# 0.22 I/o and i/s/o relationships in time and frequency

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## I/o and i/s/o representation of siso linear systems

I/O I/S/O
variables: $\left(u,y\right)$ variables: $\left(u,x,y\right)$
$\frac{d q}{d t}}y(t)=\frac{d p}{d t}}u(t),n=\mathrm{deg}(q)\ge \mathrm{deg}(p)$ $\frac{d x(t)}{d t}}=Ax(t)+Bu(t),y(t)=Cx(t)+Du(t)$
$u(t),y(t)\in \mathbb{R}$ $x(t)\in \mathbb{R}^{n},\begin{pmatrix}A & B\\ C & D\\ \end{pmatrix}\in \mathbb{R}^{(n+1\times n+1)}$
Impulse Response
$\frac{d q}{d t}}h(t)=\frac{d p}{d t}}\delta (t)$ $h(t)=D\delta (t)+Ce^{At}B,t\ge 0$
$H(s)=ℒ(h(t))=\frac{p(s)}{q(s)}$ $H(s)=D+CsI-A^{(-1)}B$
Poles - characteristic roots - eigenfrequencies
${\lambda }_{i},q({\lambda }_{i})=0,I=1,\dots ,n$ $\det ({\lambda }_{i}I-A)=0$
Zeros
$H({z}_{i})=0⇔p({z}_{i}),1,\dots ,n$ $\begin{vmatrix}{z}_{i}I-A & \mathrm{-B}\\ \mathrm{-C} & \mathrm{-D}\end{vmatrix}=0$
Matrix exponential
$(e^{At}=\sum_{k=0} )\implies$ t k k A k t A t A A t A t A
$ℒ(e^{At})=sI-A^{(-1)}$
BIBO stability
$y=(h, u)$ , requirement
$\exists \forall u\colon ({\mathrm{Norm}(u)}_{})\implies$ u
$⇔{(h)}_{1}=\int_{0} \,d t$ h t
$⇔\Re ({\lambda }_{i})< 0⇔\mathrm{poles}\in \mathrm{LHP}$
Solution in the time domain
$y(t)={y}_{\mathrm{zi}}(t)+{y}_{\mathrm{zs}}(t)$ $x(t)={x}_{\mathrm{zi}}(t)+{x}_{\mathrm{zs}}(t)$
$y(t)=\sum_{I=1}^{n} {c}_{i}e^{{\lambda }_{i}t}+\int_{{0}^{-}}^{t} h(t-\tau )u(\tau )\,d \tau$ $x(t)=e^{At}x({0}^{-})+\int_{{0}^{-}}^{t} e^{A(t-\tau )}Bu(\tau )\,d \tau$
$y(t)=Ce^{At}x({0}^{-})+\int_{{0}^{-}}^{t} (D\delta (t-\tau )+Ce^{A(t-\tau )}B)u(\tau )\,d \tau ,h(·)=D\delta (t-\tau )+Ce^{A(t-\tau )}B$
$y(t)=Ce^{At}x({0}^{-})+\int_{{0}^{-}}^{t} h(t-\tau )u(\tau )\,d \tau$
Laplace Transform: Solution in the frequency domain
$Y(s)=\frac{r(s)}{q(s)}+H(s)U(s)$ $X(s)=sI-A^{(-1)}x({0}^{-})+sI-A^{(-1)}BU(s)$
$Y(s)=CsI-A^{(-1)}x({0}^{-})+(D+CsI-A^{(-1)}B)U(s),H(s)=D+CsI-A^{(-1)}B$

## Definition of state from i/o description

Let $H(s)=D+\frac{\langle p(s)\rangle }{q(s)}$ , $\mathrm{deg}(\langle p\rangle )< \mathrm{deg}(q)$ . Define $w$ so that $\frac{d q}{d t}}w(t)=u(t)$ , $(y(t)=\frac{d \langle p\rangle }{d t}}w+Du(t))\implies ({x}^{T}=\begin{pmatrix}w & w^{1} & \dots & w^{(n-1)}\\ \end{pmatrix}\in \mathbb{R}^{n})$ , $n$ : degree of $q(s)$ .

## Various responses

Zero-input or free response
response due to initial conditions alone.
Zero-state or forced response
response due to input (forcing function) alone (zero initial condition).
Homogeneous solution
general form of free-response (arbitrary initial conditions).
Particular solution
forced response.
response obtained for large balues of time $T\to$ .
Transient response
full response minus steady minus state response.

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