Stability Analysis of Discrete Time Systems Mapping Between the splane and the zplane Mapping Between the splane and the zplane Mapping Between the splane and the zplane Mapping Between the splane and the zplane ID: 720070
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Slide1
Digital Control Systems
Stability Analysis of Discrete Time SystemsSlide2
Mapping Between the s-plane and the z-planeSlide3
Mapping Between the s-plane and the z-planeSlide4
Mapping Between the s-plane and the z-planeSlide5
Mapping Between the s-plane and the z-planeSlide6
Mapping Between the s-plane and the z-planeSlide7
Mapping Between the s-plane and the z-planeSlide8
Stability
Analysis of Discrete Time Systems
BIBO Stability
An initially relaxed (all the initial conditions of the system are zero) LTI system is said
to be BIBO stable if for every bounded input, the output is also bounded.However, the stability of the following closed loop systemcan
be determined from the location of closed loop poles in z-plane which are the roots of the characteristic equationSlide9
Stability
Analysis of Discrete Time Systems
BIBO Stability
For
the system to be stable, the closed loop poles or the roots of the characteristic equation must lie within the unit circle in z-plane. Otherwise the system would be unstable.If a simple pole lies at |z| = 1, the system becomes marginally stable. Similarly if a pair of
complex conjugate poles lie on the |z| = 1 circle, the system is marginally stable. Multiple poles on unit circle make the system unstable.Slide10
Stability
Analysis of Discrete Time Systems
BIBO Stability
Example:
Determine the closed loop stability of the system shown in Figure when K = 1.Slide11
Stability
Analysis of Discrete Time Systems
BIBO Stability
Example:Slide12
Stability
Analysis of Discrete Time Systems
BIBO Stability
Stability Tests
Applied to characteristic equationSchur-Cohn stability testJury Stability testRouth stability coupled with bi-linear transformation.Applied to state space
Lyapunov stability analysisSlide13
Stability
Analysis of Discrete Time Systems
Jury Stability Test
Jury TableSlide14
Stability
Analysis of Discrete Time Systems
Jury Stability Test
Jury TableSlide15
Stability
Analysis of Discrete Time Systems
Jury Stability Test
Jury Table
Th
e
system will be stable if:Slide16
Stability
Analysis of Discrete Time Systems
Jury Stability Test
Example:Slide17
Stability
Analysis of Discrete Time Systems
Jury Stability Test
Example:Slide18
Stability
Analysis of Discrete Time Systems
Jury Stability Test
Example:Slide19
Stability
Analysis of Discrete Time Systems
Jury Stability Test
Example:Slide20
Stability
Analysis of Discrete Time Systems
Jury Stability Test
Singular CasesSlide21
Stability
Analysis of Discrete Time Systems
Jury Stability Test
Example:Slide22
Stability
Analysis of Discrete Time Systems
Jury Stability Test
Example:Slide23
Stability
Analysis of Discrete Time Systems
Bilinear
TransformationThe bilinear transformation has the following form.
where a, b, c, d are real constants.If we consider a
= b = c = 1 and d = −1, then thetransformation
takes a form
This transformation maps the inside of the unit circle in the z-plane into the left half of the w-planeSlide24
Stability
Analysis of Discrete Time Systems
Bilinear
TransformationLet
the inside of the unit circle in z-plane can be represented by:Slide25
Stability
Analysis of Discrete Time Systems
Bilinear
TransformationLet
the inside of the unit circle in z-plane can be represented by:Slide26
Stability
Analysis of Discrete Time Systems
Routh
Stability CriterionSlide27
Stability
Analysis of Discrete Time Systems
Routh
Stability Criterion
N
ecessary
and sufficient condition for all roots of
Q(w)
to
be located in
the
left-half
plane is that all
the
are
positive and all of the coefficients in the
first
column
be
positive
.
Slide28
Stability
Analysis of Discrete Time Systems
Routh
Stability CriterionExample:
There is one sign change in the first column of the
Routh
array. Thus the system is
unstable
with one pole at right hand side of the w-plane or outside the unit circle of z-plane.Slide29
Stability
Analysis of Discrete Time Systems
Routh
Stability CriterionExample:
All elements in the first column of
Routh
array are positive. Thus the system is stable.Slide30
Stability
Analysis of Discrete Time Systems
Routh
Stability
CriterionExample:Find out the range of K for whichthe system is stable.Slide31
Stability
Analysis of Discrete Time Systems
Routh
Stability CriterionExample:Slide32
Stability
Analysis of Discrete Time Systems
Routh
Stability CriterionExample:Slide33
Stability
Analysis of Discrete Time Systems
Routh
Stability CriterionSingular casesThe first element in any row is zero→replace zero by a small number and then proceed with
the tabulation.All the elements in a single row are zeroPairs of real roots with opposite signs.Pairs of imaginary roots. Pairs of complex conjugate roots which are equidistant from the origin.
→ an auxiliary equation is formed by using the coefficients of the row just above the row of all zeros. The tabulation is continued by replacing the row of
zeros by the coefficients of