17. Linear Time-Invariant Systems

A linear time invariant system is governed by a linear ODE with constant coefficients. These equations can be recast as a first order system of ODEs,

\[\begin{split}\begin{cases} \dot{\mathbf{x}} = \mathbf{A} \mathbf{x} + \mathbf{B} \mathbf{u} \\ \mathbf{y} = \mathbf{C} \mathbf{x} + \mathbf{D} \mathbf{u} \\ \mathbf{x}(0^-) = \mathbf{x}_0 \end{cases}\end{split}\]

Exploiting the properties of matrix exponential the general expression of the state can be written as the sum of the free response to initial condition and the forced response.

\[\begin{split}\begin{aligned} \mathbf{x}(t) & = e^{\mathbf{A}t} \mathbf{x}_0 + \int_{\tau=0^-}^{t} e^{\mathbf{A}(t-\tau)} \mathbf{B} \mathbf{u} (\tau) \, d \tau \\ \mathbf{y}(t) & = \mathbf{C} e^{\mathbf{A}t} \mathbf{x}_0 + \mathbf{C} \int_{\tau=0^-}^{t} e^{\mathbf{A}(t-\tau)} \mathbf{B} \mathbf{u} (\tau) \, d \tau + \mathbf{D} \mathbf{u}(t) \\ \end{aligned}\end{split}\]

Proof in time domain

Multipying by \(e^{-\mathbf{A} t}\),

\[\begin{split}\begin{aligned} e^{-\mathbf{A}t} ( \dot{x}(t) - \mathbf{A} \mathbf{x}(t) ) & = e^{-\mathbf{A} t} \mathbf{B} \mathbf{u}(t) \\ \dfrac{d}{dt} \left( \mathbf{x} e^{-\mathbf{A}t} \right) & = e^{-\mathbf{A} t} \mathbf{B} \mathbf{u}(t) \\ \end{aligned}\end{split}\]

and integrating from \(0^-\) to a generic time value \(t\),

\[e^{-\mathbf{A}t} \mathbf{x}(t) - \mathbf{x}_0 = \int_{\tau=0^-}^{t} e^{-\mathbf{A}\tau} \mathbf{B} \mathbf{u} (\tau) \, d \tau\]

The state \(\mathbf{x}(t)\) can be written as the sum of the free response and a force response. The general expression of the state and the output as a function reads

\[\begin{split}\begin{aligned} \mathbf{x}(t) & = e^{\mathbf{A}t} \mathbf{x}_0 + \int_{\tau=0^-}^{t} e^{\mathbf{A}(t-\tau)} \mathbf{B} \mathbf{u} (\tau) \, d \tau \\ \mathbf{y}(t) & = \mathbf{C} e^{\mathbf{A}t} \mathbf{x}_0 + \mathbf{C} \int_{\tau=0^-}^{t} e^{\mathbf{A}(t-\tau)} \mathbf{B} \mathbf{u} (\tau) \, d \tau + \mathbf{D} \mathbf{u}(t) \\ \end{aligned}\end{split}\]

Laplace domain. The Laplace transform of the problem reads

\[\begin{split}\begin{cases} s \hat{\mathbf{x}} = \mathbf{A} \hat{\mathbf{x}} + \mathbf{B} \hat{\mathbf{u}} + \mathbf{x}_0 \\ \hat{\mathbf{y}} = \mathbf{C} \hat{\mathbf{x}} + \mathbf{D} \hat{\mathbf{u}} \\ \end{cases}\end{split}\]

\[(s\mathbf{I} - \mathbf{A}) \hat{\mathbf{x}} = \mathbf{B} \hat{\mathbf{u}} + \mathbf{x}_0\]

\[\begin{split}\begin{aligned} \hat{\mathbf{x}}(s) & = (s\mathbf{I} - \mathbf{A})^{-1} \mathbf{x}_0 + (s\mathbf{I} - \mathbf{A})^{-1} \mathbf{B} \hat{\mathbf{u}}(s) \\ \hat{\mathbf{y}}(s) & = \mathbf{C} (s\mathbf{I} - \mathbf{A})^{-1}\mathbf{x}_0 + \left[ \mathbf{C} (s\mathbf{I} - \mathbf{A})^{-1} \mathbf{B} + \mathbf{D} \right] \hat{\mathbf{u}}(s) \end{aligned}\end{split}\]

Performing inverse Laplace transform allows to go back to time domain (just use Laplace inverse transform of a matrix exponential, and the formula (13.1) for Laplace transform of convolution).

