2.4-2.7.pdf Dynamic Response PDF

74 Figure 2.5.3 Response for Example 2.5.2. CHAPTER 2 Dynamic Response and the Laplace Transform Method 10 8 x(t ) 6 4 2 0 –2 0 0.2 0.4 0.6 0.8 1 t 1.2 1.4 1.6 1.8 2 2.5.4 TIME CONSTANTS AND COMPLEX ROOTS The model in Example 2.5.2 has complex roots: −3 ± 4 j. These lead to the term e−3t in the response. Thus we may apply the concept of a time constant to complex roots by computing the time constant from the negative inverse of the real part of the roots. Here the model’s time constant is τ = 1/3, and thus the response is essentially at steady state for t > 4τ = 4/3 as shown in Figure 2.5.3. Since complex roots occur only in conjugate pairs, each pair has the same time constant. E X A M P L E 2.5.3 Responses for Second-Order, Imaginary Roots ■ Problem Identify the transient, steady-state, free, and forced responses of the following equation, whose characteristic roots are ±5 j: ẍ + 25x = c ■ Solution From Table 2.5.1, Entry 3, we see that the solution form is x(t) = c + C1 sin 5t + C2 cos 5t 25 steady state Because there are no terms that disappear as t → ∞, there is no transient response. Note that part of the steady-state response is oscillatory. This example shows that the steady-state response need not be constant. To identify the free and forced responses, we must obtain the expressions for C1 and C2 as functions of the initial conditions. These are C2 = x(0) − c/25 and C1 = ẋ(0)/5. Thus the 2. 5 Response Parameters and Stability solution can be expressed as ẋ(0) c sin 5t + (1 − cos 5t) 5 25 x(t) = x(0) cos 5t + free forced Since the roots here have no real part, no time constant is deﬁned for this model. This makes sense because there is no transient response here. 2.5.5 NATURAL FREQUENCY Consider the solution to the equation ẍ +25x = c given in Example 2.5.3. Suppose that the input c is zero. The free response oscillates about x = 0 with a radian frequency of 5. The period of the oscillation is 2π/5. Using our solution tables, we can generalize this result to the equation m ẍ + kx = 0 (2.5.6) Its free response is given by x(t) = x(0) cos ωn t + where ẋ(0) sin ωn t ωn (2.5.7) ωn = k m (2.5.8) √ The radian frequency of oscillation of the free response will be ωn = k/m, and is called the natural frequency. Similarly, the natural period of oscillation is given by Pn = 2π/ωn . The response is a constant-amplitude oscillation. The amplitude depends on the initial conditions x(0) and ẋ(0), but the oscillation frequency and the period are independent of the initial conditions. 2.5.6 DAMPED NATURAL FREQUENCY Suppose the equation contains a ﬁrst derivative term, as ẍ +6ẋ +25x = 0. The solution is given in Example 2.5.2 with c = 0. ẋ(0) + 3x(0) −3t e sin 4t x(t) = x(0)e−3t cos 4t + 4 The free response oscillates with a radian frequency of 4. The period of the oscillation is 2π/4 = π/2, but the amplitude decays with time. Note that the addition of the derivative term 6ẋ changes the frequency from 5 to 4, and causes the oscillations to decay. For reasons we will encounter in Chapter 4, the ﬁrst-order derivative term is called the damping term. Using our solution tables, we can generalize this result to the equation m ẍ + c ẋ + kx = 0 (2.5.9) ms 2 + cs + k = 0 (2.5.10) The characteristic equation is and the roots are s= −c ± √ c2 − 4mk = −a ± bj 2m (2.5.11) 75 76 CHAPTER 2 Dynamic Response and the Laplace Transform Method where we have assumed that the roots are complex (the only case to give oscillatory response) and k c 2 c b= − a=− 2m m 2m The solution is x(t) = e−at x(0) cos bt + ẋ(0) + ax(0) sin bt b (2.5.12) The frequency of oscillation of the free response is b, and is called the damped natural frequency. It is frequently denoted by the symbol ωd . k c 2 − ωd = (2.5.13) m 2m From this we can see that the damping term c ẋ decreases the oscillation frequency. The largest possible oscillation frequency occurs when c = 0, and in this case ωd equals ωn . If c, the coefﬁcient of the damping term, is large enough, then ωd will be zero or imaginary. In this case, both roots are real and no oscillation occurs. The value of c for which ωd = 0 and both roots are real and equal is the value √ (2.5.14) c = 2 mk This value is called the critical damping value, because it represents the dividing line √ between√ oscillatory and non-oscillatory free response. If c < 2 mk, oscillation occurs. If c > 2 mk, the response is exponential. 2.5.7 THE DAMPING RATIO If both roots are negative or have negative real parts, the free response of a second-order equation can be conveniently characterized by the damping ratio, denoted by ζ (and sometimes called the damping factor). For the characteristic equation ms 2 +cs +k = 0, the damping ratio is deﬁned as the ratio of the actual value of c to its critical value, as c (2.5.15) ζ = √ 2 mk This deﬁnition is not arbitrary but is based on the way the roots change from real to complex as the value of c is changed. Three cases can occur: 2 1. Critically √ Damped Case: Repeated roots occur if c − 4mk = 0; that is, if c = 2 mk. This value of the damping constant is the critical damping constant and when c has this value the equation is said to be critically damped. √ 2. The Overdamped Case: If c > 2 mk, two distinct, real roots exist, and the equation is overdamped. √ 3. The Underdamped Case: If c < 2 mk, complex roots occur, and the equation is underdamped. Note that 1. For a critically damped system, ζ = 1. 2. Exponential behavior occurs if ζ > 1 (the overdamped case). 3. Oscillations exist when and only when ζ < 1 (the underdamped case). 2. 