part 2 - step and impulse response graphical models

7/25/2019 Part 2 - Step and Impulse Response Graphical Models

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Linear models Step 1st order Step 2nd order Impulse 1st order Impulse 2nd order

Part II

Step and Impulse ResponseGraphical Models


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Classification

RecallTypes of modelsfrom Part I:

1 Mental or verbal models

2 Graphs and tables3 Mathematical models, with two subtypes:

First-principles, analytical models

Models from system identification

Step and impulse responses are graph models; they belong to thesecond category.


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Recall example

A hard drive read-write head, with input = motor voltage, and output =

head position.


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Classification (continued)

Step and impulse response models also belong to the class ofnonparametric models: they are functions of time that, in general,cannot be represented by a finite number of parameters.

The study of these models is called transient analysis, since it relies

in a large part on the transient regime of the response.

Note: We will in fact derive parametric models from the

nonparametric graph model.

Li d l St 1 t d St 2 d d I l 1 t d I l 2 d d


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Table of contents

1 Recap: Continuous-time linear models

2 Step response models of first-order systems

3 Step response models of second-order systems

4 Impulse response models of first-order systems

5 Impulse response models of second-order systems



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A definition of linear systems

A system islinearif it satisfies:

Superposition: If for inputu1(t)the system responds with outputy1(t); and foru2(t)the system responds withy2(t); thenfor inputu1(t) + u2(t)the system will respond withy1(t) + y2(t).

Homogeneity: If for inputu(t)the system responds with output y(t);

then for inputu(t)the system will respond withy(t).



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Transfer function representation

The transfer function is:

H(s) = Y(s)

U(s)

= bms

m + bm1sm1 +. . .+ b1s+ b0

ansn

+ an1sn

1

+. . .+ a1s+ a0, m

n

whereU(s)and Y(s)are, respectively, the Laplace transforms of theinput and output signalsu(t)and y(t).(Note: in zero initial conditions.)

TheLaplace transformof a signalf(t)is:

F(s) = L[f(t)] =

0

f(t)estdt



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Laplace transform interpretation

sis calledcomplexargument (as in complex number), and theLaplace transform can be seen as taking a function from the timedomaintto the more abstract, complex domains.

The motivation is that many signal operations common inengineering (differentiation, integration, etc.) become muchsimpler in thesdomain.

Intuitively, L can be seen as a representation off in terms of itsexponential components, like the Fourier transform is arepresentation in terms of periodic components.



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Table of contents



First order system

Step response. Determining the parameters

Example






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p p p p

First order system: Motivating example

First-order systems are common. Typical example: a thermal system.Consider an object at temperature1 (output variable) placed in anenvironment at temperature2 (input variable). Then:

C1(t) =2(t) 1(t)

R

whereCis the thermal capacitance and Ris the thermal resistance.

Applying the Laplace transform on both sides:

Cs1(s) =2(s)1(s)

R

leading to the transfer function:

H(s) =1(s)

2(s)=

1CR

s+ 1



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p p p p

First order system: General form

H(s) = K

Ts+ 1where:

Kis the gain (=1 in the example)

Tis the time constant (= CR

in the example)



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Table of contents



First order system


Example






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Ideal step input

uS(t) = 0 t 0

1 t>0



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Ideal 1st order response

Solving the differential equation fory(t)(or easier: solve forY(s)andthen apply inverse Laplace transform L1), we get:

y(t) =K(1 et/T)from where:

y(t) = K

Tet/T, y(0) =

K

Te0 =

K

T

y(T) =K(1 e1) 0.632Kand similar fort=2T, 3T, 4T(see figure).

Note: output within 2% of steady state after 4T.



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Determining the parameters

So far, everything known from: Sys. Theory, Process Modeling.

Now, consider we are given a step response of a real, unknownsystem: this response is the nonparametric model. We can use it tofind an approximate transfer function (a parametric model) of thesystem.

Algorithm

1 Read the steady-state value. That is the gainK.

2 Determine the time value where the output reaches 0.632 of its

steady state value. That is the time constantT.



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Nonzero initial conditions

In practice, we often cannot use ideal step signals: the system must

be kept around a safe/profitable operating point. Real-lifeapproximations of step inputs are rectangular signals as below. Thesystem response is therefore nonstandard.

But recall linearity properties. Our new input is

u(t) =u0+ (uss u0)uS(t)withuS(t)the ideal step input. Then,denoting the ideal step response by yS(t), we have the new output:

y(t) =y0+ (uss u0)yS(t)

simply a shifted and scaled version. (Note we assumed the system

was in steady-state aty0 with input held constant atu0.)



