Espacio estados y punto de equilibrio

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State Space representation and Equilibrium Points of 2 Differential Equations Gerardo Macías, Fellow, IEEE

Abstract— A state-space representation is a mathematical model of a physical system as a set of input, output and state variables related by differential equations. This representation is an important tool in the arsenal of today’s control system designer. The graphical interpretation of a linearization is to find the shape of the tangent line at a point of a curve function. This point or points are called equilibrium points.

T

Z! + 2Z! = b!                      (6) Z! − Z! = b!                      (6)

Applying the Cramer rule to solve system equations: AX = B b 2 Z! = ! b! −1 Z!

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I. INTRODUCTION

analysis of a dynamics systems has an important place when is required to apply a control law. Approximately since 60’s modern control theory has developed with a new approach that is based in the concept of state [1]. A state of a system is the smallest group of variables also called state variables. A state space representation can be used with linear differential equations and also with nonlinear ones. To transform linear into nonlinear equation needs to find the linear approximation to a function at a given point. Linearization is a method for assessing the local stability of an equilibrium point of a system of nonlinear differential equations [2].

X = A!! B

HE

II. DEVELOP The two nonlinear differential equations are: x! + 2x! + 3x!" + x! ! = u!                      (1) x! − x! + x! x! + x!" = u!                            (2)

And auxiliary equations: x!" = sin(x! )                                                                            (3)   x!" = x! !                                                                                          (4)

Equations number 1 and 2 has derivatives with same order. To solve this problem is necessary to do a variable change and apply Cramer Rule: Z! = x!              (5) Z! = x!              (5)

Replacing equations number 3, 4 and 5 in 1 and 2:

1 x! = 3 x! 1 3

2 3 b! −1 b! 3

Multiplying and solving where b! = −3 sin(x! )   − x! ! + u! and b! = −x! x! − x! ! + u! 1

2

2

1

2

3 1

3 1

3 1

3 1

3 1

3

3

3

3

x! = − x! ! − x! x! − x!  ! − sin(x! )   + u! + cos(u! )      (7) !

 !

x! = − x! + x! x! + x! − sin(x! )   + u! − cos(u! )      (8) 3

A. State Space Representation The make state space representation is necessary to use the eq. 7 and 8. There will be two state variables for each equation: Z! Z! Z! Z!

= x! = x!           = x! = x!

Z! = x! = Z! 1 ! 2 2 1 2 Z! = x! = − Z! − Z! Z! − Z!  ! − sin(Z! )   + u! + cos(u! ) 3 3 3 3 3                         Z! = Z! 1 1 1 1 1 Z! = x! = − Z! ! + Z! Z! + Z!  ! − sin Z! + u! − cos(u! ) 3 3 3 3 3

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Also could be represented in matrix notation. Following the form x t = A t x t + B t u(t)

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

Z! 1 Z! 2 Z! 3 Z! 4

⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= ⎥ ⎢ ⎥ ⎢ ⎦ ⎢ ⎢⎣

-

-

0 sin(Z 1 )

1 2 - Z4 3

Z1 0 sin(Z 1 )

1

Z1

3

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣

0 Z4

⎤ 2 2 1 2 ⎥⎡ - Z3 - Z4 ⎥ ⎢ 3 3 ⎥⎢ ⎥ ⎢⎢ 0 1 1 2 1 2 ⎥⎢ Z3 - Z4 ⎥ ⎣ 3 3 ⎥⎦ 0

III. CONCLUSION

0

⎤ ⎥ 2 cos(u 2 ) ⎥ ⎡⎢ u ⎥⎢ 1 3u 2 ⎥⎢ 0 0 ⎥ ⎢⎢ u 2 ⎥⎣ 1 1 cos(u 2 ) ⎥ 3 3u 2 ⎥⎦ 0 1 3

Z1 ⎤ ⎥ Z2 ⎥ ⎥+ Z3 ⎥ Z 4 ⎥⎦

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

As can be seen is a nonlinear state space representation. B. Equilibrium Points An equilibrium point of a dynamic system represents the conditions of system variables in static position. Then this x ≡ 0 [3]. means x! ≡ 0 and thus acceleration !! The equations 7 y 8 around the equilibrium points are: 1

2

2

1

3

3

3

3

x!" = − x!" ! − x!" x!" − x!"  ! − sin(x!" )   + u!" + cos(u!" )                                                                  (9) 1

1

3

3

3

1

1

3

3

x!" = − x!" ! + x!" x!" + x!"  ! − sin(x!" )   + u!" 1

− cos(u!" )                                                              (10) 3

Removing

velocities,

accelerations

and

defining

u1 = u 2 = 3 in 9 and 10: 2

2

3 1

3 1

3

3

sin(x!" )   + x!"  ! = 1 + cos(3)                              (11) sin(x!" )   + x!"  ! = 1 − cos(3)                              (12)

Isolating

x 203 from eq. 11 and replacing in 12: 2

x!"  ! = 1 + cos(3) − sin(x!" )   sin(x!" )   +

1 3

1+

Now isolating

2 3

3

cos(3) − sin(x!" )  

3 2

3 2

                        1

= 1 − cos(3)      (13) 3

x10 from eq. 13 and replacing in eq. 11 to

obtain two operation points.

1 1 − cos(3) sin(x!" )   = 6 3      (14) 3 1−6

This calculated equilibrium point could be used to linearize the main differential 1 and 2 equations by expanding in Taylor series to apply a linear control like PID although this is not always necessary because it can be applied a nonlinear control. REFERENCES

0

2

2

[1]

K. Ogata, “Análisis de sistemas de control” Dinámica de sistemas, 1st ed. Prentice Hal, 1987.

[2]

K. Ogata, “Modelado matemático de sistemas de control” Ingeniería de control moderna, 5 ed. Madrid: Pearson, 2020.

[3]

A. F. Sarasola, “Conceptos básicos de los sistemas de control” Control de los sistemas continuos, 2nd ed. Oviedo: Universidad de Oviedo, 2007.