5 Response Parameters and Stability If any root is positive or has a positive real part, the damping ratio is meaningless and therefore not deﬁned. For example, because the equation s 2 − 4s + 25 = 0 has the roots s = 2 ± 5 j, its free response will increase without bound, and the damping ratio is negative and therefore meaningless. The damping ratio can be used as a quick check for oscillatory behavior. For example, with the characteristic equation s 2 + 5ds + 4d 2 = 0, where d > 0, the √ damping ratio is ζ = 5d/(2 4d 2 ) = 5/4. Because ζ > 1, no oscillations can occur in the free response regardless of the value of d > 0 and regardless of the initial conditions. Equation (2.5.10) has three parameters, m, c, and k. We can reduce the number of parameters to two by writing the characteristic equation in terms of the parameters ζ and ωn . First divide (2.5.10) by m and use the facts that ωn2 = k/m and ⎞ ⎛ c c k ⎠= ⎝ 2ζ ωn = 2 √ m m 2 mk The characteristic equation becomes s 2 + 2ζ ωn s + ωn2 = 0 and the roots are s = −ζ ωn ± jωn 1 − ζ2 (2.5.16) where we have assumed that the roots are imaginary or complex and thus ζ ≤ 1. We thus see that the damped frequency of oscillation can be expressed in terms of ζ and ωn as ωd = ωn 1 − ζ 2 (2.5.17) When ζ ≤ 1, this equation shows that the damped frequency is always less than the undamped frequency. The real part of the root is the negative reciprocal of the time constant, and so we have 1 (2.5.18) τ= ζ ωn Remember that this formula applies only if ζ ≤ 1. Table 2.5.1 summarizes the formulas for these parameters. The solution of ẍ + 6ẋ + 25x = 0 from Example 2.5.2 for ẋ(0) = 0 and x(0) = 10 is 15 sin 4t x = 10e−3t cos 4t + 2 Table 2.5.1 Response parameters for second-order models. 1. Model: 2. Characteristic Equation: 3. Natural Frequency: 4. Damping Ratio: 5. Damped Natural Frequency: 6. Time Constant (if ζ ≤ 1): m ẍ + c ẋ + kx = f (t) cs + k = 0 ms 2 + k ωn = m c ζ = √ 2 mk ωd = ωn 1 − ζ 2 1 2m = τ= c ζ ωn 77 78 CHAPTER 2 Dynamic Response and the Laplace Transform Method This √ is plotted in Figure 2.5.3. For this differential equation the damping ratio is ζ = 6/(2 25) = 0.6 < 1, so we would expect to see oscillations with a radian frequency of 4, but we do not. Why not? The answer lies in the relation of the time constant to the oscillation period. The roots are s = −3 ± 4 j, so the time constant is τ = 1/3. The response will be essentially zero after t = 4τ = 4/3 = 1.33, but the period is 2π/4 = 1.57. Because the period is greater than 4τ , the oscillations do not have time to repeat before their amplitude becomes very small. Oscillations will not be very visible in the response if the damped period 2π/ωd is greater than four time constants. This occurs when the following ratio is greater than 1. 2π/ωd 2π 1 π ζ 2π ζ ωn R= = = = 2 4τ ωd 4/ζ ωn 2 1 − ζ2 ωn 1 − ζ 4 This ratio R will be greater than 1 if ζ > 0.537. 2.5.8 STABILITY We have seen free responses that approach 0 as t → ∞. We have also seen free responses that approach a constant-amplitude oscillation as t → ∞. The free response may also approach ∞. For example, the model ẋ = 2x has the free response x(t) = x(0)e2t and thus x → ∞ as t → ∞. Let us deﬁne some terms.1 Unstable Stable A model whose free response approaches ∞ as t → ∞ is said to be unstable. If the free response approaches 0, the model is stable. Neutral Stability The borderline case between stable and unstable. Neutral stability describes a situation where the free response does not approach ∞ but does not approach 0. The stability properties of a linear model are determined from its characteristic roots. To understand the relationship between stability and the characteristic roots, consider some simple examples. The ﬁrst-order model ẋ + ax = f (t) has the free response x(t) = x(0)e−at , which approaches 0 as t → ∞ if the characteristic root s = −a is negative. The model is unstable if the root is positive because x(t) → ∞ as t → ∞. A borderline case, called neutral stability, occurs if the root is 0. In this case, x(t) remains at x(0). Neutral stability describes a situation where the free response does not approach ∞ but does not approach 0. Now consider the following second-order models, which have the same initial conditions: x(0) = 1 and ẋ(0) = 0. Their responses are shown in Figure 2.5.4. 1. The model ẍ − 4x = f (t) has the roots s = ±2, and the free response: 1 x(t) = e2t + e−2t 2 2. The model ẍ − 4ẋ + 229x = f (t) has the roots s = 2 ± 15 j, and the free response: 2 2t sin 15t cos 15t − x(t) = e 15 1 There is some variation in terminology related to stability. A stable model is sometimes said to be asymptotically stable. A neutrally stable model is said by some to be critically stable. 2. 5 Response Parameters and Stability Figure 2.5.4 Examples of unstable and neutrally stable models. 6 2 4 2 1 3 x(t ) 0 –2 –4 –6 –8 0 0.1 0.2 0.3 0.4 0.5 t 0.6 0.7 0.8 79 0.9 1 3. The model ẍ + 256x = f (t) has the roots s = ±16 j, and the free response: x(t) = cos 16t From the plots we can see that none of the three models displays stable behavior. The ﬁrst and second models are unstable, while the third is neutrally stable. Thus, the free response of a neutrally stable model can either approach a nonzero constant or settle down to a constant amplitude oscillation. The effect of the real part of the characteristic roots can be seen from the free response form for the complex roots σ ± ωj (entry 4 in Table 2.5.1, with c = 0): x(t) = eσ t (C1 sin ωt + C2 cos ωt) Clearly, if the real part is positive (that is, if σ > 0), then the exponential eσ t will grow with time and so will the amplitude of the oscillations. This is an unstable case. If the real part is 0 (that is, σ = 0), then the exponential becomes e0t = 1, and the amplitude of the oscillations remains constant. This is the neutrally stable case. The imaginary part of the root is the frequency of oscillation; it has no effect on stability. Model 1 is unstable because of the exponential e2t , which is due to the positive root s = +2. The negative root s = −2 does not cause instability because its exponential disappears in time. If we realize that a real number is simply a special case of a complex number whose imaginary part is 0, then these examples show that a linear model is unstable if at least one root has a positive real part. We will see that the free response of any linear, constant-coefﬁcient model, of any order, consists of a sum of terms, each multiplied by an exponential. Each exponential will approach ∞ as t → ∞, if its corresponding root has a positive real part. A root is said to have a multiplicity k if it is repeated k times. For example, the roots s = 0, 0 have multiplicity 2, as do the roots s = 4 j, 4 j, −4 j, −4 j, which are the roots of s 4 + 32s 2 + 256 = 0. Consider the equation ẍ = 0. Its roots are s = 0, 0 and its solution is x(t) = x(0) + v0 t, where v0 = ẋ(0). Now, |x(t)| → ∞ as t → ∞ if v0 = 0. Thus the model ẍ = 0 exhibits unstable behavior if v0 = 0 and neutrally stable behavior if v0 = 0. Since stability is a property of the model only, and not of the initial conditions, we cannot say that the model ẍ = 0 is stable, or even neutrally stable. 