(ode:lti:response)= # LTI system response Usually the response of LTI to 3 different input are studied - integrable input, being response to non-zero initial condition equivalent to an impulsive force at time $t = 0$, see [equivalence impulsive force - istantanteous change of state](ode:lti:impulsive-force) - periodic input - stochastic input ## LTI $$\begin{cases} \dot{\mathbf{x}} = \mathbf{A} \mathbf{x} + \mathbf{B} \mathbf{u} \\ \mathbf{y} = \mathbf{C} \mathbf{x} + \mathbf{D} \mathbf{u} \end{cases}$$ with initial conditions, $\mathbb{x}(0) = \mathbf{x}_0$. (ode:lti:response:general)= ## Response of LTI systems The response of a LTI can be written as the sum of a free and forced response. $$\begin{aligned} \mathbf{x}(t) & = e^{\mathbf{A} t} \mathbf{x}_0 + \int_{\tau=0}^{t} e^{\mathbf{A}(t-\tau)} \mathbf{B} \mathbf{u}(\tau) \, d \tau \\ \mathbf{y}(t) & = \mathbf{C} \, e^{\mathbf{A} t} \mathbf{x}_0 + \int_{\tau=0}^{t} \mathbf{C} e^{\mathbf{A}(t-\tau)} \mathbf{B} \mathbf{u}(\tau) \, d \tau + \mathbf{D} \mathbf{u}(t) \\ \end{aligned}$$ (eq:lagrange:formula) ```{dropdown} Proof **todo** ``` ```{admonition} Impulse response, $ \mathbf{H}(t)$ :class: tip Functions $$\begin{aligned} \mathbf{H}_x(t) & := e^{\mathbf{A} t} \mathbf{B} \\ \mathbf{H} (t) & := \mathbf{C} e^{\mathbf{A} t} \mathbf{B} + \mathbf{D} \delta(t) \\ \end{aligned}$$ are defined **impulse response functions** for the state $\mathbf{x}$ and the output $\mathbf{y}$, as they determine with the evolution of the state and the output of a system with zero initial condition $\mathbf{x}_0 = \mathbf{0}$ as a consequence of a impulsive force $\mathbf{u}_{\delta}(t) = \mathbf{u} \delta(t)$. Using Lagrange formula {eq}`eq:lagrange:formula` $$\begin{aligned} \mathbf{x}(t) & = e^{\mathbf{A}t} \mathbf{B} \mathbf{u} && = \mathbf{H}_x(t) \mathbf{u} \\ \mathbf{y}(t) & = \left[ \mathbf{C} e^{\mathbf{A}t} \mathbf{B} + \mathbf{D} \delta(t) \right] \mathbf{u} && = \mathbf{H}(t) \mathbf{u} \\ \end{aligned}$$ (eq:lagrange:formula:impulse) ``` ```{admonition} Convolution :class: tip Integral in Lagrange formula {eq}`eq:lagrange:formula` is the convolution of the impulse response function and the input. Changing the integration variable $z = t - \tau$, $d\tau = - dz$, $\tau = 0: \, z = t$, $\tau = t, \ z = 0$ (and going back from $z$ to $\tau$ as the symbol for the dummy integration variable), it's possible to swap the arguments of the impulsive force and the forcing, $$\begin{aligned} \mathbf{x}(t) & = e^{\mathbf{A} t} \mathbf{x}_0 + \int_{\tau=0}^{t} \mathbf{H}_x(\tau) \mathbf{u}(t-\tau) \, d \tau \\ \mathbf{y}(t) & = \mathbf{C} \, e^{\mathbf{A} t} \mathbf{x}_0 + \int_{\tau=0}^{t} \mathbf{H}(\tau) \mathbf{u}(t-\tau) \, d \tau \\ \end{aligned}$$ (eq:lagrange:convol) ``` ```{admonition} Linear but not time-invariant :class: tip **todo** *The response of linear system, but not time-invariant can be written with the same expression used by Lagrange fomrula {eq}`eq:lagrange:formula`, with $\mathbf{H}(t)$ not equal to ..., but solution of the ODE...