control chap5

CONTROL SYSTEMS THEORY

Transient response stability

CHAPTER 5STB 35103

Objective To determine the stability of a system

represented as a transfer function.

Introduction In chapter 1, we learnt about 3

requirements needed when designing a control system Transient response Stability Steady-state errors

Introduction What is stability?

Most important system specification. We cannot use a control system if the system is

unstable Stability is subjective

From Chapter 1, we have learned that we can control the output of a system if the steady-state response consists of only the forced response. But the total response, c(t)

forced naturalc t c t c t

Introduction Using this concept we can summarize the

definitions for linear, time-invariant systems.

Using natural response; A system is stable if the natural response

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

approaches infinity as time approaches infinity. A system is marginally stable if the natural

response neither decays nor grows but remains constant or oscillates.

Introduction A system is stable if every bounded input

yields a bounded output. (bounded = terkawal). We call this statement the bounded-input, bounded-output (BIBO).

Using the total response (BIBO) A system is stable if every bounded input

yields a bounded output. A system is unstable if any bounded input

yields an unbounded output.

Introduction We can also determine the stability of a

system based on the system poles. 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 1 on the imaginary axis.

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

Introduction Figure 5.1 a indicates closed-loop poles for

a stable system.

Figure 5.1 a - Closed-loop poles and response for stable system

Introduction Figure 5.1 a indicates closed-loop poles for

an unstable system.

Figure 5.1 b - Closed-loop poles and response for unstable system

Stability summary

Introduction In order for us to know the stability of our

system we need to draw the system poles. To find the poles we need to calculate the roots of the system polynomials.

Try to get the system poles for the systems in Figure 5.1 a and Figure 5.1 b.

Introduction What about this system? Can you find the

root locus for this polynomial?

A method to find the stability without solving for the roots of the system is called Routh-Hurwitz Criterion.

Figure 5.2 – Close loop system with complex polynomial.

Routh-Hurwitz Criterion We can use Routh-Hurwitz criterion method

to find how many closed-loop system poles are in the LHP, RHP and on the jω-axis

Disadvantage : We cannot find their coordinates

The method requires two steps: Generate a data table called a Routh table Interpret the Routh table to tell how many close-

loop system poles are in the left half-plane, the right half-plane, and on the jω-axis

Routh-Hurwitz Criterion Example

Figure 5.3 displays an equivalent closed loop transfer function. In order to use Routh table we are only going to focus on the denominator.

Figure 5.3 – Equivalent closed-loop transfer function

Routh-Hurwitz Criterion First step (1)

Based on the denominator in Figure 5.3, the highest power for s is 4, so we can draw initial table based on this information. We label the row starting with the highest power to s0.

s4

s3

s2

s1

s0

Routh-Hurwitz Criterion Input the coefficient values for each s

horizontally starting with the coefficient of the highest power of s in the first row, alternating the coefficients.

s4 a4 a2 a0

s3 a3 a1 0

s2

s1

s0

Remaining entries are filled as follows. Each entry is a negative determinant of entries in the previous two rows divided by the entry in the first column directly above the calculated row.

Routh-Hurwitz Criterion

Routh-Hurwitz Criterion

Routh-Hurwitz Criterion

Routh-Hurwitz Criterion

Routh-Hurwitz Criterion

Routh-Hurwitz Criterion

Routh-Hurwitz Criterion

Routh-Hurwitz Criterion

Routh-Hurwitz Criterion

Routh-Hurwitz Criterion Example 6.1

Make a Routh table for the system below

Answer: get the closed-loop transfer function

Routh-Hurwitz Criterion We can multiply any row in Routh table by

a positive constant without changing the rows below.

Routh-Hurwitz Criterion Interpreting the basic Routh table

In this case, the Routh table applies to the systems with poles in the left and right half-planes.

Routh-Hurwitz criterion declares that 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.

Routh-Hurwitz Criterion

Routh-Hurwitz Criterion If the closed-loop transfer function has all poles in the

left half of the s-plane, the system is stable. The system is stable if there are no sign changes in

the first column of the Routh table. Example:

Routh-Hurwitz Criterion

Based on the table, there are two sign changes in the first column. So there are two poles exist in the right half plane. Which means the system is unstable.

+

-

-

+

Routh-Hurwitz Criterion Exercise 1

Make a Routh table and tell how many roots of the following polynomial are in the right half-plane and in the left half-plane.

