1
\$\begingroup\$

I have a discrete time system

Suppose I have the following AB matrix of my system $$A = \begin{bmatrix} 0 & 1 \\ 1 & 1.5 \\ \end{bmatrix} $$

$$B = \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix}$$

To design a state feedback stabilizer, I want relationship $$u = Kx$$

where K is my feedback gain, u is input to my discrete time system

To design this, use well known ackerman's formula as follows:

$$K = - \begin{bmatrix} 0 1 \end{bmatrix}*C^{-1}_{AB}*r(A)$$

where r(A) is my characteristic polynomial containing the pole I want to place

MATLAB directly calculates this using "acker" (and "place", but we will limit to one problem at a time)

But I keep on getting two different results and it is driving me crazy! I checked every single step!

Please help me to pin point the error in the following code:

%Approach 1: Raw formula

A = [0 1; 1 1.5]
B = [0 1]'
CAB = ctrb(A,B)
K1 = -1*[0 1]*inv(ctrb(A,B))*(A^2 - A + 0.25) %I wish to place the poles at {0.5, 0.5}

%Note (A^2 - A + 0.25) = (A - 0.5)(A - 0.5), the characteristic equation containing the two poles

%Approach 2: Using Build-in Acker

K2 = -1*acker(A,B,[0.5,0.5])

It is completely illogical to me why

K1 =

-1.2500 -0.7500

K2 =

-1.2500 -0.5000

I am using (in principle) identical way to calculate the K vector! Also note this error has never occurred to me in any previous designs before my system started blowing up all over.

Can someone who knows control theory pin point exactly what maybe the problem? MATLAB also contains bugs so that is something to be kept in mind.

Thanks

\$\endgroup\$
  • 1
    \$\begingroup\$ You need to multiply 0.25 in the characteristic equation by the identity matrix. \$\endgroup\$ – Suba Thomas Apr 16 '15 at 14:19
  • \$\begingroup\$ Thanks, why not reply as an answer? \$\endgroup\$ – Carlos - the Mongoose - Danger Apr 16 '15 at 19:45
2
\$\begingroup\$

The constant 0.25 in the characteristic equation needs to be multiplied by the identity matrix.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.