Lecture 37 - Output Feedback Design Examples - Electrical Engineering (EE) PDF Download

Clear all your Electrical Engineering (EE) doubts with EduRev Join for Free

Join for Free

Lecture 37 - Output feedback design examples, Control Systems

1 Output feedback design examples

In the last lecture, we have discussed about the incomplete state feedback design and output feedback design. In this lecture we would solve some examples to make the procedure properly understood.

Example 1: Let us consider the following system

x(k + 1) = Ax(k) + Bu(k)

y(k) = C x(k)

for which the A, B , C matrices are as follows

We know, for output feedback,

u(k) = −Gy(k)

where the matrix G has to be designed. Since C has rank 2 and the rank of B is also 2, minimum 2 eigenvalues can be placed at desired locations. Let these two be 0.1 and 0.2. The characteristic equation of A is

|zI − A| = z³ + 1

B ^∗ is written as

B^∗ = BW =

which has two independent parameters in terms of w₁ and w₂. Controllability matrix for the pair (A, B ^∗) is

It will be non singular if   Let

G^∗ =[g₁^∗ g₂^∗]C

Then

Since C has a rank of 2, G^∗C has two independent parameters in terms of g₁^∗ and g₂^∗. The closed loop characteristic equation is

φ(z) = z³ + α³z² + α₂z + α₁ = 0

Thus
G^∗C =[ 0 0 1 ] Uc^∗−1 φ(A)

or,

The last row in the above equation corresponds to the following constraint equation.

             (1)

Since 2 of the three eigenvalues can be arbitrarily placed, w1 and w2 can be arbitrary provided the condition   is satisfied. But they should be selected such that the third eigenvalue is stable. This puts an additional constraint on w₁ and w₂.

For example, the necessary condition for the closed loop system to be stable is |α| < 1.

To satisfy this condition, w₂ cannot be equal to zero.

For z = 0.1 and 0.2 to be the roots of the characteristic equation

z³ + α³z² + α₂z + α₁ = 0

the following equations must be satisfied

α₁ + 0.001 + α₃0.01 + 0.1α₂ = 0
α₁ + 0.008 + α₃0.04 + 0.2α₂ = 0

Simplifying the above equations,

α₂ + 0.3α₃ + 0.07 = 0 (2) α₁ − 0.02α₃ − 0.006 = 0 (3)

Solving equations (1), (2) and (3) together

If we set w₁ = 0 and w₂ = 1, we get α1 = 0.00133, α₂ = 0 and α₃ = −0.23333.
With the above coefficients we find the roots to be z₁ = 0.1, z₂ = 0.2 and z₃ = −0.0667.
Thus the third pole is placed within the unit circle and the closed loop system is stable.
There also exist some other combinations of w₁ and w₂ for which z₁ = 0.1, z₂ = 0.2 and the closed loop system is stable.
Putting the values of w₁ and w₂ and corresponding α₁, α₂ and α₃ in the expression of G^∗C , we get

Thus the feedback matrix can be calculated as

Hence,

Example 2: Consider the same system as in the previous example except for the fact that now

which has rank 1. This implies that

which has only one independent parameter in terms of g₁^∗ and g₂^∗. Thus

Last two rows of the above equation are constrained to be zero. Thus we can only assign w₁ or w₂ arbitrarily, not both. The constraint equations are as follows.

If we want two closed loop eigenvalues to be placed at z = 0.1 and z = 0.2, we will altogether have four equations with five unknowns. Only one of these five unknowns can be assigned arbitrarily.
But these four equations would be nonlinear in w₁ and w₂, hence difficult to solve. The simpler way would be to use the following equation

Only two coefficients can be arbitrarily assigned. Since the constant term is equal to 1, the system cannot be stabilized with output feedback.