Moulton  Lectures

 

On

Electro-Acoustics

 

 

 

Presented by:

Dave L Moulton

 

 

 

 

Location:  Thales Acoustics Harrow UK

Date:  26-February-2003

 

Content

 

Lecture 14 looked at a definition for stability. I will now take the subject  of stability a little further and look at a powerful method for determining the stability of a system by interrogating the zeros of the characteristic equation.

 

On with the lecture

 

The Characteristic Equation

First lets define the Characteristic equation. We can use our understanding of the ANR feedback system model as an example:

 

From this model we can determine the transfer function relating the ratio of the cancelled pressure P_e(s) to the incident noise pressure P₀(s), resulting in the now familiar expression:

 

 

Equation [15.1] is an example of what is commonly known as an error feedback function.

 

The denominator (1 – kG(s)H(s)) is known as the Characteristic Equation, whereas the whole expression is known as the system feedback function.

 

Relationship between the Characteristic Equation and system stability.

 

The stability of [15.1] is determined by the location of its poles on the s-plane.  Looking at the feedback function it should be clear that the poles are going to be a determined by the Characteristic Equation.

 

Let us look at this in a little more detail. We can write the functions G(s) and H(s) as fractions with zeros (z) in the numerator and poles (p) in the denominator.

 

Thus:

     and   

Putting these expressions in to [15.1] we get:

 

Rearranging:

 

Expressing the numerator  in a more general form showing polynomials:

 

 

The polynomial fraction in the numerator is a representation of the Characteristic Equation in terms of generalized polynomials.

 

Thus:

 

We can go back to [15.3]  and rewrite our feedback function as:

 

 

Thus:

 

What we have done is take the Characteristic Equation and express it in terms of ‘pole and zero’ polynomials.  Putting these expressions into the feedback function results in the Characteristic Equation zeros becoming the poles of the feedback function.

 

Thus:

 

In order to investigate the stability of the feedback function we need to investigate the location of the zeros of the Characteristic Equation.

 

The zeros of the Characteristic equation are in fact the poles of the system feedback function.

 

Thus we are interested in analyzing the polynomial:

 

 

If we could easily factorize this function it will take the form:

 

 

Where the location of the roots r on the s-plane determine the degree of stability of the system.

 

For example, a 3^rd order polynomial could take the form:

 

 

In this case we have:   A₃ = 1, A₂= 4, A₁= -21,  A₀ = -30,

 

Fortunately [15.10] can be easily factorized to give:

 

 

revealing poles at:  s = -2,  s = 3 and s = -5.

 

These poles can be plotted onto the s-plane,

Indicating that one of the poles is located in the Right Hand s-plane, thus making the system unstable.

 

Stability Analysis Using The Routh Hurwitz Method

 

What I want to show now is a short hand method of determining the location of the closed loop system poles from an analysis of the Characteristic Equation  zero polynomial coefficients.  Thus we are interested in analyzing A_n,  A_n-1,  A_n-2,  A_n-3, etc……. From equation  [15.8]

 

Lets have a closer look at the factorized version of [15.8] represented in [15.9]

 

 

Expanding [15.9] we result in:

 

 

 

……………….[15.13]

 

This expression can be explained as::

 

a_nSⁿ - a_nS^n-1(Sum of all the roots) + a_nS^n-2(Sum of the product of the roots taken 2 at a time) -  a_nS^n-2(Sum of the product of the roots taken 3 at a time) +…….+ a_nS^n-2(-1)ⁿ(Product of  all n roots) = 0

 

 

Now looking at [15.9] it is clear that the format for a pole to lie in the left hand plane is: (s – r)=0 where r is –ve, thus if r =(-r_p) then  (s-(-r_p))=0, leading to a pole r_p being located in the left hand plane.

 

Thus if any pole lies in the left hand plane it will add a negative contribution to each of the coefficients in [15.12] and [15.13]. 