80 CHAPTER 2 Dynamic Response and the Laplace Transform Method When a model has roots on the imaginary axis that are of multiplicity k, these roots generate a term in the solution of the form t k−1 . So if k ≥ 2, the term t k−1 will grow with time and thus the model will be unstable. For example, a model with the characteristic equation s 5 + 3s 4 + 32s 3 + 96s 2 + 256s + 768 = 0 has the roots s = −3, 4 j, 4 j, −4 j, −4 j. So the model will have solutions x(t) containing the terms C1 e−3t , C2 t sin 4t, C3 t cos 4t If the initial conditions are such that either C2 or C3 is nonzero, then |x(t)| → ∞. Neutral stability is associated with roots having zero real parts, but these examples show that a model is not neutrally stable if it has roots of multiplicity 2 or greater on the imaginary axis. Such models are said to be marginally stable. Thus we can make the following statement about linear models. Stability Test for Linear Constant-Coefﬁcient Models A constant-coefﬁcient linear model is stable if and only if all of its characteristic roots have negative real parts. The model is neutrally stable if one or more roots have a zero real part with no roots on the imaginary axis of multiplicity 2 or greater, and the remaining roots have negative real parts. The model is unstable if any root has a positive real part. If a linear model is stable, then it not possible to ﬁnd a set of initial conditions for which the free response approaches ∞ as t → ∞. However, if the model is unstable, there might still be certain initial conditions that result in a response that disappears in time. For example, the model ẍ − 4x = 0 has the roots s = ±2, and thus is unstable. However, if the initial conditions are x(0) = 1 and ẋ(0) = −2, then the free response is x(t) = e−2t , which approaches 0 as t → ∞. Note that the exponential e2t corresponding to the root at s = +2 does not appear in the response because of the special nature of these initial conditions. Because the time constant is a measure of how long it takes for the exponential terms in the response to disappear, we see that the time constant is not deﬁned for neutrally stable and unstable cases. 2.5.9 A PHYSICAL EXAMPLE A physical example illustrating the meaning of stability is shown in Figure 2.5.5. Suppose that there is some slight friction in the pivot point of the pendulum. In part (a) Figure 2.5.5 Pendulum illustration of stability properties. m L ␪ L ␪ m (a) (b) 2. 5 Response Parameters and Stability 81 the pendulum is hanging and is at rest at θ = 0. If something disturbs it slightly, it will oscillate about θ = 0 and eventually return to rest at θ = 0. We thus see that the system is stable. If the friction is large enough, the pendulum will not oscillate before coming to rest. In both cases, however, the system is stable because the pendulum eventually returns to its rest position at θ = 0. Now suppose that there is no friction. The pendulum will oscillate with a constant amplitude forever about θ = 0 if disturbed. This is an example of neutral stability. If the pendulum is perfectly balanced at θ = π, as in Figure 2.5.5b, it will never return to θ = π if it is disturbed. So, this equilibrium is an unstable one. 2.5.10 THE ROUTH-HURWITZ CONDITION The characteristic equation of many systems has the form ms 2 + cs + k = 0. A simple criterion exists for quickly determining the stability of such a system. This is proved in homework problem 2.2.5. The condition states that the second-order system whose characteristic polynomial is ms 2 + cs + k is stable if and only if m, c, and k have the same sign. This requirement is called the Routh-Hurwitz condition. 2.5.11 STABILITY AND EQUILIBRIUM An equilibrium is a state of no change. The pendulum in Figure 2.5.5 is in equilibrium at θ = 0 and when perfectly balanced at θ = π . The equilibrium at θ = 0 is stable, while the equilibrium at θ = π is unstable. From this we see that the same physical system can have different stability characteristics at different equilibria. So we see that stability is not a property of the system alone, but is a property of a speciﬁc equilibrium of the system. When we speak of the stability properties of a model, we are actually speaking of the stability properties of the speciﬁc equilibrium on which the model is based. Figure 2.5.6 shows a ball on a surface that has a valley and a hill. The bottom of the valley is an equilibrium, and if the ball is displaced slightly from this position, it will oscillate forever about the bottom if there is no friction. In this case, the equilibrium is neutrally stable. The ball, however, will return to the bottom if friction is present, and the equilibrium is stable in this case. If we displace the ball so much to the left that it lies outside the valley, it will never return. Thus, if friction is present, we say that the valley equilibrium is locally stable but globally unstable. An equilibrium is globally stable only if the system returns to it for any initial displacement. The equilibrium on the hilltop is globally unstable because, if displaced, the ball will continue to roll down the hill. For linear models, stability analysis using the characteristic roots gives global stability information. However, for nonlinear models, linearization about an equilibrium gives us only local stability information. Figure 2.5.6 Surface illustration of stability properties. Hill equilibrium Valley equilibrium 82 CHAPTER 2 Dynamic Response and the Laplace Transform Method 2.6 TRANSFER FUNCTIONS The complete response of a linear ODE is the sum of the free and the forced responses. For zero initial conditions the free response is zero, and the complete response is the same as the forced response. Thus we can focus our analysis on the effects of the input only by taking the initial conditions to be zero temporarily. When we have ﬁnished analyzing the effects of the input, we can add to the result the free response due to any nonzero initial conditions. The concept of the transfer function is useful for analyzing the effects of the input. Consider the model ẋ + ax = f (t) (2.6.1) and assume that x(0) = 0. Transforming both sides of the equation gives s X (s) + a X (s) = F(s) Then solve for the ratio X (s)/F(s) and denote it by T (s): T (s) = X (s) 1 = F(s) s+a The function T (s) is called the transfer function of (2.6.1). The transfer function is the transform of the forced response divided by the transform of the input. It can be used as a multiplier to obtain the forced response transform from the input transform; that is, X (s) = T (s)F(s). The transfer function is a property of the system model only. The transfer function is independent of the input function and the initial conditions. The transfer function concept is extremely useful for several reasons. 