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Nonzero initial conditions (continued)

Then the parameters are recovered easily:

K = yss y0uss u0

y(T) =y0+ 0.632(yss

y0)

(read T where the output raises 0.632 of the difference)

The start time of the step may also be different from 0, solved triviallyby shifting the time axis.



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Table of contents



First order system


Example






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Example: Thermal system

Consider the thermal system in the figure. The input is the voltage V

applied to the lamp, the output is the temperature read by athermocouple at the back of the steel plate.



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Thermal system: Experimental data

The data is obtained from the Daisy database(http://homes.esat.kuleuven.be/smc/daisy/).

Note: the presence ofnoisein the data! This is virtually always true inidentification experiments.

We use the first step for identification, and keep the others for

validation.

http://homes.esat.kuleuven.be/~smc/daisy/http://homes.esat.kuleuven.be/~smc/daisy/http://homes.esat.kuleuven.be/~smc/daisy/http://homes.esat.kuleuven.be/~smc/daisy/http://homes.esat.kuleuven.be/~smc/daisy/


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Thermal system: Model and parameters

The nonparametric model is the graph. We use it to estimate atransfer function model.

We haveyss 192 C, y0 129 C. Also, the inputuss=9 V and(from the experiment) we know that u0 =6 V. Therefore:

K = yss y0uss u0

192 1299 6 21

Further,y(T) =y0+ 0.632(yss

y0)

169, and identifying this point

on the graph we getT 60.



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Thermal system: Transfer function model

K =21T =60H(s) = K

T s+ 1

= 21

60s+ 1



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Thermal system: Validation of transfer function model

The simulation does not take into account the non-zero initialcondition of the system, hence the first part is different.

The fit is not great the cooling dynamics are quite slower than theheating dynamics, for example, so in reality this is not a simplefirst-order system.

Nevertheless, the transfer function is sufficient for a rough initial

model: this is the typical use of transient analysis.



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Thermal system: Validation (continued)

Mean squared error (MSE) on the validation data (second and furthersteps):

J= 1

N

Nk=1

e2(k) = 1

N

Nk=1

(y(k) y(k))2 62.21Note that although we treated the data as continuous-time so far, it isof course sampled in discrete time, so a meaningful MSE can be

computed. The sampling time isTs =2 s.



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Table of contents




Second order system


Example

Other issues


5

Impulse response models of second-order systems



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Second order system: Motivating example

Second-order systems are also quite common.

Consider a massmtied to a spring, to which we apply a force f (the

input) away from the string. We measure the position xof the mass(output). From Newtons second law:

mx(t) =f(t) kx(t)wherekis the spring constant.

Applying the Laplace transform on both sides:ms2X(s) =F(s) kX(s)

leading to the transfer function:

H(s) = X(s)

F(s)

= 1

ms2

+ k



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Second order system: General form

H(s) = K2n

s2 + 2ns+2n

where:

Kis the gain (= 1k

in the example)

is the damping (=0 in the example)

nis the natural frequency (= k/min the example)



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Table of contents




Second order system


Example

Other issues


5

Impulse response models of second-order systems



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2nd order step response shapes

Damping factoris crucial in determining step response shape.

=0, undamped

(0, 1), underdamped



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2nd order step response shapes (continued)

=1, critically damped

>1, overdamped


U d d d 2 d d


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Underdamped 2nd order step response

We are mostly interested in the underdamped case ( (0, 1))

Solving fory(t)we get:

y(t) =K

1 1

1 2ent sin(n

1 2t+ arccos )


R h t i ti


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Response characteristics

Steady-state valueK

To get the peaks and valleys, we solve for zero derivative, y(t) =0:

y(t) = Kn1 2 ent sin(n1 2t) =0

tm= mn

1 2

y(tm) =K[1 + (

1)mMm], whereovershoot M=e

12


N t i d l


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Nonparametric model

Now, consider we are given a step response of a real, unknownsystem: this response is the nonparametric model. Using the insightdeveloped above, we can find an approximate transfer function (aparametric model) of the system.


D t i i th t


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Algorithm

1 Determine steady-state output valueyss. That is the gainK.2 Determine overshootM, (a) from the first peak: M= y(t1)yss

yss, or

(b) from ratio of first valley and first peak: M=

yss

y(t2)

y(t1)yss .3 SolveM=e

12 , leading to= log1/M

2+log2 M

4 Read period of oscillation T0=t3 t1= 2n

12, getting

n= 2

T012

, or equivalentlyn= 2T02 + log

2 M.




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Similar to 1st order case: new input u(t) =u0+ (uss u0)uS(t), sothe new output is again just a shifted and scaled version of the idealstep responseyS(t): y(t) =y0+ (uss u0)yS(t).The modified formulas are:

1 GainK = yssy0ussu0 .

2 Overshoot (a)M= y(t1)yssyssy0 (we need to subtracty0), or (b)

M= yssy(t2)y(t1)yss (no change in this formula).

