Chapter 6

Modeling and Control of Dynamic Systems MESB323

Chapter 7

Stability

7.1. Introduction

Stability is a very important aspect of control system. One of the main objectives of control system is to achieve stability. If a system is not stable then the response will become uncontrollable and the system will be unstable. In this topic we will learn about system stability and how to determine whether a system is stable or unstable.

7.2. Definitions

There are two definitions of stability:

First Definition:Based on total response of the system which we call 'Bounded Input Bound Output' (BIBO).

Second Definition:Based on a natural response of the system due to initial condition.

Based on the first definition:(a)A Linear time invariant (LTI) system is considered to be stable if a bounded input yields a bounded otput. Bounded mean the signal does not go out bounded.

(b)A LTI system is considered to be unstable if a bounded input yields an unbounded output. This mean the output goes beyond bound.

Based on the second definition:

(a)A system is said to be stable, if the natural response of the system approaches zero as time (t) approaches infinity ().

(b)A system is said to be unstable if the natural response of the system approaches infinity as time (t) approaches infinity ().

Let us examine the two cases below:

Recalling basic chapter 5, output response of system consists of two parts:

Forced response + natural responses

C(t) = Cf (t) + Cn (t)

Total response = forced response + natural response

(i)Transfer function(pole is 2)y(t) = output and u(t)= unit step input. The initial condition y(0) = 0.

Suppose u(t) =1, then U(s)=,then the output is

y (t) = A + B e2t

Force responseNatural response

As . Alternatively, output y(t) is bounded and the unit step input u(t) is also bounded. Hence the system is stable.

(ii) Transfer function (pole is +2)

For the same input condition as in case (i),

The input u(t) is bounded but the output is unbounded due to positive exponential term. Hence the system is unstable.

Case (i) above where as only happens if the closed loop poles of the system are on the left half of the s plane, basically if the poles have negative real values.

Closed loop poles on the right half plane yields instability as shown in case (ii)

A system is stable if the closed loop poles are on the left hand half of the splane

7.3. Necessary Condition of Stability

The closed loop poles are all on the left half plane if the coefficients of the characteristic equation are all present and having the same sign. Characteristic equation (C.E) is the denominator of the closed loop system's transfer function. If the coefficients of the C.E are all having the same sign it does not mean that all the poles are on the left half plane. On the other hand, if the poles are on the left half plane then all the coefficients of the C.E must have the same sign.

Example 7.1For the closed loop transfer function given below, determine the stability of the system.

(a)

(b)

(c)

Solution:

(a) The C.E is

All coefficients are positive

Since all coefficients are same sign then the system satisfies the necessary condition for stability. However we cannot surely state that the system is stable without locating the closed loop poles.

(b) The C.E is

In this case all the coefficients are not the same sign as the coefficient for s2 is negative whereas all the other coefficients are all positive. Therefore this system does not satisfy the necessary condition and it is not a stable system.

(c)

The C.E is

In this case all the coefficients are having the same sign but not all the coefficients are present. Coefficient for s2 is 0, therefore this is not a stable system. In order to check for stability, we need to find the exact location of the closed loop poles. It is easy to find if the characteristic equation is in the order of 2 or 3. However, it is very tedious when it comes to higher order equation. In this case RouthHurwitz Criterion becomes very useful.

7.4. RouthHurwitz Criterion

RouthHurwitz Criterion states that, the roots of the equation are all in the left half plane (LHP) if all the elements of the first column of the Routh's tabulation are having the same sign. The number of sign changes in the first column of the Routh table equals to the number of roots with positive real parts on the right half plane (RHP).

Routh's Table:

Let the transfer function of closed loop system be T(s) where

The characteristic equation in this case:-

C.E = = 0

(a) Start a column with the highest power of s in the C.E to s0.

Table 7.1

(b) Fill the first row starting with the coefficient of the highest power of s followed by every other coefficient in the C.E equation.

Table 7.2

(c) Do the same for the next row starting with next highest power of s.

0

No more coefficients

Table 7.3

(d) Each entries of the column are actually the negative determinant of entries in the previous two rows divided by the entry in the first column directly above the calculated row. The left hand column of the determinant is always the first column of the two previous rows. The right hand column of the determinants is the elements column above and to the right.

0

Table 7.4

Note: To find the determinant, we do a cross multiply as shown below:

Example 7.2

Given a closed loop transfer function:

(a)Indicate how many poles are on the right half s plane and how many poles are on the left half plane.

(b)Is the system stable?

Solution:

The characteristic equation:

CE = s3- 4s2 + s + 6 =0 By observation, we know that the system is unstable as the coefficients of the C.E are not having the same signal. First we solve it the normal way where we locate the closed loop poles and then we will verify this with Routh Hurwitz method. In this case we can do both ways as the order of the C.E is only 3.

