Lecture 1 Lecture 1: The Nyquist Criterionsudhoff/ee631/dcst_lecture1.pdfLecture 1 1 Lecture 1: The Nyquist Criterion S.D. Sudhoff Energy Sources Analysis Consortium ESAC DC Stability

Lecture 1

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Lecture 1: The Nyquist Criterion

S.D. SudhoffEnergy Sources Analysis ConsortiumESAC DC Stability Toolbox Tutorial

January 4, 2002Version 2.1

Lecture 1

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Lecture 1 Outline

• Motivation: Negative Impedance Instability• Analysis Objectives• The Cauchy Principal• The Nyquist Criterion• Application to DC Systems

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Motivation: Negative Impedance Instability

• High bandwidth regulation leads to constant power loads

• Constant power loads appear as negative incremental resistance

• Negative resistance is highly destabilizing

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Motivation: Example System

SM IMTurbineMechanical

Load

IMControls

Vol. Reg./Exciter

iabcsvdci

vdcsvr

ωrm,imTe,desvdcs

*

3 - Uncontrolled

Rectifier

φ 3 - Fully C ontrolled

Inverter

φLCFilter

C apacitiv eFilter

Tie Line

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Motivation: System Performance

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Analysis Objectives

• All methods based on Nonlinear Average Value Model (NLAM) so state-space description is non time-varying

• Most straightforward method: eigenanalysis• We’re not really interested in stability

analysis though, we really are interested in driving design specs. An approach to this end is through the use of Nyquist techniques

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Contour Evaluation of Complex Functions

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-2

-1.5

-1

-0.5

0

0.5

1

1.5

2A Nyquist Contour and its Corresponding Nyquist Contour Evaluation

131)( 2 ++

+=

ssssH

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The Cauchy Principal

A contour evaluation of a complex function will only encirclethe origin if the contour contains a singularity of that function...

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Cauchy Principal: Example 1

-1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1A Nyquist Contour and its Corresponding Nyquist Contour Evaluation

)1)(1(5.0)(+−

+=

ssssH

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Cauchy Principal: Example 2

-1.5 -1 -0.5 0 0.5 1 1.5 2 2.5-1.5

-1

-0.5

0

0.5

1

1.5A Nyquist Contour and its Corresponding Nyquist Contour Evaluation

)1)(1(5.0)(+−

+=

ssssH

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Justification

L

L

))()(())(()(

321

21pspsps

zszssH−−−

−−=

( )( )L

L

+−∠+−∠+−∠−+−∠+−∠=∠

)()()()()()(

321

21pspsps

zszssH

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Justification: Example 1

-1-0.5

00.5

11.5

22.5

33.5

4

-2-1.5

-1-0.5

00.5

11.5

2-60

-40

-20

0

20

40

60

)1)(1(1)(

jsjsssH

−++++

=

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Justification: Example 2

-4-3.5

-3-2.5

-2-1.5

-1-0.5

00.5

1

-2-1.5

-1-0.5

00.5

11.5

2-400

-300

-200

-100

0

100

200

300

400

500

)1)(1(1)(

jsjsssH

−++++

=

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The Nyquist Contour

Assumption:Traverse the Nyquist contourin CW direction

Observation #1:Encirclement of a poleforces the contour to gain 360degrees so the Nyquist evaluationencircles origin in CCW direction

Observation #2Encirclement of a zeroforces the contour to loose 360degrees so the Nyquist evaluationencircles origin in CW direction

∞j

∞− j

∞

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Nyquist Theory

)(1)()(sGsGsH

+=

)()()(sdsnsG =

)()()()(1

sdsnsdsG +

=+

Clearly, stability of determined by roots of)()(

)()(sdsn

snsH+

=

Transfer function of a closedloop plant

Break G into numerator anddenominator

Also for your consideration...

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Implications...

= Net number of CW encirclements of originof Nyquist contour of

= Net number of CW encirclements of -1of Nyquist contour of

= Number of unstable open loop poles

= Number of unstable closed loop poles

=

cweN

cweN

uolpN

uclpN

cweN uolpuclp NN −

)(sG

)(1 sG+

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Summary

• Perform Nyquist evaluation of• Let the net number of CW encirclements of

-1 be denoted • Let the number of unstable open loop poles

be denoted• The number of unstable closed loop poles is

given by• Stability of closed loop system determined

by evaluation of open loop plant

cweN

)(sG

uolpN

uolpcwe NN +

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Application to DC Systems

Z

v

ii

vv+

_+ +

Z

s l

ss

ll

lls

ss

ls

l vZZ

Zv

ZZZ

v+

++

=

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Application to DC Systems

lls

ss

ls

l vZZ

Zv

ZZZ

v+

++

=

s

ss D

NZ =

l

ll D

NZ =

lssl

llssslDNDNvDNvDN

v++

=

)1( lssl

llssslYZDNvDNvDN

v++

=

Transfer function

Impedance definitions

Rearranging a little

Rearranging some more

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Application to DC Systems

)1( lssl

llssslYZDNvDNvDN

v++

= From previous page

Assumption #1: Load stable when fed from ideal source.Implication #1: has no roots in the right-half plane.

Assumption #2: Source stable when supplying constant current load.Implication #2: has no roots in the right- half plane.

lN

sD

Conclusion: System will be stable provided does not have zeros in the right-half plane.

lsYZ+1

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Aside...

• Mathematical definition of a load– Component which obeys assumption #1

• Mathematical definition of a source– Component which obeys assumption #2

• These definitions are not mutually exclusive• Often, but not always, these definitions

correspond to components which use and produce power, respectively

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Applications to DC Systems

From previous page, system will be stable provideddoes not have any zeros in the right-half plane

Observation: does not have any unstable open loop poles since does not have any unstable roots and does not have

any unstable roots (Assumptions 1 and 2, previous slide)

1+lsYZ

lN sDlsYZ

Conclusion: Number of unstable closed loops poles is equalto the number of clockwise encirclements of -1 by the Nyquistcontour of . We do not need to consider the possibilityof unstable open loop poles in our analysis.

lsYZ

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Summary

The source load system is stable provided that the evaluationof along the Nyquist contour does not encircle -1lsYZ