

 State space systems

(Blank Abstract)
I/o and i/s/o representation of siso linear systems
I/O 
I/S/O 
variables:
$\left(u,y\right)$ 
variables:
$(u,x,y)$ 
$\frac{d q}{d t}y(t)=\frac{d p}{d t}u(t),n=\mathrm{deg}(q)\ge \mathrm{deg}(p)$ 
$\frac{d x(t)}{d t}=Ax(t)+Bu(t),y(t)=Cx(t)+Du(t)$ 
$u(t),y(t)\in \mathbb{R}$ 
$x(t)\in \mathbb{R}^{n},\begin{pmatrix}A & B\\ C & D\\ \end{pmatrix}\in \mathbb{R}^{(n+1\times n+1)}$ 
Impulse Response 
$\frac{d q}{d t}h(t)=\frac{d p}{d t}\delta (t)$ 
$h(t)=D\delta (t)+Ce^{At}B,t\ge 0$ 
$H(s)=\mathcal{L}(h(t))=\frac{p(s)}{q(s)}$ 
$H(s)=D+CsIA^{(1)}B$ 
Poles 
characteristic roots  eigenfrequencies 
${\lambda}_{i},q({\lambda}_{i})=0,I=1,\dots ,n$ 
$\det ({\lambda}_{i}IA)=0$ 
Zeros 
$H({z}_{i})=0\iff p({z}_{i}),1,\dots ,n$ 
$\begin{vmatrix}{z}_{i}IA & \mathrm{B}\\ \mathrm{C} & \mathrm{D}\end{vmatrix}=0$ 
Matrix exponential 
$(e^{At}=\sum_{k=0} )\implies $∞
t
k
k
A
k
t
A
t
A
A
t
A
t
A 
$\mathcal{L}(e^{At})=sIA^{(1)}$ 
BIBO stability 
$y=(h, u)$ , requirement 
$\exists \forall u\colon ({\mathrm{Norm}(u)}_{})\implies $∞ ∞
u ∞ ∞ 
$\iff {(h)}_{1}=\int_{0} \,d t$∞
h
t ∞ 
$\iff \Re ({\lambda}_{i})< 0\iff \mathrm{poles}\in \mathrm{LHP}$ 
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 $ 

$y(t)=Ce^{At}x({0}^{})+\int_{{0}^{}}^{t} (D\delta (t\tau )+Ce^{A(t\tau )}B)u(\tau )\,d \tau ,h(\xb7)=D\delta (t\tau )+Ce^{A(t\tau )}B$ 

$y(t)=Ce^{At}x({0}^{})+\int_{{0}^{}}^{t} h(t\tau )u(\tau )\,d \tau $ 
Laplace
Transform: Solution in the frequency domain 
$Y(s)=\frac{r(s)}{q(s)}+H(s)U(s)$ 
$X(s)=sIA^{(1)}x({0}^{})+sIA^{(1)}BU(s)$ 

$Y(s)=CsIA^{(1)}x({0}^{})+(D+CsIA^{(1)}B)U(s),H(s)=D+CsIA^{(1)}B$ 
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
$\frac{d q}{d t}w(t)=u(t)$ ,
$(y(t)=\frac{d \langle p\rangle}{d t}w+Du(t))\implies ({x}^{T}=\begin{pmatrix}w & w^{1} & \dots & w^{(n1)}\\ \end{pmatrix}\in \mathbb{R}^{n})$ ,
$n$ : degree of
$q(s)$ .
Various responses
Zeroinput or free response
 response due to initial conditions alone.
Zerostate or forced response
 response due to input (forcing function) alone (zero
initial condition).
Homogeneous solution
 general form of
freeresponse (arbitrary initial conditions).
Particular solution
 forced response.
Steadystate response
 response obtained
for large balues of time
$T\to $∞ .
Transient response
 full response minus steady
minus state response.
Questions & Answers
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CYNTHIA
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Uday
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what is system testing?
AMJAD
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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AMJAD
what is system testing
AMJAD
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Stotaw
In this morden time nanotechnology used in many field .
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Azam
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Prasenjit
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Damian
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Azam
I'm interested in Nanotube
Uday
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Prasenjit
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how did you get the value of 2000N.What calculations are needed to arrive at it
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Source:
OpenStax, State space systems. OpenStax CNX. Jan 22, 2004 Download for free at http://cnx.org/content/col10143/1.3
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