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Introduction to System Modeling

1 Introduction

A mathematical model is an equation or set of equations which adequately describes thebehavior of a system.

System modeling is a subject in its own right. Essentially there are two approaches to findingthe model, [2].

In the first, the system is broken down into smaller elements. For each element a mathe-matical description is then established by working from the physical laws which describethe systems behavior. The simplest such technique is lumped-parameter modeling, whichis considered in this course.

The second approach is known as system identification in which it is assumed that anexperiment can be carried out on the system, and that a mathematical model can be foundfrom the results. This approach can clearly only be applied to existing plants, whereaslumped-parameter modeling can additionally be applied to a plant yet to be built, workingpurely from the physics of the proposed plant components.

System modeling is a specialization of the more general area of mathematical modeling.In a system model, the important relationship is that between the manipulated inputs andmeasurable outputs. Ideally this relationship should be linear and capable of being describedby an expression of low-order (that is, an equation or set of equations containing as few termsas possible). An usual graphical representation of single input, single output (SISO) dynamicalsystem is shown in Figure 1. A model of such a system is a mathematical relationship betweenthe inputu(t) and the output y(t).

Dynamic Systeminput

u(t)

output

y(t)

Figure 1: SISO dynamic system

Most commonly, a low-order linear differential equation model is used.

When fundamental physical laws are applied to the lumped-parameter model, the resultingequations may be nonlinear in which case further assumptions may have to be made in order toproduce an ordinary linear differential equation model which is soluble. In such cases, it is not

unusual to assume that system operation will be restricted to small perturbations about a given

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operating condition. If the assumed operating region is small enough, most nonlinear plantsmay be described by a set of linear equations. If these assumptions cannot be made, nonlinearcontrol techniques may be appropriate.

2 Lumped-parameter models

The systems studied in this course are:

Linear- It must obey the principle of superposition that is, if an input u1(t) causes an outputy1(t) and an input u2(t) causes an output y2(t), then an input u1(t)+u2(t) causes anoutput y1(t)+y2(t) (see next section Linear Approximations of Physical systems)

In the case of the resistor, if an input of 1A produces an output of 3V, and an input of 2Aproduces an output of 6V then an input of 3A produces an output of 9V.

Stationary(or time invariant) - The parameters inside the element must not vary with time.In other words, an input applied today must give the same result as the same input appliedyesterday or tomorrow.

A vehicle which burns large masses of fuel, such as racing cars or space vehicles, is anexample of system which is not stationary. Its dynamic behavior will alter significantlyas its mass decreases, [2].

Deterministic- The outputs of the system at any time can be determined from a knowledgeof the systems inputs up to that time. In other words there is no random (or stochastic)behavior in the system, since its outputs are always a specific function of the inputs. Theterm causalis also used for such systems, [2].

Examples. Consider an element representing an idealized component such as:

The resistor. If the resistor is linear then its model is a relationship between the voltagev(t) and the currenti(t):

i(t) = 1

Rv(t)

The inductor is described by the equation:

i(t) = 1

L

v(t)dt or v(t) =L

di(t)

dt

The capacitor is described by:i(t) =C

dv(t)

dt

The spring-mass-damper mechanical system (Figure 2). Assuming the system is linear,a model of the system can be derived from the Newtons second law of motion. It is arelationship between the (input) forcer(t) and the (output) displacementy(t). The systemcould represent for example an automobile shock absorber. The

Md2y(t)

dt2 + f

dy(t)

dt + ky(t) =r(t)

where: fis the friction coefficient, M - the mass, k - the stiffness of the linear spring.

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k

displacement

Frictionf

Force

y(t)

r(t)

MassM

Figure 2: Spring-mass-damper system

3 Linear Approximation of Physical Systems

A great majority of physical systems are linear within some range of variables. However, allsystems ultimately become nonlinear as the variables are increased without limit.