3 : same as before.4 T0: same as before, the time axis is unchanged.


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2nd order example


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2nd order example

Simulation example, 500 samples with sampling time 0.047.

Note again the measurement noise. Also, while the experiment haszero initial condition (u0 =y0 =0), the steps still have nonstandardvalues (different from 1).

We will use step 1 for identification, steps 3-5 for validation (using the

fact that step 2 returns the system to zero initial condition).


Example: step response model


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Example: step response model

The nonparametric model is the graph. We use it to estimate atransfer function model.

Since the output is quite noisy, we determine the steady state valueby averaging several samples, namely numbers 90 to 100, betweent4andt5:

yss 111

100k=90

y(k) 1.00

We read on the graph: t1

0.65,t2

1.35,t3

1.96,y(t1)

1.37,

y(t2) 0.86. Finally,uss=4.


Example: Determining the parameters


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Example: Determining the parameters

1 GainK = yssy0ussu0 0.25.

2 Overshoot M= y(t1)yssyssy0 0.36.

3 Damping= log 1/M2+log2 M

0.31.4 PeriodT0 =t3 t1 1.31, so natural frequency

n= 2

T0

12 5.05.


Example: Transfer function model


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K =0.25

=0.31

n=5.05H(s) = K2ns2 + 2ns+2n = 6.38s2 + 3.09s+ 25.51

(Note: Actual calculations done in double precision with Matlab, so

using the two-decimal precision numbers given in the slides can leadto slightly different results. This remark applies to all the examples.)


Example: Validation of transfer function model


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Example: Validation of transfer function model

Very good fit (not surprising since this is synthetic data).

Mean squared error (MSE):

J= 1

N

Nk=1

e2(k) = 1

N

Nk=1

(y(k) y(k))2 9.66 105


Table of contents


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Table of contents




Second order system


Example

Other issues




Choosing the order


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Choosing the order

Even when it is overdamped orcritically-damped, att=0 a 2ndorder system response will have aderivative of 0: it will betangent tothe time axis. In contrast, thetangent slope isK/Tfor 1st ordersystems.


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Table of contents


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Table of contents

1

Recap: Continuous-time linear models


3

Step response models of second-order systems


Impulse signal. Relation between step and impulse responses

First-order impulse response. Determining the parametersExample



Ideal impulse input


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dea pu se put

The ideal impulse is the Dirac delta. An informal definition:

uI(t) =

t=00 t=0

with the additional condition uI(t)dt=1.(In fact, the ideal impulse is not a function and requires the notion ofdistributions to be formally defined.)


Practical impulse realization


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p

In practice, we cannot create signals of infinite amplitude.So, an approximate impulse is realized with the rectangular signal:

uIR(t) =

1 t [0, )0 otherwise

where (much smaller than the time constants in the system).Note the signal still obeys

uIR(t)dt=1 (rectangle has area 1).

So there will be approximation error, but for small the error is notlarge. We develop the analysis in the ideal case, while the examples

use the practical realization.


Useful property of impulse response


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p p y p p

In the Laplace domain:

step inputUS(s) = 1

s, impulse inputUI(s) =1

Recall that the time-domain response of a system can be expressedas: y(t) = L1 {Y(s)}, andY(s) =H(s)U(s). So:

YS(s) = 1

sYI(s), YI(s) =sYS(s)

yS(t) = t0

yI()d, yI(t) = yS(t)

The impulse response is the derivative of the step response.


Table of contents


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Recall: First order system


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y

H(s) = K

Ts+ 1

where:

K is the gain

Tis the time constant


Ideal 1st order impulse response


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Using the relation to the step response, and the derivative of the stepresponse we already computed, we have the impulse response:

yI(t) = K

Tet/T, t 0

from where:

yI(0) = KT

=ymax

yI(T) = K

Te1 =ymaxe

1 0.368ymax

Note: yI(4T) =0.0183ymax, so like for the step response, the output

is roughly in steady state after 4T.




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When the initial conditions are non-zero, the impulse is shifted alongthe vertical axis.

From linearity, given the shiftedu(t) =u0+ uI(t), we have a shiftedy(t) =y0+ yI(t). (We assumed the system was in steady-state at y0with input held constant atu0.)

So the behavior is:

ymax=y0+K

T

y(T) =y0+ 0.368(ymax y0)

Noteu0 =uss, y0 =yss.