Setting C. E = 0 (We can locate the closed loop poles)

s3 4s2 + s + 6 = (s 2)(s + 1)(s 3)=0

Therefore the poles are at

s = 2, s = 1 and s = 3

There are two poles on the right half s plane and one pole on the left half splane. This system is definitely unstable.

Second Method: Routh Hurwitz Method

C.E=s34s2+s+6 =0s311

s2-46

s1a1=2.5 a2=0

s0b1=60

Table 7.5

Basically the table will be

s311

s2-4 (1st sign change)6

s12.5 (2nd sign change)0

s06

Table 7.6

Routh Hurwitz criterion states the number of sign changes in the first column of the Routh table indicates the number of poles on the right half splane. In the table above we can observe there are two sign changes. Therefore there are two poles on the right half s plane and another pole is on the left half splane. This is the same result that we found by solving the C.E to obtain the closed loop poles. This system is unstable

Example 7.3:

Given the closed loop transfer function of a system is T(s) where

T(s) =

(i)Indicate how many poles on the right half s plane and how many poles on the left half s plane.

(ii)Indicate whether the system is stable or unstable.

Simplified the table will be: Two sign change

4620

2100

-1420

0

20

Table 7.8

(a) From the Table 7.8, we can see that there are two sign changes therefore 2 poles on the right half splane. Since this system supposed to have 4 poles (the denominator of the system is 4th order) the other two poles are on the left half s plane.

(b) The system is unstable as the system has poles on the right half splane.

7.3.1Special Cases for Routh Table

There are two special cases that we need to consider while tabulating the Routh table.

1. The first element in anyone row of Routh's table is zero but others are not.

This case will be explained using Example 7.4

Example 7.4:

Given the closed loop transfer of a system

(a) Tabulate the Routh table.

(b) Find how many poles are on the right half splane and how many poles on the left half splane. Solution:

(a) CE = s4 +s3 +4s2 +4s+3

s4143

s3140

s2

s1?

s0

Table 7.9

As we can see in the table above, the value of bl is undefined as a1 is zero. For cases like this above where the first element in any one row is zero, the table can be completed by a replacing a very small value which is called epsilon (). This can be taken as a very small positive value or a very small negative value.

(b) (very small positive value)

s4143

s3140

s2

3

s1

0

s03

Table 7. 10

Since is a very small positive value, the sign in the first row of Table 6.10 is

+

+

+

-

+

1 sign change

2 sign change

Table 7. 11

(a)From the table above, the two sign changes. Therefore there are two poles on the right half splane and two poles on the left half splane. We will get the same result if we take to be a very small negative value as shown below.

(b) (very small negative value)

+

+

-

+

+

Table 7.12

The results are just the same as in (ii) where a positive small value is used.

2. The second special case occurs when all elements in a row are zeros.

This indicates one or more of the following conditions:

(a) The equations has one pair of real roots with equal magnitude but opposite signs. (b)The equation has one or more roots on the imaginary axis. (c)The equation has complex conjugate roots forming symmetry at the origin.

Example 7.5:

Given a closed loop transfer function of a system

(a) Tabulate the Routh table.

(b) Indicate the location of the closed loop poles solution.

Solution:

(a) CE = s5110169

110169

000

All element in the row are zero

Table 7.13

The situation with all zeros in the row like above is solved using auxiliary equation of P(s) = 0. Auxiliary equation P(s) is formed from the coefficients of the row just before the row of zeros in the Routh tabulation. For this case the row above the row of zeros have coefficient 1 for s4, 10 for s2 and 169 for s0.

Therefore:

P(s) = s4 + 10s2 +169s0

Take the differentiation of P(s) with respect to s

The coefficients of are filled replacing the row of zeros and the other elements are found like usual using these values.

s5110169

110169

420

5169

-115.2

169

Table 7.14

(b) For this case, we have two consider the tables in two different parts in determining the location of the closed loop transfer function. The first part will be from the beginning of the table till the row before the row of zeros (ROZ). The second part will be a test of even polynomial from the row before the row of zeros (ROZ) till the end of the table.

(i)For the first part we have:

+

+

1 pole*Note this is order 1 as only 1 pole

Table 7.15

There are no sign changes indicating 1 pole on the left hand side of the s-plane.

The second part, we have

+

1 pole

+

2 pole

+

3 pole

-

4 pole

+

Note: This is orderfour so there are 4poles with 2 sign changes

Table 7.16

In this case there is 2 sign changes indicating 2 poles of the right hand side of the 2plane and 2 poles on the left hand side of the splane because of the requirement of symmetry. Basically we can summarize in a table form as shown below:

LocationOtherOrder 1EvenFourth OrderTotalFifth Order

Right hand side022

Left hand side123

000

Table 7.17

The system is unstable since there are two poles on the right hand side of the splane.

Note: If there are no sign changes in the second part then there is no poles on the right half plane which means there are no poles on left half plane as well (symmetry condition). Therefore all the 4 poles will be on the imaginary axis.

10 CYL