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If I transform wave equation for vibrating string Mx′′+Cx′+Kx=b(t) in linear system using $x_1(t)=x(t)$ and $x_2(t)=x_1^{'}(t)$ vibrating string equation becomes $Md_tx_2(t)+Cx_2(t)+Kx_1(t)=b(t)$. That is:

\begin{align} \frac{d}{dt}\left[ \begin{array}{c} x_1 \\ x_2 \\ \end{array} \right]&= \left[ \begin{array}{cc} 0 & 1 \\ -M^{-1}K& -M^{-1}C \\ \end{array} \right]\left[ \begin{array}{c} x_1 \\ x_2 \\ \end{array} \right]+\left[\begin{array}{c} 0 \\ M^{-1} \end{array}\right]b(t) \end{align}

We think of $x(t)$ as the output, $b(t)$ as the input and define output equation as:

$y=\left[\begin{array}{cc} 1&0 \end{array}\right]\left[\begin{array}{c} x_1 \\ x_2 \end{array}\right]$

So now, in the standard control theory notation of first equation:

$x'=Ax+Bu$ $y=Cx$

We make the following identifications:

$A= \left[ \begin{array}{cc} 0 & 1 \\ -M^{-1}K& -M^{-1}C \\ \end{array} \right], $ a $B=\left[\begin{array}{c} 0 \\ M^{-1} \end{array}\right]$

$C=\left[\begin{array}{cc} 1&0 \end{array}\right]$ $y(t)=x_1(t);u(t)=b(t).$ Shouldn't this be a MIMO system ( I conclude it from dimension of B)? I'm asking because I got information that this is an SISO system, but I can't figure it out?

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  • $\begingroup$ Yairs answer Is right. In nature there really are no strictly defined inputs and outputs. That's really up to the person modeling, to decide what's an input and output. But for your specific example, it's SISO. The B and C coupling vectors each have only one non zero term. $\endgroup$ – docscience Jan 29 '16 at 21:16
  • $\begingroup$ I understand how it looks in nature, but it's hard for me to connected it with theory. From the theory I use, B would be of dimension $2n \times n.$ Maybe, my linearization wasn't so good? I'm making algorithms for computation of reduced system and for me is important if something is scalar, vector or matrix. $\endgroup$ – Efe Jan 29 '16 at 21:31
  • $\begingroup$ You missed my point. But in a control application if you have only one actuator and one sensor, the system, in practical considerations, is SISO regardless if you can couple other outputs or inputs. The ones that do not connect to sensors or actuators are useless in the control. $\endgroup$ – docscience Jan 29 '16 at 21:37
  • $\begingroup$ I think I understand now. When I multiply $Bu$, I'll get vector that has only one non zero term? $\endgroup$ – Efe Jan 29 '16 at 21:45
  • $\begingroup$ No, in order to couple in two actuators, your input coupling, $B$ needs to be 2 X 2 or the input itself, u needs to be 2 X 1. $\endgroup$ – docscience Jan 29 '16 at 21:52
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I don't understand why are you basing your definition as either SISO or MIMO on the dimensionality of $B$.

The same physical system (viberating string in your case), can be either SISO or MIMO depending on your configuration.

The question of classifying a system as SISO or MIMO depends on your control parameters, and the parameters which you "read out" or sample.

From the way I understand your question your control system scheme includes one control parameter $b(t)$, which indicates applied force, and one output parameter $x$ which indicates displacement.

Hence this is a SISO system.

If for instance you would have been interested also in the velocity of the string, you could have defined a second output variable $\dot{x}$. In this case you would've had:

$$ \begin{bmatrix} y_1 & y_2 \end{bmatrix}= \begin{bmatrix} 1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} $$

In this case, the same system wouldn't have been SISO any more. The way to understand whether your SISO or MIMO isn't according to the dimensionality, but according to the non-zero entries in $B$ and $C$.

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  • $\begingroup$ I understand what you mean but the definition of state-space system I have says that for SISO system dimension of matrix B is $n \times 1$ and $1\times n$ for C but in my equations these dimensions are different. $\endgroup$ – Efe Jan 29 '16 at 21:12
  • $\begingroup$ I don't know where did you get this definition from, but as I wrote the classification cannot be done according to dimensionality alone. That being said, in your case you have exactly the dimensions you wrote in your comment, with $n=2$ $\endgroup$ – Yair M Jan 29 '16 at 21:20
  • $\begingroup$ Does it mean that I should "see" matrices M, C and K as scalars? Definiton is from A.C.Antoulas, Approximation of Large-Scale Dynamical Systems. $\endgroup$ – Efe Jan 29 '16 at 21:42
  • $\begingroup$ The exact same equation with "scalar" $M$, $C$, and $K$, could in principle describe one mass connected to a string and damper, right? Would you agree then that this is a SISO system? As I wrote, and as @docscience wrote, the physical system in question doesn't cllasify the system. Only your control scheme specifying inputs, and outputs does. $\endgroup$ – Yair M Jan 29 '16 at 21:51
  • $\begingroup$ Then I should distinguish the physical system from control scheme, right? Would you agree that for this control scheme trasfer function $C(sI-A)^{-1}B$ is scalar? $\endgroup$ – Efe Jan 29 '16 at 22:06
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Although it can be generalized to accommodate an n-dimensional non-linear state space system, I'll describe for you a 2 dimensional linear state space system as you have posed in your specific example. This way to make matters simpler and more illustrative.

For this system, the $B$ vector is a 2 X 1 vector and $b(t)$ is a scalar. The $B$ vector is properly called the input coupling vector and its purpose is to couple the input, $b(t)$ into the state equations.

Similarly, the $C$ vector is a 1 X 2 vector and $y(t)$ is a scalar. The $C$ vector is properly called the output coupling vector and its purpose is to couple the state equations into the output, $y(t)$.

Since $b(t)$ and $y(t)$ are both scalars, the system is properly called Single Input, Single Output (SISO).

If $b(t)$ were rather a vector there would be 2 inputs thus making the system a multiple input system. Although $B$ does not directly determine whether the system is single or multiple input, it must conform to the dimensionality of $b(t)$ and in your example it would have to be 2 X 2 with the dimension of $b(t)$ taken as 2 X 1.

Likewise If $y(t)$ were rather a vector there would be 2 outputs thus making the system a multiple output system. Although $C$ does not directly determine whether the system is single or multiple output, it must conform to the dimensionality of $y(t)$ and in your example it would have to be 2 X 2 with the dimension of $y(t)$ taken as 2 X 1.

To summarize it's the dimensions of the input and output vectors that strictly define a single or multiple description of the system. The coupling matrices (or vectors) however must conform to their dimensions and properly describe how they couple to the state equations within.

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