自动控制原理与设计[李中华译]课后答案

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自动控制原理与设计(第五版)李中华译 课后答案
13 k,(X-X K,x, m k(x1-x2) k(x2-y) n k a-k( )-b元 k X ⅹ1-Ⅹ k, m k, (x2-y) X -X (X1-X kk( F-k(x-x)-b(-元) 2. Write the equations of motion of a pendulum consisting of a thin, 2-kg stick of length l suspended from a pivot. How long should tI he rod be in order for the period to be exactly 2 secs?(The inertia I of a thin stick about an endpoint is -mL. Assume 0 is small enough that sin 0 8) Let’ s use eq(2.14 M=l 14 Moment about point O 19 sin 0=log as we assumed o is small 3g The frequency only depends on the length of the rod 2丌 3 1.49m 2丌 <Notes> 15 Figure 2.39: Double pendulum (a)Compare the formula for the period, T=2TVd with the well known formula for the period of a point mass hanging with a string with length l.T= (b) Important In general, Eg.(2. 14 is valid only when the reference point for the moment and the moment of inertia is the mass center of the body. However, we also can use the formular with a reference point other than mass center when the point of reference is fixed or not accelerating, as was the case here for point O 3. Write the equations of motion for the double-pendulum system shown in Fig. 2. 39. Assume the displacement angles of the pendulums are small enough to ensure that the spring is always horizontal. The pendulum rods are taken to be massless, of length l, and the springs are attached 3/4 of the way down 16 b If we write the moment equilibrium about the pivot point of the left pen- dulen from the free body diagram M=-mgl sin0-k-l(sin 0 sin 0)cose ml 8 +mgl sin 0 +ihl cos 0(sin 8 - 0)=0 Similary we can write the equation of motion for the right pendule mgl sin 0 +k-l(sin 0 - sin 0)cos 0 - l=ml 0 As we assumed the angles are small, we can approximate using sin g 6,sin6≈b,cos≈1, and cos≈1. Finally the linearized equations of motion becomes 9 mlg +mg6 16 k(0-6) 9 m8+m9+gkl(6-6) O 17 +20 7+16m 0-6 9 k 0+90+ 0) 16m 4. Write the equations of motion for a body of mass M suspended from a fixed point by a spring with a constant k. Carefully define where the body' s displacement is zero Some care needs to be taken when the spring is suspended vertically in the presence of the gravity. We define =0 to be when the spring is unstretched with no mass attached as in(a). The static situation in(b results from a balance between the gravity force and the spring k y=0(x=-mg/k) g From the free body diagram in(b), the dynamic equation results m2元=-kx-m9 We can manipulate the equation 72 n k so it we replace a using y =2+ F9, n -ke i+ ky 18 The equilibrium value of a including the effect of gravity is at I=-kg and y represents the motion of the mass about that equilibrium point An alternate solution method, which is applicable for any problem involving vertical spring motion, is to define the motion to be with respect to the static equilibrium point of the springs including the effect of gravity and then to proceed as if no gravity was present. In this problem, we would define y to be the motion with respect to the equilibrium point then the FBD in(c)would result directly in 5. For the car suspension discussed in Example 2.2 (a) write the equations of motion(Eqs. (2.10)and(2. 11))in state-variable form. Use the state vector a i y y1 (b) Plot the position of the car and the wheel after the car hits a"unit bump(i.e, r is a unit step) using MATLAB. Assume that m 10kg,m=350kg,k=500,000N/m,ks=10,000N/m.Find the value of b that you would prefer if you were a passenger in the can (a) We can arrange the equations of motion to be used in the state- variable form y y m n k b xbm k —+—y+ 0/+ k j=-8x+ y So, for the given sate vector of = the state-space form will be 0 k b k 0 0 y b b (b)Note that b is not the damping ratio, but damping. We need to find the proper order of magnitude for 6, which can be done by trial and error. What passengers feel is the position of the car. Some general requirements for the smooth ride will be, slow response with small overshoot and oscillation From the figures, b a 3000 would be acceptable. There is too much overshoot for lower values, and the system gets too fast(and harsh) for larger values Problem 2.5 b clear all. close all 1=10 m2=350 kW=500000 ks=10000; B=[1000200030004000] t=0:0.01:2; for i= 1: 4 B(i); F=[0100;-(ks/m1+kw/m1)-b/m1ks/m1b/m1; 0001;ks/m2b/m2-ks/m2-b/m2] G=[0;kw/m1;0;0 H=[1000;0010] y (F,G,H,J,1,t); subplot(2, 2, i) plot( t, y(:, 1), (:,2),); legend(Wheel, Car) ttl= sprintf('Response with b=%4.lf, b) title(ttl) end 6. Automobile manufacturers are contemplating building active suspension systems. The simplest change is to make shock absorbers with a change- able damping, b(u). It is also possible to make a device to be placed in parallel with the springs that has the ability to supply an equal force, u in opposite directions on the wheel axle and the car body (a) Modify the equations of motion in Example 2.2 to include such con- trol input (b) Is the resulting system linear? (c) Is it possible to use the forcer, u. to completely replace the springs and shock absorber? Is this a good idea a) The Fbd shows the addition of the variable force, u, and shows b as in the fbd of fig. 2.5. however. here b is a function of the control variable. u. The forces below are drawn in the direction that would result from a positive displacement of a

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fwchina_2310 答案与《自动控制原理与设计》第六版的习题有些不同,需要自己对应相应的习题来看。
2020-05-15
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骑猫赶集 还是有帮助的~
2019-08-21
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qq_31738481 答案很有帮助,对应第六版的话要自己去对应章数和题号,赞一个
2015-10-17
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zj151948 还不错,挺好的
2015-09-27
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gezhongzawu 很好,英文版
2015-09-27
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