matlab与控制系统仿真课件-未知版本1ch

1、Chapter3 STATE VARIABLE MODEL1.INTRODUCTION2. TWO SIMPLE PHYSICAL SYSTEM3. SUMMARYEXAMPLE 1Consider the system shown in Figure 3.2. The variables of interest are noted on the figure and defined as M1, M2=mass of cart, P, q=position of cart, u =external force acting on system, k1, k2=spring constant,

2、 b1, b2=damping coefficient.A variety of methods is available for developing the equations of motion for the two cars. Here we use the free-body diagram approach. The free-body diagram of mass M1 is shown in Figure 3.3, where velocity of M1, and M2, respectively,We assume that the cars have negligib

3、le rolling friction. We consider any existing rolling friction as lumped into the damping coefficients, b1 and b2. When using free-body diagrams we can encounter difficulty when assigning direction to the spring force, k1(p-q), and damping force, b1, . Let us consider the mass M1 first. The position

4、 of M1 is denoted by p. The positive direction of p is specified by the pointing direction of the arrow (see Figure 3.2). Which direction we choose to point the arrow is irrelevant; but once we have specified a direction, the applied forces must be consistent with that direction.Suppose for example,

5、 that we could hold M2 so that it cannot move (that is , q=0 ). Then we move the mass M1 in a positive direction. In moved to the right, we obtain a reaction force from the spring and damper that resists the motion. The direction of the forces is in the negative direction, or pointing to the left in

6、 the free-body diagram. Thus the spring force is ,and the damping force isOf course, if ,the spring force is affected. In fact, if p = q, then the spring is not compressed or stretched, and the spring force is zero. The above discussion applies to mass M2 as well. Now ,given the free-body diagram wi

7、th forces and directions appropriately applied, we use Newtons second law (sum of the forces equals mass of the object multiplied by its acceleration) to obtain the equation of motion-one equation for each mass. For mass M1 we have ,or, (3.1)Similarly, for mass M2 we have ,or (3.2)We now have a mode

8、l given by the two second-order ordinary differential equations in Eqs. (3.1)and (3.2). We can start developing a state-space model by definingSimilarly, for mass M2 we have ,or (3.2)We now have a model given by the two second-order ordinary differential equations in Eqs. (3.1)and (3.2). We can star

9、t developing a state-space model by definingDenoting the derivatives of x1 and x2 as x3 and x4 , respectively, it follows that ,(3.3) . (3.4)Eqs. (3.3), (3.4), (3.7), and (3.8) can be written as where , and and u is the external force acting on the system (see Figure 3.2). If we choose p as our outp

10、ut, then .The Matlab script used to simulate the step response of the rolling car system is shown if Figure 3.4. The parameter values used in the simulation are M1=1(slug), M2=1(slug), k1=10(Ib/ft), k2=50(Ib/ft), b1=1(1b-sec/ft), b2=2(1b-sec/ft).When the input is a unit step-that is , u = 1 Ib-the s

11、teady-state position of cart is approximately 1.4 inch. To view the position time-history of cart 2, we can change the output matrix to: C = ( 0 1 0 0 ).STEP ReSPONSE:k1=10;k2=50;b1=1;b2=2;m1=1;m2=1;a=0 0 1 0; 0 0 0 1;-k1/m1 k1/m1 -b1/m1 b1/m1;k1/m2 -(k1+k2)/m2 b1/m2 -(b1+b2)/m2;b=0 0 1/m1 0;c=1 0 0

12、 0;d=0;sys = ss(a,b,c,d)step(sys)Example 2. Electrical :RLC Series CircuitConsider the RLC series eircuit shown in Figure 3.5, where vin = input voltage (volts), L = inductance (H), R = resistance (), C = capacitance (F), vc = voltage across the capacitor (volts), i = current (anps).Using Kirchhoffs

13、 voltage law, we obtain , (3.9)where . (3.10)Then taking the derivative of xi and using Eq. (3.10) yields .(3.11)Also, taking the time derivative of x2 and using Eq. (3.9) yields .(3.12)We can write Eqs. (3.11) and (3.12) in matrix form as where ,and .With , and , we have .If we can measure ,then we have ,where and .We can also computer the transfer function as So fo