A system is defined as linear in terms of the system excitation and response. In the caseof an electrical network, the excitation is the input current and the response is the voltage. Ingeneral a necessary condition for a linear system can be determined in terms of and excitationx(t) and a response y(t). When a system at rest is subjected to an excitation x1(t), it provides aresponse y1(t). Furthermore, when the system is subjected to an excitation x2(t), it provides acorresponding responsey2(t). For a linear system, it isnecessarythat the excitation x1(t)+ x2(t)result in a response y1(t) + y2(t). This is usually called the principle of superposition.

Furthermore, it is necessary that the magnitude scale factor be preserved in a linear system.Again, consider a system with an inputxand an outputy. Then it is necessary that the responseof a linear system to a constant multiple of an input x be equal to the response to the inputmultiplied by the same constant so that the output is equal to y. This is called the propertyofhomogeneity.

Example 1. A system characterized by the relation y = x2 is not linear because thesuperposition property is not satisfied. A system represented by the relation y = mx+ b isnot linear because it does not satisfy the homogeneity property. However, this device maybe considered linear about an operating point x0, y0 for small changes x and y. Whenx= x0+ xand y = y0+ y, we have

y= mx + b

ory0+ y= mx0+ mx + b

and thereforey= mx

which satisfies the necessary conditions.

Consider a general element with an excitation x(t) and a responsey(t). The relationship ofthe two variables is written as:

y(t) =g(x(t))

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where g(x(t)) indicates y(t) is a function of x(t). The normal operating point is designatedby x0. Because the curve (function) is continuous over the range of interest, a Taylor seriesexpansion about the operating point may be utilized. Then we have:

y = g(x) =g(x0) + dgdx|x=x0 x x01! +

d2gdx2

|x=x0 (x x0)22! + ...

The slope at the operating point,dg

dx|x=x0,

is a good approximation to the curve over a small range of ( x x0), the deviation from theoperating point. Then, as a reasonable approximation, the expression from example 1 becomes:

y = g(x0) +dg

dx|x=x0(x x0) =y0+ m(x x0),

wherem is the slope at the operating point. Finally, this equation can be rewritten as the linearequation

(y y0) =m(x x0)or

y= mx.

If the dependent variable y depends upon several excitation variablesx1, x2,...,xn, then thefunctional relationship is written as:

y= g(x1, x2,...,xn).

The Taylor series expansion about the operating point x10, x20,...,xn0 is useful for a linear

approximation to the nonlinear function. When the higher-order terms are neglected, the linearapproximation is written as

y= g(x10, x20,...,xn0) + g

x1|x=x0(x1 x10) +

g

x2|x=x0(x2 x20) + ...+

g

xn|x=x0(xn xn0)

where x0 is the operating point. An example will clearly illustrate this method.

Example 2. Pendulum oscillator model.

Consider the pendulum oscillator shown in Figure 3. The torque on the mass is

T =MgLsin(x)

where g is the gravity constant. The equilibrium condition for the mass is x0 = 0o. The first

derivative evaluated at equilibrium provides the linear approximation, which is:

T T0=M gL sinxx |x=x0(x x0),

where T0= 0. Then, we have

T =M gL(cos0o)(x 0o) =MgLxThe approximation is reasonably accurate for/4 x /4. For example, the response ofthe linear model for the swing through

30o is within 2% of the actual nonlinear pendulum

response.

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Mass M

Length L

angle x

Figure 3: Pendulum oscillator

4 The Laplace Transform

The ability to obtain linear approximations of physical systems allows the analyst to considerthe use of Laplace transformation. The Laplace transform method substitutes the relativelyeasily solved algebraic equations for the more difficult differential equations. The basic Laplacetransform of a time signal f(t) is defined as:

F(s) =

0

f(t)estdt

and written symbolically as: F(s) [f(t)]It is common practice to use a capital letter Fwhich is a function of the new variable s for

the transform of the time signal f(t). Also it is assumed thatf(t) is zero for all times beforet=0. In the definition of Laplace transform the exponent st must be dimensionless otherwisethe expression est is meaningless. Thus the variable s has dimension of 1/time which is thedimension of frequency. Since s is also a complex quantity it is often referred to as the thecomplex frequency.