* $$\begin{aligned} \boldsymbol{\Phi} \left( \dot{\mathbf{x}} - \mathbf{A}\mathbf{x} \right) & = \boldsymbol{\Phi} \mathbf{B} \mathbf{u} \\ \dfrac{d}{dt} \left( \boldsymbol{\Phi} \mathbf{x} \right) & = \boldsymbol{\Phi} \mathbf{B} \mathbf{u} \ , \end{aligned}$$ with $\dot{\boldsymbol{\Phi}} = - \boldsymbol{\Phi} \mathbf{A}$, and proper initial conditions $\boldsymbol{\Phi}(t,0) = \mathbf{I}$. ... $$\mathbf{x}(t) = \boldsymbol{\Phi}(t,0) \mathbf{x}_0 + \int_{\tau=0}^{t} \boldsymbol{\Phi}(t, \tau) \mathbf{B}(\tau) \mathbf{u}(\tau) \, d \tau \ .$$ ``` (ode:lti:response:equilibria)= ## Equilibria Equilibrium solution of a ODE is a stationary solution, independent from time, $\dot{\overline{\mathbf{x}}} = \mathbf{0}$. With a steady forcing $\overline{\mathbf{u}}$, $$\begin{cases} \mathbf{0} = \mathbf{A} \overline{\mathbf{x}} + \mathbf{B} \overline{\mathbf{u}} \\ \mathbf{y} = \mathbf{C} \overline{\mathbf{x}} + \mathbf{D} \overline{\mathbf{u}} \ . \end{cases}$$ **If $\mathbf{A}$ is not singular**, for every steady forcing $\overline{\mathbf{u}}$ only one equilibrium exists for the system, $\overline{\mathbf{x}} = \mathbf{A}^{-1} \mathbf{B} \overline{\mathbf{u}}$. **If $\mathbf{A}$ is singular**...**todo** *rely on [Linear algebra](math:linear-algebra:intro)* (ode:lti:response:stability)= ## Stability **Asymptotic stability to initial perturbations.** A system is stable under perturbation of initial conditions if the free response goes to zero as $t \rightarrow + \infty$, $$\lim_{t \rightarrow +\infty} \mathbf{x}_{free}(t) = \mathbf{0} \ .$$ **todo** *for all perturbations*(?). If matrix $\mathbf{A}$ is diagonalizable, a system is asymptotically stable if all the eigenvalues of the matrix $\mathbf{A}$ have negative real part, $\text{re}\{\lambda_i(\mathbf{A})\} < 0$. **todo** *rely on [Linear algebra](math:linear-algebra:intro)* (ode:lti:response:laplace)= ## Response to integrable input - and Laplace transform $$\begin{aligned} \mathbf{x}(s) & = \left(s \mathbf{I} - \mathbf{A} \right)^{-1} \mathbf{x}_0 + \mathbf{H}_x(s) \mathbf{u}(s) \\ \mathbf{y}(s) & = \mathbf{C} \, \left( s \mathbf{I} - \mathbf{A} \right)^{-1} \mathbf{x}_0 + \mathbf{H}(s) \mathbf{u}(s) \end{aligned}$$ (eq:lagrange:formula:laplace) (ode:lti:response:periodic)= ## Response to periodic input of stable systems - and Fourier transform Forced response to periodic input of an stable system can be represented in Fourier domain replacing $s$ with $j \omega$ (and dropping $j$ once it's clear we're dealing with Fourier analysis) $$\begin{aligned} \mathbf{x}(\omega) & = \mathbf{H}_x(\omega) \ \mathbf{u}(\omega) \\ \mathbf{y}(\omega) & = \mathbf{H} (\omega) \ \mathbf{u}(\omega) \end{aligned}$$ (eq:lagrange:formula:fourier) (ode:lti:response:stochastic)= ## Response to stochastic input of stable systems ### Assumptions ... ### Expected value $$\boldsymbol{\mu}_{\mathbf{x}}(t) := \mathbb{E}[ \mathbf{x}(t) ]$$ Taking expected value of LTI equations $$\begin{cases} \mathbb{E}[\dot{\mathbf{x}}] = \mathbf{A} \mathbb{E}[\mathbf{x}] + \mathbf{B} \mathbb{E}[\mathbf{u}] \\ \mathbb{E}[ \mathbf{y} ] = \mathbf{C} \mathbb{E}[\mathbf{x}] + \mathbf{D} \mathbb{E}[\mathbf{u}] \\ \end{cases}$$ **todo** - *is it possible to swap time derivative and expected value operators?