Unreachable systems Consider the system shown in Figure 63 W

(Unreachable systems) Consider the system shown in Figure 6.3. Write the dynamics of the two systems as

dx/dt = Ax + Bu, dz/dt = Az + Bu.

If x and z have the same initial condition, they will always have the same state regardless of the input that is applied. Show that this violates the definition of reachability and further show that the reachability matrix Wr is not full rank.

Solution

Given, the complete system is then described by

dX/dt = AX+BU

dZ/dt = AZ+BU

The reachability matrix is Wr = · [1 1 :1 1] ¸

This matrix is singular and the system is not reachable. One implication of this is that if X and Zstart with the same value, it is never possible to find an input which causes them to have different values. Similarly, if they start with different values, no input will be able to drive them both to zero.

The system consists of two identical systems with the same input. Clearly, we cannot separately cause the first and the second systems to do something different since they have the same input. Hence we cannot reach arbitrary states, and so the system is not reachable. More subtle mechanisms for nonreachability can also occur. For example, if there is a linear combination of states that always remains constant, then the system is not reachable. To see this, suppose that there exists a row vector H such that 0 = d dt Hx = H(Ax+Bu), for all u.

Then H is in the left null space of both A and B and it follows that HWr = H [B AB ··· A ⁿ¹B ] = 0. Hence the reachability matrix is not full rank. In this case, if we have an initial condition x0 and we wish to reach a state x_f for which Hx₀ \\= Hx_f , then since Hx(t) is constant, no input u can move from x0 to x f .