5 Signals

A few commonly used signals will be introduced in this section:

1. The unit step is defined by (Figure 4 a):

u(t) =

0, t

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Table 1: Laplace transform operations

1 Transform integral f(t)0 f(t)e

stdt or [f(t)]

2 Linearity f 1(t) f2(t) F1(s) F2(s)

3 Constant multiplication af(t) aF(s)

4 Complex shift theorem eatf(t) F(sa)

5 Real shift theorem f(t-T) eTsF(s), T

0

6 Scaling theorem f( ta) aF(as)

7 First derivative ddt f(t) sF(s) - f(0)

8 n-th derivative dn

dtn f(t) snF(s)nr=1 dr1dtr1 f(0)snr

9 First integralt0f(t)dt

1sF(s)

Table 2: Laplace transforms of common functionsf(t) F(s)

1 Unit impulse (Dirac)(t) 1

2 Unit step u(t)=1 1s

3 Unit ramp v(t)=t 1s2

4 eat 1sa

5 cost ss2+2

6 sint s2+2

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Figure 4: Unit step signal

Figure 5: The ramp signal

2. The unit ramp is shown in Figure 5. It is defined by:

v(t ) =

0, t

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6 Transfer function models

The transfer function of a linear system is defined as the ratio of the Laplace transform of

the output variable to the Laplace transform of the input variable, with all the initial conditionsassumed to be zero.

A transfer function may be defined only for a linear stationary system. A transfer functionis an input-output description of the behavior of a system.

A linear differential equation that describes such a system can be:

a0r(t) + a1dr(t)

dt + ...+ am

dmr(t)

dtm =b0y(t) + b1

dy(t)

dt + ...+ bn

dny(t)

dtn

where r(t) and y(t) are the input and output variables, as shown in Figure 7. Applying the

H(s)R(s) Y(s)

Figure 7: Block diagram of a system system

Laplace transform for the initial conditions zero, the differential equation becomes:

(a0+ a1s + ...+ amsm)R(s) = (b0+ b1s + ...+ bns

n)Y(s)

and the transfer function will then be:

H(s) =Y(s)

R(s)=

a0+ a1s + ...+ amsm

b0+ b1s + ...+ bnsn

Example. Spring-mass-damper.The transfer function of the spring-mass-damper sys-tem, shown in Figure 2, is obtained from the original equation model by applying the Laplacetransform:

Md2y(t)

dt + f

dy(t)

dt + ky(t) =r(t)

M s2Y(s) + f sY(s) + kY(s) =R(s)

H(s) =Y(s)

R(s) =

1

M s2 + f s + k

The block diagram now contains all the information given in the transfer function model,that is, output = contents x input. H(s) is called a transfer function. It shows how the inputis transferred to the output.

Example. Electrical system. The transfer function for the system shown in Figure 8 can

be derived as follows: Write the fundamental relation for each element:

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L

C Rvinv

out

iL

iC

iRN

Figure 8: Electrical system

Inductor: diL

dt =

1

LvL (8)

Capacitor: dvC

dt =

1

CiC (9)

Resistor: vR= RiR (10)Write the Kirchhoffs current law at node N:

iL = iC+ iR (11)

Write the Kirchhoffs voltage law:

vin = vL+ vC (12)

vC = vR = vout (13)

Assume the initial conditions zero, replace (13) and apply the Laplace transform to the relations(8) - (12):

sIL(s) = 1

LVL(s) (14)

sVout(s) = 1

CIC(s) (15)

Vout(s) = RIR(s) (16)

IL(s) = IC(s) + IR(s) (17)

Vin(s) = VL(s) + Vout(s) (18)

Calculate to eliminateIL, IC, IR, VL from (14) - (18):

Vin= sLIL+Vout = sL(IC+IR)+ Vout= sL(sCVout+1R

Vout)+ Vout= (LCs2+ LR

s+1)Vout (19)

The transfer function is:

H(s) =Vout(s)

Vin(s) =

1

LCs2 + LRs + 1=

R

RLCs2 + Ls + R (20)

For a real physical system the function H(s) is a rational polynomial such that

H(s) =

N(s)

D(s)

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with the order of the denominatorD(s) being greater than or equal to the order of the numeratorpolynomial N(s). Such a system is called proper. In a proper system, the system order isthe degree of the denominator polynomial.