1. Transfer Functions and Software. Software packages such as MATLAB do not accept system descriptions expressed as single, higher-order differential equations. Such software, however, does accept a description based on the transfer function. In Section 2.10 we will see how MATLAB does this. In Chapter 5 we will see that the transfer function is the basis for a graphical system description called the block diagram, and block diagrams are used to program the Simulink dynamic simulation software. So the transfer function is an important means of describing dynamic systems. 2. ODE Equivalence. It is important to realize that the transfer function is equivalent to the ODE. If we are given the transfer function we can reconstruct the corresponding ODE. For example, the transfer function 5 X (s) = 2 F(s) s + 7s + 10 corresponds to the equation ẍ + 7ẋ + 10x = 5 f (t). You should develop the ability to obtain transfer functions from ODEs and ODEs from transfer functions. This process is easily done because the initial conditions are assumed to be zero when working with transfer functions. From the derivative property, this means that to work with a transfer function you can use the relations L(ẋ) = s X (s), L(ẍ) = s 2 X (s), and so forth. Examples 3.4.1 and 3.4.2 will show how straightforward this process is. 3. The Transfer Function and Characteristic Roots. Note that the denominator of the transfer function is the characteristic polynomial, and thus the transfer 2. 6 Transfer Functions 83 function tells us something about the intrinsic behavior of the model, apart from the effects of the input and speciﬁc values of the initial conditions. In the previous equation, the characteristic polynomial is s 2 + 7s + 10 and the roots are −2 and −5. The roots are real, and this tells us that the free response does not oscillate and that the forced response does not oscillate unless the input is oscillatory. Because the roots are negative, the model is stable and its free response disappears with time. 2.6.1 MULTIPLE INPUTS AND OUTPUTS Obtaining a transfer function from a single ODE is straightforward, as we have seen. Sometimes, however, models have more than one input or occur as sets of equations with more than one dependent variable. It is important to realize that there is one transfer function for each input-output pair. If a model has more than one input, a particular transfer function is the ratio of the output transform over the input transform, with all the remaining inputs ignored (set to zero temporarily). Two Inputs and One Output ■ Problem Obtain the transfer functions X (s)/F(s) and X (s)/G(s) for the following equation. 5ẍ + 30ẋ + 40x = 6 f (t) − 20g(t) ■ Solution Using the derivative property with zero initial conditions, we can immediately write the equation as 5s 2 X (s) + 30s X (s) + 40X (s) = 6F(s) − 20G(s) Solve for X (s). X (s) = 5s 2 20 6 F(s) − 2 G(s) + 30s + 40 5s + 30s + 40 When there is more than one input, the transfer function for a speciﬁc input can be obtained by temporarily setting the other inputs equal to zero (this is another aspect of the superposition property of linear equations). Thus, we obtain 6 X (s) = 2 F(s) 5s + 30s + 40 X (s) 20 =− 2 G(s) 5s + 30s + 40 Note that the denominators of both transfer functions have the same roots: s = −2 and s = −4. We can obtain transfer functions from systems of equations by ﬁrst transforming the equations and then algebraically eliminating all variables except for the speciﬁed input and output. This technique is especially useful when we want to obtain the response of one or more of the dependent variables in the system of equations. E X A M P L E 2.6.1 84 CHAPTER 2 Dynamic Response and the Laplace Transform Method E X A M P L E 2.6.2 A System of Equations ■ Problem a. Obtain the transfer functions X (s)/V (s) and Y (s)/V (s) of the following system of equations: ẋ = −3x + 2y ẏ = −9y − 4x + 3v(t) b. Obtain the forced response for x(t) and y(t) if the input is v(t) = 5u s (t). ■ Solution a. Here two outputs are speciﬁed, x and y, with one input, v. Thus there are two transfer functions. To obtain them, transform both sides of each equation, assuming zero initial conditions. s X (s) = −3X (s) + 2Y (s) sY (s) = −9Y (s) − 4X (s) + 3V (s) These are two algebraic equations in the two unknowns, X (s) and Y (s). Solve the ﬁrst equation for Y (s): s+3 Y (s) = X (s) (1) 2 Substitute this into the second equation. s+3 s+3 X (s) = −9 X (s) − 4X (s) + 3V (s) 2 2 Then solve for X (s)/V (s) to obtain s 6 X (s) = 2 V (s) s + 12s + 35 (2) Now substitute this into equation (1) to obtain Y (s) s + 3 X (s) s+3 6 3(s + 3) = = = 2 2 V (s) 2 V (s) 2 s + 12s + 35 s + 12s + 35 (3) The desired transfer functions are given by equations (2) and (3). Note that denominators of both transfer functions have the same factors, s = −5 and s = −7, which are the roots of the characteristic equation: s 2 + 12s + 35. b. From equation (2), X (s) = 6 6 5 30 V (s) = 2 = s 2 + 12s + 35 s + 12s + 35 s s(s 2 + 12s + 35) The denominator factors are s = 0, s = −5, and s = −7, and thus the partial-fraction expansion is C2 C3 C1 X (s) = + + s s+5 s+7 where C1 = 6/7, C2 = −3, and C3 = 15/7. The forced response is 6 15 − 3e−5t + e−7t 7 7 From (1) we have y = (ẋ + 3x)/2. From (4) we obtain x(t) = y(t) = 9 30 + 3e−5t − e−7t 7 7 (4) 2. 