The evolution of a physical system can be modelled in the time domain using a state-space representation. To do this, we choose several variables as the state variables, which together specify the state of the system at any given time. The state variables must be linearly independent. If a system is represented by an nth-order differential equation, then generally we require n state variables. Several other variables make up the input to the system, and the output is represented as a set of variables that can be expressed in terms of the state and input variables.

State Equations Edit

What it comes down to is that we end up with a set of two matrix/vector equations in \mathbb{R}^n (where n is the order of the system). x is the vector of state variables, u is the input vector, and y is the output vector.

\mathbf{\dot{x}} = \mathbf{Ax} + \mathbf{Bu}

\mathbf{y} = \mathbf{Cx} + \mathbf{Du}

Laplace Transform Solution Edit

We can work out the transient response of a system to an input by solving the state equations, using the Laplace transform. (Yes, we are allowed to use the Laplace transform on a vector; we do this by applying the transform to each component of the vector).

\mathcal{L}\{\mathbf{\dot{x}} = \mathbf{Ax} + \mathbf{Bu}\} \Rightarrow s \mathbf{X}(s) - \mathbf{x}(0) = \mathbf{AX}(s) + \mathbf{BU}(s)

which rearranges into

\mathbf{X}(s) = (s\mathbf{I} - \mathbf{A})^{-1}[\mathbf{x}(0) + \mathbf{BU}(s)]

We can now sub this value into the transform of the output equation:

\mathbf{Y}(s) = \mathbf{CX}(s) + \mathbf{DU}(s)

Equivalent Transfer Function Edit

We can obtain the the equivalent transfer function for the system as a special case of the Laplace transform solution, by letting the initial conditions be equal to 0.

\mathbf{X}(s) = (s\mathbf{I} - \mathbf{A})^{-1}\mathbf{BU}(s)

\mathbf{Y}(s) = \mathbf{C}(s\mathbf{I} - \mathbf{A})^{-1}\mathbf{BU}(s) + \mathbf{DU}(s) = [\mathbf{C}(s\mathbf{I} - \mathbf{A})^{-1}\mathbf{B} + \mathbf{D}]\mathbf{U}(s)

G(s) = \frac{Y(s)}{U(s)} = \mathbf{C}(s\mathbf{I} - \mathbf{A})^{-1}\mathbf{B} + \mathbf{D}

Transfer Function Poles Edit

As a shortcut to getting the poles of the transfer function, it turns out that they are equal to the eigenvalues of the matrix A. If you don't remember what eigenvalues are, they are the solutions to the equation \det ( s \mathbf{I} - \mathbf{A} ) = 0.

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