 

Routh and Hurwitz found a method of determining how many of these negative contributions are occurring by using a method of creating an array filled with numbers derived from the determinants of successive coefficients.

 

The resulting array gives an insight into how many of the poles actually lie in the Right Hand plane by looking at the number of sign changes in the first column of the array. 

 

The Routh Hurwitz Array

 

Consider the characteristic ‘zero’ polynomial shown in [15.15]:

 

 

To create the array we represent alternate coefficients on two rows.

 

The zero row contains the coefficients:  A_n,  A_n-2,  A_n-4,  A_n-6,  etc….

 

The first row contains the coefficients:  A_n-1,  A_n-3,  A_n-5,  A_n-7,  etc….

 

Row 2 is derived from the coefficients in row 1 and row 0 rows.

 

Row 3 is derived from the coefficients in the  row 2 and row 1 etc….

The values of the coefficients in rows 2 onwards are calculated using a method of determinants:

 

,  ,

 

,   ,   

 

Where the determinants are calculated in the usual way:

 

For example:

 

Consider the polynomial:

 

 

Setting up the zero and first rows of the array:

 

Now calculating each of the coefficients

 

In this array the numbers in the first column are:  1, 6,  32/3, 8.  All of these numbers are positive so there are no sign changes and no poles in the right hand plane.

 

The Routh-Hurwitz Criterion States that:

 

The number of zeros of the characteristic equation with positive real parts is equal to the number of sign changes in the first column of the coefficients in the Routh-Hurwitz array.

 

Lets take another example:

 

Just to prove the validity of the Routh Hurwitz method [15.18] can be factorized as follows:

 

 

We can see that there are two poles in the Right Hand s-plane at s=3 and s=6, thus the system is unstable.  Now lets see what the Routh Hurwitz Array reveals.

 

Here we see two sign changes in the first column. From row 0 to row 1 the number changes from positive to negative (First sign change). From row 2 to row3 the number changes from negative to positive (Second sign change).  Thus the Routh-Hurwitz Array has two sign changes and agrees with the factorized equation indicating that there are two roots in the Right hand s-plane.

 

 

Special Cases For The Routh Hurwitz Array

 

Case 1:  A zero appears in the first column before the array is completed

In constructing the Routh-Hurwitz array, there is one problem that is often encountered which prevents us from completing the array. We can see that if any of the coefficients in the first column goes to zero, except the last coefficient, then the array cannot be evaluated.

 

Consider the example:  

 

If we try and form the array we get:

 

Clearly the zero in the first column of row 2 prevents any further computation of the array.

 

To get around this problem we substitute the zero for a symbol, usually e which we define as tending to zero but remaining positive (on the positive side of zero).

 

Thus:      e  ® 0+

 

Substituting e into the array allows us to complete the array:

 

 

Since  e  is very small and positive, (8e - 20)/ e is negative. This means that there are two sign changes in the first column and therefore the closed loop function is unstable.

 

 

Case 2:  All the Coefficients of a row go to zero.

The method shown in case 1 cannot be used if all of the coefficients in a row (except the last row) go to zero. 

 

For example:

 

Forming the Array we see that  all of the coefficients in row 2 have gone to zero.

 

 

When ever this happens we must the  coefficients of the row above with the coefficients of an auxiliary equation derived from that row.  The auxiliary equation is basically the resulting coefficients of the row after differentiation.  In the example above the auxiliary equation is created from row 1 and substituted into row 1.

 

We can write row 1 as:

 

differentiating w.r.t s we get:

 

Substituting the coefficients 12, 8 and –4  into row 1 we get:

 

 

The meaning of a row of zeros.

 

Whenever all the coefficients in a row go to zero, it indicates the existence of two roots having the same magnitude but opposite sign. 

A pair of purely imaginary roots, or four roots symmetrically located about the origin of the s-plane would cause a vanishing row.

 

 

End of Lecture