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V. ROOTS OF SECOND-ORDER AND THIRD-ORDER POLYNOMIALS

We consider conditions that second- and third-order polynomials have roots in the left half of the complex s-plane.

1/ Second-order polynomials

Second-order polynomials with real coefficients have either real or complex roots of the form

( s + a ) ( s + b ) = 0 or ( s + a + jc ) ( s + a - jc ) = 0 size 12{ \( s+a \) \( s+b \) =" 0 " matrix { {} # {}} "or" matrix { {} # {}} \( s+a+"jc" \) \( s+"a - jc" \) " = 0"} {}

where a>0 and b>0. The polynomials can be expressed as

s 2 + ( a + b ) s + ab or s 2 + 2 as + a 2 + c 2 = 0 size 12{s rSup { size 8{2} } + \( a+b \) s+ ital "ab" matrix { {} # {}} ital "or" matrix { {} # {}} s rSup { size 8{2} } +2 ital "as"+a rSup { size 8{2} } +c rSup { size 8{2} } =0} {}

Thus, both polynomials have the form

s 2 + αs + β = 0 size 12{s rSup { size 8{2} } +αs+β=0} {}

where α>0 and β>0. These conditions are both necessary and sufficient.

2/ Third-order polynomials

Third-order polynomials must have one real root and either a pair of real or complex roots of the form

( s + a ) ( s + b ) ( s + c ) = 0 or ( s + b ) ( s + a + jd ) ( s + a - jd ) = 0 size 12{ \( s+a \) \( s+b \) \( s+c \) =" 0 " matrix { {} # {}} "or" matrix { {} # {}} \( s+b \) \( s+a+"jd" \) \( s+"a - jd" \) " = 0"} {}

where a>0, b>0, and c>0. The polynomials can be expressed as

s 3 + ( a + b + c ) s 2 + ( ab + ac + bc ) s + abc = 0 or s 3 + ( 2a + b ) s 2 + ( a 2 + d 2 + 2 ab ) s + ( a 2 + d 2 ) b = 0 alignl { stack { size 12{s rSup { size 8{3} } + \( a+b+c \) s rSup { size 8{2} } + \( ital "ab"+ ital "ac"+ ital "bc" \) s+ ital "abc"=0 matrix {{} # {} } } {} #ital "or" matrix { {} # {}} s rSup { size 8{3} } + \( 2a+b \) s rSup { size 8{2} } + \( a rSup { size 8{2} } +d rSup { size 8{2} } +2 ital "ab" \) s+ \( a rSup { size 8{2} } +d rSup { size 8{2} } \) b=0 {} } } {}

Thus, both can be put in the form

s 3 + αs 2 + βs + γ = 0 size 12{s rSup { size 8{3} } +αs rSup { size 8{2} } +βs+γ=0} {}

Note that α>0, β>0, and γ>0. But in addition, β>γ/α. These conditions are both necessary and sufficient.

VI. ROOT LOCUS PLOTS FOR POSITION CONTROL SYSTEMS

1/ Proportional controller

H ( s ) = K s 2 + 101 s + 100 + K size 12{H \( s \) = { {K} over {s rSup { size 8{2} } +"101"s+"100"+K} } } {}

Recall the step response of the position control system with proportional controller.

The poles of the closed-loop system function are at

s 1,2 = 101 2 ± 101 2 2 100 K 1 / 2 size 12{s rSub { size 8{1,2} } = - { {"101"} over {2} } +- left [ left [ { {"101"} over {2} } right ] rSup { size 8{2} } - "100" - K right ]rSup { size 8{1/2} } } {}

The root locus plot is

2/ Integral controller

H ( s ) = K s 3 + 101 s 2 + 100 s + K size 12{H \( s \) = { {K} over {s rSup { size 8{3} } +"101"s rSup { size 8{2} } +"100"s+K} } } {}

Recall the step response of the position control system with integral controller.

The poles are the roots of the polynomial

s 3 + 101 s 2 + 100 s + K = 0 size 12{s rSup { size 8{3} } +"101"s rSup { size 8{2} } +"100"s+K=0} {}

The root locus plot is

Note that all the poles are in the lhp for K>0 and 100>K/101 or 0<K<10100. For K>10100 two poles move into the rhp and the system is unstable.

