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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

1

Stability

• How to determine the stability of a system represented as a transfer function

• How to determine the stability of a system represented in state space

• How to determine system parameters to yield stability


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lntroduction

Definitions of stability for linear, time-invariant systems using the natural response:

1. A system is stable if the natural response approaches zero as time approaches infinity.

2. A system is unstable if the natural response approaches infinity as time approaches infinity.

3. A system is marginally stable if the natural response neither decays nor grows but remains constant or oscillates.


3

Bounded-input bounded-output (BIBO) definitions of stability for linear, time-invariant systems using the total response

1. A system is stable if every bounded input yields a bounded output.

2. A system is unstable if any bounded input yields an unbounded output.

Stable systems have closed-loop transfer functions with poles only in the left half-plane

Unstable systems have closed loop transfer functions with at least one pole in the right half-plane and/or poles of multiplicity greater than one on the imaginary axis


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Marginally stable systems have closed loop transfer functions with only imaginary axis poles of multiplicity 1 and poles in the left half-plane

Stable System Unstable System


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Routh-Hurwitz Criterion

• yields stability information without the need to solve for the closed-loop system poles

Two steps:

(1) Generate a data table called a Routh table

(2) Interpret the Routh table to tell how many closed-loop system poles are inthe left half-plane and in the right half-plane.


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Generating a Basic Routh Table

Note: any row of the Routh table can be multiplied by a positiveconstant without changing the values of the rows below.


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Interpreting the Basic Routh Table

• the number of roots of the polynomial that are in the right half-plane is equal to the number of sign changes in the first column.

- two sign changes, hence two poles in the right half-plane


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Problem: Find range of gain K for which the system is stable, K>0

Solution:

System is stable for K<1386


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Stability in State Space

• we can determine the stability of a system represented in state space by finding the eigenvalues (poles) of the system matrix, A, and determining their locations on the s-plane

• we can find the poles using equation:


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Steady-State Errors• How to find the steady-state error for a unity feedback system

• How to specify a system's steady-state error performance

• How to find the steady-state error for disturbance inputs

• How to find the steady-state error for nonunity feedback systems

• How to design system parameters to meet steady-state error performance specifications

• How to find the steady-state error for systems represented in state space


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Steady-State ErrorsSteady-state error is the difference between the input and the output for a prescribed test input as ∞→t

Test waveforms for evaluating steady-state errors of control systems


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Evaluating Steady-State Errors


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Sources of Steady-State Errors

Error:

- zero error in the steady state for a step input


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Steady-State Error for Unity Feedback Systems

Steady-State Error in Terms of T(s) – closed loop transfer function

… final value theorem


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Steady-State Error in Terms of G(s) – open loop transfer function


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Response to step, ramp and parabolic input

Step input

Thus if n ≥ 1 then 0)( and )(lim0

=∞∞=→

esGs

Ramp input

Thus if n ≥ 2 then 0)( and )(lim0

=∞∞=→

essGs

if (at least one pole must be at origin)


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Parabolic input

Thus if n ≥ 3 then 0)( and )(lim 2

0=∞∞=

→esGs

s


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Static Error Constants and System TypeStatic Error Constants

Thus, for a step input:

for a ramp input:

for a parabolic input:


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System Type

n=0 … Type 0 systemn=1 … Type 1 systemn=2 … Type 2 system


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Problem Given the control system, find the value of K so that there is 10% error in the steady state.

Solution


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Steady-State Error for Disturbances (1)

)()()( sCsRsE −=


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Steady-State Error for Disturbances (2)

- let us assume a step disturbance

- hence error can be reduced by increasing the DC gain of G1(s)or decreasing the dc gain of G2(s)


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Steady-State Error for Nonunity Feedback Systems (1)

• Control systems often do not have unity feedback because of the compensation used to improve performance or because of the physical model for the system.

• When nonunity feedback is present, the plant's actuating signal is not the actual error or difference between the input and the output.

)(/)()()()()(

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sGsHsHsGsGsG

==

G1(s) … input transducerG2(s) … plantH1(s) … feedback

Ea(s) … actuating signal (not the actual error)


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Let us modify, the block diagram to show explicitly the actual error E(s)=R(s)-C(s)

Ea(s) … actuating signal(not the actual error)

E(s)=R(s)-C(s)actual error



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If we consider step input and step disturbance, then R(s)=D(s)=1/s and


- both disturbance and nonunity feedback


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For zero error:

Above equation can be satisfied if:

1. System is stable

2. G1 is a Type 1system

3. G2 is a Type 0 system

4. H is a Type 0 system with DCgain=1


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Problem Determine the system type, error constant, and the steady state errorfor a unit step

Solution

E(s)=R(s)-C(s)actual error


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Sensitivity

Sensitivity - the degree to which changes in system parameters affect system transfer functions, and hence performance.

Sensitivity is the ratio of the fractional change in the function to the fractionalchange in the parameter as the fractional change of the parameter approaches zero.


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Problem Find the sensitivity of the steady-state error to changes in parameter Kand parameter a for the system with a step input

Solution


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Steady-State Error for Systems in State Space(1) analysis via final value theorem(2) analysis via input substitution

Analysis via Final Value Theorem

closed-loop transfer function


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Analysis via Input Substitution – avoids taking inverse of (sI - A)

Step input (r = 1)

If the input is a unit step r = 1 then a steady-state solution xSS for x is:

where Vi is constant.

Thus, for a unit step input:

BCACV 1−−==ssy


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Ramp inputs (r = t)

We balance the equation:

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