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Consider now that we are given the impulse response of a real,unknown system: this response is the nonparametric model. As in

the step case, we can use this response to find an approximatetransfer function (a parametric model) of the system.

We consider first non-zero initial conditions because that is actually afavorable situation: we have a reliable way to find the gainK.

Algorithm

1 Read the steady-state (or initial) output yss=y0 and inputuss=u0. Then,K =yss/uss.

2 Readymaxand determine the time where the output decreases to0.368 of the differenceymax

y0. That is the time constantT.


Determining the parameters in zero initial conditions


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We can estimate the gain by using ymax= K

T, but in practice this will

not be as accurate (e.g. because the impulse is not ideal).

Algorithm

1 Readymaxand determine the time where the output decreases to0.368 of the differenceymax y0. That is the time constant T.

2 FindK =ymaxT.


Table of contents


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1st order example


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Simulation example, 330 samples with sampling time Ts =0.28 (30samples are the initial steady-state regime, then 100 for each impulse

response). The practical impulses are realized with = Ts =0.28,amplitude 1/ 0.37.

Note the measurement noise and the non-zero initial condition.

We will use impulse 1 for identification, impulses 2-3 for validation.


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Example: Model and parameters (continued)


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The maximum output value isymax 0.19, reached att1 8.86.Valuey0+ 0.368(ymax y0) 0.15 is reached att2 12.60.Therefore:

1 K =yss/uss 0.25.2 T =t2 t1 3.92.




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K =0.25

T =3.92H(s) = KT s+ 1 = 0.253.92s+ 1


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State space model of annth order system


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Astate space modelof a linear system is a representation of the

system in the following form:

x(t) =Ax(t) + Bu(t)

y(t) =Cx(t) + Du(t)

where:xstate vector,x Rn withnthe order of the systemuandyare the usual input and output. They can be vectors ifthe system has several inputs or outputs, but for us here, a scalarinput and output are enough.

Astate matrix,B input matrix,Coutput matrix, andDfeedthrough matrix. They have appropriate dimensions:A Rnn,B Rn1 (a vector, due to scalar input), C R1n (avector, due to scalar output), D R (a scalar).


State space model of a general 1st order system


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Starting from transfer function model:

H(s) = K

Ts+ 1=

Y(s)

U(s)

and moving back to the time domain we get:

y(t) =1T y(t) + KTu(t)

By simply takingx=y(recall that the system has order 1 so a singlestate suffices), we can write:

x(t) = 1T

x(t) + KT

u(t)

y(t) =x(t)

so our state space model hasA = 1T

,B= KT

, C=1, D=0.


Back to example: (Approximate) state space model


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x(t) = 1Tx(t) +KTu(t) = 0.26x(t) + 0.06u(t)

y(t) =x(t)


Example: Validation with correct initial condition


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To take the initial condition into account, we simply set x(0) =y0when starting the simulation.

Mean squared error (MSE) on the validation data:

J= 1

N

Nk=1

e2(k) 3.74 106


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Recall: Second order system


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H(s) = K2n

s2 + 2ns+2n

where:

K is the gain

is the damping factor

nis the natural frequency


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Underdamped impulse response


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Using the derivative of the step response we already computed, wehave the impulse response:

yI(t) = Kn1 2 ent sin(n1 2t)

Already note that the oscillation period is unchanged, soT0 =t3 t1 =2(t2 t1).


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Nonzero initial conditions: estimatingK


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In non-zero initial conditions, the impulse is shifted,u(t) =u0+ uI(t),leading to a shiftedy(t) =y0+ yI(t). Noteu0 =uss, y0 =yss.

From the steady-state values we can estimate thegain: K = yssuss

.There is no change in T0, but the areas must now be found relative tothe steady-state value:

A+= t0,1

0

(y() y0)d =K+ KM

A= t0,2

t0,1

(y0 y())d =KM+ KM2


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Table of contents


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5

Impulse response models of second-order systemsSecond-order impulse response. Determining the parameters

Example


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Example: Steady-state values and gain


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We use the graph (nonparametric model) to estimate a transferfunction (parametric model). We haveu0 =uss=2.

We determine the steady state output (equal to the initial output) byaveraging the last 11 samples:

yss=y0 111

130

k=120

y(k) 0.5


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Back to example: (Approximate) state space model


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x(t) =

0 1

26.68

3.22

x(t) +

0

6.64u(t)

y(t) = 1 0 x(t) + 0u(t)wherexis now the whole state vector, x(t) =

x1(t)x2(t)

.


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part 2 - step and impulse response graphical models

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