If the denominator polynomial is extracted and set equal to zero, that is: D(s) = 0, theresulting equation is called the systems characteristic equation, since it can be shownto characterize the systems dynamics. Its roots are the poles of H(s). The roots of thenumerator polynomial are the zeros of the transfer function. Both the poles and zeros of H(s)can be complex values, s = +j .

For many design applications it is desirable to present the transfer function in a factoredform:

H(s) = k(s z1)(s z2)...(s zm)sr(sp1)(s p2)...(s pn)

where m n,pi and zi are the poles and zeros of the transfer function, r - the number of polesat the origin, n + r - the order of the system.

Other forms

H(s) = k

sr

m1j=1(Tjs + 1)

m2j=1(

12nj

s2 + 2jnj

s + 1)n1j=1(Tjs + 1)

n2j=1(

12nj

s2 + 2jnj

s + 1)

where k - the gain factor, nj - the natural frequencies, Tj - the time constants, j - thedamping factors.

7 System response

The transfer function can be easily used for the calculation of system response to an input signal.Given the input signal r(t) and the transfer function model of a system H(s), we are interestedin finding the system behavior, that is the output signal y(t) (see Figure 7). From the definitionof the transfer function:

Y(s) = H(s) R(s) (21)By applying the inverse Laplace transform we obtain:

y(t) =1[H(s) R(s)]. (22)

Example. Spring-mass-damper response. Consider the spring-mass-damper system

from the previous example and Figure 2. The transfer function of this system is given by:

H(s) =Y(s)

R(s) =

1

M s2 + f s + k.

We shall obtain the system response or output variable, when the input r(t) is a Dirac impulseand the system constants M , f , k are:

1. M= 1, f= 3, k= 2

2. M= 1, f= 1, k= 3

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Ifr(t) is a unit impulse, the Laplace transform will be R(s) = 1 and the output variable y(t)is:

y(t) =1[H(s)]

1. In the first case the transfer function has two real poles

Y(s) = 1

(s + 1)(s+ 2)

Expanding Y(s) in a partial fraction expansion, we obtain

Y(s) = k1s + 1

+ k2s + 2

where k1 and k2 are the coefficients of the expansion, or the residues, and are evaluatedby multiplying through by the denominator factor corresponding to ki and setting s equalto the root.

k1 = 1s + 2

|s=1= 1

k2 = 1

s + 1|s=2= 1

The inverse Laplace transform ofY(s) is then

y(t) =1

1

s + 1

1

1

s + 2

Using a Laplace transforms table, we find that

y(t) =et e2t

The responsey(t) is called overdamped and is shown in Figure 9.

2. In this caseH(s) has complex poles and can be written as:

H(s) = K

12n

s2 + 2n s + 1

where K= 1/k, 1/2n =M/k and 2/n = f /k. It can be easily proved that when >1the roots are real, when < 1 the roots are complex and conjugates. When = 1 theroots are real and repeated and the condition is called critical damping. In this case, thedamping factor = 1/2

(3) < 1 so the system response will be called underdamped as

the poles of the transfer function are complex.

y(t) =1[H(s)]

Using a table with Laplace transform properties, and writing Y(s) as

Y(s) = 2

11

112

(s + 12)2 + (

112 )

2

the system response is:

y(t) = 2

11et/2sin(

11

2 t)

The underdamped response is shown in Figure 9.

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0

0.05

0.1

0.15

0.2

0.25

0 1 2 3 4 5 6-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

0 2 4 6 8 10 12

Figure 9: Spring-mass-damper response. Overdamped case (left). Underdamped case (right)

8 Block Diagram Models

Since control systems are concerned with the control of specific variables, the interrelationshipof the control variables to the controlling variables is required. This relationship is typicallyrepresented by the transfer function of the subsystem relating the input and output variables.Therefore one can correctly assume that the transfer function is an important relation for controlengineering.

