To solve a set of differential equations we have two choices, solve them numerically or symbolically. For a symbolic solution the system of differential equations must be manipulated into a single differential equation. In this chapter we will look at methods for manipulating differential equations into useful forms.
The differential operator 'd/dt' can be written in a number of forms. In this book there have been two forms used thus far, d/dt x and x-dot. For convenience we will add a third, 'D'. The basic definition of this operator, and related operations are shown in Figure 6.1 General properties of the differential operator. In basic terms the operator can be manipulated as if it is a normal variable. Multiplying by 'D' results in a derivative, dividing by 'D' results in an integral. The first-order axiom can be used to help solve a first-order differential equation.
Figure 6.1 General properties of the differential operator
Figure 6.2 Proof of the first-order axiom
Figure 6.3 Example: A simplification with the differential operator contains an example of the manipulation of a differential equation using the 'D' operator. The solution begins by replacing the 'd/dt' terms with the 'D' operator. After this the equation is rearranged to simplify the expression. Notice that the manipulation follows the normal rules of algebra.
Figure 6.3 Example: A simplification with the differential operator
An example of the solution of a first-order differential equation is given in Figure 6.4 Example: A a solution for a first-order system. This begins by replacing the differential operator and rearranging the equation. The first-order axiom is then used to obtain the solution. The initial conditions are then used to calculate the coefficient values.
Figure 6.4 Example: A a solution for a first-order system
A typical system will be described by more than one differential equation. These equations can be solved to find a single differential equation that can then be integrated. The basic technique is to arrange the equations into an input-output form, such as that in Figure 6.5 Developing input-output equations. These equations will have only a single output variable, and these are always shown on the left hand side. The input variables (there can be more than one) are all on the right hand side of the equation, and act as the non-homogeneous forcing function.
Figure 6.5 Developing input-output equations
A sample derivation of an input-output equation from a system of differential equations is given in Figure 6.6 Example: An input output equation. This begins by replacing the differential operator and combining the equations to eliminate one of the output variables. The solution ends by rearranging the equation to input-output form.
Figure 6.6 Example: An input output equation
Figure 6.7 Drill problem: Find the second equation in the previous example
In some instances we will want to numerically integrate an input-output equation. The example starting in Figure 6.8 Example: Writing an input-output equation as a differential equation shows the development of an input-output equation for two freely rolling masses joined by a spring. The final equation has a derivative on the right hand side that would prevent it from being analyzed in many cases. In particular if the input force 'F' was a step function the first derivative would yield an undefined (infinite) value that could not be integrated.
Figure 6.8 Example: Writing an input-output equation as a differential equation
The equation is then converted to state variable form, including a step to calculate a second derivative of the input, as shown in Figure 6.9 Example: Writing state equations for equations with derivatives.
Figure 6.9 Example: Writing state equations for equations with derivatives
Input-output equations can also be written as a transfer function. The basic form for a transfer function is shown below in Figure 6.10 Example: A transfer function. The general form calls for output over input on the left hand side. The right hand side is comprised of constants and the 'D' operator. In the example 'x' is the output, while 'F' is the input.
Figure 6.10 Example: A transfer function
If both sides of the example were inverted then the output would become 'F', and the input 'x'. This ability to invert a transfer function is called reversibility. In reality many systems are not reversible.
There is a direct relationship between transfer functions and differential equations. This is shown for the second-order differential equation in Figure 6.11 The relationship between transfer functions and differential equations for a mass-spring-damper example. The homogeneous equation (the left hand side) ends up as the denominator of the transfer function. The non-homogeneous solution ends up as the numerator of the expression.
Figure 6.11 The relationship between transfer functions and differential equations for a mass-spring-damper example
The transfer function for a first-order differential equation is shown in Figure 6.12 A first-order system response. As before the homogeneous and non-homogeneous parts of the equation becomes the denominator and the numerator of the transfer function.
Figure 6.12 A first-order system response
A transfer function is in a form suitable for using normal integration techniques. If the non-homogeneous part does not includes derivatives, then the techniques presented in previous chapters can be used. An example of explicitly solving such an equation is shown in Figure 6.13 Example: Integrating an input-output equation, Figure 6.14 Example: Integrating an input-output equation, and Figure 6.15 Example: Integrating an input-output equation (cont'd).
Figure 6.13 Example: Integrating an input-output equation
Figure 6.14 Example: Integrating an input-output equation
Figure 6.15 Example: Integrating an input-output equation (cont'd)
The classic mass-spring-damper system is shown in Figure 6.16 Example: A transfer function for a mechanical system. In this example the forces are summed to provide an equation. The differential operator is replaced, and the equation is manipulated into transfer function form. The transfer function is given in two different forms because the system is reversible and the output could be either 'F' or 'x'.