7 The Impulse and Numerator Dynamics 85 2.7 THE IMPULSE AND NUMERATOR DYNAMICS In our development of the Laplace transform and its associated methods, we have assumed that the process under study starts at time t = 0−. Thus the given initial conditions, say x(0−), ẋ(0−), . . . , represent the situation at the start of the process and are the result of any inputs applied prior to t = 0−. That is, we need not know what the inputs were before t = 0− because their effects are contained in the initial conditions. The effects of any inputs starting at t = 0− are not felt by the system until an inﬁnitesimal time later, at t = 0+. If the dependent variable x(t) and its derivatives do not change between t = 0− and t = 0+, the solution x(t) obtained from the differential equation will match the given initial conditions when x(t) and its derivatives are evaluated at t = 0. The results obtained from the initial value theorem will also match the given initial conditions. However, we will now investigate the behavior of some models for which x(0−) = x(0+), or ẋ(0−) = ẋ(0+), and so forth for higher derivatives. The initial value theorem gives the value at t = 0+, which for some models is not necessarily equal to the value at t = 0−. In these cases the solution of the differential equation is correct only for t > 0. This phenomenon occurs in models having impulse inputs and in models containing derivatives of a discontinuous input, such as a step function. 2.7.1 THE IMPULSE An input that changes at a constant rate is modeled by the ramp function. The step function models an input that rapidly reaches a constant value, while the rectangular pulse function models a constant input that is suddenly removed. The impulse is similar to the pulse function, but it models an input that is suddenly applied and removed after a very short time. The impulse, which is a mathematical function only and has no physical counterpart, has an inﬁnite magnitude for an inﬁnitesimal time. Consider the rectangular pulse function shown in Figure 2.7.1a. Its transform is M(1 − e−s D )/s. The area A under the pulse is A = M D and is called the strength of the pulse. If we let this area remain constant at the value A and let the pulse duration D approach zero, we obtain the impulse, represented in Figure 2.7.1b. Because M = A/D, the transform F(s) is A 1 − e−s D Ase−s D = lim =A D→0 D D→0 s s after using L’Hopital’s limit rule. If the strength A = 1, the function is called a unit impulse. The unit impulse, called the Dirac delta function δ(t) in mathematics literature, often appears in the analysis of dynamic systems. It is an analytically convenient F(s) = lim Figure 2.7.1 (a) The rectangular pulse. (b) The impulse. A A 5 MD M 0 t D (a) t 0 (b) 86 CHAPTER 2 Dynamic Response and the Laplace Transform Method approximation of an input applied for only a very short time, such as when a highspeed object strikes a another object. The impulse is also useful for estimating the system’s parameters experimentally and for analyzing the effect of differentiating a step or any other discontinuous input function. In keeping with our interpretation of the initial conditions, we consider the impulse δ(t) to start at time t = 0− and ﬁnish at t = 0+, with its effects ﬁrst felt at t = 0+. E X A M P L E 2.7.1 Impulse Response of a Simple First-Order Model ■ Problem Obtain the unit-impulse response of the following model in two ways: (a) by separation of variables and (b) with the Laplace transform. The initial condition is x(0−) = 3. What is the value of x(0+)? ẋ = δ(t) ■ Solution a. Integrate both sides of the equation to obtain x(t) ẋ(t) d x = x(0−) t δ(t) dt = 0− 0+ t δ(t)dt + 0− δ(t)dt = 1 + 0 0+ because the area under a unit impulse is 1. This gives x(t) − x(0−) = 1 or x(t) = x(0−) + 1 = 3 + 1 = 4 This is the solution for t > 0 but not for t = 0−. Thus, x(0+) = 4 but x(0−) = 3, so the impulse has changed x(t) instantaneously from 3 to 4. b. The transformed equation is s X (s) − x(0−) = 1 X (s) = or 1 + x(0−) s which gives the solution x(t) = 1 + x(0−) = 4. Note that the initial value used with the derivative property is the value of x at t = 0−. The initial value theorem gives x(0+) = lim s X (s) = lim s s→∞ s→∞ 1 + x(0−) =1+3=4 s which is correct. E X A M P L E 2.7.2 Impulse Response of a First-Order Model ■ Problem Obtain the unit-impulse response of the following model. The initial condition is x(0−) = 0. What is the value of x(0+)? X (s) 1 = F(s) s+5 ■ Solution Because f (t) = δ(t), F(s) = 1, and the response is obtained from X (s) = 1 1 F(s) = s+5 s+5 2. 7 The Impulse and Numerator Dynamics 87 The response is x(t) = e−5t for t > 0. This gives x(0+) = lim x(t) = lim e−5t = 1 t→0+ t→0+ So the impulse input has changed x from 0 at t = 0− to 1 at t = 0+. This same result could have been obtained from the initial value theorem: x(0+) = lim s X (s) = lim s s→∞ s→∞ 1 =1 s+5 Impulse Response of a Simple Second-Order Model E X A M P L E 2.7.3 ■ Problem Obtain the unit-impulse response of the following model in two ways: (a) by separation of variables and (b) with the Laplace transform. The initial conditions are x(0−) = 5 and ẋ(0−) = 10. What are the values of x(0+) and ẋ(0+)? ẍ = δ(t) (1) ■ Solution Let v(t) = ẋ(t). Then the equation (1) becomes v̇ = δ(t), which can be integrated to obtain v(t) = v(0−) + 1 = 10 + 1 = 11. Thus ẋ(0+) = 11 and is not equal to ẋ(0−). Now integrate ẋ = v = 11 to obtain x(t) = x(0−) + 11t = 5 + 11t. Thus, x(0+) = 5 which is the same as x(0−). So for this model the unit-impulse input changes ẋ from t = 0− to t = 0+ but does not change x. b. The transformed equation is a. s 2 X (s) − sx(0−) − ẋ(0−) = 1 or sx(0−) + ẋ(0−) + 1 5s + 11 5 11 = = + 2 s2 s2 s s which gives the solution x(t) = 5 + 11t and ẋ(t) = 11. Note that the initial values used with the derivative property are the values at t = 0−. The initial value theorem gives X (s) = x(0+) = lim s X (s) = lim s s→∞ and because L(ẋ) = s X (s) − x(0−), s→∞ 5s + 11 =5 s2 5s + 11 −5 ẋ(0+) = lim s[s X (s) − x(0−)] = lim s s→∞ s→∞ s = 11 as we found in part (a). Impulse Response of a Second-Order Model ■ Problem Obtain the unit-impulse response of the following model. The initial conditions are x(0−) = 0, ẋ(0−) = 0. What are the values of x(0+) and ẋ(0+)? X (s) 1 = 2 F(s) 2s + 14s + 20 E X A M P L E 2.7.4 88 CHAPTER 2 Dynamic Response and the Laplace Transform Method ■ Solution Because f (t) = δ(t), F(s) = 1, and X (s) = 2s 2 1 1 1 1 1 1 F(s) = 2 = − + 14s + 20 2s + 14s + 20 6s+2 6s+5 The response is x(t) = (e−2t − e−5t )/6. This gives x(0+) = lim x(t) = lim t→0+ t→0+ and ẋ(0+) = lim ẋ(t) = lim t→0+ t→0+ e−2t − e−5t 6 −2e−2t + 5e−5t 6 =0 = 1 2 So the impulse input has not changed x between t = 0− and t = 0+ but has changed ẋ from 0 to 1/2. These results could have been obtained from the initial value theorem: x(0+) = lim s X (s) = lim s s→∞ s→∞ 1 =0 2s 2 + 14s + 20 and, noting that x(0−) = 0, ẋ(0+) = lim s[s X (s) − x(0−)] = lim s s→∞ s→∞ 1 s = 2s 2 + 14s + 20 2 In summary, be aware that the solution x(t) and its derivatives ẋ(t), ẍ(t), . . . will match the given initial conditions at t = 0− only if there are no impulse inputs and no derivatives of inputs that are discontinuous at t = 0. If X (s) is a rational function and if the degree of the numerator of X (s) is less than the degree of the denominator, then the initial value theorem will give a ﬁnite value for x(0+). If the degrees are equal, then initial value is undeﬁned and the initial value theorem is invalid. The latter situation corresponds to an impulse in x(t) at t = 0 and therefore x(0+) is undeﬁned. When the degrees are equal the transform can be expressed as a constant plus a partial-fraction expansion. For example, consider the transform 23 9s + 4 =9− X (s) = s+3 s+3 The inverse transform is x(t) = 9δ(t) − 23e−3t and therefore x(0+) is undeﬁned. 