The importance of the cause-effect relationship of the transfer function is evidenced by thefacility to represent the relationship of the system variables by schematic means. The blockdiagram representation of the system relationship is prevalent in control system engineering.Block diagrams consist of unidirectional, operational blocks that represent the transfer functionsof the variables of interest, [1].

The block diagram representation of a given system can often be reduced, using block di-agram reduction techniques. The resulting diagram obeys the law of superposition because ofthe linear behavior of each subsystem.

In control engineering, the block diagram is a primary tool that together with transferfunctions can be used to describe cause-and-effect relationships throughout a dynamic system.The manipulation of block diagrams adheres to a mathematical system of rules often known asblock diagram algebra [3].

In general, the block diagram of a linear time invariant system consists of four components,namely signal, block (with transfer function), summing point and pickoff point as shown inFigure 10, [3].

Figure 10: The components of a block diagram

In Figure 11 are shown the three basic connections: series (cascade), parallel and feedback.The equivalent transfer function for each of these cases can be determined as follows:

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Y2(s)

Y(s)

Y1(s)

R2(s)

R1(s)

R(s)H

1(s) H

2(s)

Y1(s)

Y(s)

Y2(s)

R(s)

H1(s)

H2(s)

`

Y(s)

Hd(s)

Hr(s)

R(s) E(s)

Figure 11: Three basic connections. 1-series, 2-parallel, 3-feedback

1. Series connection.

H1(s) = Y1(s)

R1(s); H2(s) =

Y2(s)

R2(s); Y1(s) =R2(s)

The equivalent transfer function can be calculated from:

H(s) =

Y(s)

R(s) =

Y2(s)

R1(s)=

Y2(s)

Y1(s)

R1(s) R2(s) =H1(s) H2(s)For n systems connected in cascade the equivalent transfer function will be:

H(s) =n

j=1

Hj(s)

2. Parallel connection.

H1(s) = Y1(s)

R(s); H2(s) =

Y2(s)

R(s); Y(s) = Y1(s) Y2(s)

The equivalent transfer function is:

H(s) = Y(s)

R(s) = H1(s)H2(s)

For n systems connected in parallel the equivalent transfer function is:

H(s) =n

j=1

Hj(s)

3. Feedback connection. The error signalE(s) can be calculated from:

E(s) =R(s)Hr(s) Y(s)13

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The output is:Y(s) =Hd(s) E(s)

IfE(s) is eliminated between the previous two relations, the overall transfer function will

be:H(s) =

Y(s)

R(s)=

Hd(s)

1Hd(s) Hr(s)

Diagram transformations

1. Point behind a block - Figure 12

X2

X1X

3

GX

1G

X2

X3

G

Figure 12: Point behind a block

2. Moving a pickoff point ahead of a block - Figure 13

X2

X1

X2

G

X2

X1 X2

G

G

Figure 13: Moving a pickoff point ahead of a block

3. Moving a pickoff point behind a block - Figure 14

X1

X1

X2

G

X1

X1 X2

G

1/G

Figure 14: Moving a pickoff point behind a block

4. Moving a summing point ahead of a block - Figure 15

8.1 The overlap of signals

If a system has more than one input, than the output can be considered a result of all effects ofthese signals.

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Consider the system shown in Figure 16. A transfer function relating each input and theoutput can be calculated, when the other input is considered zero.

Y(s) =R1(s)

H01(s)

|R2(s)=0+ R2(s)

H02(s)

|R1(s)=0

Y(s) = H1H2

1 + H1H2H3 R1(s) + H2

1 + H1H2H3R2(s)

8.2 Example

Consider a system represented by the block diagram shown in Figure 17. Determine the closed-loop transfer function T(s) = Y(s)/R(s).

The block diagram will be transformed so the basic rules of block diagram reduction can beapplied. First, the summing points encircled in Figure 18 (left) will be considered. Note that

the signal Zcan be calculated as:

Z=R (X+ Y) =R X Y (23)

and the two summing points can be reduced to a single one, as shown in Figure 18 (right)

From now on, the basic rules apply. The inner loop is reduced first to one block and theequivalent transfer function is:

Gin(s) =10s+5

1 + 10s+5 5=

10

s + 55 (24)

This resulting block, is in a series connection with 1/s and the equivalent of these two is theproductGin/s. The new block diagram is presented in Figure 19.

