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# ANALYSIS OF DIFFERENTIAL EQUATIONS

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1. ANALYSIS OF DIFFERENTIAL EQUATIONS1.1 Introduction1.2 Explicit Solutions1.3 Responses1.3.1 First-order1.3.2 Second-order1.3.3 Other Responses1.4 Response Analysis1.5 Non-Linear Systems1.5.1 Non-Linear Differential Equations1.5.2 Non-Linear Equation Terms1.5.3 Changing Systems1.6 Case Study1.7 Summary1.8 Problems1.9 Problems Solutions1.10 Challenge Problemspage 1721. ANALYSIS OF DIFFERENTIAL EQUATIONS1.1 IntroductionIn the previous chapter we derived differential equations of motion for translating systems. These equations can be used to analyze the behavior of the system and make design decisions. The most basic method is to select a standard input type (a forcing func-tion) and initial conditions, and then solve the differential equation. It is also possible to estimate the system response without solving the differential equation as will be discussed later.Figure 1.1 shows an abstract description of a system. The basic concept is that the system changes the inputs to outputs. Say, for example, that the system to be analyzed is an elevator. Inputs to the system would be the mass of human riders and desired elevator height. The output response of the system would be the actual height of the elevator. For analysis, the system model could be developed using differential equations for the motor, elastic lift cable, mass of the car, etc. A basic test would involve assuming that the elevator starts at the ground floor and must travel to the top floor. Using assumed initial values and input functions the differential equation could be solved to get an explicit equation for ele-vator height. This output response can then be used as a guide to modify design choices (parameters). In practice, many of the assumptions and tests are mandated by law or by groups such as Underwriters Laboratories (UL), Canadian Standards Association (CSA) and the European Commission (CE).Topics:Objectives:• To develop explicit equations that describe a system response.• To recognize first and second-order equation forms.• First and second-order homogeneous differential equations• Non-homogeneous differential equations• First and second-order responses• Non-linear system elements• Design casepage 173Figure 1.1 A system with and input and output responseThere are several standard input types used to test a system. These are listed below in order of relative popularity with brief explanations.• step - a sudden change of input, such as very rapidly changing a desired speed from 0Hz to 50Hz.• ramp - a continuously increasing input, such as a motor speed that increases con-stantly at 10Hz per minute.• sinusoidal - a cyclic input that varies continuously, such as wave height that is continually oscillating at 1Hz. • parabolic - an exponentially increasing input, such as a motor speed that is 2Hz at 1 second, 4rad/s at 2 seconds, 8rad/s at 3 seconds, etc.After the system has been modeled, an input type has been chosen, and the initial conditions have been selected, the system can be analyzed to determine its behavior. The most fundamental technique is to integrate the differential equation(s) for the system.1.2 Explicit SolutionsSolving a differential equation results in an explicit solution. This equation pro-vides the general response as a function of time, but it can also be used to find frequencies and other characteristics of interest. This section will review techniques used to integrate first and second-order homogenous differential equations. These equations correspond to systems without inputs, also called unforced systems. Non-homogeneous differential equations will also be reviewed.The basic types of differential equations are shown in Figure 1.2. Each of these equations is linear. On the left hand side is the integration variable ’x’. If the right hand side is zero, then the equation is homogeneous. Each of these equations is linear because each of the terms on the left hand side is simply multiplied by a linear coefficient.inputssystemoutputsdifferentialequationsfunction functionNote: By convention inputs are on the left, and outputs are on the right. forcing responsepage 174Figure 1.2 Standard equation formsA general solution for a first-order homogeneous differential equation is given in Figure 1.3. The solution begins with the solution of the homogeneous equation where a general form is ’guessed’. Substitution leads to finding the value of the coefficient ’Y’. Following this, the initial conditions for the equation are used to find the value of the coef-ficient ’X’. Notice that the final equation will begin at the initial displacement, but approach zero as time goes to infinity. The e-to-the-x behavior is characteristic for a first-order response.Ax··Bx·Cx++ Df t()=second-order non-homogeneousAx··Bx·Cx++ 0=second-order homogeneousAx·Bx+ Cf t()=first-order non-homogeneousAx·Bx+0=first-order homogeneouspage 175Figure 1.3 General solution of a first-order homogeneous equationAx·Bx+0=Given the general form of a first-order homogeneous equation,xXeYt–=Guess a solution form and solve.x·YX– eYt–=AYX– eYt–()BXeYt–()+0=AY–()B+0=YBA---=andx0() x0=initial conditionxhXeBA--- t–=Next, use the initial conditions to find the remaining unknowns.x0XeBA--- 0–=xhXeBA--- t–=Therefore the general form is,x0X=xt() x0eBA---t–=Therefore the final equation is,Note: The general form below is useful for finding almost all homogeneous equationsxht() XeYt–=page 176Figure 1.4 Drill Problem: First order homogeneous differential equationThe general solution to a second-order homogeneous equation is shown in Figure 1.5. The solution begins with a guess of the homogeneous solution, and the solution of a quadratic equation. There are three possible cases that result from the solution of the qua-dratic equation: different but real roots; two identical real roots; or two complex roots. The three cases result in three different forms of solutions, as shown. The complex result is the most notable because it results in sinusoidal oscillations. It is not shown, but after the homogeneous solution has been found, the initial conditions need to be used to find the remaining coefficient values.x·2x+0=Solve the following differential equation given the initial condition.x0() 3=xt() 3e2t–=ans.page 177Figure 1.5 Solution of a second-order homogeneous equationAs mentioned above, a complex solution when solving the homogeneous equation results in a sinusoidal oscillation,

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