Automation Systems Lecture 5 - Stability of linear dynamic systems Jakub Mozaryn

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Automation Systems

Lecture 5 - Stability of linear dynamic systems

Jakub Mozaryn

Institute of Automatic Control and Robotics, Department of Mechatronics, WUT

Warszawa, 2017

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Introduction

Stability

A stable system is a dynamic system witch a bounded response to a bounded input

Stable system, if pushed out a state of equilibrium (considered operating point P) returns to the state of equlibrum (to some state K ) after the termination of factors (disturbaces d ) that pushed the system from a state of equilibrum.

In the case of linear systems, the behavior of the system after the termination of an action, that pushed the system out of the equilibrum position, is characteristic feature of the system and does not depend on the type of the action before it’s termination. (simple analysis) In the case of nonlinear systems, their behavior as effect of an action, that pushed the system out of the equilibrum point, can depend on the type and magnitude of the action before it’s termination. (complicated analysis)

Jakub Mozaryn Automation Systems

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Stability

There are three types of behavior after pushing a system out of the equilibrum point:

1 A system returns to equilibrium state in the operating point prior to action that pushed system out of balance - asymptotical stability,

2 A system returns to equilibrium state in the operationg point other than one present when action pushed system out of balance - non-asymptotical stability, neutral, marginal

3 A system doesn’t reach a state of equilibrym - ustability,

instability; a special case of such behavior are sustained oscillations with constant amplitude - system on the border of stability.

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Impulse response

Impulse response is the simplest case, that allows to determine the stability of the linear time invariant dynamical system (LTI).

Using the following formula

y (t)|_u(t)=δ= g (t) = L⁻¹{G (s)} (1) Some inverse Laplace transforms for impulse responses, useful for further analysis are:

L⁻¹ 1 s

= 1(t) (2)

L⁻¹ 1 s²

= t (3)

L⁻¹

1

s ± α

= e^∓αt (4)

L⁻¹

1

(s ± α)²

= te^∓αt (5)

L⁻¹

As + B s²+ Cs + D

= Ae^C²^tcos(t r

D −C²

4 )+ 2B − AC

√

4D + C²e^C²^tsin(t r

D −C² 4 ) (6) if: C²− 4D < 0.

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Stability

Figure 2 : Impulse response examples for: 1, 2 non-asymptoticaly stable systems, 3, 4 - asymptoticaly stable systems, 5, 6 unstable systems, 7 system on the border of stability (sustained oscillations)

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Stability

Equation of motion of a LTI system:

a_nd⁽ⁿ⁾y (t)

dt⁽ⁿ⁾ + · · · + a₁dy (t)

dt + a₀= b_md^(m)tu(t)

dt^(m) + · · · + b₁du(t)

dt + b₀ (7) Transfer function:

G (s) = Y (s)

U(s) =bms^m+ · · · + b2s²+ b1s + b0

ansⁿ+ · · · + a2s²+ a1s + a0

(8) Impulse response:

y (t)|_u(t)=δ= g (t) = L⁻¹{G (s)} (9)

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Stability - asymptotical stability

Case 1: Characteristic equation has only single roots with nonzero negative real parts.

G (s) = L(s)

a_n(s − s₁)(s − s₂) . . . (s − s_n) (10) After the partial fraction decomposition

G (s) = C1

s − s1

+ C2

s − s2

+ · · · + Cn

s − sn

(11) g (t) = L⁻¹{G (s)} = C1e^s¹^t+ C2e^s²^t+ ... + Cne^sⁿ^t (12)

t→∞lim g (t) = 0 if s1, . . . , sn< 0 (13)

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Stability - asymptotical stability

Figure 3 : Placement of roots of the system on the s-plane

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Stability - asymptotical stability

Case 2: Characteristic equation has one double root with nonzero real part, and the rest of roots are unique with nonzero negative real parts.

G (s) = L(s)

an(s − s1)²(s − s3) . . . (s − sn) (14) After the partial fraction decomposition

G (s) = C1

s − s₁+ C2

(s − s₁)²+ C3

s − s₃+ · · · + Cn

s − s_n (15) g (t) = L⁻¹{G (s)} = C1e^s¹^t+ C₂te^s¹^t+ C₃e^s³^t+ ... + C_ne^sⁿ^t (16)

t→∞lim g (t) = 0 if s₁, . . . , s_n< 0 (17) because, in case of s1< 0, expotential function e^s¹^t decreases faster than coefficient t increases.

