线性控制系统课件Module4(免费)

∑τ = J ω
•
τ m −τ1 − cmωm = Jm ωm
•
•
(for the motor shaft )
τ 2 − clωl = Jl ωl (for the load shaft ) ωl = ωm N τ 2 = Nτ1 (for the gearbox )
Jl • cl τ m = Jm + 2 ωm + cm + 2 ωm N N
1 −2 c(t ) = 1− e − te 2
− t 2 t
Homework: P71: 4.3 P72: 4p between overshoot and damping ratio when system is underdamped.
Sample Problem 4.1
A closed-loop transfer function has the form
C 9 = 2 R s + 4.5s + 9
The damped natural frequency is given by
ωd = ωn 1−ζ 2 = 1.98rad/s
If the system is subject to a unit step input, the output may be written as
C(s) =
9R(s) 9 = 2 s2 + 4.5s + 9 s s + 4.5s + 9
τ m = Jm ωm + cmωm
Taking the Laplace transform
•
Ω(s) 1 = Tm (s) Js + c
R c > L J
⇒
K p Ka Km NR θl G = = 2 θd 1+ GH Js + cs + Kp Ka Km NR
Compare with the standard form:
When ζ<1 —the underdamped case
2 s2 + 2ζωn s + ωn = 0
s1,2 = −ζωn ± ωn 1−ζ
2
c(t ) = 1−
e
−ζωnt
1−ζ 2
sin (ωd t +φ ) ωd = ωn 1−ζ 2
tan φ =
1−ζ 2
ζ
We can recognize from here (P65) 1. The system oscillates at frequency ωd 2. The response decays due to the exponent –ζωn 3. The overshoot depends on the value of ζ
Comparing the closed-loop transfer function with the generalized form
2 ωn C 9 = 2 = 2 2 R s + 4.5s + 9 s + 2ζωns + ωn
It is seen that
ωn = 3rad/s
ζ=
4.5 = 0.75 2ωn
R(s)
2 ωn 2 s2 + 2ζωns + ωn
C(s)
Fig.4.14 Generalized closed-loop transfer function
Open-loop transfer function
R(s)
2 ωn s2 + 2ζωns
C(s)
DC motor position control system
Linear Control System Engineering
Module 4
Second-order System
Second-order system: a system is represented by a second order differential equation.
The behavior of second-order system is different from that of firstorder system, and it may cause oscillation response (振荡相应) and overshoot(系统超调)
If the system is subject to a unit step input, the output may be written as
C(s) = 14 1 1 12 = − − 2 s (s +1 2) (s +1 2)2 s(s +1 2)
The time-domain response is therefore
−ωnt
(1−ωnt )
When ζ>1 —the overdamped case
2 s2 + 2ζωn s + ωn = 0
s1,2 = −ζωn ± ωn ζ −1
2
c(t ) = 1−
e−ζωnt
ζ −1
2
sinh (ωd t +φ )
ζ 2 −1 ωd = ζ 2 −1 tan φ = ζ
For the R-L-C circuit, Compare with the standard form:
1 Uo (s) 1 LC = = 2 Ui (s) LCs + RCs +1 s2 + R s + 1 L LC
R 1 2 ωn = 2ζωn = L LC
1 R C ωn = ζ= LC 2 L
The block diagram of this system
Fig.4.13 position control system block diagram
1. Motor field winding electrical circuit
di υa = iR + L dt
Taking the Laplace transform yields
Solution
The closed-loop transfer function is
C K s(s +1) K = = 2 R 1+ K s(s +1) s + s + K
Comparing with the generalized second-order system, we get
ωn = K and
Determine the undamped natural frequency, the damping ratio, and the damped natural frequency. What is the steady-state output for a unit step input?
Solution
(
)
Applying the final-value theorem yields
lim c(t ) = lim sC(s) = 1
t →∞ s→0
Sample Problem 4.3
A control system with unity feedback is shown in Fig.SP4.3.1. Determine the value of K necessary to make the damping ratio ζ=1. For this value of K calculate the output c(t) when the input r(t) is the unit step function.
Second-order electrical system example:
The UO(s) is written as:
Ui Uo
Uo (s) =
1 1 Cs R + Ls + Cs ⋅
Ui (s)
The transfer function is:
1 Uo (s) 1 LC = = 2 Ui (s) LCs + RCs +1 s2 + R s + 1 L LC
ζ= 1 1 = 2ζωn 2 K 2ζωn =1
If it required that ζ=1, then from the above
K= 1 4
When K takes this value, the transfer function becomes
C 14 14 = 2 = R s + s +1 4 (s +1 2)2
Standard form of the second-order system:
2 ωn C(s) = 2 2 R(s) s + 2ζωns + ωn
There are two characteristic parameters here: ζ－ the damping ratio(阻尼比) ωn － the undamped natural frequency (无阻尼自然频率)
c(t ) = 1−
e−ζωnt 1−ζ 2
sin (ωd t +φ ) tan φ = 1−ζ 2
ωd = ωn 1−ζ 2
ζ
The characteristic equation and the roots
s + 2ζωns + ω = 0
2 2 n
s1,2 = −ζωn ± ωn ζ 2 −1
The step response for the four cases of ζ
When ζ=0 —the undamped case