1
$\begingroup$

I have a MIMO system:

$\begin{bmatrix} y_1\\ y_2 \end{bmatrix} = \frac{1}{1+\frac{6}{s(s+5)}}\begin{bmatrix} \frac{6}{s(s+5)} & 1\\ 1&-1 \end{bmatrix} \begin{bmatrix} u_1\\ u_2 \end{bmatrix}$

in which the transfer function matrix has a zero at $s=0$. I have found the zero input direction for ths zero at the origin, and I have found that it is:

$v_z=\begin{bmatrix} 0\\ -0.8863 \end{bmatrix}$

I know that the zeros have the property that is I send an input of the form:

$u(s)=v_ze^{-zt}$

where $z$ is the zero starting from the state zero direction, $x(0)=\xi$, I obtain an output which is equal to zero, so $y(t)=0$. (zeros)

So, they have a blocking ptroperty.

One thing that I noticed is that the zero input direction has one component which is equal to zero. I have tried to go deeper on this, but I have not understood how to interpret this.

I know that the input zero direction vector specify the proportions in which the exponential $e^{-zt}$ should be present at the corresponding inputs to ensure that, if the initial condition is chosen as the state zero direction $\xi$, this exponential appears in none of the outputs.

So, given this interpretation I was thinking that, since the first component of the zero input direction is equal to zero, it means that $e^{-zt}$ is not present in the corresponding input, and so the first input has no influence in rendering the output equal to zero.

But this seems to forced and I am not really sure I am reasoning correctly. Moreover, I have not very well understood what is $e^{-zt}$.

I have found slides at the last slide that a zero at the origin gives problems in controlling the system at stationarity. I don't know why. Is this related to the fact that the zero input vector has this particular form? (this concept is also written here : here, where it says that there is a singular value which is constant, and in my case, if I use the Matlab command sigma(G) I see that also in my case one singular value is constant. I guess because I have also a zero at the origin).

[EDIT] I have just understood that \$e^{-zt}\$ is the mode given by the zero, so if we consider $u(t)=u_0e^{-zt}$ is an exponential input , and the output will be:

$y(t)=G(s)u_0e^{-zt}$

where $G(s)$ is the transfer function.

So, I think I can try to apply this reasoning as follows:

I can consider an input in the direction $v_z$, and this imput is:

$u(t)=v_ze^{-zt} = \begin{bmatrix} 0\\ -0.8863 \end{bmatrix}e^{-zt}$

and I know that a zero has a blocking property, so I will have $y(t)=0$ if I apply this input.

So, an idea that came into my mind is that I could have controllabilitu broblems, but I am really not sure. I am confused about this topic.

The concept is also explained here: zeros.

So, does the fact that the zero input direction has the form $v_z=\begin{bmatrix} 0\\ -0.8863 \end{bmatrix}$ implies some particular situation?

My doubts came from a just theoretical issue. The fact that I have a zero input vector, looks to me that the mode$ e^{z1t}$ cannot be excited by, for example, an impulse in the first input. So, this form of the zero input direction gave me doubts that the system may have controllability issues. I tried on Matlab to use the command ctrb(A,B), but the output tells me the system is controllable. But to me it looks like I cannot control the the state $x1$, since I cannot send an input in the first channel. I would like to say that mine is only a reasoning.

I have fpund here https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-30-feedback-control-systems-fall-2010/lecture-notes/MIT16_30F10_lec10.pdf, that if the zero input direction is $0$, then the system is not controllable. Has this something to do with my case?

Can somebody please help me?

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.