# Another Control System Question

Discussion in 'Homework Help' started by Kayne, Jun 23, 2010.

1. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
For a discrete time system

$\frac{Y(z)}{R(z)}=\frac{z^-1+z^-2}{1+1.3z^-1+0.4z^-2}$

Rewritting into a state space form

$X_{n+1}=\Phi X_{n} + \Gamma U_{n}, y_{n}=CX_{n}$

I have found the following to the question

$\begin{array}{1}X_{1}\\ X_{2}\end{array} =\begin{array}{1}0&1\\ -0.4&-1.3\end{array}\begin{array}{1}X_{1(k)}\\ X_{2(k)}\end{array}+\begin{array}{1}0\\ 1\end{array}uk$

$y(k+1)=\begin{array}{1}1&1\end{array}\begin{array}{1}X_{1(k+1})\\ X_{2(k+2)}\end{array}$

Using state variable technique design a controler with z=±0.5

This is where I am not sure if I times $\begin{array}{1}0&1\\ -0.4&-1.3\end{array}$ by ±0.5 to get the answer....

2. ### tskaggs New Member

Jun 17, 2010
26
3
The transfer function in the z-domain is

$\T(z)=C*(zI-A)^{-1}*B+D$

I think the purpose of this problem is to find a controller so that when you apply it to the given system, the poles are located at the given values of Z.

I'm not familiar with the state-variable method... is that meaning state-feedback?

For the sake of this post let's assume you want to design a controller that implements full state-feedback. The following wiki link gives a little more info on the subject:
http://en.wikipedia.org/wiki/Full_state_feedback

So basically with state-feedback you are measuring the states of your system and feeding them back so that your input is equal to the states (u=-K*x, where x is the state vector and K is your control gain matrix). If you substitute the input equation back into the state-space equations your new transfer function looks like -

$\T(z)=C*[zI-(A-B*K)]^{-1}*B+D$

This method is otherwise known as the "pole-placement method," because given knowledge of every state of the system you can create the control gain matrix K to place the poles of the system anywhere you want.

Now we are only interested in the poles of the transfer function. Since D in our case is the empty matrix then denominator of the transfer function is equal to

$\det[zI-(A-B*K)]$

Since you want the poles to be at z=$\pm$0.5, then you need to solve the following equation for the elements of K

$\(z-0.5)*(z+0.5)=det[zI-(A-B*K)]$

Last edited: Jun 23, 2010
3. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
I am a little confused to what you said so I have used both the formulas that you said and worked though them which is below.

Using the formula that you have stated
$C \times (zI-A)^-1 \times B$
and working though this am I correct with the answer below.

$= \left(\begin{array}{cc}1&1\end{array}\right) \times (\left(\begin{array}{cc}z&0\\0&z\end{array}\right)-\left(\begin{array}{cc}0&1\\-0.4&1.3\end{array}\right))^-1 \times \left(\begin{array}{cc}0\\1\end{array}\right)$

$= \left(\begin{array}{cc}1&1\end{array}\right) \times (\left(\begin{array}{cc}z&-1\\0.4&1.3\end{array}\right))^-1 \times \left(\begin{array}{cc}0\\1\end{array}\right)$

$= \left(\begin{array}{cc}1&1\end{array}\right) \times \left(\begin{array}{cc}z+1.3 &1\\-0.4&z\end{array}\right) \times \left(\begin{array}{cc}0\\1\end{array}\right)$

$=\left(\begin{array}{cc}0&z+1\end{array}\right)$

Also I have done this using $\det [zI- (A-B \times k]$ formula that you have said

$\det [(\left(\begin{array}{cc}z&0\\0&z\end{array}\right)-(\left(\begin{array}{cc}0&1\\-0.4&1.3\end{array}\right)- \left(\begin{array}{cc}0\\1\end{array}\right) \times \left(\begin{array}{cc}k_{1}&k_{2} \end{array}\right) )]$

$\det [(\left(\begin{array}{cc}z&0\\0&z\end{array}\right)-(\left(\begin{array}{cc}0&1\\-0.4&1.3\end{array}\right)- \left(\begin{array}{cc}0\\k_{2}\end{array}\right)]$

Have I got this correct so far to continue with the poles

Thanks again

Last edited: Jun 24, 2010
4. ### tskaggs New Member

Jun 17, 2010
26
3
No, remember you are working with matrices, so all the dimensions must match up if you want to add, subtract, or multiply.

The brilliance of this solution is that when you take the inverse of (zI-A+BK) the denominator is the determinant. And since we are only worried about the poles we only have to work with the denom.

I've attached a pdf to show you how to compute the solution, because I think it will help you understand.

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5. ### Kayne Thread Starter Active Member

Mar 19, 2009
105
0
I will go back and re do the answer, I new that the matrices had to be the same but with the K component was unsure how to do this. After I do it again I will compare with the answer you have said and see where I have gone wrong.