Neutral Stability

Lecture 9Feedback Control Systems

President University Erwin Sitompul FCS 9/1

Dr.-Ing. Erwin SitompulPresident University

http://zitompul.wordpress.com2 0 1 3


Neutral Stability Root-locus technique determine the stability of a closed-

loop system, given only the open-loop transfer function KG(s), by inspecting the denominator in factored form.

Frequency response technique determine the stability of a closed-loop system, given only the open-loop transfer function KG(jω) by evaluating it and performing a test on it.

The principle will be discussed now.

Chapter 6 The Frequency-Response Design Method


Neutral Stability Suppose we have a system defined by

• System is unstable if K > 2.• The neutrally stable points lie

on the imaginary axis, K = 2 and s = ±j.

1( )( ) 180

KG sKG s

All points on the locus fulfill:1( )

( ) 180KG jKG j



Neutral Stability The frequency response of the

system for various values of K is shown as follows.

At K = 2, the magnitude response passes through 1 at the frequency ω = 1 rad/sec (remember s = ±j), at which the phase passes through 180°.

After determining “the point of neutral stability”, we know that: K < Kneutral stability stable, K > Kneutral stability unstable.

At (ω = 1 rad/sec, G(jω) = –180°), |KG(jω)| < 1 stable K, |KG(jω)| > 1 unstable K.



Neutral Stability A trial stability condition can now be stated as follows

If |KG(jω)| < 1 at G(jω) = –180°, then the system is stable.

This criterion holds for all system for which increasing gain leads to instability and |KG(jω)| crosses magnitude = 1 once.

For systems where increasing gain leads from instability to stability, the stability condition is

If |KG(jω)| > 1 at G(jω) = –180°, then the system is stable.

For other more complicated cases, the Nyquist stability criterion can be used, which will be discussed next.

While Nyquist criterion is fairly complex, one should bear in mind, that for most systems a simple relationship exists between the closed-loop stability and the open-loop frequency response.

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The Nyquist Stability Criterion The Nyquist stability criterion is based on the argument

principle. It relates the open-loop frequency response KG(s) to the

number of closed-loop poles of the system (roots of the characteristic equation 1+ KG(s)) that are in the RHP.

It is useful for stability analysis of complex systems with more than one resonance where the magnitude curve crosses 1 several times and/or the phase crosses 180° several times.

Advantage: From the open-loop frequency response (in the form of Bode plot), we can determine the stability of the closed-loop system without the need to determine the closed-loop poles.



The Nyquist Stability Criterion Consider the transfer function H1(s) with poles and zeros as

indicated in the s-plane in (a) below. We wish to evaluate H1 for values of s on the clockwise

contour C1, with s0 as test point,1 0 1 2 1 2( ) .H s

As s travels along C1 in the clockwise direction starting at s0, the angle α increases and decreases, and finally returns to the original value as s returns to s0, without rotating through 360°, see (b).



The Nyquist Stability Criterion α changes, but did not undergo a net change of 360° because

there are no poles or zeros within C1. Thus, none of the angles θ1, θ2, Φ1, or Φ2 go through a net

revolution. Since there is no net revolution, the plot of H1(s) in (b) will not

encircle the origin.


• Magnitude of H is equivalent to distance from origin to s0

• Phase of H is equivalent to α


The Nyquist Stability Criterion Now, consider the transfer function H2(s) with poles and zeros

as indicated in the s-plane in (c). We wish to evaluate H2 for values of s on the clockwise

contour C1, noting that there is a pole within C1. As s travels in the clockwise direction around C1, the

contributions from θ1, θ2, and Φ1 change, but return to their original values as s returns to s0.

In contrast, Φ2 undergoes a net change of –360° after one full trip around C1 H2 encircles the origin in ccw direction.



The Nyquist Stability Criterion The essence of the argument principle:

A contour map of a complex function will encircle the origin Z – P times, where Z is the number

of zeros and P is the number of poles of the function inside the contour.