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Stability - asymptotical stability

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Stability - neutral

Case 3: Characteristic equation has one zero root, and the rest of roots are unique with nonzero real parts.

G (s) = L(s)

a_n(s)(s − s₂) . . . (s − s_n) (18) After the partial fraction decomposition

G (s) = C1

s + C2

s − s2

+ · · · + Cn

s − sn

(19) g (t) = L⁻¹{G (s)} = C1+ C2e^s²^t+ ... + Cne^sⁿ^t (20)

t→∞lim g (t) = C1 if s2, . . . , sn< 0 (21)

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Stability - asymptotical stability

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Stability - instability

Case 4: Characteristic equation has two (or more) zero roots, and the rest of roots are unique with nonzero real parts.

G (s) = L(s)

an(s)²(s − s3) . . . (s − sn) =C₁ s + C₂

(s)² + C₃ s − s3

+ · · · + C_n s − sn

(22) After the partial fraction decomposition

G (s) = L(s)

an(s)²(s − s3) . . . (s − sn) =C1

s + C2

(s)² + C3

s − s3

+ · · · + Cn

s − sn

(23) g (t) = L⁻¹{G (s)} = C₁+ C₂t + C₃e^s³^t+ ... + C_ne^sⁿ^t (24)

t→∞lim g (t) = ∞ (25)

system is unstable

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Stability - asymptotical stability

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Stability - asymptotical stability

Case 5: Characteristic equation has unique real roots with nonzero parts, and complex roots with negative nonzero imaginary parts.

x1= a + jb, x2= a − jb (26)

G (s) = L(s)

an(s²− 2as + a²+ b²)(s − s3) . . . (s − sn) (27) After the partial fraction decomposition

G (s) = C₁s + C₂

(s²− 2as + a²+ b²)+ C₃ s − s3

+ · · · + C_n s − sn

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g (t) = L⁻¹{G (s)} = C₁e^atcos(bt)+C2+ aC1

b e^atsin(bt)+C₃e^s³^t+...+C_ne^sⁿ^t (29)

t→∞lim g (t) = 0 if a < 0, and Re(s3), . . . , Re(sn) < 0 (30)

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Stability - asymptotical stability

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Stability - sustaining oscillations

Case 6: Characteristic equation has unique real roots with nonzero parts, and complex roots with zero real parts.

x₁= jb, x₂= −jb (31)

G (s) = L(s)

(s²+ b²)(s − s₃) . . . (s − s_n) (32) After the partial fraction decomposition

G (s) = C1s + C2

(s²+ b²)+ C3

s − s3

+ · · · + Cn

s − sn

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g (t) = L⁻¹{G (s)} = C1cos(bt) +C₂

b sin(bt) + C3e^s³^t+ ... + Cne^sⁿ^t (34) If s3, . . . , sn < 0, then system is on the boundary of stability and has sustainig oscillations.

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Stability - asymptotical stability

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Stability

Summary:

System is asymptoticaly stable, if it’s chatacteristic equation has roots with nonzero negative real parts.

System is neutral, if it’s chatacteristic equation has one zero root and the rest of roots with nonzero negative real parts.

System is unstable if it’s chatacteristic equation has more than one zero root or unique roots with positive real parts.

System is on the boundary of stability (generates sustaining oscillations) if it’s chatacteristic equation has one zero root and doesn’t have unique roots with positive real parts, but has complex roots with zero real parts.

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Stability - excercises

Determine the stability of systems described by following transfer functions

G (s) = 1

2s + 1 (35) G (s) = 1

2s − 1 (36)

G (s) = 1

2s²+ 3s + 1 (37) G (s) = kp

1 + 1

T_is

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and system with following block diagram model

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Stability - stability criteria

To determine the stability of linear (LTI) systems there is sufficient knowledge of the distribution of roots of characteristic equation on the complex plane (s-plane).

The problems that arise within this method:

Calculation of roots of higher order algebraic equations is complicated.

Characteristic equation of the system is sometimes unknown.

Other methods of determining stability - the so-called: stability criteria do not require the determination of the roots of the characteristic equation:

analytic criteria (Routh-Hurwitz, Routh), graphic criteria ( Michajlov,Evans-root locus), graphical-analytical criterion: ( Nyquist).