Note that the feedback path has a transfer function equal to 1, therefore it is a unity feedbackloop. The equivalent transfer function from the inputR(s) to the output Y(s) is calculated asfollows:

T(s) = Y(s)

R(s) =

Gin(s) 1s1 + Gin(s) 1s 1

=

10s(s+55)

1 + 10s(s+55)=

10

s2 + 55s+ 10 (25)

9 The Transfer Matrix

Because of the great number of systems with multiple inputs and outputs (MIMO systems),calculus techniques specific for this kind of systems were developed. In case of linear systemsthe mathematical techniques use matrix calculus instead of transfer functions.

In Figure 20 the block diagram of a system with m inputs and n outputs is shown.

If the signal transfer is unidirectional and each input influences each output, as shown inFigure 21, we can write the following equation system:

Y1 = H11R1+ H12R2+ . . . H 1mRmY2 = H21R1+ H22R2+ . . . H 2mRm

...

Yn= Hn1R1+ Hn2R2+ . . . H nmRm

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where R and Yare the Laplace transforms of the inputs r(t) and outputs y (t), and

Hjk = YjRk

is the transfer function from the input k to the output j .Thus, a multiple linear system is composed by nxm elements having one input and one

output. The equation system can be written in a matrix form:

Y= H Rwhere

R= [R1(s)R2(s)... Rm(s)]T

is the input vector (m x 1),Y= [Y1(s)Y2(s)... Yn(s)]

T

is the output vector (n x 1), and

H=

H11 H12 . . . H 1mH21 H22 . . . H 2m. . . . . . . . . . . .

Hn1 Hn2 . . . H nm

is called the transfer matrix.

In case of connecting linear MIMO systems, we can use transfer matrices instead of transferfunctions, obeying the rules of matrix calculus.

For example consider the three basic connections shown in Figure 22.

1. Series connection.

Y1 = H1 R; Y= H2 R2; R2= Y1; Y= H2 H1 RThe equivalent transfer matrix will be:

H= H2 H1For n linear systems connected in series, the equivalent transfer matrix is:

H=1

j=n

Hj

2. Parallel connection.

Y1 = H1 R; Y2= H2 R; Y= Y1 Y2The equivalent transfer matrix will be:

H= H1 H23. Feedback connection.

E= R Hr Y; Y= Hd E; Y= Hd (RHr Y)(1Hd Hr) Y= Hd R; Y= (1Hd Hr)1 Hd R

The equivalent transfer matrix will be:

H= (1 Hd Hr)1

Hd

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References

[1] Richard C. Dorf and Robert H. Bishop, Modern Control Systems, Addison-Wesley, 1995.

[2] Ken Dutton, Steve Thompson, and Bill Barraclough, The Art of Control Engineering,Addison-Wesley, 1997.

[3] C. Mei,On teaching the simplification of block diagrams, International Journal of EngineeringEducation 18 (2002), no. 6, 697703.

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X1

G

X2

X3

X2

X1

X3

G

1/G

Figure 15: Moving a summing point ahead of a block

R1

YR

2

H1 H2

H3

H01

H02

Y

R2

R1

Figure 16: Two-input system

Figure 17: Block diagram example exercise.

Figure 18: Block diagram transformation.

Figure 19: Block diagram transformation.

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.

..

linear multipleinput multipleoutput (MIMO)

system.

..

r1(t)

r2

(t)

rm(t)

y1(t)

y2

(t)

yn(t)

Figure 20: MIMO system

...

...

R1(s)

R2(s)

Rm(s)

Y1(s)

Y2(s)

Yn(s)

...

H11

H22

Hnm

Hn1H

n2

H12

Figure 21: Transfer functions between inputs and outputs

Y(s)Y1(s)

R2(s)

R(s)

H1(s) H

2(s)

Y1(s)

Y(s)

Y2(s)

R(s)

H1(s)

H2(s)

Y(s)

Hd(s)

Hr(s)

R(s) E(s)

Figure 22: MIMO system

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