INGENIERÍA DE
CONTROL
UNIDAD I: FUNDAMENTOS DE LA
INGENIERÍA DE CONTROL
Prof. Ing. Jack Cutipa
Docente de Ing. Mecatrónica e Ing. Electrónica
vidis.cutipa@upn.pe
Introducción a los sistemas de control.
SESIÓN 7
LOGRO DE LA SESION
Al finalizar la sesión, el estudiante analiza el error en estado estable
utilizando la función de transferencia y señales estandar de prueba,
empleando los conocimientos adquiridos en clase, con claridad y
criterio.
https://www.youtube.com/watch?v=qhXPb-cJ94c
ERROR en ESTADO ESTACIONARIO
CONTENIDOS
1. Error en estado estable
Introduction
• Any physical control system inherently suffers
steady-state error in response to certain types of
inputs.
• A system may have no steady-state error to a step
input, but the same system may exhibit nonzero
steady-state error to a ramp input.
• Whether a given system will exhibit steady-state
error for a given type of input depends on the type
of open-loop transfer function of the system.
Classification of Control Systems
• Control systems may be classified according to
their ability to follow step inputs, ramp inputs,
parabolic inputs, and so on.
• The magnitudes of the steady-state errors due
to these individual inputs are indicative of the
goodness of the system.
Classification of Control Systems
• Consider the unity-feedback control system
with the following open-loop transfer function
• It involves the term sN in the denominator,
representing N poles at the origin.
• A system is called type 0, type 1, type 2, ... , if
N=0, N=1, N=2, ... , respectively.
Classification of Control Systems
• As the type number is increased, accuracy is
improved.
• However, increasing the type number
aggravates the stability problem.
• A compromise between steady-state accuracy
and relative stability is always necessary.
Steady State Error of Unity Feedback Systems
• Consider the system shown in following figure.
• The closed-loop transfer function is
Steady State Error of Unity Feedback Systems
• The transfer function between the error signal E(s) and the
input signal R(s) is
E( s ) 1
=
R( s ) 1 + G( s )
• The final-value theorem provides a convenient way to find
the steady-state performance of a stable system.
• Since E(s) is
• The steady state error is
Static Error Constants
• The static error constants are figures of merit of
control systems. The higher the constants, the
smaller the steady-state error.
• In a given system, the output may be the position,
velocity, pressure, temperature, or the like.
• Therefore, in what follows, we shall call the output
“position,” the rate of change of the output
“velocity,” and so on.
• This means that in a temperature control system
“position” represents the output temperature,
“velocity” represents the rate of change of the
output temperature, and so on.
Static Position Error Constant (Kp)
• The steady-state error of the system for a unit-step input is
• The static position error constant Kp is defined by
• Thus, the steady-state error in terms of the static position
error constant Kp is given by
Static Position Error Constant (Kp)
• For a Type 0 system
• For Type 1 or higher systems
• For a unit step input the steady state error ess is
Static Velocity Error Constant (Kv)
• The steady-state error of the system for a unit-ramp input is
• The static position error constant Kv is defined by
• Thus, the steady-state error in terms of the static velocity
error constant Kv is given by
Static Velocity Error Constant (Kv)
• For a Type 0 system
• For Type 1 systems
• For type 2 or higher systems
Static Velocity Error Constant (Kv)
• For a ramp input the steady state error ess is
Static Acceleration Error Constant (Ka)
• The steady-state error of the system for parabolic input is
• The static acceleration error constant Ka is defined by
• Thus, the steady-state error in terms of the static acceleration
error constant Ka is given by
Static Acceleration Error Constant (Ka)
• For a Type 0 system
• For Type 1 systems
• For type 2 systems
• For type 3 or higher systems
Static Acceleration Error Constant (Ka)
• For a parabolic input the steady state error ess is
Summary
Example#1
• For the system shown in figure below evaluate the static
error constants and find the expected steady state errors
for the standard step, ramp and parabolic inputs.
100( s + 2)( s + 5)
R(S) C(S)
2
s ( s + 8)( s + 12)
-
Example#1 (evaluation of Static Error Constants)
100( s + 2)( s + 5)
G( s ) =
s 2 ( s + 8)( s + 12)
K p = lim G( s )
s →0 K v = lim sG( s )
s →0
100( s + 2)( s + 5)
K p = lim 2 100s( s + 2)( s + 5)
s →0 s ( s + 8)( s + 12) K v = lim 2
s →0 s ( s + 8)( s + 12)
Kp =
Kv =
K a = lim s 2 G( s ) 100s 2 ( s + 2)( s + 5)
K a = lim 2
s →0
s →0
s ( s + 8 )( s + 12 )
100( 0 + 2)(0 + 5)
K a = = 10.4
( 0 + 8)(0 + 12)
Example#1 (Steady Sate Errors)
Kp = Kv = K a = 10.4
=0
=0
= 0.09
Bibliografia
Katsuhiko Ogata, Ingeniería de control moderna, 5ta Edición
Benjamin Kuo, Sistemas de control automatico, 7ma Edición