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Routh-Hurwitz criterion

Routh-Hurwitz criterion allows to verify that algebraic equation of any degree has only the roots with negative real parts. The use of

Routh-Hurwitz criterion is limited to LTI systems with the transfer function in the analytical form.

Algebraic equation of the degree n with constant, real coefficients a_nd⁽ⁿ⁾y (t)

dt⁽ⁿ⁾ + a_n−1d⁽ⁿ⁻¹⁾y (t)

dt⁽ⁿ⁻¹⁾ + · · · + a₁dy (t)

dt + a₀ (39) has only the roots with negative real parts, when there are fulfilled folowing conditions (Hurwitz conditions).

1 CONDITION I: All coeficients a0, a1, ..., an, of this equation are non-zero, and have the same signs,

2 CONDITION I: All determinants of principal minors of so-called Hurwitz matrix ∆_n are positive.

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Routh-Hurwitz criterion

Hurwitz amtrix

Hurwitz matrix has following form

∆n=

an−1 an 0 − 0 0 0

an−3 an−2 an−1 − 0 0 0 a_n−5 a_n−4 a_n−3 − 0 0 0

− − − − − − −

0 0 0 − a₂ a₃ a₄

0 0 0 − a₀ a₁ a₂

0 0 0 − 0 0 a₀

_n×n

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Routh-Hurwitz criterion

Example: determine Hurwitz matrix for the 4th degree equation

a₄s⁴+ a₃s³+ a₂s²+ a₁s + a₀= 0 (41)

∆₄=

a₃ a₄ 0 0 a1 a2 a3 a4

0 a0 a1 a2

0 0 0 a0

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It’s primary determinants are:

∆₂=

a3 a1

a4 a2

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∆₃=

a₃ a₁ 0 a₄ a₂ a₀

0 a₃ a₁

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Routh-Hurwitz criterion - example H1

Determine the value of gain kp that will ensure stable work of the system.

Gs =

1 (Ts+1)⁴

1 +_(TS+1)¹ 4

= 1

(Ts + 1)⁴+ k_p (45)

Characteristic equation:

(Ts + 1)⁴+ kp= 0 (46)

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Routh-Hurwitz criterion - example H1

Characteristic equation:

(Ts + 1)⁴+ kp= 0 (47)

therefore

T⁴s⁴+ 4T³s³+ 6T²s²+ 4Ts + 1 + k_p= 0 (48) a₄= T⁴, a₃= 4T³, a₂= 6T², a₁= 4T , a₀= 1 + k_p (49) I. Hurwitz condition will be fulfilled if

a0= 1 + kp> 0, which gives kp> −1 (50)

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Routh-Hurwitz criterion - example H1

Hurwitz matrix

∆4=

a3 a4 0 0 a1 a2 a3 a4

0 a0 a1 a2

0 0 0 a0

=

4T³ 4T 0 0

4T 6T² 4T³ T⁴

0 1 + kp 4T 6T²

0 0 0 1 + kp

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II. Hurwitz condition will be fulfilled if det(∆2) = det

4T³ 4T 4T 6T²

> 0 (52)

det(∆₃) = det





4T³ 4T 0

4T 6T² 4T³ 0 1 + k_p 4T



> 0 (53)

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Routh-Hurwitz criterion - example H1

det(∆2) = 24T⁵− 4T⁵= 20T⁵> 0 (54) det(∆3) = 96T⁶− 16T⁶− 16T⁶kp− 16T⁶= 64T⁶− 16T⁶kp> 0 (55) therefore, from II Hurwitz condition

k_p< 4 (56)

I. and II. Hurwitz conditions will be fulfilled if

−1 < kp< 4 (57)

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Routh-Hurwitz criterion

Remaks to Routh-Hurwitz criterion

REMARK 1: Nonasymptotic stability occurs when in the

characteristic equation of degree n coefficient a₀= 0 (equation has one zero root), while the remaining coefficients are positive.

After dividing both sides of the equation by s yields the equation of order n − 1, for which the Routh-Hurwitz criterion should be applied, in order to check the sign of the other elements.

If this equation satisfies the Hurwitz conditions it will mean that the system has one zero root and the remaining roots have negative real parts and the system is nonasymptoticaly stable.

REMARK 2: Routh-Hurwitz criterion can’t be used to determine the stability of system with delays.

