现代控制工程-第十四章

ω=+ ∞
ω=0
Example 7.5 （2 order type I system）Please discuss system） the stability of the CL system.
R(s) + -
K ) s(τs +1
Y(s)
consider the upper imaginary axis, the mapping contour is the polar plot as:
| G( jω) |=1 ∠G( jω) = −180o
-1
0
1
2. System margins In frequency domain, stable system has 2 ways to collapse to unstable. | G( jω) |=1 Consider point a ∠G( jω) ≠ −180o If a rotates clockwise the angle r , then | G( jω) |=1, ∠G( jω) = −180o Consider point b
i =1 M
=0
F( s) = 1 + L( s) = 1 + G( s)H( s) = 0
where L(s) = G(s)H(s) is the corresponding openopenloop transfer function.
Nyquist criterion can be stated as: 1. if the open-loop transfer function has no poles in the right half of the s-plane, and the mapped contour ΓL does not encircle the point (−1, jo) , the closed-loop system will be stable. 2. if the open-loop transfer function has P poles in the right half of the s-plane, and the mapped contour ΓL encircle the point (−1, jo) P times in the counterclockwise direction, the closed-loop system will be stable.
K G(s) = s(s +1)(0.1s +1)
γ (ωcm ) =ϕ (ωcm ) +1800 =120 (K = 5);
−120 (K = 20)
ϕ(ω) = −900 −tg−1ω −tg−10.1 = −1800 ω
Kg = −L(ωcp ) = 6db( K = 5) , −6db( K = 20)
∠G( jω) = −180o | G( jω) |≠1
r
j
1/h
-1
b a
0
1
If b shift leftward to -1, then | G( jω) |=1, ∠G( jω) = −180o
a. Phase crossing frequency ωcp and the gain margin Kg Phase crossing frequency ωcp
denote when
φ( jω) as the CL frequency characteristics
ω > ωb
20lg φ( jωb ) < 20lg φ( jo) −3
For RC network,
1 20lg φ( jωb ) = 20lg = 20lg 2 2 2 1+T ωb
1 The bandwidth is ωb = ，it is the low pass filter. T
Kg, Gain margin in dB
phase crossing frequency ωcp
b. Magnitude crossing frequency ωcm and the phase margin φm Unity circle Magnitude crossing frequency ωcm
Example 7.1 the transfer function of unit negative feedback system is Determine the gain margin and phase margin when K=5 and K=20 . Hints：determine the crossing frequencies （0dB） and （-180 degrees）first. (ω) = 20lg K − 20lg ω − 20lg 1+ω2 − 20lg 1+ (0.1ω)2 = 0 L
∠GH( jωcp ) = −180
gain margin
1 Kg = GH( jωcp )
Kg >1
20lg(Kg ) = −L(ωcp ) > 0
(-1,0)
GH ( jωcp )
We need
GH ( jωcp ) <1
else, the system is unstable.
Gain margin on Bode diagram
So, the bandwidth is proportional to the nature frequency(+), and increase when ξ decrease. It is favorable to the swiftness of the system response. This is also true for most systems.
resonant frequency ωr Maximum magnitude occurs at resonant frequency when ξ < 0.707 in the underdamped systems.
(2) Maximum magnitude
Mr and
Mr =
1 2 ξ 1−ξ
7.2 Nyquist Criterion
The start point to determine the stability is the characteristic equation of the control system
F(s) =
or
k∏(s + si )
n
∏(s + s )
k =1 k
2
the resonant frequency is:
ωr = ωn 1− 2ξ
2
the magnitude crossing frequency, phase margin can be related with time domain parameters ωn and ξ as:
G( jωc ) = 1
GH( jωcm) = 1
Phase margin
φm = ∠GH( jωcm) − −180
= 180 + ∠GH( jωcm )
We need
(
)
(-1,0)
φm
பைடு நூலகம்
φm > 0
else, the system is unstable.
Phase margin on Bode diagram
Magnitude crossing frequency ωcm Phase margin
Hope: establish exact relationship between methods and indices in the time and frequency domain.
In the case of underdamped systems, we can try to make it. In general situation, only empirical formula are available.
1
For underdamped system thus
2 ωn φ(s) = 2 2 S + 2ξωns +ωn
2 2 ωn 2 ωb (1− 2 ) + 4ξ 2 2 = 2 ωn ωn
And we have
ωb = ωn (1− 2ξ 2 ) + (1− 2ξ 2 )2 +1

1 2
1. Marginally stable in frequency domain
Minimum phase system is the system without OL pole in the right half s plain, or p=0 for Nyquist criterion.
j
Thus, when the minimum phase system is marginal stable, the polar plot G(jw) will cross the point (-1,j0), and satisfies
Chapter 7 Stability in the Frequency Domain
7.1 Mapping Contours in the s-plane 7.2 The Nyquist Criterion 7.3 Relative Stability and the Nyquist Criterion 7.4 Performance Indices in the Frequency Domain
(-1,0)
ω=+ ∞
−Kτ1τ2 τ1 +τ2