

 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
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
sure. what is your question?
ninjadapaul
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X6)^2
so it's 20 divided by X6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
if A not equal to 0 and order of A is n prove that adj (adj A = A
rolling four fair dice and getting an even number an all four dice
Differences Between Laspeyres and Paasche Indices
No. 7x 4y is simplified from 4x + (3y + 3x) 7y
J, combine like terms 7x4y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)1/7 (x1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
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Porter
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Yasmin
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Cesar
I'm interested in nanotube
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
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AMJAD
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Stotaw
In this morden time nanotechnology used in many field .
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and may other field for details you can check at Google
Azam
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Prasenjit
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maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
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Prasenjit
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Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
I'm interested in Nanotube
Uday
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Prasenjit
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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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