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Nyquist criterion

Nyquist criterion enables to determine the stability of a closed system based on the frequency characteristics of the open system.

Closed system transfer function

GZ(s) = G1(s)

1 + G₁(s)G₂(s) (58)

Open system transfer function

G0(s) = G1(s)G2(s) (59)

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Simplified Nyquista criterion

In the case when the characteristic equation of the open system does not have roots with positive real parts (may have any number of zero value roots), a closed system is stable where amplitude-phase characteristic of open system does not include point {−1, j 0}.

’does not include’ means, that moving along the characteristics toward higher frequencies, point {−1, j 0} stays to the left side of the

characteristics.

REMARK: Simplified Nyquist criterion doesn’t include cases when characteristic equation of open system, besides negative or zero value roots, has roots with positive real parts.

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Nyquist criterion

Nyquist criterion properties

frequency response of the open system, on the basis of which stability of the closed system is determined, can be easily determined analytically or experimentally.

criterion allows not only a stability check, but also allows designing the system with specified dynamic properties,

criterion allows stability check of systems with delays.

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Nyquist criterion

Figure 9 : The amplitude-phase characteristics of open system for 1) stable closed-loop system, 2) unstable closed-loop system

Nyquist conditions

M(ω_−π) < 1; where ω_−π : ϕ(ω_−π) = −π (60)

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Nyquist criterion - characteristics examples

Figure 10 : The amplitude-phase characteristics of open-loop systems, corresponding to: stable closed-loop systems - characteristics doesn’t include point {−1, j 0}

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Nyquist criterion - characteristics examples

Figure 11 : The amplitude-phase characteristics of open-loop systems, corresponding to: unstable closed-loop systems - characteristics include point {−1, j0}

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Nyquist criterion

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Nyquist criterion - Bode charakteristics

Nyquist conditions for amplitude and phase characteristics

L(ω_−π) = 20 log M(ω_−π) < 0;

(62) ϕ(ωp) > −π; gdzie L(ωp) = 0 (63)

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Nyquist criterion - gain margin and phase margin

Gain margin

∆M = 1

M(ω_−π) (64)

∆L = −20 log M(ω_−π) (65) Phase margin

∆ϕ = π + ϕ(ωp) (66) Gain margin and phase margin of the stable system have positive values.

INDUSTRIAL PRACTICE

30 deg < ∆ϕ < 60 deg (67) 2 ≤ ∆M ≤ 4 → 6dB ≤ ∆L ≤ 12dB

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Nyquist criterion - example N1

Using Nyquist criterion determine the stability of the following system:

G₀(s) = 1

s³+ 3s²+ s + 1 (69)

G₀(j ω) = 1

−i ω³− 3ω²+ j ω + 1 = 1

1 − 3ω²+ j (ω − ω³)

1 − 3ω²− j(ω − ω³ 1 − 3ω²− j(ω − ω³) = 1 − 3ω²

(1 − 3ω²)²+ (ω − ω³)² + j −(ω − ω³) (1 − 3ω²)²+ (ω − ω³)²

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Real part and imaginary part P(ω) = 1 − 3ω²

(1 − 3ω²)²+ (ω − ω³)²; Q(ω) = −(ω − ω³) (1 − 3ω²)²+ (ω − ω³)²

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Nyquist criterion - example N1

Real part

P(ω) = 1 − 3ω²

(1 − 3ω²)²+ (ω − ω³)² (72) Imaginary part

Q(ω) = −(ω − ω³)

(1 − 3ω²)²+ (ω − ω³)² (73)

ω 0 p1/3 1 ∞

P(ω) 1 0 -0.5 0

Q(ω) 0 -2.6 0 0

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Real shape of Nyquist diagram - MATLAB

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Nyquist criterion - example N2

Using Nyquist criterion determine the stability and gain/phase margins of the following system:

G₀(s) = 10

(0.1s + 1)(0.001s + 1) 1

0.3s (74)

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Nyquist criterion - example N2

G0(s) = G1(s)G2(s)G3(s)G4(s) (75) therefore

G0(s) = 10 1 0.1s + 1

1 0.01s + 1

1 0.3s

(76) In the case of serial connection the Bode characteristics of each elements are summed.

L0(ω) = L1(ω) + Lω+ L3(ω) + L4(ω) (77) ϕ0(ω) = ϕ1(ω)+ϕω+ϕ3(ω)+ϕ4(ω) (78)

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