[eBook] Feedback Control of Dynamic Systems 5E (Franklin) Solution Manual

[eBook] Feedback Control of Dynamic Systems 5E (Franklin) Solution Manual

(Parte 2 de 8)

(b) Is the resulting system linear?

(c) Is it possible to use the forcer, u2, to completely replace the springs and shock absorber? Is this a good idea?

Solution:

(a) The FBD shows the addition of the variable force, u2, and shows b as in the FBD of Fig. 2.5, however, here b is a function of the control variable, u1. The forces below are drawn in the direction that would result from a positive displacement of x.

(b) The system is linear with respect to u2 because it is additive. But b is not constant so the system is non-linear with respect to u1 because the control essentially multiplies a state element. So if we add controllable damping, the system becomes non-linear.

are now available on some carswhere the driver chooses between

(c) It is technically possible. However, it would take very high forces and thus a lot of power and is therefore not done. It is a much better solution to modulate the damping coefficient by changing oriÞce sizes in the shock absorber and/or by changing the spring forces by increasing or decreasing the pressure in air springs. These features as ofto r stiff ride.

7. Modify the equation of motion for the cruise control in Example 2.1, Eq(2.4), so that it has a control law; that is, let

This is a proportional control law where the difference between vr and the actual speed is used as a signal to speed the engine up or slow it down.

Put the equations in the standard state-variable form with vr as the input and v as the state. Assume that m = 1000 kg and b =5 0 N·s/m, and

Þnd the response for a unit step in vr using MATLAB. Using trial and error, Þnd a value of K that you think would result in a control system in which the actual speed converges as quickly as possible to the reference speed with no objectional behavior.

Solution:

úv + b

m u substitute in u = K (vr − v) œv + b m u = K

A block diagram of the scheme is shown below where the car dynamics are depicted by its transfer function from Eq. 2.7.

m bs m

Thes tate-variablef ormo ft he equationsi s,

m vr so that the matrices for Matlab are

Time (sec.)

A m p lit u d e

Step Response

We can see that the larger the K is, the better the performance, with no objectionable behaviour for any of the cases. The fact that increasing K also results in the need for higher acceleration is less obvious from the plot but it will limit how fast K canb ei nt he real situationb ecauset he engine has only so much poop. Note also that the error with this scheme gets quite large with the lower values of K. You will Þnd out how to eliminate this error in chapter 4 using integral control, which is contained in all cruise control systems in use today. For this problem, a reasonable compromise between speed of response and steady state errors would be K = 1000, where it responds is 5 seconds and the steady state error is 5%.

%P roblem 2.7 clear all, close all

% Overlay the step response hold on for i=1:length(k)

Problems and Solutions for Section 2.2

8. In many mechanical positioning systems there is ßexibility between one part of the system and another. An example is shown in Figure 2.6 where there is ßexibility of the solar panels. Figure 2.40 depicts such a situation, where a force u is applied to the mass M and another mass m is connected to it. The coupling between the objects is often modeled by a spring constant k with a damping coefficient b, although the actual situation is usually much more complicated than this.

(a) Write the equations of motion governing this system, identify appropriate state variables, and express these equations in state-variable form.

(b) Find the transfer function between the control input, u, and the output, y.

Solution: (a) The FBD for the system is

25 Figure 2.40: Schematic of a system with ßexibility

which results in the equations

Let the state-space vector x =[ x œxy œy ]T m y+ b

m œy

M x+ b

M œx − k M y − b M œy + 1

M u

œx ¤x œy ¤y

− km − bm km b m kM bM − kM − b M y œy

(b) We have the complete state-variable form from part a. We will learn the systematic way to convert from the state-variable form to transfer function later in chapter 7. If we make Laplace Transform of the equations of motion m X+ b m sX − k m

M Y + b

26 CHAPTER 2. DYNAMIC MODELS From Cramer s Rule,

det

Finally,

9. For the inverted pendulum, Eqs. (2.34),

(a) Try to put the equations of motion into state-variable form using the state vector x =[ θ úθ x œx ]T. Why is it not possible?

(b) Write the equations in the descriptor form and deÞne values for E, F0,a nd G0 (note that E is a 4 × 4m atrix). Then show how you would compute F and G for the standard statevariable description of the equations of motion.]

Solution: (a) It is impossible because the acceleration terms are coupled.

x œx

10. The longitudinal linearized equations of motion of a Boeing 747 are given in Eq. (9.28). Using MATLAB or other computer aid:

(a) Determine the response of the altitude h for a 2-sec pulse of the elevator with a magnitude of 2◦. Note that, since Eq. (9.28) represents a set of linearized equations, the state variables actually represent the deviation of the state from the nominal operating point. For example, h represents the amount the altitude of the aircraft differs from 20,0 ft.

(b) Consider using the feedback law δe = Khh + δe,ext (127) where the elevator input angle is the sum of a term proportional to the error in altitude h plus an external input (a disturbance or command input). Note from part (a) that a positive change in elevator causes a negative change in altitude, so that the proposed proportional feedback law has the logical sign to anticipate a stable system provided Kh > 0. By trial and error, try to Þnd a value for the feedback gain Kh such that a 2◦ pulse of 2 sec on δe,ext yields a more stable altitude response.

(Parte 2 de 8)

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