* - *does time derivative of a stochastic process always exist? What's the right way to interpret time derivative in stochastic equations? As an increment?* $$d \mathbf{x} = \mathbf{A} \mathbf{x} \, d t + \mathbf{B} \mathbf{u} \, d t$$ so that if $\mathbf{u}(t) \sim \boldsymbol{\xi}(t)$, then $\mathbf{u} \, dt \sim d \mathbf{W}$ ## Analysis of response of LTI in Fourier domain **Energy or power.** For - time integrable signals: don't average in time; energy - periodic or stochastic non-integrable signals: average in time; power **Stationary stochastic process.** Strong-sense (joint probability independent from time) or wide-sense (moments up to order $m$ are independent on time) **Ergodic process.** Average over random process can be replaced with average over time. ```{admonition} Example of stationary non-ergodic process As an example of stationary, but not ergodic process is a random process whose realizations are $$X_t = \theta \ , \qquad \forall t \in [0,T]$$ with $\theta$ drawn from a random variable $\Theta$, with average $\mu$. Average in time over a realization gives $\theta$, while average over realizations gives $\mu$ for all $t$, $$\begin{aligned} \text{Avg. in time over a realization: } & \quad \dfrac{1}{T} \int_{t=0}^{T} X_t \, dt = \dfrac{1}{T} \int_{t=0}^{T} \theta \, dt = \theta \\ \text{Avg. in probability over realizations: } & \quad \mathbb{E}[X_t] = \mathbb{E}[\Theta] = \mu \ . \end{aligned}$$ ``` ```{admonition} White noise... ``` ### Stationary Ergodic processes **Auto-correlation** $$\mathbf{R}_{XX}(t_1, t_2) = \mathbb{E}\left[ \mathbf{X}_{t_1} \, \mathbf{X}_{t_2}^* \right]$$ **Auto-covariance** $$\mathbf{K}_{XX}(t_1, t_2) = \mathbb{E}\left[ (\mathbf{X}_{t_1} - \boldsymbol{\mu}_{t_1} ) (\mathbf{X}_{t_2} - \boldsymbol{\mu}_{t_2} )^* \right]$$ For stationary processes, both auto-correlation and auto-covariance don't depend on two time instant $t_1$, $t_2$ but only on the time-shift between them, $t_2 = t_1 + \tau$, $$\mathbf{R}_{XX}(\tau) = \mathbb{E}\left[ \mathbf{X}_{t} \, \mathbf{X}_{t + \tau}^* \right] \ , \qquad \forall t \ .$$ ```{dropdown} Autocorrelation and ESD of time integrable processes :open: With causal, $\mathbf{x}(t \le 0) = 0$, and integrable processes, autocorrelation is defined as $$\tilde{\mathbf{R}}_{xx}(\tau) = \int_{t = 0}^{+\infty} \mathbf{x}(t) \mathbf{x}^*(t+\tau) \, d t \ ,$$ **Energy Spectral Density (ESD)** is defined as its Fourier transform $$\tilde{\boldsymbol{\Phi}}_{xx}(f) = \mathscr{F} \left\{ \mathbf{R}_{xx}(\tau) \right\} (f) \ ,$$ ``` ```{dropdown} Autocorrelation and PSD of infinite-time processes :open: Using average over time as