2.7.2 NUMERATOR DYNAMICS The following model contains a derivative of the input g(t): 5ẋ + 10x = 2ġ(t) + 10g(t) Its transfer function is X (s) 2s + 10 = G(s) 5s + 10 Note that the input derivative ġ(t) results in an s term in the numerator of the transfer function, and such a model is said to have numerator dynamics. So a model with input derivatives has numerator dynamics, and vice versa, and thus the two terms describe the same condition. With such models we must proceed carefully if the input is discontinuous, as is the case with the step function, because the input derivative produces an impulse when 2. 7 The Impulse and Numerator Dynamics 89 acting on a discontinuous input. To help you understand this, we state without rigorous proof that the unit impulse δ(t) is the time-derivative of the unit-step function u s (t); that is, d (2.7.1) δ(t) = u s (t) dt This result does not contradict common sense, because the step function changes from 0 at t = 0− to 1 at t = 0+ in an inﬁnitesimal amount of time. Therefore its derivative should be inﬁnite during this time. To further indicate the correctness of this relation, we integrate both sides and note that the area under the unit impulse is unity. Thus, 0+ 0+ d u s (t) dt = u s (0+) − u s (0−) = 1 − 0 = 1 δ(t) dt = 0− 0− dt Thus an input derivative will create an impulse in response to a step input. For example, consider the model 5ẋ + 10x = 2ġ(t) + 10g(t) If the input g(t) = u s (t), the model is equivalent to 5ẋ + 10x = 2δ(t) + 10u s (t) which has an impulse input. Numerator dynamics can signiﬁcantly alter the response, and the Laplace transform is a convenient and powerful tool for analyzing models having numerator dynamics. A First-Order Model with Numerator Dynamics ■ Problem Obtain the transfer function and investigate the response of the following model in terms of the parameter a. The input g(t) is a unit-step function. 5ẋ + 10x = a ġ(t) + 10g(t) x(0−) = 0 ■ Solution Transforming the equation with x(0−) = 0 and solving for the ratio X (s)/G(s) gives the transfer function: X (s) as + 10 = G(s) 5s + 10 Note that the model has numerator dynamics if a = 0. For a unit-step input, G(s) = 1/s and as + 10 1 a−5 1 X (s) = = + s(5s + 10) s 5 s+2 Thus the response is a − 5 −2t e x(t) = 1 + 5 From this solution or the initial-value theorem we ﬁnd that x(0+) = a/5, which is not equal to x(0−) unless a = 0 (which corresponds to the absence of numerator dynamics). The plot of the response is given in Figure 2.7.2 for several values of a. The initial condition is different for each case, but for all cases the response is essentially constant for t > 2 because the term e−2t becomes small. E X A M P L E 2.7.5 90 CHAPTER 2 Figure 2.7.2 Plot of the response for Example 2.7.5. Dynamic Response and the Laplace Transform Method 2 1.8 a = 10 1.6 1.4 x(t ) 1.2 a=5 1 0.8 a=1 0.6 0.4 a=0 0.2 0 E X A M P L E 2.7.6 0 0.2 0.4 0.6 0.8 1 t 1.2 1.4 1.6 1.8 2 A Second-Order Model with Numerator Dynamics ■ Problem Obtain the transfer function and investigate the response of the following model in terms of the parameter a. The input g(t) is a unit-step function. 3ẍ + 18ẋ + 24x = a ġ(t) + 6g(t) x(0−) = 0 ẋ(0−) = 0 ■ Solution Transforming the equation with zero initial conditions and solving for the ratio X (s)/G(s) gives the transfer function: as + 6 X (s) = 2 G(s) 3s + 18s + 24 Note that the model has numerator dynamics if a = 0. For a unit-step input, G(s) = 1/s and X (s) = s(3s 2 11 a−3 1 3 − 2a 1 as + 6 = + + + 18s + 24) 4s 6 s+2 12 s + 4 Thus the response is x(t) = 1 a − 3 −2t 3 − 2a −4t + e + e 4 6 12 From this solution or the initial-value theorem we ﬁnd that x(0+) = 0, which is equal to x(0−), and that ẋ(0+) = a/3, which is not equal to ẋ(0−) unless a = 0 (which corresponds to the absence of numerator dynamics). The plot of the response is given in Figure 2.7.3 for several values of a. Notice that a “hump” in the response (called an “overshoot”) does not occur for smaller values of a and the height of the hump increases as a increases. However, the value of a does not affect the steady-state response. 2. 8 Additional Examples 91 Figure 2.7.3 Plot of the response for Example 2.7.6. 0.5 a = 10 0.4 a=5 x(t ) 0.3 a=3 0.2 a=0 0.1 0 0 0.2 0.4 0.6 0.8 1 t 1.2 1.4 1.6 1.8 2 2.8 ADDITIONAL EXAMPLES This section contains additional examples of solving dynamic models. A Mixing Process E X A M P L E 2.8.1 ■ Problem A mixing tank is shown in Figure 2.8.1. Pure water ﬂows into the tank of volume V = 600 m3 at the constant volume rate of 5 m3 /s. A solution with a salt concentration of si kg/m3 ﬂows into the tank at a constant volume rate of 2 m3 /s. Assume that the solution in the tank is well mixed so that the salt concentration in the tank is uniform. Assume also that the salt dissolves 2 m3/s Solution si Figure 2.8.1 A mixing process. 