transient response stability

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Transient response stability

System is analyzed for its:The transient response Steady state response Stability

Concept of stability Stability is the most important system specification. Many physical systems are inherently openloop unstable. Feedback control is introduced by engineer to stabilize the unstable plant

For open-loop stable plant, we still need feedbackto adjust performance to meet the design specification. What is stability ?

Example

Tacoma Narrows Bridge (a) as oscillation begins (b) at catastrophic failure.

Example

The M2 robot is more energy-efficient but less stable than many other designs that are well-balanced but consume much more power.

Stability Stability is the most important system specification. If the system is unstable, transient response and steady state errors are moot points. An unstable system can not be designed for a specific transient response or steady state error requirement

Stable and unstable system response Stable state: A linear , time invariant system is stable if the natural response approaches zero as time approaches infinity. Unstable: A linear , time invariant system is unstable if the natural response approaches infinity as time approaches infinity.

Critically stable: If the natural response neither decays nor grows but remains constant or oscillates.

A system is stable if every bounded yields a bounded outputThis statement the bounded inputBounded output (BIBO) defination of stability

Test for stability1.Preliminary test: Check the roots of the characteristic equation to lie on the left half of s-plane. 2. Routh-Hurwitz criterion

Preliminary testBy pole configuration

Stable system-In terms of pole configuration Stable :A linear system stable , if all poles of its transfer function lie in the left-half plane

Unstable system-In terms of pole configuration Instability A linear system is only unstable, if atleast one pole of its transfer function lies in the right-half plane, or, if at least one multiple pole (multiplicity ) is on the imaginary axis of the plane

Marginally stable system-In terms of pole configuration Marginally stable A linear system is critically stable, if at least one single pole exists on the imaginary axis, no pole of the transfer function lies in the right-half plane, and in addition no multiple poles lie on the imaginary axis

All cases

The concept of stability Pole location and stability

Routh-Hurwitz Stability Criterion

The Routh-Hurwitz Stability Criterion The Routh-Hurwitz stability criterion provides a method to examine the system stability without the need to solve for poles of a system. Using Routh-Hurwitz Criterion one can find how many poles are in the left half-plane,right halfplane, and on the imaginary axis. However, using this method, one cannot find the exact coordinates of the poles.

The Routh-Hurwitz Stability Criterion The method requires two steps: - Generate data table called as Routh table - Interpret the data table to find how many poles are in the right half-plane of the complex plane

The Routh-Hurwitz stability Criterion The closed loop gain/Transfer functionY (s) G (s) N (s) Z T (s) ! ! ! ! R (s) 1 G (s)H (s) ( (s) P N(s) is the numerator, its roots are called as zeros ( (s) (or G(s)) is the denominator or characteristic equation, its roots are called as poles. The characteristic as:

function in the Laplace variable is writtenn1 n3

D(s) ! ans an1s an2s

n

a1s a0

Routh Table Consider the characteristic equation

D ( s ) ! a n s n a n 1 s n 1 a n 2 s n 3 a1 s a 0

Create Routh Table Consider closed loop system with transfer functionN ( s) a4 s 4 a3 s 3 a2 s 2 a1s1 a0Initial Routh Table

s43 S

s2 s1 s0

a4 a3 b1 c1 d1

a2 a1 b2 c2

a0 0

0 0

To find other terms Routh Table

From the Rouths criterion If all the elements of the first column of the Rouths array are of the same sign, the roots of the polynomial are all in the left half of the splane. The system is stable. If there are changes of sign in the first column, the number of sign changes indicates the number of roots with positive real part, i.e. the number of roots that lie on the right half of the splane. The system is unstable

Example: D(s) = s3 + 2s2 + 4s + 3 Initial Routh tableAll the elements of the 1st column Are of the same sign, Hence all roots lie on the left half of s plane. System is said to be stable.

Calculate the other coefficients

Example: D(s) = s4 + 5s3+5s2 + 10s + 12 The Rouths array

The 1st column has two sign changes and so there are two roots of the characteristic Equation that lie on the right half of the s-plane. The system is said to be unstable.

Example:

Two sign change=2 poles on right half Plane, system unstable

Exercise 1: Test the stability of the closedloop system

Rouths array

Take the common factor

The 1st column is

The two sign changes, poles on right half of s-plane Unstable system

Special case.A zero element in the first column

A zero element in the first column If 1st element of the column is zero, replce with null element with small positive number epsilon( I ) and proceed with the construction of the array

Equation P ( s ) ! s 4 s 3 2 s 2 2 s 5 ! 0s4 1 2 5 s3 1 2 0 s2 (0)I 5 0 s1 (2I 5) / I 0Now assume, Epsilon = -ve ,means there are two sign changes-s3-s2 &s2-s1,so unstable epsilon = +ve=If it is 1 & 2 , there two sign changes =+ve =if it is > 2, it becomes stable

s0 5

Epsilon = -ve, 2 sign changes , unstable Epsilon = -ve, check what value the column does changes sign=unstable

Example Example

Example Rouths array S4 1 11 S3 6 6 s2 60-6/6= 10 6k-0/6=k S1 60-6k/10 0 S0 [(60-6k)/10]*K/((60-6k)/10)=k K 0 0 0

To make system stable,1st column should not have any sign change If K=0,Last row S0=0 If K=10, s1 row =0 IfK>0 and K

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