VII. STABILIZATION OF UNSTABLE SYSTEMS

1/ Many common systems are unstable

Some common systems are annoyingly unstable (adapted from Figure 11.7 in Oppenheim&Willsky, 1983).

We can model the audio feedback system with SIMULINK as follows.

Another example of an unstable system is an inverted pendulum. For example, balancing a broomstick in your hand is an example of an inherently unstable system that is stabilized by your motor control system.

Figure adapted from Figure 11.2 in Oppenheim&Willsky, 1983.

2/ Inverted pendulum

We will analyze an inverted pendulum attached to a cart. A schematic diagram is shown on the left and a free-body diagram showing the forces on the pendulum is shown on the right.

The forces of attachment of the cart and pendulum on the pendulum are obtained from the equations of rectilinear motion.

The equation of rotational motion about the center of mass is

J d 2 θ ( t ) dt 2 = mgl sin θ ( t ) ml cos θ ( t ) d 2 x ( t ) dt 2 size 12{J { {d rSup { size 8{2} } θ \( t \) } over { ital "dt" rSup { size 8{2} } } } = ital "mgl""sin"θ \( t \) - ital "ml""cos"θ \( t \) { {d rSup { size 8{2} } x \( t \) } over { ital "dt" rSup { size 8{2} } } } } {}

where J is the moment of inertia of the mass about the mass, m is the mass, and g is acceleration of gravity. For small θ, this differential equation is linearized by noting that sinθ≈θ and cos θ ≈ 1. Therefore,

J d 2 θ ( t ) dt 2 mgl sin θ ( t ) = ml d 2 x ( t ) dt 2 size 12{J { {d rSup { size 8{2} } θ \( t \) } over { ital "dt" rSup { size 8{2} } } } - ital "mgl""sin"θ \( t \) = - ital "ml" { {d rSup { size 8{2} } x \( t \) } over { ital "dt" rSup { size 8{2} } } } } {}

The system function is

H ( s ) = θ ( s ) X ( s ) = ( ml / J ) s 2 s 2 ( mgl / J ) size 12{H \( s \) = { {θ \( s \) } over {X \( s \) } } = { { - \( ital "ml"/J \) s rSup { size 8{2} } } over {s rSup { size 8{2} } - \( ital "mgl"/J \) } } } {}

Therefore, the poles occur at

s 1,2 = ± mgl J 1 / 2 size 12{s rSub { size 8{1,2} } = +- left [ { { ital "mgl"} over {J} } right ] rSup { size 8{1/2} } } {}

Hence, the system is inherently unstable since one of its poles (natural frequencies) is in the rhp.

3/ Stabilization of the inverted pendulum

To stabilize the inverted pendulum, a rotary potentiometer is used to measure θ(t). A current proportional to θ(t) − θo, where θo is the desired angle, drives a motor so as to increase x(t) (adapted from Figure 6.4-1, Siebert, 1986).

This inverted pendulum is connected in a feedback configuration as follows.

M(s) represents the system function of the motor dynamics, 1 + (a/s) is the system function of the proportional plus integral controller for θ(t), and c + bs is the system function of the proportional plus derivative controller for x(t).

Demo of stabilization of an inverted pendulum.

4/ Stabilize by pole cancellation?

Why not cascade an unstable system with another system that cancels the unstable pole?

  • Perfect cancellation is very difficult.
  • The unstable pole can be excited by other inputs.

VIII. CONCLUSIONS

  • Interconnection of systems requires attention to their interactions.
  • Feedback is a powerful method to improve the performance of systems. However, feedback systems have the capacity to become unstable.
  • Instability can be determined by examining the poles of the closed-loop transfer function. It may also be important to examine the stability of components of the closed-loop system. Inherently unstable systems can be stabilized with feedback.

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Source:  OpenStax, Signals and systems. OpenStax CNX. Jul 29, 2009 Download for free at http://cnx.org/content/col10803/1.1
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