7 6 5 4 3 23 9 6 4 7 8 2 6P s s s s s s s s

Routh-Hurwitz CriterionAnswer

Routh-Hurwitz CriterionAnswer

Routh Hurwitz Exercise 2

Routh Hurwitz Solution

Routh Hurwitz Exercise 3

Solution

Routh Hurwitz Exercise 4

Routh Hurwitz Solution

Routh-Hurwitz Criterion: Special cases Two special cases can occur:

Routh table has zero only in the first column of a row

Routh table has an entire row that consists of zeros.

s3 1 3 0

s2 3 4 0

s1 0 1 2

s3 1 3 0

s2 3 4 0

s1 0 0 0

Routh-Hurwitz Criterion: Special cases Zero only in the first column

There are two methods that can be used to solve a Routh table that has zero only in the first column.

1. Stability via epsilon method

2. Stability via reverse coefficients

Routh-Hurwitz Criterion: Special cases Zero only in the first column

Stability via epsilon method

Example 6.2Determine the stability of the closed-loop

transfer function

5 4 3 2

10

2 3 6 5 3T s

s s s s s

Routh-Hurwitz Criterion: Special cases Solution: We will begin forming the Routh table

using the denominator. When we reach s3 a zero appears only in the first column.

s5 1 3 5

s4 2 6 3

s3 0 7/2 0

s2

s1

s0Zero in first column

Routh-Hurwitz Criterion: Special cases If there is zero in the first column we

cannot check the sign changes in the first column because zero does not have ‘+’ or ‘-’.

A solution to this problem is to change 0 into epsilon (ε).

s5 1 3 5

s4 2 6 3

s3 0 7/2 0

s2

s1

s0

ε

Routh-Hurwitz Criterion: Special cases We will then calculate the determinant for

the next s values using the epsilon.

Routh-Hurwitz Criterion: Special cases If we all the columns and rows in the

Routh table we will get

Routh-Hurwitz Criterion: Special cases We can find the number of poles on the

right half plane based on the sign changes in the first column. We can assume ε as ‘+’ or ‘-’

Routh-Hurwitz Criterion: Special cases There are two sign changes so there are

two poles on the right half plane. Thus the system is unstable.

Routh-Hurwitz Criterion: Special cases Zero only in the first column

Stability via reverse coefficients

Example 6.3Determine the stability of the closed-loop

transfer function

5 4 3 2

10

2 3 6 5 3T s

s s s s s

Routh-Hurwitz Criterion: Special cases Solution: First step is to write the denominator in

reverse order (123653 to 356321)

We can form the Routh table using D(s) values.

5 4 3 23 5 6 3 2 1D s s s s s s

Routh-Hurwitz Criterion: Special cases The Routh table indicates two signal

changes. Thus the system is unstable and has two right-half plane poles.

Routh-Hurwitz Criterion: Special cases Entire row is zero

the method to solve a Routh table with zeros in entire row is different than only zero in first column.

When a Routh table has entire row of zeros, the poles could be in the right half plane, or the left half plane or on the jω axis.

Routh-Hurwitz Criterion: Special cases Example 6.4

Determine the number of right-half-lane poles in the closed-loop transfer function

5 4 3 2

10

7 6 42 8 56T s

s s s s s

Routh-Hurwitz Criterion: Special cases Solution: Start with forming the initial Routh table

Routh-Hurwitz Criterion: Special cases We can reduce the number in each row

Routh-Hurwitz Criterion: Special cases We stop at the third row since the entire

row consists of zeros.

When this happens, we need to do the following procedure.

Routh-Hurwitz Criterion: Special cases Return to the row immediately above the

row of zeros and form the polynomial.

The polynomial formed is

4 26 8P s s s

Routh-Hurwitz Criterion: Special cases Next we differentiate the polynomial with

respect to s and obtain

We use the coefficient above to replace the row of zeros. The remainder of the table is formed in a straightforward manner.

34 12 0dP s

s sds

Routh-Hurwitz Criterion: Special cases The Routh table when we change zeros

with new values

Routh-Hurwitz Criterion: Special cases Solve for the remainder of the Routh table

There are no sign changes, so there are no poles on the right half plane. The system is stable.

Example 1- normal

Example 2 - special case

Example 3 – special case

Example 4 – special case

Exercise Given that G(s) is the open loop transfer

function for a unity feedback system, find the range of K to yield a stable system

Solution

Exercise Find the range of K to yield a stable

system given the closed loop transfer function below

Solution