5 m3/s Water V 5 600 m3 so qvo Mixer 92 CHAPTER 2 Dynamic Response and the Laplace Transform Method completely so that the volume of the mixture remains the same. The salt concentration so kg/m3 in the outﬂow is the same as the concentration in the tank. The input is the concentration si (t), whose value may change during the process, thus changing the value of so . (a) Obtain a dynamic model of the concentration so . (b) Suppose that si is a step function of magnitude S. Find the steady state value of so and estimate how long it will take to reach steady state. ■ Solution Two mass species are conserved here: water mass and salt mass. The tank is always full, so the mass of water m w in the tank is constant, and thus conservation of water mass gives dm w = 5ρw + 2ρw − ρw qvo = 0 dt where ρw is the mass density of fresh water, and qvo is the volume outﬂow rate of the mixed solution. This equation gives qvo = 5 + 2 = 7 m3 /s. The salt mass in the tank is so V , and conservation of salt mass gives d (so V ) = 0(5) + 2si − so qvo = 2si − 7so dt or, with V = 600, dso 600 = 2si − 7so (1) dt This is the model. The time constant for the mixing process is 600/7 = 85.7 s. Thus, if si is initially zero and then becomes a nonzero constant value S, the salt concentration so in the outﬂow will eventually become constant at the value 2S/7 after approximately 4(85.7) = 343 s. E X A M P L E 2.8.2 An Example of Subsystems: Two Coupled Tanks ■ Problem Consider two brine tanks each containing 500 L (liters) of brine connected as shown in Figure 2.8.2. At any time t,the ﬁrst and the second tank contain x1 (t) and x2 (t) kg of salt, respectively. The brine concentration in each tank is kept uniform by continuous stirring. Brine containing r kg of salt per liter is entering the ﬁrst tank at a rate of 15 L/min, and fresh water is entering the second tank at a rate of 5 L/min. The incoming brine density r (t) can be changed to regulate the process, so r (t) is an input variable. Brine is pumped from the ﬁrst tank to the second one at a rate of 60 L/min and from the second tank to the ﬁrst one at a rate of 45 L/min. Brine is discharged from the second tank at a rate of 20 L/min. Figure 2.8.2 Mixing in two brine tanks. Brine 15 L/min r kg/L Fresh Water 5 L/min 45 L/min X1(t) X2(t) 500 L 500 L 60 L/min Tank 1 Tank 2 Brine 20 L/min 2. 8 Additional Examples a. Obtain the differential equations, in terms of x1 and x2 , that describe the salt content in each tank as a function of time. b. Obtain the transfer functions X 1 (s)/R(s) and X 2 (s)/R(s). c. Suppose that r (t) = 0.2 kg/L. Determine the steady state values of x1 and x2 , and estimate how long it will take to reach steady state. ■ Solution a. We assume that the liquid volume does not change when the salt is dissolved in it. We also observe that the total volume of the brine in each tank remains constant at 500 L because the incoming and the outgoing volume ﬂow rates for each tank are equal. Therefore, each liter of brine in the ﬁrst tank contains x1 /500 kg of salt, and salt leaves the ﬁrst tank at a rate of 60(x1 /500) kg/min. Considering that each liter of brine in the second tank contains x2 /500 kg of salt, the rate at which salt leaves the second tank and enters the ﬁrst one is 45(x2 /500) kg/min. In addition, new brine enters the ﬁrst tank at a rate of 15r kg/min because each liter of the new brine contains r kg of salt. A similar argument can be given for the second tank. Then the rates of change of the salt content of each tank, in kg/min, can be expressed as d x1 x1 x2 = 15r (t) − 60 + 45 (1) dt 500 500 x1 x2 d x2 = 0 + 60 − 65 (2) dt 500 500 b. Multiplying each equation by 100 gives d x1 100 = 1500r (t) − 12x1 + 9x2 dt d x2 = 12x1 − 13x2 100 dt Transforming each equation, using zero initial conditions, and collecting terms gives (100s + 12)X 1 (s) − 9X 2 (s) = 1500R(s) −12X 1 (s) + (100s + 13)X 2 (s) = 0 Solving these equations gives the transfer functions. 1500(100s + 13) X 1 (s) = 4 2 R(s) 10 s + 2500s + 48 1.8 × 104 X 2 (s) = 4 2 R(s) 10 s + 2500s + 48 c. (3) (4) Note that (3) has numerator dynamics, whereas (4) does not. The characteristic roots are the roots of the denominator, which are s = −0.021 and s = −0.029. Thus the system is stable and the time constants are τ = 1/0.021 = 47.62 and τ = 1/0.229 = 4.37. The dominant time constant is 47.62 minutes, so the time to reach steady state is approximately 4(47.62) = 190.5 minutes. The salt content x1 may actually take longer than this estimate because of the numerator dynamics in its transfer function. Determination of the actual value requires solving (1) and (2) or (3) and (4). Applying the Final Value Theorem to (3) and (4) with gives the steady state values. x1ss = lim s s→0 x2ss = lim s s→0 1500(100s + 13) 0.2 104 s 2 + 2500s + 48 s 1.8 × 104 0.2 4 2 10 s + 2500s + 48 s = 1500(13)(0.2) = 81.25 kg 48 = 1.8 × 104 0.2 = 75 kg 48 These are the steady state values regardless of the initial values of x1 and x2 . 93 94 CHAPTER 2 Dynamic Response and the Laplace Transform Method This example illustrates a typical problem in system dynamics. The system consists of two subsystems, the two tanks. To model the entire system, we must ﬁrst understand the dynamics of each subsystem. This was illustrated in Example 2.8.1, where we derive the model of a single tank. Only then can we build the model of the system having two tanks. Building a model of a more complex system, say one containing many tanks, would then proceed in a similar way. E X A M P L E 2.8.3 Transform of A sin(ωt + φ) ■ Problem (a) Derive the Laplace transform of the function A sin(ωt + φ). (b) Generalize the answer from part (a) to ﬁnd the Laplace transform of Ae−at sin(ωt + φ). ■ Solution a. Because we already have the transforms of sin ωt and cos ωt, we can use the following trigonometric identity to obtain our answer: sin(ωt + φ) = sin ωt cos φ + cos ωt sin φ From the linearity property of the transform, L[A sin(ωt + φ)] = A ∞ sin ωt cos φe −st ∞ dt + A 0 cos ωt sin φe−st dt 0 = A cos φL(sin ωt) + A sin φL(cos ωt) ω s = A cos φ 2 + A sin φ 2 s + ω2 s + ω2 Combining these terms gives the answer: s sin φ + ω cos φ s 2 + ω2 Following the same procedure and using the fact that L[A sin(ωt + φ)] = A b. L e−at sin ωt = we see that ω (s + a)2 + ω2 L e−at cos ωt = s+a (s + a)2 + ω2 L Ae−at sin(ωt + φ) = A cos φL e−at sin ωt + A sin φL e−at cos ωt = A cos φ ω s+a + A sin φ (s + a)2 + ω2 (s + a)2 + ω2 Combining these terms gives the answer: L Ae−at sin(ωt + φ) = A E X A M P L E 2.8.4 s sin φ + a sin φ + ω cos φ (s + a)2 + ω2 Response in the Form A sin(ωt + φ) ■ Problem Given that 3s + 10 s 2 + 16 obtain f (t) in the form f (t) = A sin(ωt + φ), where A > 0. F(s) = 2. 