Solving for the state-space response

State response

The state equation can be solved by a synthesis of familiar techniques, as follows. First, we rearrange: \[\begin{aligned} \label{ssresp_eq:state-ready-for-integrating-factor} \frac{d\bm{x}}{d t} - A \bm{x} = B \bm{u}. \end{aligned}\] An integrating factor would be clutch, but what should it be? It looks analogous to a scalar ODE that would use the natural exponential \(\exp(-a t)\) (for positive constant \(a\)), but we have a vector ODE. We need a matrix-version of the exponential. Recall that a series definition of the scalar exponential function \(\exp:\mathbb{C}\rightarrow\mathbb{C}\) is \[\begin{aligned} \exp z = \sum_{k=0}^\infty \frac{1} {k!} z^k.\end{aligned}\] We define the matrix exponential \(\exp:\mathbb{C}^n\times\mathbb{C}^n \rightarrow \mathbb{C}^n \times \mathbb{C}^n\) (we use the same symbol) to be, for \(n\times n\) complex matrix \(Z\), \[\begin{aligned} \label{ssresp_eq:matrix_exponential_def} \exp Z = \sum_{k=0}^\infty \frac{1} {k!} Z^k. \end{aligned}\] because why not? For the hell of it, let’s see if the matrix exponential \[\begin{aligned} \exp(-A t) \end{aligned}\] works as an integrating factor, if for no other reason than it was constructed to be a sort of matrix-analog of \(\exp(-a t)\), which would work for the scalar case. Premultiplying¹ on both sides: $$\begin{align*} \exp(-A t) \frac{d \bm{x}}{d t} - \exp(-A t) A \bm{x} &= \exp(-A t) B \bm{u} \quad \Rightarrow \\ \frac{d} {d t} \left( \exp(-A t) \bm{x} \right) &= \exp(-A t) B \bm{u}. \end{align*}$$ Rearranging and integrating over the interval \((0,t)\), $$\begin{align*} d\left( \exp(-A t) \bm{x} \right) &= \exp(-A t) B \bm{u} dt \quad \Rightarrow \\ \int_0^t d\left( \exp(-A \tau) \bm{x}(\tau) \right) &= \int_0^t \exp(-A \tau) B \bm{u}(\tau) d\tau \quad \Rightarrow \\ \exp(-A t) \bm{x} - \bm{x}(0) &= \int_0^t \exp(-A \tau) B \bm{u}(\tau) d\tau \tag{$\exp(0) = I$}. \end{align*}$$ This last expression can be solved for \(\bm{x}\), the state response solution. Before we do this, however, let’s define the matrix function called the state transition matrix \(\Phi\) to be the matrix-valued function \[\begin{aligned} \label{ssresp_eq:state_transition_matrix} \Phi(t) = \exp(A t), \end{aligned}\] Substituting \(\Phi\) and solving, $$\begin{align}\label{ssresp_eq:state_response} \bm{x} &= \Phi(t) \bm{x}(0) + \Phi(t) \int_0^t \Phi(-\tau) B \bm{u}(\tau) d\tau \\ &= \Phi(t) \bm{x}(0) + \int_0^t \Phi(t - \tau) B \bm{u}(\tau) d\tau. \end{align}$$ Note that the first term of each version of is the free response (due to initial conditions) and the second term is the forced response (due to inputs).

State transition matrix

The state transition matrix \(\Phi\) introduced in wound up being a key aspect of the response, which is why we call it that. We used two of its properties (in matrix exponential form) during that derivation: the initial-value $$\begin{align} \Phi(0) = I \tag{where $I$ is the identity matrix} \end{align}$$ and the inverse \[\begin{aligned} \Phi^{-1}(t) = \Phi(-t). \end{aligned}\]

There is a third property that might be called the bootstrapping property: for time intervals \(\Delta t_i\), \[\begin{aligned} \Phi(\Delta t_1 + \Delta t_2 + \cdots) = \Phi(\Delta t_1) \Phi(\Delta t_2) \cdots. \end{aligned}\] This allows one to compute the state transition matrix² incrementally, from one previously computed.

A final property we’ll consider is the special-case of a diagonal \(A\) with diagonal elements \(a_{11}, a_{22}, \cdots, a_{nn}\), which yields a diagonal state transition matrix \[\begin{aligned} \Phi(t) = \begin{bmatrix} e^{a_{11} t} & & & 0 \\ & e^{a_{22} t} & & \\ & & \ddots & \\ 0 & & & e^{a_{nn} t} \end{bmatrix}.\end{aligned}\]

The last property turns out to be quite convenient for deriving $\Phi$ for a given system, as we will see in . For now, we must rely on the definition of \(\Phi\) from and the series definition of the matrix exponential from . This requires us to derive the first several terms of the series solution and attempt to divine the corresponding scalar exponential series, a rather tedious task. Other than to familiarize ourselves with the definition through exercises, we prefer the derivation method of .

Output response

The output response \(\bm{y}(t)\) requires little additional solution: assuming we have solved for the state response \(\bm{x}(t)\), the output is given in the output equation . Through direct substitution, we find the output response solution $$\begin{align} \label{ssresp_eq:output_response} \bm{y}(t) &= C \bm{x}(t) + D \bm{u}(t) \\ &= C \Phi(t) \bm{x}(0) + C \int_0^t \Phi(t - \tau) B \bm{u}(\tau) d\tau + D \bm{u}(t). \end{align}$$