an approximation of the probability average $$\mathbb{E}[ f(t) ] \sim \lim_{T \rightarrow +\infty} \dfrac{1}{T} \int_{t=0}^{T} f(\tau) \, d \tau \ ,$$ **Power Spectral Density (PSD)**, defined as the Fourier transform of the auto-correlation, reads $$\begin{aligned} \boldsymbol{\Phi}_{YY}(f) & = \mathscr{F}[ \mathbf{R}_{YY}(\tau) ](f) = \\ & = \int_{\tau = -\infty}^{+\infty} \mathbb{E} \left[ \mathbf{y}(t) \mathbf{y}^*(t+\tau) \right] \, e^{-i 2 \pi f \tau} \, d \tau \ . \end{aligned}$$ For asymptotically stable systems subject to periodic forcing, response to non-homogenoeus initial conditions vanishes while the forced response can be written either in Fourier or in time domain, exploiting the property of the transform of convolution {prf:ref}`transform:fourier:convolution`, as $$\begin{aligned} \mathbf{y}(f) = \mathbf{H}(f) \, \mathbf{u}(f) \qquad , \qquad \mathbf{y}(t) = \mathbf{H} \ast \mathbf{u}(t) = \int_{\tau=-\infty}^{+\infty} \mathbf{H}(\tau) \mathbf{u}(t - \tau) \, d \tau \ . \end{aligned}$$ Inserting harmonic response into the expression of PSD, it becomes $$\begin{aligned} \boldsymbol{\Phi}_{YY}(f) & = \int_{\tau = -\infty}^{+\infty} \mathbb{E} \left[ \left( \int_{\xi=-\infty}^{+\infty} \mathbf{H}(\xi) \, \mathbf{u}(t - \xi) \, d\xi \right) \left( \int_{\chi=-\infty}^{+\infty} \mathbf{H}(\chi) \, \mathbf{u}(t+\tau-\chi) \, d \chi \right)^* \right] \, e^{-i 2 \pi f \tau} \, d \tau = && (1) \\ & = \int_{\tau = -\infty}^{+\infty} \, \int_{\xi=-\infty}^{+\infty} \, \int_{\chi=-\infty}^{+\infty} \mathbf{H}(\xi) \mathbb{E} \left[ \mathbf{u}(t - \xi) \mathbf{u}^*(t+\tau-\chi) \right] \mathbf{H}^*(\chi) \, e^{-i 2 \pi f \tau} \, d \tau \, d \chi \, d \xi = && (2) \\ & = \int_{\tau = -\infty}^{+\infty} \, \int_{\xi=-\infty}^{+\infty} \, \int_{\chi=-\infty}^{+\infty} \mathbf{H}(\xi) \mathbb{E} \left[ \mathbf{u}(t ) \mathbf{u}^*(t+\tau-\chi+\xi) \right] \mathbf{H}^*(\chi) \, e^{-i 2 \pi f \tau} \, d \tau \, d \chi \, d \xi = && (3) \\ & = \int_{z = -\infty}^{+\infty} \, \int_{\xi=-\infty}^{+\infty} \, \int_{\chi=-\infty}^{+\infty} \mathbf{H}(\xi) \mathbb{E} \left[ \mathbf{u}(t ) \mathbf{u}^*(t+z) \right] \mathbf{H}^*(\chi) \, e^{-i 2 \pi f \left( z - \xi + \chi \right)} \, d z \, d \chi \, d \xi = \\ & = \int_{\xi = -\infty}^{+\infty} \mathbf{H}(\xi) \, e^{i 2 \pi f \xi} \, d \xi \, \int_{z=-\infty}^{+\infty} \mathbf{R}_{uu}(z) \, e^{-i 2 \pi f z } \, dz \, \int_{\chi=-\infty}^{+\infty} \mathbf{H}^*(\chi) \, e^{-i 2 \pi f \chi } \, d \chi = (4) \\ & = \overline{\mathbf{H}}(f) \, \boldsymbol{\Phi}_{uu}(f) \, \mathbf{H}^T(f) \ . \end{aligned}$$ having used $(1)$ the fact that the impulse response function is deterministic, $(2)$ time shift in the autocorrelation for a **stationary** random process $\mathbf{u}(t)$ as an input, and $(3)$ the change of coordinates $(z, \xi, \chi) = (\tau-\chi+\xi, \xi, \chi)$, and $(4)$ a slight notation abuse ndicating Fourier transform with the same symbols of the functions in time domain. ```