Figure 6.16 Example: A transfer function for a mechanical system
Mass-spring-damper systems are often used when doing vibration analysis and design work. The first stage of such analysis involves finding the actual displacement for a given displacement or force. A system experiencing a sinusoidal oscillating force is given in Figure 6.17 Example: Explicit analysis of a mechanical system. Numerical values are substituted and the homogeneous solution to the equation is found.
Figure 6.17 Example: Explicit analysis of a mechanical system
The solution continues in Figure 6.18 Example: Explicit analysis of a mechanical system (continued) where the particular solution is found and put in phase shift form.
Figure 6.18 Example: Explicit analysis of a mechanical system (continued)
The system is assumed to be at rest initially, and this is used to find the constants in the homogeneous solution in Figure 6.19 Example: Explicit analysis of a mechanical system (continued). Finally the displacement of the mass is used to find the force exerted through the spring on the ground. In this case there are two force frequency components at 1.392rad/s and 6rad/s. The steady-state force at 6rad/s will have a magnitude of .2932N. The transient effects have a time constant of 4 seconds (1/0.25), and should be negligible within a few seconds of starting the machine.
Figure 6.19 Example: Explicit analysis of a mechanical system (continued)
A decision has been made to reduce the vibration magnitude transmitted to the ground to 0.1N. This can be done by adding a mass-spring isolator, as shown in Figure 6.20 Example: Vibration isolation system. In the figure the bottom mass-spring-damper combination is the original system. The mass and spring above have been added to reduce the vibration that will reach the ground. Values must be selected for the mass and spring. The design begins by developing the differential equations for both masses.
Figure 6.20 Example: Vibration isolation system
For the design we are only interested in the upper spring, as it determines the force on the ground. An input-output equation for that spring is developed in Figure 6.21 Example: Developing an input output equation. The given values for the mass-spring-damper system are used. In addition a value for the upper mass is selected. This is arbitrarily chosen to be the same as the lower mass. This choice may need to be changed later if the resulting spring constant is not practical.
Figure 6.21 Example: Developing an input output equation
This particular solution of the differential equation will yield the steady-state displacement of the upper mass. This can then be used to find the needed spring coefficient.
Figure 6.22 Example: Developing an input output equation
Figure 6.23 Example: Finding the particular solution
Figure 6.24 Example: Finding the particular solution (cont'd)
Finally the magnitude of the particular solution is calculated and set to the desired amplitude of 0.1N. This is then used to calculate the spring coefficient.
Figure 6.25 Example: Calculation of the spring coefficient
· The differential operator can be manipulated algebraically
· Equations can be manipulated into input-output forms and solved as normal differential equations
1. Develop the input-output equation and transfer function for the mechanical system below. There is viscous damping between the block and the ground. A force is applied to cause the mass to accelerate.
2. Find the input-output form for the following equations.
3. Find the transfer function for the systems below. Here the input is a torque, and the output is the angle of the second mass.
4. Find the input-output form for the following equations.
5. The following differential equations were converted to the matrix form shown. Use Cramer's rule to find an input-output equation for `y'.
6. Find the input output equation for y2. Ignore the effects of gravity.
7. Find the input-output equations for the systems below. Here the input is the torque on the left hand side.
8. Write the input-output equations for the mechanical system below. The input is force `F', and the output is `y' or the angle theta (give both equations). Include the inertia of both masses, and gravity for mass `M'.
9. The applied force `F' is the input to the system, and the output is the displacement `x'.
a) Find x(t), given F(t) = 10N for t >= 0 seconds.
b) Using numerical methods, find the steady-state response for an applied force of F(t) = 10cos(t + 1) N ?
c) Solve the differential equation to find the explicit response for an applied force of F(t) = 10cos(t + 1) N ?
d) Set the acceleration to zero and find an approximate solution for an applied force of F(t) = 10cos(t + 1) N. Compare the solution to the previous solutions.
10. Find the transfer function for the system below.
11. For the system below find the a) state and b) input-output equations. The cable always remains tight, and all deflections are small. Assume that the value of J2 is negligible. The input is the force F and the output is the angle `theta'.
12. Find the input-output equations for the differential equations below if both 'x' and 'y' are outputs.
13. For the system pictured below find the input-output equation for y2.
1. Solve the following differential equation assuming a unit step input.
2. Solve the following differential equation assuming a unit step input.
Irwin, J.D., and Graf, E.R., Industrial Noise and Vibration Control, Prentice Hall Publishers, 1979.
Close, C.M. and Frederick, D.K., "Modeling and Analysis of Dynamic Systems, second edition, John Wiley and Sons, Inc., 1995.