A MIMO system with 2 input and 2 output decoupling method to a SISO system is described in many articles and books. How about m*n size transfer functions systems? How can we generalize the method for example to 3*3 or 3*7 MIMO systems?

Here is a 2*2 MIMO system description:

with $\mathrm{D_{11}(s)=D_{22}(s)=1}$ to the form

$$\mathrm{D(s)}=\begin{bmatrix} D_{11}(s) & D_{12}(s) \\ D_{21}(s) & > D_{22}(s) \\ \end{bmatrix}$$

Here we specify a decoupled response and the decoupler with the structure in Equation

$$G_p(s)D(s)=\begin{bmatrix} G_{11}(s) & 0 \\0 & G_{22}(s) > \end{bmatrix} \\ \begin{bmatrix} G_{11}(s) & G_{12}(s) \\ G_{21}(s) & > G_{22}(s) \end{bmatrix} \begin{bmatrix} 1 & D_{12}(s) \\ D_{21}(s) & 1 > \end{bmatrix} > = \begin{bmatrix} G^*_{11}(s) & 0 \\ 0 & G^*_{22}(s) \end{bmatrix}$$

And we can solve four equations in four unknowns to find

$$D_{12}(s) = -\frac{G_{12}(s)}{G_{11}(s)} \\ D_{21}(s) = > -\frac{G_{21}(s)}{G_{22}(s)} \\ G_{l1}(s) = G_{11}(s) = -\frac{G_{12}(s)G_{21}(s)}{G_{22}(s)} \\ G_{l2}(s) = G_{22}(s) = -\frac{G_{21}(s)G_{12}(s)}{G_{11}(s)} $$

  • $\begingroup$ You need to be looking in possibly a network analysis and synthesis textbooks, eg Kuo, or Brian D.O. Anderson & Sumeth Vongpanitlerd. It's not a subject that is taught much these days. $\endgroup$ Commented Feb 25, 2016 at 8:40
  • $\begingroup$ I think that you are looking for the state space form. $\endgroup$ Commented Jul 17, 2016 at 1:53
  • $\begingroup$ This topic on the math stackexchange might help math.stackexchange.com/questions/1297659/… $\endgroup$
    – jos
    Commented Feb 27, 2017 at 14:05

1 Answer 1


I cannot give you the solution by using transfer functions. However I can give you a general form by using the state space representation. I will do it for a square system, i.e. the number of inputs and outputs is equal. For a system with $n$ inputs and $m$ outputs it is getting more messy and a lot harder to solve the problem.

The system $$\dot{x}=f(x)+ g_1(x)u_1 + \ldots + g_m(x)u_m$$ with outputs $$y_1 = h_1(x), \ldots, y_m = h_m(x)$$

First introducing the Lie Derivative. The Lie Derivative of $h$ with respect to $f$ or along $f$ is $$L_{f}h(x) = \frac{\partial h}{\partial x}f(x)$$ For example, the following notation is used: $$\begin{split} L_g L_f &= \frac{\partial (L_f h)}{\partial x}g(x)\\ L_f^2 h(x) &= L_f L_f h(x) &= \frac{\partial (L_f h)}{\partial x}f(x)\\ L_f^kh(x) &= L_f L_f^{k-1}h(x) &= \frac{\partial (L_f^{k-1})}{\partial x}f(x) \end{split}$$

Introducing the notion of relative degree with respect to each output. Consider the $i$-th output and differentiate it with respect to time: $$\dot{y}_i = L_fh_i(x) +L_{g_1}h_i(x)u_1 +\ldots L_{g_m}h_i(x)u_m$$ This expression depends explicitly on at least one input if (for all $x$): $$\left(L_{g_1}h_i(x),\ldots,L_{g_m}h_i(x) \right)\neq (0,\ldots,0)$$ If so, the $i$-th output has relative degree $k_i = 1$.

In general the relative degree $k$ per output $i$ if $$\left(L_g,L_f^{k_i-1}h_i(x),\ldots,L_{g_m}L_f^{k_i-1}h_i(x)\right) \neq (0,\ldots,0)$$ for all $x$.

The system is now Input-Output Linearised (hence decoupled) when applying the following feedback $$u(x) = -A^{-1}(x)N(x)+A^{-1}(x)v$$ with the decoupling matrix $A(x)$, vector $N(x)$ and new input vector $v$. Where $$A(x) = \begin{pmatrix} L_{g_1}L_f^{k_1-1}h_1(x) & \ldots & L_{g_m}L_f^{k_1-1}h_1\\ \vdots & & \vdots\\ L_{g_1}L_f^{k_m-1}h_m(x) & \ldots & L_{g_m}L_F^{k_m-1}h_m\\ \end{pmatrix},\quad N(x) = \begin{pmatrix} L_f^{k_1}h_1(x)\\ \vdots\\ L_f^{k_m}h_m(x) \end{pmatrix} $$.

Hence $A(x)$ needs to be invertible for all $x$. If you want the transfer functions, simply apply Laplace.


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