8 Additional Examples 95 ■ Solution From the results of part (a) of Example 2.8.3, we see that the form f (t) = A sin(ωt + φ) has the transform s sin φ + ω cos φ A s 2 + ω2 Comparing this with F(s) we see that ω = 4 and A(s sin φ + ω cos φ) = 3s + 10. Therefore, A sin φ = 3 and Aω cos φ = 4A cos φ = 10, or 3 10 sin φ = cos φ = (1) A 4A Because A > 0, these equations reveal that sin φ > 0 and cos φ > 0. Therefore, φ is in the ﬁrst quadrant (0 ≤ φ ≤ π/2 rad) and can be computed as follows: φ = tan−1 sin φ 3/A 3 = tan−1 = tan−1 = 0.876 rad cos φ 10/4A 10/4 To ﬁnd A, we use the fact that sin2 φ + cos2 φ = 1 for any angle φ. From equations (1) we have 2 2 3 10 sin φ + cos φ = + =1 A 4A 2 2 Because A was speciﬁed to be positive, we take the This gives A2 = 9 + 100/16 = 244/16. √ positive square root to obtain A = 61/2. The solution is 1√ f (t) = 61 sin(4t + 0.876) 2 Sine Form of the Response ■ Problem Obtain the solution to the following problem in the form of a sine function with a phase angle: 3ẍ + 12ẋ + 87x = 5 x(0) = 2 ẋ(0) = 7 ■ Solution Applying the Laplace transform we obtain 3 s 2 X (s) − sx(0) − ẋ(0) + 12[s X (s) − x(0−)] + 87X (s) = 5 s which gives X (s) = 6s 2 + 45s + 5 2s 2 + 15s + 5/3 = 3s(s 2 + 4s + 29) s (s + 2)2 + 25 (1) Because the denominator roots are s = 0 and s = −2 ± 5 j, we know that the form of the solution is x(t) = C1 + C2 e−2t sin(5t + φ) Using the results of Example 2.8.3 with a = 2 and ω = 5, we can write the transform of x(t) as C1 s sin φ + 2 sin φ + 5 cos φ X (s) = + C2 s (s + 2)2 + 25 This can be expressed as a single fraction as follows: X (s) = = C1 (s + 2)2 + 25 + C2 s(s sin φ + 2 sin φ + 5 cos φ) s (s + 2)2 + 25 (C1 + C2 sin φ)s 2 + (4C1 + 2C2 sin φ + 5C2 cos φ)s + 29C1 s (s + 2)2 + 25 (2) E X A M P L E 2.8.5 96 CHAPTER 2 Dynamic Response and the Laplace Transform Method Comparing the numerators of equations (1) and (2) we see that C1 + C2 sin φ = 2 4C1 + 2C2 sin φ + 5C2 cos φ = 15 29C1 = 5 3 The third equation gives C1 = 5/87 = 0.0575, and the ﬁrst equation gives 169 87 From the second equation, we have C2 cos φ = 947/435. Taking C2 to be positive, we see that both sin φ and cos φ are positive, and thus φ is in the ﬁrst quadrant. Therefore C2 sin φ = 2 − C1 = −1 φ = tan sin φ cos φ −1 = tan Because sin2 φ + cos2 φ = 1, sin φ + cos φ = 2 2 169 87C2 169/87 947/435 2 + = 0.729 rad 947 435C2 2 =1 which gives C2 = 2.918. Thus the solution is x(t) = 0.0575 + 2.918e−2t sin(5t + 0.729) The shifting property, entry 6 in Table 2.2.2, can be used to invert transforms containing the function e−Ds . This enables us to obtain the response to inputs composed of shifted step or ramp functions. E X A M P L E 2.8.6 Pulse Response of a First-Order Model ■ Problem Suppose a rectangular pulse P(t) of unit height and duration 2 is applied to the ﬁrst-order model ẋ + 4x = P(t) with a zero initial condition. Use the Laplace transform to determine the response. ■ Solution The problem to be solved is ẋ + 4x = P(t) x(0−) = 0 Taking the transform of both sides of the equation and noting that the initial condition is zero, we obtain s X (s) + 4X (s) = P(s) = 1 − e−2s s Solve for X (s). X (s) = 1 − e−2s P(s) = s+4 s(s + 4) Let Y (s) = Then 1 s(s + 4) X (s) = 1 − e−2s Y (s) 2. 8 Additional Examples 0.3 97 Figure 2.8.3 Plot of the solution for Example 2.8.6. 0.25 x(t ) 0.2 0.15 0.1 0.05 0 0 0.5 1 1.5 2 t 2.5 3 3.5 4 and from the shifting property, x(t) = y(t)u s (t) − y(t − 2)u s (t − 2) (1) To ﬁnd y(t), note that the denominator roots of Y (s) are s = 0 and s = −4. Thus we can express Y (s) as follows: C1 C2 1 Y (s) = + = s s+4 4 1 1 − s s+4 Therefore, y(t) = and from equation (1), x(t) = 1 1 −4t − e 4 4 1 1 −4t − e 4 4 − 1 1 −4(t−2) − e u s (t − 2) 4 4 So for 0 ≤ t ≤ 2, x(t) = 1 1 −4t − e 4 4 and for t ≥ 2, 1 1 −4t 1 1 −4(t−2) 1 − e − + e = e−4(t−2) − e−4t 4 4 4 4 4 The function’s graph is shown in Figure 2.8.3. x(t) = Series Solution Method ■ Problem Obtain an approximate, closed-form solution of the following problem for 0 ≤ t ≤ 0.5: ẋ + x = tan t x(0) = 0 (1) E X A M P L E 2.8.7 98 CHAPTER 2 Dynamic Response and the Laplace Transform Method ■ Solution If we attempt to use separation of variables to solve this problem we obtain dx = dt tan t − x so the variables do not separate. In general, when the input is a function of time, the equation ẋ + g(x) = f (t) does not separate. The Laplace transform method cannot be used when the Laplace transform or inverse transform either does not exist or cannot be found easily. In this example, the equation cannot be solved by the Laplace transform method, because the transform of tan t does not exist. An approximate solution of the equation ẋ + x = tan t can be obtained by replacing tan t with a series approximation. The number of terms used in the series determines the accuracy of the resulting solution for x(t). The Taylor series expansion for tan t is t3 2t 5 17t 7 π + + + ··· |t| < 3 15 315 2 The more terms we retain, the more accurate is the series. Also, the series becomes less accurate as the absolute value of t increases. To demonstrate the series solution method, let us use a series with two terms: tan t = t + t 3 /3. At the largest value of t, t = 0.5, the two-term series gives 0.5417 versus 0.5463 for the true value of tan 0.5. So the two-term series is accurate to at least two decimal places over the range of t we are interested in (0 ≤ t ≤ 0.5). Using the two-term series we need to solve the following problem: tan t = t + ẋ + x = t + t3 3 x(0) = 0 Using the Laplace transform we obtain s X (s) + X (s) = 1 1 3! + 2 s 3 s4 or X (s) = s2 + 2 s 4 (s + 1) The inverse transform was obtained in Example 2.4.7 and is 1 3 t − t 2 + 3t − 3 + 3e−t 3 We can expect this approximate solution of equation (1) to be accurate to at least two decimal places for 0 ≤ t ≤ 0.5. Of course, greater accuracy can be achieved by retaining more terms in the Taylor series for tan t. x(t) = 2.9 COMPUTING EXPANSION COEFFICIENTS WITH MATLAB You can use MATLAB to easily compute the coefﬁcients in the partial-fraction expansion. The appropriate MATLAB function is residue. Let X (s) denote the transform. In the terminology of the residue function, the expansion coefﬁcients are called the residues and the factors of the denominator of X (s) are called the poles. The poles include the characteristic roots of the model and any denominator roots introduced by the input function. If the order m of the numerator of X (s) is greater than the order n of the denominator, the transform can be represented by a polynomial K (s), called the direct term, plus a ratio of two polynomials where the denominator degree is greater

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