UNIVERSITY OF CINCINNATIDept. of Aerospace Engineering and Engineering Mechanics

20-EGFD-615 Introduction to Modern Control Theory Autumn, 2002Solution to Homework 1

Problem 1:

The equations of motion are:

Note that the positions p1 , p2 , and p3 appear in the spring force equations. Therefore, if we use thevelocities v1 , v2 , and v3 as states, we will also need their integrals, the positions p1 , p2 , and p3, asstates. In fact, only the position differences p2 - p1 and p3 - p2, appear in the equations of motion, andthe given output is the position difference p3 - p1 = (p3 - p2) + (p2 - p1), so the model requires knowledgeonly of these two position differences, not necessarily all three positions.

Define the states as:

Differentiating each state and using the equations above to express the time derivatives of thevelocities, we get for the state equations:

and the output equation is just y = x5 + x4. In vector-matrix form:

and .

For the linear graph solution to this problem, start with the linear graph shown at the left below.This graph has 8 branches, 4 nodes (left, top, right,and bottom reference node), and 1 source (F branch). So, we should have 15 unknowns, including thevelocity difference across the force source branch. We can eliminate many of the velocity differenceunknowns by immediately using three of thecompatibility (or “loop”) equations for the upper left,upper right, and center loops (which yield )vB1 =)vK1, )vB2 = )vK2, and )vF = v2, respectively) andthen using the left and right loops to get )vB1 = )vK1

= v2 - v1 and )vB2 = )vK2 = v3 - v2, respectively. Now, all the branch velocity differences are in terms

of the velocities v1 , v2 , and v3. The 7 constitutive equations are:

Finally, the node equations at the left, right, and top nodes, respectively, are:

Define the states as the three velocities v1, v2, and v3 and the two integrated quantities that appear inthe spring force constitutive relationships, namely the integrals of (v2 - v1 ) and of (v3 - v2), which ofcourse are the position differences used as states above. Substituting into the nodal equations yieldsthree of the state equations, with the other two coming from the fact that the time derivatives of the

integrated states are just the integrands (here, the velocity differences). The results are exactly thesame as above.

Problem 2: The circuit diagram (which is also essentially the linear graph once the voltage drops andcurrents are defined for each element):

The graph has 8 branches, 6 nodes (numbered), and 2 sources. So, there will be 14 unknowns, 2 ofwhich are the current through the voltage source iv and the voltage across the current source vi. Thestate variables are already chosen above as the energy storage variables for the two capacitors(voltage drops) and the inductor (current). There are 6 elemental or constitutive equations:

There are 5 node equations at nodes 1 through 5:

Now we need 3 independent compatibility (or “loop”) equations to complete the set of 14 equationsnecessary to solve for all the unknowns. There are many ways to choose these 3 equations, one ofwhich is to use the 3 successively larger loops that all involve the voltage source. Using these 3:

There are an infinite number of ways to proceed with the algebra at this point. The idea is toeliminate all of the unknowns except the states x1, x2, and x3 from the equations and then to solve forthe first time derivatives of the states using the last three constitutive equations. ONE way to do thisis as follows.

Use the third and fifth node equations to eliminate the variables iC1 and iC2 in favor of i2 and -is,respectively. Notice also that the first node equation expresses iv in terms of i1, and that iv does notappear anywhere else, so we have eliminated it. Now, use the first three constitutive equations (for

the resistors) to eliminate the resistor voltages v1, v2, and v3 in favor of the currents. The remainingconstitutive, node, and compatibility equations after these eliminations are:

Solve the two remaining node equations for i2 and i3, solve the second compatibility equation for vL,and drop the last compatibility equation because it merely gives the value of the voltage across thecurrent source. After the substitutions, that leaves:

Solve the last equation for i1 to get:

and substitute it into the first three equations here to get:

and except for dividing each equation by the coefficient of the state derivative terms, these equationsare in state space form. The matrix version is: