Bài giảng Control system design - Chapter IX: Stability in the frequency domain - Nguyễn Công Phương
Stability in the Frequency Domain
1. Mapping Contours in the s – Plane
2. The Nyquist Criterion
3. Relative Stability and the Nyquist Criterion
4. Time – Domain Performance Criteria in the
Frequency Domain
5. System Bandwidth
6. The Stability of Control Systems with Time
Delays
7. PID Controllers in the Frequency Domain
8. Stability in the Frequency Domain Using
Control Design Software
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Nguyễn Công Phương
CONTROL SYSTEM DESIGN
Stability in the Frequency Domain
Contents
I. Introduction
II. Mathematical Models of Systems
III. State Variable Models
IV. Feedback Control System Characteristics
V. The Performance of Feedback Control Systems
VI. The Stability of Linear Feedback Systems
VII. The Root Locus Method
VIII.Frequency Response Methods
IX. Stability in the Frequency Domain
X. The Design of Feedback Control Systems
XI. The Design of State Variable Feedback Systems
XII. Robust Control Systems
XIII.Digital Control Systems
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Stability in the Frequency Domain
1. Mapping Contours in the s – Plane
2. The Nyquist Criterion
3. Relative Stability and the Nyquist Criterion
4. Time – Domain Performance Criteria in the
Frequency Domain
5. System Bandwidth
6. The Stability of Control Systems with Time
Delays
7. PID Controllers in the Frequency Domain
8. Stability in the Frequency Domain Using
Control Design Software
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Mapping Contours
Ex. 1 in the s – Plane (1)
jω jv
s – plane F( s ) F(s) – plane
j2 D j2 A
D A
j1 j1
−2 −1 0 1 2 σ −2 −1 0 1 2 u
− j1 − j1
C B
− j2 − j2 B
C
u =2σ + 1
F() s= 2 s + 1 =2(σ +j ω )1 + =(2σ + 1) + j 2 ω =u + jv →
v = 2ω
=+ =σ + ω → =+ =σ ++ ω = ×++ ×=+
As 1 j 1 j AujvF (2 1) j (2 ) (2 1 1)j (2 1) 3 j 2
=− → = ×++ ×− =−
BjBs1 1 F (211) j [2(1)]3 j 2
=−− → = ×−++ ×− =−−
CjCs1 1 F [2(1)1] j [2(1)] 1 j 2
=−+ → = ×−++ ×=−+
DjDs1 1 F [2(1)1] j (21) 1 j 2
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Mapping Contours
Ex. 2 in the s – Plane (2)
s
F( s ) =
s + 2
s-plane F(s)-plane
1.5 1.5
D A D
1 1
0.5 0.5
A
ω
0 jv 0
j
B
-0.5 -0.5
-1 -1
C B C
-1.5 -1.5
-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5
σ u
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Mapping Contours
in the s – Plane (3)
s-plane F(s)-plane
2 2
1 F() s= 2 s + 1 1
ω
jv
j 0 0
-1 -1
-2 -2
Cauchy’s -2theorem-1 : 0If a contour1 2 Γ3s in the s-plane encircles-2 -1 Z zeros0 and1 P2 poles3 of F(s)
and does not pass throughσ any poles or zeros of F(s) and the traversalu is in the
clockwise direction s-planealong the contour, the corresponding contourF(s)-plane ΓF in the F(s)-plane
encircles1.5 the origin of the F(s)-plane N = Z – P1.5times in the clockwise direction.
1 1
0.5 0.5
ω
jv
j 0 0
-0.5 s-0.5
F( s ) =
-1 s +-12
-1.5 -1.5
-2 -1 0 1 -2 -1 0 1
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Mapping Contours
Ex. 3 in the s – Plane (4)
s
F( s ) =
+
s 0.5 s-plane F(s)-plane
6 0.15
4 0.1
2 0.05
ω
jv
j 0 0
-2 -0.05
-4 -0.1
-6
-6 -4 -2 0 2 4 6 0.9 0.95 1 1.05 1.1 1.15
σ u
If a contour Γs in the s-plane encircles Z zeros and P poles of F(s) and does not pass
through any poles or zeros of F(s) and the traversal is in the clockwise direction along
the contour, the corresponding contour ΓF in the F(s)-plane encircles the origin of the
F(s)-plane N = Z – P times in the clockwise direction.
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Mapping Contours
Ex. 2 in the s – Plane (5)
s s+ z
F( s ) = = =Fs() ∠= Fs () Fs ()( ∠−φ φ )
s + 2 s+ p z p
If a contour Γs in the s-plane encircles Z zeros and P poles of F(s) and does not pass
through any poles or zeros of F(s) and the traversal is in the clockwise direction along
the contour, the corresponding contour ΓF in the F(s)-plane encircles the origin of the
F(s)-plane N = Z – P times in the clockwise direction.
s-plane F(s)-plane
1.5 1.5
1 1
φ φ− φ
0.5 z 0.5 z p
ω
0 jv 0
j φ
p
-0.5 -0.5
-1 -1
-1.5 -1.5
-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5
σ sites.google.com/site/ncpdhbkhn u 8
Stability in the Frequency Domain
1. Mapping Contours in the s – Plane
2. The Nyquist Criterion
3. Relative Stability and the Nyquist Criterion
4. Time – Domain Performance Criteria in the
Frequency Domain
5. System Bandwidth
6. The Stability of Control Systems with Time
Delays
7. PID Controllers in the Frequency Domain
8. Stability in the Frequency Domain Using
Control Design Software
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The Nyquist Criterion (1)
• F(s) = 1 + L(s) = 0
• A feedback system is stable if and only
jω s – plane
if the contour ΓL in the L(s) – plane
does not encircle the (–1, 0) point when
the number of poles of L(s) in the right r → ∞
– hand s – plane is zero ( P = 0).
0 σ
• (when the number of poles of L (s) in the Γ
right – hand s – plane is other than s
zero ) A feedback system is stable if and
only if, for the contour ΓL , the number Nyquist contour
of counterclockwise encirclements of
the (–1, 0) point is equal to the number
of poles of L(s) with positive real parts.
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The Nyquist Criterion (2)
Ex. 1
1 1 R( s ) 1 1 Y( s )
.
τs+1 τ s + 1 τ s +1 τ s + 1
T( s ) = 1 2 (− ) 1 2
1 1
1+ K .
τ+ τ +
1s1 2 s 1 K
1 1 Γ 1 1
→+A feedback1K system . is stable ==+ if 01() and only L if s the→ contourL( s ) = L Kin the L(s) . – plane does not encircle
the (–1, 0)τ point+ when τ the + number of poles of L(s) in the rightτ –+ hand τ s – + plane is zero ( P = 0).
1s1 2 s 1 1s1 2 s 1
= ω
As j ω→∞
jω jv
A= L( j ω ) = 0 s – plane
L ω→∞ A
= σ
Bs σ →∞ ω → ∞
r → ∞
B= L (σ ) = 0
L σ →∞ D B A, B,C D
= ω 0 σ 0 u
Cs j ω→−∞ − 1 − 1
τ τ ω = 0
=ω = 1 2
CL L( j )ω→−∞ 0
Γ
= + C s
Ds 0 j 0
= =
DL L(0) K L(s) – plane
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The Nyquist Criterion (3)
Ex. 2
jω
K s – plane
L( s ) = D
s(τ s + 1)
C r → ∞
A= j ω B
s ω<0, ω → 0 E
0 σ
− 1 ε
τ
A= L( j ω ) A
L ω<0, ω → 0
Γ
F s
= K
ω τω +
(j )( j 1) ω<0, ω → 0
Kτ K
= − − j =−Kτ + j ∞
τω22+1 ωτω ( 22 + 1)
ω<0, ω → 0
Cauchy’s theorem : If a contour Γs in the s-plane encircles Z zeros and P poles of L(s)
and does not pass through any poles or zeros of L(s) and the traversal is in the
clockwise direction along the contour, the corresponding contour ΓF in the L(s)-plane
encircles the origin of the L(s)-plane N = Z – P times in the clockwise direction.
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The Nyquist Criterion (4) K = 2, τ = 1
Ex. 2
ω 20
j A
K s – plane 15
L( s ) = D
s(τ s + 1) 10
C r → ∞
A=− Kτ + j ∞ 5
L B E
jv 0
0 σ D, F B
− 1 ε
B = σ τ
s σ →0 A -5
Γ -10
K F s
B= L (σ ) = -15
L σ →0 σ( σ + 1) C
σ →0 -20
= ∞ 0 10 20
K u
C= j ω →=CLj(ω ) = =−−∞ Kj τ
s ω>0, ω → 0 L ω>0, ω → 0 ω τω +
(j )( j 1) ω>0, ω → 0
K
D= j ω →=D L( j ω ) = = 0
s ω→∞ L ω→∞ ω τω +
(j )( j 1) ω→∞
K
F= j ω →=F L( j ω ) = = 0
s ω→−∞ L ω→−∞ ω τω +
(j )( j 1) ω→−∞
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The Nyquist Criterion (5)
jω jv
K s – plane
L( s ) = A
(τs+ 1)( τ s + 1) ω → ∞
1 2 r → ∞
D B A, B,C D
• It is sufficient to 0 σ 0 u
− 1 − 1
construct the contour τ τ ω = 0
Γ for the frequency 1 2
L Γ
range 0< ω<∞ in order C s
to investigate the τ
stability. K = 2, = 1
jω 20
s – plane A
D 15
• The magnitude of L(s) 10
as s = re jϕ and r →∞ C r → ∞
5
will normally approach B E
zero or a constant. jv 0 D, F B
− 1 0 ε σ
τ -5
A
-10
K Γ
L( s ) = F s -15
s(τ s + 1) C
-20
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u
jω
s – plane
The Nyquist Criterion (6) D
Ex. 3 C r → ∞
B
= K E
L( s ) 0 σ
τ+ τ + − 1 ε
s(1 s 1)( 2 s 1) τ
A
K
L( j ω ) =
ω ωτ+ ωτ + Γ
j( j1 1)( j 2 1) F s
−+−ττ ω − ωττ2
= K(12 ) jK (1/ )(1 12 )
+ωτ22 + τ 2 + ωττ 422
1 (1 2 ) 12
K − −
= ∠−−−[tan(1ωτ ) tan( 1 ωτ ) π /2]
ωτ4+ τ 22 + ω − ωττ 22 1 2
(12 ) (1 12 )
=→==ω ω K ∠− π
Cjsω> ω → C L lim()lim Lj (/2)
0, 0 ω→0 ω → 0 ωτ4+ τ 22 + ω − ωττ 22
(12 ) (1 12 )
K
Dj=ω →= Dlim( Lj ω ) = lim ∠− (3/2) π
sω→∞ L ω→∞ ω →∞ τ τ ω 3
1 2
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jω
s – plane
The Nyquist Criterion (7) D
Ex. 3 C r → ∞
B
= K E
L( s ) 0 σ
τ+ τ + − 1 ε
s(1 s 1)( 2 s 1) τ
K K A
C=∠−(π /2); D =∠− (3/2) π
L L ∞ Γ
0 F s
−ττ + ωωττ − 2
ω =K(1 2 ) − K (1/ )(11 2 )
L( j ) 22 2 422 j 22 2 422
+ωτ ++ τ ωττ + ωτ ++ τ ωττ K = 1, τ = 5, τ = 5
1()12 12 1() 12 12 1 2
6
K(1/ω )(1− ωττ2 )
Im{()}0L j ω = →1 2 = 0
+ωτ22 + τ 2 + ωττ 422 4
1 (1 2 ) 12
− τ τ 2
→ω = 1 → = K 1 2
Re(L ) ω= 1
τ τ τ τ τ+ τ
1 2 1 2 1 2
jv 0
-2
−Kττ ττ +
→12 ≥−→1 K ≤ 12
ττ+ ττ -4
12 12
-6
-6 -4 -2 0 2 4
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The Nyquist Criterion (8)
Ex. 3
K τ+ τ
L( s ) = K ≤ 1 2
τ+ τ + τ τ
s(1 s 1)( 2 s 1) 1 2
K = 1, τ = 1, τ = 1 K = 2, τ = 1, τ = 1 K = 3, τ = 1, τ = 1
1 2 1 2 1 2
1.5 1.5 1.5
1 1 1
0.5 0.5 0.5
jv
jv 0 jv 0 0
-0.5 -0.5 -0.5
-1 -1 -1
-1.5 -1.5 -1.5
-2 -1 0 1 -2 -1 0 1 -2 -1 0 1
u u u
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jω
s – plane
The Nyquist Criterion (9) D
Ex. 4 C r → ∞
K B E
L( s ) = 0 σ
2 τ + − 1 ε
s( s 1) τ
A
K K −
L() j ω= = ∠−− [tan()] π1 ωτ
2 4 2 6 Γ
−ω(j ωτ + 1) ω+ τ ω F s
C= j ω
s ω>0, ω → 0 2
→=ω =K ∠− π 1.5
CL lim( L j ) lim ( )
ω→0 ω → 0 ω2
1
= ω 0.5
Ds j ω→∞
jv 0
→=ω =K ∠− π
DL lim L ( j ) lim3 (3/2)
ω→∞ ω →∞ τω -0.5
-1
B= ε e jφ
s ε→0
-1.5
→ =ε jφ = K −2 j φ
BL lim Le ( ) lim e -2
ε→0 ε → 0 ε 2 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2
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The Nyquist Criterion (10)
Ex. 5 2.5
jω 2
Y( s ) s – plane
R( s ) 1 1 1.5
K1 D
s −1 s 1
−
( ) C r → ∞ 0.5
B E jv 0
0 σ -0.5
ε 1
K -1 K1
L( s ) = 1 A
s( s − 1) -1.5
Γ -2
F s
-2.5
-3 -2 -1 0
u
Ex. 6
jω 2
Y( s ) s – plane
R( s ) 1 1 D 1.5
K1
− 1
(− ) s 1 s
− r → ∞
( ) K C 0.5
2 B
E jv 0
− 1 0 1 σ -0.5
K2 ε
K( K s + 1) -1
L( s ) = 1 2 A
− -1.5
s( s 1) Γ
F s -2
-5 -4 -3 -2 -1 0
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The Nyquist Criterion (11)
Ex. 6
jω 2
Y( s ) s – plane
R( s ) 1 1 D 1.5
K1
− 1
(− ) s 1 s
− r → ∞
( ) K C 0.5
2 B
E jv 0
− 1 0 1 σ -0.5
K2 ε
K( K s + 1) -1
L( s ) = 1 2 A
− -1.5
s( s 1) Γ
F s -2
-5 -4 -3 -2 -1 0
u
KKj(ω+− 1) KK ω2 ( + 1) K ωω (1 − K 2 )
→=L( jω ) 12 = 12 + j 12
jωω( j − 1) ωω42 + ωω 42 +
Kω(1− K ω 2 ) 1
Im{()}0L j ω=→1 2 =→= 0 ω 2
ω4+ ω 2
K2
−ω2 +
ω =K1( K 2 1) = −
Re{Lj ( )} ω2 = 4 2 KK1 2
1/ K2 ω+ ω
ω2 =
1/ K2
− →=+=−+=→
IfKK12 1 KK 12 1 ZNP 110stable
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The Nyquist Criterion (12)
Ex. 6
jω 2
Y( s ) s – plane
R( s ) 1 1 D 1.5
K1
− 1
(− ) s 1 s
− r → ∞
( ) K C 0.5
2 B
E jv 0
− 1 0 1 σ -0.5
K2 ε
> A -1
K1 K 2 1 -1.5
Γ
F s -2
-5 -4 -3 -2 -1 0
u
K = 0.5, K = 3, K K = 1.5 K = 0.5, K = 2, K K = 1 K = 0.5, K = 1, K K = 0.5
1 2 1 2 1 2 1 2 1 2 1 2
2 2 2
1 1 1
jv
jv 0 jv 0 0
-1 -1 -1
-2 -2 -2
-4 -2 0 -4 -2 0 -4 -2 0
u u
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Stability in the Frequency Domain
1. Mapping Contours in the s – Plane
2. The Nyquist Criterion
3. Relative Stability and the Nyquist Criterion
4. Time – Domain Performance Criteria in the
Frequency Domain
5. System Bandwidth
6. The Stability of Control Systems with Time
Delays
7. PID Controllers in the Frequency Domain
8. Stability in the Frequency Domain Using
Control Design Software
sites.google.com/site/ncpdhbkhn 22
Relative Stability
Ex. 1 and the Nyquist Criterion (1)
τ = 1; τ = 1
1 2
K 0.5
L( s ) =
τ+ τ +
s(1 s 1)( 2 s 1)
K
L( j ω ) =
ωτω+ τ ω +
j(1 j 1)( 2 j 1) 0
−ω2 τ + τ
= K (1 2 )
ωτ4(+ τ ) 22 + ω (1 − ττω 22 ) jv τ τ
12 12 K 1 2
2
Kω(1− ττω ) -0.5 τ+ τ
− j 1 2 1 2
ωτ4+ τ 22 + ω − ττω 22
(12 ) (1 12 )
K = 2.5
2
ω− ττω K = 1.2
ω = →K (11 2 ) =
Im{L ( j )} 04 22 22 0 K = 0.3
ωτ(+ τ ) + ω (1 − ττω ) -1
12 12 -1.5 -1 -0.5 0
u
→ω = 1
τ τ
1 2
−Kωττ2 ( + ) − K ττ
ω =12 = 12
Re{L ( j )} ω= τ τ 4 22 22
1/ 1 2 ωττ++− ω ττω ττ +
(12 ) (1 12 ) ω= τ τ 12
1/ 1 2
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Relative Stability
and the Nyquist Criterion (2)
τ = 1; τ = 1
1 2
0.5
Phase difference
before instability
0
α
jv d
= = 1
Gain marginGM 20log
d -0.5
K = 2.5
K = 1.2
= Φ = α K = 0.3
Phase margin M -1
-1.5 -1 -0.5 0
u
Gain difference before instability
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Relative Stability
Ex. 2 and the Nyquist Criterion (3)
6 0.5
Given L( s ) =
(s+ 2)( s2 + 2 s + 2) Phase
Find its gain & phase margin? difference
before
6 0
L( j ω ) = instability α
(jω+ 2)( −+ ω2 2 j ω + 2)
jv
−ω 2 d
= 24(1 )
22 2 22
16(1−ω ) + ω (6 − ω ) -0.5
6(6ω− ω 2 )
− j
16(1−ω22 ) + ω 2 (6 − ω 22 )
2
6(6ω− ω ) -1
Im{L }= =→= 0ω 2.45 rad/s -1.5 -1 -0.5 0
16(1−ω22 ) + ω 2 (6 − ω 22 ) u
Gain difference before instability
24(1−ω 2 )
d=Re{ L } = = 0.3
ω=2.45 −ω22 + ω 2 − ω 22
16(1 ) (6 ) ω =2.45
1 1
Gain margin==G 20log = 20log = 10.46 dB
M d 0.3
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Relative Stability
Ex. 2 and the Nyquist Criterion (4)
6 0.5
Given L( s ) =
(s+ 2)( s2 + 2 s + 2) Phase
Find its gain & phase margin? difference
before
ω = 0
L( j ) 1 instability α
2
24(1−ω 2 )
→ jv
22 2 22 d
16(1−ω ) + ω (6 − ω )
2 -0.5
6(6ω− ω 2 )
+ = 1 →ω = 1.253 rad/s
16(1−ω22 ) + ω 2 (6 − ω 22 )
→ ∠ω = Im{L }
L( j )ω = atan
1.253 -1
Re{L } ω=1.253 -1.5 -1 -0.5 0
u
6(6ω− ω 2 ) Gain difference before instability
16(1−ω22 ) + ω 2 (6 − ω 22 )
=atan = − 112.3 o
24(1−ω 2 )
−ω22 + ω 2 − ω 22
16(1 ) (6 ) ω=1.253
→ =Φ==−α o −− o = o
PhasemarginM 112.3 ( 180) 67.7
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Relative Stability
and the Nyquist Criterion (5)
τ = 1; τ = 1
1 2
0.5
= = 1
Gain marginGM 20log Phase
d difference
The gain margin is the increase in the before 0
system gain when phase = –180 ° that instability α
will result in a marginally stable jv d
system with intersection of the
–1 + j0 point on the Nyquist diagram. -0.5
K = 2.5
K = 1.2
K = 0.3
Phase margin = Φ = α -1
M -1.5 -1 -0.5 0
u
The phase margin is the amount of phase Gain difference before instability
shift of the L(jωt) at unity magnitude that
will result in a marginally stable system
with intersection of the – 1 + j0 point on
the Nyquist diagram.
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Relative Stability
Ex. 2 and the Nyquist Criterion (6)
6 Gain margin=G = 10.46 dB,ω = 2.45 rad/s
Given L( s ) = M
(s+ 2)( s2 + 2 s + 2)
Phasemargin=Φ= 67.7,o ω = 1.253rad/s
Find its gain & phase margin? M
Bode Diagram
5
0
-5
GM
-10
Magnitude (dB) Magnitude -15
-20
-25
-45 1 1.5 2 2.5 3 3.5 4
-90
-135 Φ
M
Phase (deg) Phase
-180
-225
1 1.5 2 2.5 3 3.5 4
Frequency (rad/s)
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Relative Stability
Ex. 3 and the Nyquist Criterion (7)
1 1
Compare Ls()= &() Ls = .
1 ss(+ 1)(0.2 s + 1)2 ss ( + 1) 2
Bode Diagram
10
0
GM 2
-10 GM 1
-20
Magnitude (dB) Magnitude
> -30
GM1 G M 2
Φ > Φ -40
M1 M 2 -90
L
1
L
-135 2
Φ
M 1
-180 Φ
M 2
Phase (deg) Phase
-225
-270
0.5 0.6 0.7 0.8 0.9 1 2 3 4
Frequency (rad/s)
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Relative Stability
and the Nyquist Criterion (8)
ω 2 ω2 ω
L( j ω ) = n =n ω2 +4 ζ 22 ω ∠−− 90 o atan
ω ω+ ζω ωn ζω
j( j 2n ) 2 n
ω
o
→Φ =−90 − atan 1
M ζω
2 n Exact
0.9 Linear approximation
ω 2
n ω2+ ζ 2 ω 2 = 0.8
c4 n 1
ω 0.7
c
ζ
ω2 0.6
c 4 2
→ =4ζ +− 1 2 ζ 0.5
ω2
n 0.4
Damping ratio, Damping
→ Φ = 2 0.3
M atan
4+1/ζ 4 − 2 0.2
0.1
0
ζ= Φ ζ ≤ 0 10 20 30 40 50 60 70 80 90 100
0.01 , 0.7 Phase margin, Φ ( o )
M M
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Stability in the Frequency Domain
1. Mapping Contours in the s – Plane
2. The Nyquist Criterion
3. Relative Stability and the Nyquist Criterion
4. Time – Domain Performance Criteria in the
Frequency Domain
5. System Bandwidth
6. The Stability of Control Systems with Time
Delays
7. PID Controllers in the Frequency Domain
8. Stability in the Frequency Domain Using
Control Design Software
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Time – Domain Performance Criteria
in the Frequency Domain (1)
R( s ) Y( s )
Y( j ω ) G( jω )( Gj ω ) Gc ( s ) G( s )
T( j ω ) = = c
ω+ ω ω (− )
Rj()1 GjGjc ()()
φ ω
= M(ω ) e j ( ) H( s )
ω ω = +
Gjc ( )( Gj ) u jv
ω ω 1.5
Gc ( j )( Gj ) = M = 0.7
→M (ω ) = M 1.5 M =1
+ ω ω 1
1Gc ( j )( Gj )
ujv+ uv2 + 2 0.5
= = =
1++ujv (1 ++ u ) 2 v 2 0 M 2 M = 0.5
-0.5
-1
2 2 2
M 2 M
→−u += v -1.5
1−M2 1 − M 2 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5
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Time – Domain Performance Criteria
in the Frequency Domain (2)
1 4.5
K
M 1
1 K
M 2 2
4
0.5
3.5
0 M1 3
2.5
jv -0.5
Magnitude 2
M2
-1 1.5
K> K
1 2 1
-1.5
0.5
K1
K2
-2 0
-2.5 -2 -1.5 -1 -0.5 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
u ω
Polar plot of Frequency response of
ω ω
ωω= ω = ω Gc ( j )( Gj )
GjGjc ()() Lj () KPj () T( j ω ) =
1+ G ( jω )( Gj ω )
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Time – Domain Performance Criteria
in the Frequency Domain (3)
6 R( s ) Y( s )
φ = o
10 Gc ( s ) G( s )
(− )
4
φ = o
20 H( s )
2
Y( j ω ) G( jω )( Gj ω ) φ ω
Tj()ω= =c = Me () ω j ( )
o ω+ ω ω
φ = 30 Rj()1 GjGjc ()()
1.5
0 = M = 0.7
M 1.5 M =1
φ = − 30 o 1
-2 0.5
0 M = 2 M = 0.5
φ = − 20 o
-4 -0.5
-1
φ = − 10 o
-6 -1.5
-4 -3 -2 -1 0 1 2 3 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5
2 2 2 2 2 2
1 111 M 2 M
u+ +− v = 1 + u− + v =
2 2φ 4 φ 2 1−M2 1 − M 2
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Time – Domain Performance Criteria
Ex. in the Frequency Domain (4)
R( s ) Y( s )
=0.47 = G( s ) G( s )
GsGsc ()() ,()1 Hs c
s( s2 + s + 1) (− )
H( s )
Nichols Chart
20
1 dB
15
3 dB
10
6 dB
5
0
-5
Open-Loop Gain (dB) GainOpen-Loop
-10
-15
-20
-270 -225 -180 -135 -90
Open-Loop Phase (deg)
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Stability in the Frequency Domain
1. Mapping Contours in the s – Plane
2. The Nyquist Criterion
3. Relative Stability and the Nyquist Criterion
4. Time – Domain Performance Criteria in the
Frequency Domain
5. System Bandwidth
6. The Stability of Control Systems with Time
Delays
7. PID Controllers in the Frequency Domain
8. Stability in the Frequency Domain Using
Control Design Software
sites.google.com/site/ncpdhbkhn 36
System Bandwidth (1)
Ex. 1
0
T
=1 = 1 1
Ts1() , Ts 2 () T
s+1 5 s + 1 -3 2
-6
-8
-10
20log|T|, dB 20log|T|,
-18
0 0.2 1 2
ω
Step Response
1
0.8
0.6
Amplitude 0.4
T
1
0.2 T
2
0
0 2 4 6 8 10
Time (seconds)
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System Bandwidth (2)
Ex. 2
100 900
Ts()= ,() Ts =
1ss2++10 100 2 ss 2 ++ 30 900
10
Step Response
T 1.2
1
T
T 1
2
0 T
2
-3
1
-10
0.8
-20
-30
0.6
20log|T|, dB 20log|T|,
Amplitude
-40
0.4
-50
0.2
-60
-70 0
0 1 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6
10 10 10
ω sites.google.com/site/ncpdhbkhn Time (seconds) 38
Stability in the Frequency Domain
1. Mapping Contours in the s – Plane
2. The Nyquist Criterion
3. Relative Stability and the Nyquist Criterion
4. Time – Domain Performance Criteria in the
Frequency Domain
5. System Bandwidth
6. The Stability of Control Systems with Time
Delays
7. PID Controllers in the Frequency Domain
8. Stability in the Frequency Domain Using
Control Design Software
sites.google.com/site/ncpdhbkhn 39
The Stability of Control Systems
with Time Delays (1)
• Time delay : the time interval between the start
of an even at one point in a system and its
resulting action at another point in the system.
