控制原理Chapter 3-1

Usually, the inverse-transform operation can be performed by direct reference to transform tables or by use a digital computer program.
G (s) Y zs ( s ) X (s) bm s an s
y zs ( t ) L
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[ G ( s ) X ( s )]
Chapter 3-1
If initial condition not zero, can we get the time response? How to do?
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The solution of differential equation: LT method (see P97-106)
• We will develop valuable relationships between the performance specifications and the natural frequency and damping ratio for second-order systems.
• Relying on the notion of dominant poles, we can extrapolate the ideas associated with second-order systems to those of higher order. • A general format for the complementary solution in terms of the state transition matrix (STM) is introduced too.
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The solution of differential equation: LT method (see P97-106) Popularly, a nth system expressed by differential equation in time domain
Steady-state component of the solution: which has the same form as input Transient component of the solution: which is the solution of the corresponding homogeneous equation The form of the transient response depends only on the roots of the characteristic equation.
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The solution of differential equation: classical method
Complete solution (response) of a linear differential equation:
m n
b m 1 s a n 1 s
m 1
b1 s b 0 a1 s a 0
n 1
G (s)
Y zs ( s ) X (s) Y zs ( s ) X (s)

Y (s) ( s s1 )( s s 2 ) ( s s n ) A1 s s1 A2 s s2 An s sn
There are two methods to obtain the solution (time response) of a linear differential equation.
One is solving differential equation directly, getting the particular solution part, and complementary solution part respectively. Then add these two parts got the whole solution. (see P66-73). Another method is using Laplace transform (see P97106).----review it!
Solution of linear differential equations Standard inputs to control system Steady-state responses Transient response
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The solution of differential equation
CHAPTER 3
Time Domain Analysis of Control System
and Its Characteristics
By Hui Wang
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Chapter 3-1
1
Outline of this chapter
Introduction Solution of linear differential equations Dynamic of first and second order system Time response specifications(indices) Routh’s stability criterion Steady-state error analyzing Solution of the state Equation ………
• Based on Routh’s stability criterion, we will discuss the stability of control systems
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Solution of linear differential equations
G (s)



The key is how to get the partial-fraction expansions?
There are four classes, depending on the denominator X(s)(see 4.8).
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an y
(n)
( t ) a n 1 y
( n 1 )
(t ) a 0 y ( tb0 x ( t )
Take Laplace transfome(zero initial condition), it will be
G (s) Y zs ( s ) X (s) bm s an s
Control System
y(t)
Output response Transient response
c ( t ) c ( t ) ss c ( t ) t c ss c t
Mathematical model
Stable? Satisfied? …….
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Chapter 3-1
m n
b m 1 s a n 1 s
m 1
b1 s b 0 a1 s a 0
n 1
where usually nm.
注意：此式与P98(4.32)式 中的符号与系数标注不同！
To get the system time response, it is required to perform the inversetransform operation.
The steady state response is the one that persists after transients have elapsed (i.e, when the system reaches steady state) Design specifications: normally include several time-response indices (pl. of index) for a specified input command as well as a desired steady-state accuracy.
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Introduction
• The ability to adjust the transient and steady-state response of a feedback control system is a beneficial outcome of the design of control systems. • One of the first steps in the design process is to specify the measures of performance for the controlled system. • In the first, we review how to get the general solution of a linear differential equation and its components. • Then, we will introduce the common time-domain specifications such as percent overshoot, settling time, time to peak, time to rise, and steady-state tracking error. • To observe responses of the systems, we will use selected input signals such as the step and ramp to test the response of the control system.