC = b The only way this equation can be true is if s + a = 0 and aC = b. Thus s = −a and C = b/a. The remaining constant, D, can be determined from the initial value x(0) as follows. Substituting t = 0 into the solution form gives x(0) = C + De0 = C + D. Thus D = x(0) − C = x(0) − b/a, and the solution can be written as   b −at b (2.1.5) x(t) = + x(0) − e a a The exponential coefficient s is called the characteristic root, and its equation s + a = 0 is called the characteristic equation. Characteristic roots are of great use in determining the form of the trial solution. This insight leads to the trial-solution method for solving equations. It can be used to solve higher-order equations as well. The method is useful for quickly obtaining solutions of common ODEs whose solution forms are already known from experience. 2.1.6 ROOTS AND COMPLEX NUMBERS Many of the equations of system dynamics are second-order or higher, and thus we will need to work with equations, such as roots of characteristic polynomials, that involve complex numbers. Table 2.1.1 gives some properties of complex numbers. You should 2. 1 Differential Equations 35 Table 2.1.1 Roots and complex numbers. The quadratic formula The roots of as 2 + bs + c = 0 are given by √ −b ± b2 − 4ac s= 2a For complex roots, s = σ ± jω, the quadratic can be expressed as   as 2 + bs + c = a (s + σ )2 + ω2 = 0 Complex numbers Rectangular representation: √ z = x + j y, j = −1 Complex conjugate: z = x − jy Magnitude and angle: |z| = θ =  z = tan−1 x 2 + y2 y (See Figure 2.1.2) x Polar and exponential representation: z = |z| θ = |z|e jθ Equality: If z 1 = x1 + j y1 and z 2 = x2 + j y2 , then z 1 = z 2 if x1 = x2 and y1 = y2 Addition: z 1 + z 2 = (x1 + x2 ) + j (y1 + y2 ) Multiplication: z 1 z 2 = |z 1 ||z 2 | (θ1 + θ2 ) z 1 z 2 = (x1 x2 − y1 y2 ) + j (x1 y2 + x2 y1 ) Complex-Conjugate Multiplication: (x + j y)(x − j y) = x 2 + y 2 Division: 1 1 x − jy = = 2 z x + yj x + y2 z1 |z 1 |  (θ − θ ) = 1 2 z2 |z 2 | z1 x1 + j y1 x1 + j y1 x2 − j y2 (x1 + j y1 )(x2 − j y2 ) = = = z2 x2 + j y2 x2 + j y2 x2 − j y2 x22 + y22 Figure 2.1.2 Polar representation of a complex number z = x + j y . Imaginary z y |z| ␪ x Real 36 CHAPTER 2 Figure 2.1.3 Polar representation of the complex number z1 = 3 + 4 j . Imaginary Dynamic Response and the Laplace Transform Method z1 5 3 1 4j 4 ␪1 5 tan21 4 3 5 0.927 rad 5 53.18 5 ␪ Real 3 review these properties if you have not used complex numbers in a while. For example, the complex number z 1 = 3 + 4 j can be represented in polar form by computing its magnitude |z 1 | using the Pythagorean theorem (see Figure 2.1.3). |z 1 | = 32 + 42 = 5 The angle θ1 is found from trigonometry. θ1 = tan−1 4 3 = 0.927 rad = 53.1◦ So the polar representation of z 1 is z 1 = 5 0.927 rad = 5 53.1◦ and is illustrated in Figure 2.1.3. Using trigonometry with Figure 2.1.2, we see that x = |z| cos θ y = |z| sin θ and thus z = |z| cos θ + |z| sin θ j = |z|(cos θ + j sin θ ) Using the Euler identity (Table 2.1.2), e jθ = cos θ + j sin θ Table 2.1.2 The exponential function. Taylor series x3 xn x2 + + ··· + + ··· 2 6 n! Euler’s identities e jθ = cos θ + j sin θ e− jθ = cos θ − j sin θ ex = 1 + x + Limits lim xe−x = 0 x→∞ lim e t→∞ −st =0 if x is real. if the real part of s is positive. If a is real and positive, e−at < 0.02 if t > 4/a. e−at < 0.01 if t > 5/a. The time constant is τ = 1/a. 2. 2 The Laplace Transform Method Imaginary z2 5 23 1 4j 4 5 ␪2 23 4 ␪ 5 ␲ 2 tan21 3 5 2.214 rad 5 126.98 37 Figure 2.1.4 Polar representation of the complex number z2 = −3 + 4 j . Real we have z = |z|(cos θ + j sin θ ) = |z|e jθ This is the complex exponential representation of the complex number z. Suppose that z 2 = −3 + 4 j. Then from Figure 2.1.4, z 2 = −3 + 4 j = 5 2.214 rad = 5 126.9◦ Addition and subtraction of two complex numbers is most easily done with the rectangular representation, as z 1 + z 2 = 3 + 4 j + (−3 + 4 j) = 0 + 8 j = 8 j z 1 − z 2 = 3 + 4 j − (−3 + 4 j) = 6 + 0 j = 6 However, multiplication and division are best done using the polar or complex exponential forms, if they are already available. z 1 z 2 = (3 + 4 j)(−3 + 4 j) = 3(−3 + 4 j) + 4 j (−3 + 4 j) = −9 + 12 j − 12 j − 16 = −25 z 1 z 2 = (5 53.1◦ )(5 126.9◦ ) = 25 (53.1◦ + 126.9◦ ) = 25 180◦ or z 1 z 2 = 25(cos 180◦ + j sin 180◦ ) = 25(−1 + 0 j) = −25 When dividing with the rectangular form, first multiply the numerator and denominator by the complex conjugate of the denominator (complex conjugate numbers have the same real part but opposite-signed imaginary parts). 3 + 4j 3 + 4 j −3 − 4 j 7 − 24 j z1 = = = 0.28 − 0.96 j = z2 −3 + 4 j −3 + 4 j −3 − 4 j 25 5 53.1◦ z1 = = 1 (53.1◦ − 126.9◦ ) = 1 (−73.8◦ ) z2 5 126.9◦ or z1 = 1[cos(−73.8◦ ) + j sin(−73.8◦ )] = 0.28 − 0.96 j z2 Note that polar angles add with multiplication and subtract with division. 2.2 THE LAPLACE TRANSFORM METHOD The Laplace transform provides a systematic and general method for solving linear ODEs, and is especially useful either for nonhomogeneous equations whose right-hand side is a function of time or for sets of equations. Another advantage is that the transform converts linear differential equations into algebraic relations that can be handled easily.