• A pure time delay, without attenuation:
–sT
Gd(s) = e
• A pure time delay does not change the
magnitude of the transfer function.
• The Nyquist criterion remains valid for a
system with a time delay.
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The Stability of Control Systems
T
Ex. with Time Delays (2) −s + 1
e−sT ≈ 2
31.5 −
T( s ) = e s T
+ +2 ++ s +1
(s 1)(30 ss 1)[( / 3) s / 3 1] 2
6
4
2
0
-2
20log10|T|, dB 20log10|T|,
-4
-6
-0.3 -0.2 -0.1 0 0.1
10 10 10 10 10
ω
-100
With time delay
Without time delay
-150
)
o
( -180 Φ
φ without time delay
-200
Φ
with time delay
-250
-0.3 -0.2 -0.1 0 0.1
10 10 10 10 10
ω
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Stability in the Frequency Domain
1. Mapping Contours in the s – Plane
2. The Nyquist Criterion
3. Relative Stability and the Nyquist Criterion
4. Time – Domain Performance Criteria in the
Frequency Domain
5. System Bandwidth
6. The Stability of Control Systems with Time
Delays
7. PID Controllers in the Frequency Domain
8. Stability in the Frequency Domain Using
Control Design Software
sites.google.com/site/ncpdhbkhn 42
PID Controllers
Ex. in the Frequency Domain (1)
Td ( s )
Kω 2
R( s ) K ng Y( s )
K+I + K s
Ps D (τs+ 1)( s2 + 2 ζω s + ω 2 )
(− ) ng n g
H( s )
K = –7000, τ = 5s. Design the PID controller so that
o o
GM ≥ 6dB, 30 ≤ ΦM ≤ 60 , rise time Tr < 4s, time to peak TP < 10s.
s2 +(/ KKs )(/ + KK )
Ls( ) = Kω 2 K PD ID
ng D ss(τ+ 1)( s2 + 2 ζω s + ω 2 )
ng n g
ζ = Φ= ×o =
0.01M 0.01 30 0.3
2.16ζ + 0.6 2.16 × 0.3 + 0.6
T = = 4ω 0.31
r ω ω n
n n
2.16× 0.3 + 0.6 π π
ω =→=0.4T == 3s, T = = 8s
n r 0.4 P ω− ζ 2 − 2
n 1 0.4 1 0.3
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PID Controllers
Ex. in the Frequency Domain (2)
K = –7000, τ = 5s. Design the PID controller so that
o o
GM ≥ 6dB, 30 ≤ ΦM ≤ 60 , rise time Tr < 4s, time to peak TP < 10s.
s2 +(/ KKs )(/ + KK )
= ω2 PD ID ζ= ω =
Ls( ) Kn K D , 0.3,n 0.4
g ss(τ+ 1)( s2 + 2 ζω s + ω 2 )
ng n g
Bode Diagram
20
10
0
GM
-10
-20
Magnitude (dB) Magnitude
-30
-40
-45
-90
-135
Phase (deg) Phase Φ
-180 M
-225
-1 0 1
10 10 10
Frequency (rad/s)
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Stability in the Frequency Domain
1. Mapping Contours in the s – Plane
2. The Nyquist Criterion
3. Relative Stability and the Nyquist Criterion
4. Time – Domain Performance Criteria in the
Frequency Domain
5. System Bandwidth
6. The Stability of Control Systems with Time
Delays
7. PID Controllers in the Frequency Domain
8. Stability in the Frequency Domain Using
Control Design Software
sites.google.com/site/ncpdhbkhn 45
Stability in the Frequency Domain
Using Control Design Software
• nyquist
• nichols
• margin
• pade
• ngrid
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