In this lecture we define the stabilization and pole placement problems and then provide an algorithm that solves these problems.


Stabilization Problem: Given the system

Find the vector K so that all the the roots of have negative real part, or equivalently the closed loop system

is stable.


Pole Placement Problem: Given a single input system:

which is controllable, and given a set of symmetric set complex numbers (symmetric means that the complex number and its conjugate must be in the set)

find K so that

n is the number of state variables. The eigenvalues of the closed loop system (new A ) are equal to the above complex numbers.

Pole Placement Algorithm (Ackermann’s formula): This algorithm solves the above pole placement problem. Follow the steps below to apply the algorithm:

Step 1: Compute the controllability matrix

.

Step 2: Compute the inverse of the controllability matrix .

Step 3: Compute

Step 4: Compute the state feedback gain vector

where,

Example 1: Consider the following system. Design a state feedback controller to place the poles (eigenvalues) of the system at -1,-2,-3.

We first evaluate the stability of the system by calculating and checking the roots.

The system is unstable since there is a root at 1 which is nonnegative. We now design our state feedback to move all three of our system poles to s = -1,-2,-3. We apply Ackermann’s formula:

Step 1:

Note that the system us controllable and thus we can use Ackermann’s formula.




Step 2:

Step 3: From our desired poles location, we can construct our desired characteristic equation

We now evaluate the same polynomial expression of our system with s replaced by the A matrix.

Step 4: Finally, we may solve for our gain vector K .

Click here to download a Matlab code that solves this example.

Example 2: Evaluate the stability of the following system and use a feedback control law of the from to place the closed loop poles at [-1,-1,-1].

We first evaluate the stability of the system by calculating and checking the roots.

This means that we have three system poles at s = 0, hence the system is unstable. We now design our state feedback to move all three of our system poles to s = -1. We apply Ackermann’s formula:

Step 1:

Note that the system us controllable and thus we can use Ackermann’s formula.


Step 2:

Step 3: From our desired poles location, we can construct our desired characteristic equation

We now evaluate the same polynomial expression of our system with s replaced by the A matrix.

Step 4: Finally, we may solve for our gain vector K .

Therefore, our control law for this feedback design is:

Now, let us find the new system representation: