Control System

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INTRODUCTION

Open- and closed-loop transfer functions have certain basic characteristics that permit transient and steady-state analyses of the feedback-controlled system. Five factors of prime importance in feedback-control systems are stability, the existence and magnitude of the steady-state error, controllability, observability, and parameter sensitivity. The stability characteristic of a linear time-invariant system is determined from the system’s characteristic equation. Routh’s stability criterion provides a means for determining stability without evaluating the roots of this equation. The steady-state characteristics are obtainable from the open-loop transfer function for unity feedback systems (or equivalent unity-feedback systems), yielding figures of merit and a ready means for classifying systems.

ROUTH HURWITZ STABILITY CRITERION
Stability
The response of control systems consists of (1) natural response and forced response, or (2) zero input response and zero state response. For natural (zero input) response, a system is,

1. stable if the natural response approaches zero as time approaches infinity. 2. unstable if the natural response approaches infinity as time approaches infinity. 3. marginally stable if the natural response neither decays nor grows but remians constant or oscillates.

For the total response, a system is
1. stable if every bounded input yields a bounded output.
2. unstable if any bounded input yields an unbounded output.

The Routh Hurwitz Stability Criterion
The Routh Hurwitz stability criterion is a tool to judge the stability of a closed loop system without solving for the poles of the closed loop system. Generating a Routh-Hurwitz table


SOLVED PROBLEMS

1. Given the characteristic equation, is the system described by this characteristic equation stable?

Answer:
One coefficient (-2) is negative.
Therefore, the system does not satisfy the necessary condition for stability.

2. Given the characteristic equation, is the system described by this characteristic equation stable? Answer:
All the coefficients are positive and nonzero.
Therefore, the system satisfies the necessary condition for stability.
We should determine whether any of the coefficients of the first column of the Routh array are negative.

The elements of the 1st column are not all positive: the system is unstable

3.

Any row of the Routh table may be multiplied by a positive number. 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. In the above example,

s3 1 31 0
s2 1 103 0
s1 −72 0 0
s0 103 0 0

There are two sign changes in the first column.
1 =) −72 and − 72 =) 103
Therefore, two of the four roots of s3 +10s2 +31s+1030 = 0 are in the right half plane of s-plane.

4. Special case: Zero in the first column
10
s5 + 2s4 + 3s3 + 6s2 + 5s + 3

When zero is resulted in the first column, replace 0 with a small є.

Examine the first column by assuming є is a either positive or negative number.
Since there are two sign changes, the polynomial has two poles in the right half plane.

5. Special case: An entire row is zero
10
s5 + 7s4 + 6s3 + 42s2 + 8s + 56

The row if s3 is entirely zero. Let the polynomial above (row of s4) be P(s). Then, differentiate P(s).

P(s) = s4 + 6s2 + 8
= 4s3 + 12

Then, replace the row of s3 with the coefficients of dP(s)/ds. Then, proceed normally.
There are no sign changes, so the system T(s) is not unstable.

6. Q(s) = s3 + s2 +2s +8 ( = (s + 2)(s2 – s + 4))
Routh array
s312
s218
s1-6
s08

Two sign changes in the first column, therefore there are two roots in right half plane and the system is unstable.

1 -6 81 ± j√15¬¬
22

7.
Q(s) = s5 + 2s4 + 2s3 + 4s2 + 11s + 10
Routh...
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