I do know what the difference in meaning is between a transfer function (output over the input) and a state transition matrix \$\Phi\$ (describes the unforced response of the system). Yet when looking closer at the mathematics to me it seems like they are both the same. Could someone clarify?

$$\dot{q} = Aq(t)+Bu(t)$$ where \$A\$ is the state matrix, \$q\$ the state vector, \$B\$ the input matrix and u the input vector.

$$sQ(s) = AQ(s) + BU(s)$$ $$ sQ(s)- AQ(s) = BU(s)$$ $$ Q(s)(sI-A) = BU(s) $$ $$ Q(s) = (sI-A)^{-1}BU(s) $$

where \$(sI-A)^{-1} = \Phi\$ ie the state transition matrix.

$$ Q(s) = \Phi BU(s)$$ $$ \Phi = \frac{Q(s)}{BU(s)}$$

\$\frac{Q(s)}{BU(s)}\$ looks to me like the representation of a transfer function and based on the math it looks like the state transition matrix in fact equals a transfer function, yet that doesn't correspond to the interpretation I have of both of those things. Could somebody please elaborate a bit on that?


1 Answer 1


A transfer function relates output variables to input variables. In the equation you have shown you only consider state variables (q) and inputs (u).

This model assumes that state variables are completely accessible from the outside.

A more comprehensive model would comprise an output equation such as:

$$ y(t) = C \cdot q(t) + D \cdot u(t) $$

which transformed becomes:

$$ Y(s) = C \cdot Q(s) + D \cdot U(s) $$

A transfer function (or, better, a transfer function matrix, since we are modeling MIMO systems, apparently) takes into account also the C and D matrices.

  • \$\begingroup\$ regarding the more comprehensive model: y/u would represent the transfer function and q/u the state transition matrix. Correct? \$\endgroup\$ Commented Nov 19, 2017 at 14:04
  • \$\begingroup\$ this message as a reminder of my quesion just above. That way I can accept your answer. \$\endgroup\$ Commented Nov 20, 2017 at 20:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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