digital control systems stability analysis of discrete time systems

Digital Control Systems

Stability Analysis of Discrete Time Systems

Mapping Between the s-plane and the z-plane


BIBO StabilityAn 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 system

can be determined from the location of closed loop poles in z-plane which are the roots of the characteristic equation


BIBO Stability

1. 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.

2. 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.


BIBO StabilityExample:Determine the closed loop stability of the system shown in Figure when K = 1.


BIBO StabilityExample:


BIBO StabilityStability Tests

Applied to characteristic equation• Schur-Cohn stability test• Jury Stability test• Routh stability coupled with bi-linear transformation.

Applied to state space • Lyapunov stability analysis


Jury Stability Test

Jury Table


Jury Stability Test

Jury Table

The system will be stable if:


Jury Stability TestExample:


Jury Stability TestSingular Cases


Jury Stability TestExample:


Bilinear Transformation

The 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-plane


Bilinear Transformation

Let the inside of the unit circle in z-plane can be represented by:


Routh Stability Criterion


Routh Stability Criterion

𝑄 (𝑤)=𝑎❑0𝑤𝑛+𝑎1𝑤𝑛−1+…+𝑎𝑛−1𝑤+𝑎𝑛

¿¿¿¿𝑤1¿ 𝑓 1 ¿¿¿¿¿𝑤0 ¿𝑔0 ¿¿ ¿¿¿¿

Necessary 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.

iaia


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.



All elements in the first column of Routh array are positive. Thus the system is stable.



Find out the range of K for whichthe system is stable.


Routh Stability CriterionSingular cases• The 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 zero

o Pairs of real roots with opposite signs.o Pairs of imaginary roots.o 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 (w)dA

dw

(w) 0A

digital control systems stability analysis of discrete time systems

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