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Application to Control Design To apply the argument principle to

control design, we let the contour C1 to encircle the entire RHP in the s-plane.

As we know, RHP is the region in the s-plane where a pole would cause an unstable system.

The resulting evaluation of H(s) will encircle the origin only if H(s) has a RHP pole or zero.



Application to Control Design As the stability is decided by the roots of the characteristic

equation 1 + KG(s), the argument principle must be applied to the function 1 + KG(s).

If the evaluation contour, which encloses the entire RHP of the s-plane, contains a zero or pole of 1 + KG(s), then the evaluated argument of 1 + KG(s) will encircle the origin.

Furthermore, if the argument of 1 + KG(s) encircles the origin, then it is equivalent with if the argument of KG(s) encircles –1 on the real axis.

≡



Application to Control Design Therefore, we can plot the contour evaluation of the open-

loop KG(s), examine its encirclements of –1, and draw the conclusions about the origin encirclements of the closed-loop functions 1 + KG(s).

Presentation of the argument evaluation of KG(s) in this manner is often referred to as a Nyquist plot, or polar plot.

≡



Application to Control Design 1 + KG(s) is now written in terms of poles and zeros of KG(s),

( )1 ( ) 1( )b sKG s Ka s

( ) ( ) .

( )a s Kb s

a s

The poles of 1 + KG(s) are also the poles of G(s) The poles of closed-loop transfer function are also the

poles of the open-loop transfer function. It is safe to assume that the poles of G(s) are known. Since in most of the cases, the open-loop stability can be inspected in advance.

The zeros of 1 + KG(s) are roots of the characteristic equation If any of these zeros lies in the RHP, then the system will be unstable. An encirclement of –1 by KG(s) or, equivalently, an

encirclement of 0 by 1 + KG(s) indicates a zero of 1 + KG(s) in the RHP, and thus an unstable root of the closed-loop system.



Application to Control Design This basic idea can be generalized by noting that:

A clockwise contour C1 enclosing a zero of 1 + KG(s) —that is, an unstable closed-loop pole— will result in KG(s) encircling the –1 point in a clockwise direction.

A clockwise contour C1 enclosing a pole of 1 + KG(s) —that is, an unstable open-loop pole— will result in KG(s) encircling the –1 point in a counter-clockwise direction.

Furthermore, if two poles or two zeros are in the RHP, KG(s) will encircle –1 twice, and so on.

The net number of clockwise encirclements, N, equals the number of zeros in the RHP, Z, minus the number of open-loop poles in the RHP, P.N Z P



Application to Control Design Procedure for Plotting the Nyquist Plot:

1. Plot KG(s) for –j∞ ≤ s ≤ +j∞. Do this by first evaluating KG(jω) for ω = 0 to ω = +∞, which is exactly the frequency response of KG(jω). Since G(–jω) is the complex conjugate of G(jω), the evaluation for ω = 0 to ω = –∞ can be obtained by reflecting the 0 ≤ s ≤ +j∞ portion about the real axis. The Nyquist plot will always be symmetric with respect to the real axis.

2. Evaluate the number of clockwise encirclements of –1, and call that number N. If encirclements are in the counterclockwise direction, N is negative.

3. Determine the number of unstable (RHP) poles of G(s), and call that number P.

4. Calculate the number of unstable closed-loop roots Z: Z = N + P.

If the system is stable, then Z = 0; that is, no characteristic equation roots in the RHP



Application to Control Design

Root locus of G(s) = 1/(s+1)2

Bode plot for KG(s) = 1/(s+1)2, K = 1

( )Y s+–

( )R sK 2

1( 1)s



Application to Control Design• Nyquist plot of the evaluation

of KG(s) = 1/(s+1)2.

• The evaluation contour is C1, with K = 1.

• No value of K would cause the plot to encircle –1 ( N = 0 ).

• No poles of G(s) in the RHP ( P = 0 ).

• The closed-loop system is stable for all K > 0 ( Z = N + P = 0 ).

Bode plot for KG(s) = 1/(s+1)2, K = 1



Application to Control Design( )Y s

+–

( )R sK 2

1( 1)s s

Root locus of G(s) = 1/[s(s+1)2]

Bode plot for KG(s) = 1/[s(s+1)2], K = 1



If there is a pole on the imaginary axis, the C1 contour can be modified to take a small detour around the pole to the right.

The contour is now divided into 4 sections, I-II, II-III, III-IV, and IV-I.


I

II

III

IV



Application to Control Design Evaluation of path I-II can be obtained

from the Bode plot, corresponds to the frequency response of KG(jω) for K = 1 and ω = 0 to ω = +∞.

I

II

IIIIV

I

II



Application to Control Design( ) ( )( )G j G jG j

12

1 90 2 tan( 1)

2 2

1 1( 1) ( 1)j j j j

• Nyquist plot is nothing but the polar magnitude of G(jω).• As the frequency traverses from ω = 0 to ω = +∞, the

corresponding magnitude and phase is plotted on the polar coordinate.



Application to Control Design2

1( )( 1)

G ss s

2

1( )( 1)

G jj j

2

1(1 2 )j j

2 2

12 (1 )j

2 2

2 2

2 (1 )2 (1 )

jj

2 2

4 2 2 2 4 2 2 2

2 (1 )4 (1 ) 4 (1 )

j

2

2 2 2 2 2 2

2 (1 )( )4 (1 ) 4 (1 )

G j j

• At ω = 0–, G(jω) is asymptotic at –2 – j∞.• At ω = +∞, G(jω) is asymptotic at 0– + j0+.• Plotting G(jω) for ω = 0 to ω = +∞ can be

performed by plotting the complex coordinate given by the last equation for several frequencies.



Application to Control Design Evaluation of path III-IV is the

reflection of Path I-II about the real axis.

We know that this path corresponds to the frequency response of KG(jω) for K = 1 and ω = 0 to ω = –∞.

I

II

IIIIV

III

IV



Re ( )G s

Im ( )G s

Application to Control Design Path II-III is a half circle with a very large radius,

routing from positive imaginary axis to negative imaginary axis in clockwise direction.

This path can be evaluated by replacing,js R e

90 90 , clockwise.

I

II

IIIIV

R : very large number

2

1( )

( 1)j j js R eG s

R e R e

Performing it,

3 3

1jR e

3 3jr e

r : very small number

• Evaluation of G(s) forms a 1.5-circle with a very small radius.

• Inserting the value of θ, the circle starts at –270° and ends at 270°, routing in ccw direction.

III

II



Application to Control Design Path IV-I is a half circle with a very small radius,

routing from negative imaginary axis to positive imaginary axis in counterclockwise direction.

This path can be evaluated by replacing,js r e

90 90 , counterclockwise.

I

II

IIIIV

2

1( )

( 1)j j js r eG s

r e r e

1jr e

jR e • Evaluation of G(s) forms a half circle

with a very large radius.• Inserting the value of θ, the half

circle starts at 90° and ends at –90°, routing in clockwise direction.


Performing it,



Combining all four sections, we will get the complete Nyquist plot of the system.

I

II

IIIIV

III

IV

I

II



Application to Control Design The plot crosses the real axis at ω = 1 at –0.5. No unstable poles of G(s), P = 0.

No encirclement of –1, N = 0. Z = N + P = 0 → the system is stable.

If K > 2, the plot will encircle –1 twice in cw direction, N = 2. Z = N + P = 2 → the system is unstable with 2 roots in RHP.



Homework 9Draw the Nyquist Plot for

( 1)( )( 2)( 3)

sKG s Ks s

Choose K = 1. Based on the plot, determine the range of K for which the system is stable.Hint: Get representative frequency points before you sketch the Nyquist plot. The easiest way is to choose them is to choose:The 10n points (0.001, 0.001, …, 1, 10, 100, …)The break points ωbThe points of corrections, 1/5ωb and 5ωb


Due: Wednesday, 04.12.2013.

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Neutral Stability