Stability in the Frequency Domain

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Stability in the Frequency Domain

• Modern Control Systems
• Lecture 21

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Outline

• The issue of stability
• Mapping contours in the s-plane
• Nyquist criterion
• Examples to illustrate Nyquist criterion

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The issue of stability

• A stable system is a dynamic system with a
  bounded response to a bounded input.
• The Routh-Hurwitz criterion is used to
  investigate stability (and relative stability)
  for a system with its characteristic equation
  expressed in terms of complex variable ssj?.
• The frequency response is useful for determining
  system stability in the real frequency domain
  based on the Nyquist criterion.
• To study the Nyquist criterion, we need to
  understand the mapping of contours in the s-plane.

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Mapping contours in the s-plane

• Let F(s)0 represent a systems characteristic
  equation. To ensure stability, all the zeros of
  F(s) must lie in the left hand s-plane. Nyquist
  thus proposed a mapping of the right hand s-plane
  into the F(s)-plane.
• A contour map is a contour or trajectory in one
  plane mapped or translated into another plane by
  relation F(s).
• Since s is a complex variable, ssj?, F(s) is
  itself complex, defined as F(s)ujv, and can be
  represented on a complex F(s)-plane with
  coordinates u and v.

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Mapping a square contour by F(s)2s12(s1/2)

• Consider F(s)2s12(s1/2). We map the s-plane
  unit square contour to the F(s)-plane, obtained
  by
• ujvF(s)2s12(sj?)1(2s1)j(2?)

The contour in the s-plane encircles the zero
s-1/2 of F(s). The contour in the F(s)-plane
encircles the origin once.
The points A,B,C,D in the s-plane contour map
into the points A,B,C,D in the F(s)-plane. A
direction of traversal of the contours is shown
by arrows. Clockwise traversal of the contour is
assumed to be positive. The area within a contour
to the right of the traversal of the contour is
considered to be the area enclosed by the contour.
Rule clockwise eyes right
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Mapping a square contour by F(s)s/(s2)
We can calculate values of F(s), as shown in the
table above, as s traverses the square contour.
The contour in the s-plane encircles the zero s0
of F(s), but not the pole s-2 of F(s). The
contour in the F(s)-plane encircles the origin
once.
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Mapping a square contour by F(s)s/(s1/2)
The contour in the s-plane encircles the zero s0
and the pole s-1/2 of F(s). The contour in the
F(s)-plane does not encircle the origin.
What is the relationship between the encirclement
of the zeros and poles of F(s) by the s-plane
contour and the encirclement of the origin in the
F(s)-plane?
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Cauchys theorem

• If a contour Gs 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, then
  the corresponding contour GF in the F(s)-plane
  encircles the origin of the F(s)-plane NZ-P
  times in the clockwise direction.

NZ-P3-12
s-plane
F(s)-plane
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Nyquist contour

• Consider the characteristic equation F(s)0. For
  a system to be stable, all zeros of F(s) must lie
  in the left hand s-plane. Thus, we choose the
  so-called Nyquist contour Gs in the

s-plane that encloses the entire right hand
s-plane, and we use Cauchys theorem to determine
if any zeros of F(s) lie within Gs. That is, we
plot GF in the F(s)-plane and determine the
number of encirclement of the origin. Then the
number of zeros of F(s) within the Gs contour
i.e., unstable zeros of F(s) is ZNP.
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Nyquist criterion

• Consider the characteristic equation
  F(s)1GH(s). We decide to use
• GH(s)F(s)-1
• because the open-loop TF GH(s) is often available
  in factored form. Then the mapping of Gs in the
  s-plane will be through the function GH(s) into
  the GH(s)-plane.
• Thus, the encirclement of the origin of the
  F(s)-plane becomes the encirclement of the (-1,0)
  point in the GH(s)-plane.

A feedback system is stable if and only if the
contour GGH in the GH(s)-plane does not encircle
the (-1,0) point when the number of poles of
GH(s) in the right hand s-plane is zero (P0).
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Nyquist criterion (contd)
A feedback system is stable if and only if, for
the contour GGH, the number of counterclockwise
encirclements of the (-1,0) point is equal to
the number of poles of GH(s) with positive real
parts.

• The basis of Nyquist criterion is the fact that,
  for the GH(s) mapping, the number of zeros of
  1GH(s) in the right hand s-plane is represented
  by
• ZNP
• and it is the zeros of 1GH(s) that are the
  characteristic roots of the system, which
  determine the system stability.

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Example system with two real poles

• Consider the system shown in the figure, where

For t11,t21/10, K100, we calculate GH(j?)
and the phase of GH(j?) for selected values of ?.
The j?-axis is mapped into the solid line. The
j?-axis is mapped into the dashed line.
The semicircle with r?8 in the s-plane is mapped
into the origin of GH(s)-plane. As P0 and N0,
the closed-loop system is stable.
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Example system with a pole at the origin

• Consider the system shown in the figure, where

For this example, we have to take an
infinitesimal detour around the pole at the
origin, effected by a small semicircle of radius
??0. This detour is required as a result of the
Cauchys theorem.
We now consider each portion of Gs and find the
corresponding portion of GGH.

• The origin of s-plane.

The semicircle detour is represented by
setting -900 at ?0- to 900 at ?0. The mapping
is
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Example system with a pole at the origin
(contd)

• Consider the system shown in the figure, where

• The portion from ?0 to ?8.

This portion of Gs is mapped by GH(s) as the real
frequency polar plot as sj?. When ??8, GH(j?)
approaches zero at an angle of -1800, because
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Example system with a pole at the origin
(contd)

• Consider the system shown in the figure, where

• The portion from ?8 to ?-8.

This portion of Gs is mapped into the origin of
GH(s)-plane.
As F changes from 900 at ?8 to -900
The system is stable for all Kgt0. Why?
at ?-8, GGH moves from -1800 at ?8 to 1800 at
?-8.

• The portion from ?-8 to ?0-. This portion of
  GGH is symmetrical to the polar plot from ?8 to
  ?0.