17.1. Impulsive force

The effect of an impulsive force at time \(t=0\) is equivalent to an instantaneous change in the initial state, from time \(0^-\) before the impulse to time \(0^+\) after the impulse. Splitting the input \(\mathbf{u}(t)\) as the sum of impulsive input and regular input,

\[\begin{split}\begin{aligned} \mathbf{u}(t) & = \mathbf{u}_r(t) + \mathbf{u}_\delta \delta(t) \\ \hat{\mathbf{u}}(s) & = \hat{\mathbf{u}}_r(s) + \mathbf{u}_\delta \\ \end{aligned}\end{split}\]

the solution in time and Laplace domain reads

\[\begin{split}\begin{aligned} \mathbf{x}(t) & = e^{\mathbf{A}t} \mathbf{x}_0 + \int_{\tau=0^-}^{t} e^{\mathbf{A}(t-\tau)} \mathbf{B} \left( \mathbf{u}_r (\tau) + \mathbf{u}_\delta \delta(\tau) \right) \, d \tau = \\ & = e^{\mathbf{A}t} \left( \mathbf{x}_0 + \mathbf{B} \mathbf{u}_\delta \right) + \int_{\tau=0^-}^{t} e^{\mathbf{A}(t-\tau)} \mathbf{B} \mathbf{u}_r (\tau) \, d \tau \\ \mathbf{y}(t) & = \mathbf{C} e^{\mathbf{A}t} \left( \mathbf{x}_0 + \mathbf{B} \mathbf{u}_\delta \right) + \int_{\tau=0^-}^{t} \mathbf{C} e^{\mathbf{A}(t-\tau)} \mathbf{B} \mathbf{u}_r (\tau) \, d \tau + \mathbf{D} \mathbf{u}_r(t) + \mathbf{D} \mathbf{u}_\delta(t) \\ \end{aligned}\end{split}\]

\[\begin{split} \hat{\mathbf{x}}(s) & = (s\mathbf{I} - \mathbf{A})^{-1} \mathbf{B} \hat{\mathbf{u}}_r(s) + (s\mathbf{I} - \mathbf{A})^{-1} \left( \mathbf{x}_0 + \mathbf{B} \mathbf{u}_\delta \right) \\ \hat{\mathbf{y}}(s) & = \left[ \mathbf{C} (s\mathbf{I} - \mathbf{A})^{-1} \mathbf{B} + \mathbf{D} \right] \hat{\mathbf{u}}_r(s) + \mathbf{C} (s\mathbf{I} - \mathbf{A})^{-1} \left( \mathbf{x}_0 + \mathbf{B} \mathbf{u}_\delta \right) + \mathbf{D} \mathbf{u}_{\delta} \end{split}\]

17.2. Properties

Matrix exponential.

\[e^{\mathbf{A} t} = \sum_{k = 0}^{+\infty} \frac{\mathbf{A}^k t^k}{k!} \ .\]

Assuming it’s possible swap derivative operator and summation (when?), it’s possible to write

\[\frac{d}{dt} e^{\mathbf{A}t} = \dfrac{d}{dt} \sum_{k = 0}^{+\infty} \frac{\mathbf{A}^k t^k}{k!} = \sum_{k=1}^{+\infty} k t^{k-1} \frac{\mathbf{A}^k t^{k-1}}{k!} = \mathbf{A} e^{\mathbf{A} t} \ .\]

Laplace transform of exponential matrix.

\[\begin{split}\begin{aligned} \mathscr{L}\left\{ e^{\mathbf{A}t} \right\}(s) & := \int_{t=0^-}^{+\infty} e^{\mathbf{A} t} e^{-s t} \, dt = \\ & = \int_{t=0^-}^{+\infty} e^{(-s\mathbf{I} + \mathbf{A}) t} \, dt = \\ & = (-s\mathbf{I} + \mathbf{A})^{-1} \left.e^{(-s\mathbf{I} + \mathbf{A}) t}\right|_{t=0^-}^{+\infty} = \\ & = (s \mathbf{I} - \mathbf{A})^{-1} \ , \end{aligned}\end{split}\]

for all the values of \(s\) for which \(-s\mathbf{I} + \mathbf{A}\) is asymptotically stable, i.e. has all the eignevalues (thus, assuming that the matrix \(\mathbf{A}\) can be diagonalizable. What happens if not? Exploit other matrix decompositions to draw conclusions) with negative real parts, and thus for all the values of \(s > \max \text{re}\{ s_k(\mathbf{A}) \}\), as it’s shown in Example 17.1

Example 17.1 (Asymptotic stability of a matrix \(\ \mathbf{A}\))

An \(N \times N\) diagnonalizable matrix \(\mathbf{A}\),

(17.1)\[\mathbf{A} \mathbf{v}_k = \mathbf{v}_k \, s_k\]

has all the eigenvalues with negative real part, \(\text{re}\left\{ s_k \right\} < 0\), \(\forall k=1:N\).

The eigenvalues of a matrix \(a \mathbf{I} + \mathbf{A}\) are \(a + s_k\), while the eigenvectors are the same as those of the matrix \(\mathbf{A}\). This can be easily proved adding \(a \mathbf{I} \mathbf{v}_k\) to both sides of equation (17.1),

\[\left( a \mathbf{I} + \mathbf{A} \right) \mathbf{v}_k = \mathbf{v}_k (a + s_k) \ .\]

Transform of the convolution.