Question on residuez results

I wanted to partial fract the following

3x^2+x-2/(((x-2)^2)(1-2x))

I wrote the following

a=conv([-2 1],[-2 1]);
a=conv(a,[1 -2]);
[R P K]=residuez([3 1 -2],flipslr(a))

I am not getting the expected residues of

1/3, -5/3 and -4.  Why?

Answer 1

[Roger Stafford]

The actual partial fraction expansion of 3*x^2+x-2/(((x-2)^2)(1-2x)) is:

   3*x^2 + x - (4/9)/(x-2) + (2/3)/(x-2)^2 + (4/9)/(x-1/2)

I don't see how that agrees with your "expected" values. What are the values you obtained in [R P K]?

[Seetha Rama Raju Sanapala]

Since the denominator was not given descending powers of the x, I am just forming the denominator polynomial using the conv operation on factors of the denominator. I think your answer is wrong. The fraction is proper it can not have direct terms. Only residues.

The answer I obtained was

R = [0.333 -2 0.1666], P=[2 2 0.5], K=[];

The expected answers are

R = [-2.666 -4 0.1666], P=[2 2 0.5], K=[];

[Roger Stafford]

As you wrote the expression, Seetha, the partial fraction expansion I gave is actually correct. However, I believe you meant to write the following for your expression:

   (3*x^2+x-2)/(((x-2)^2)(1-2*x))

Note the all-important parentheses around 3*x^2+x-2. With this modification the revised partial fraction expansion would be:

   (-5/3)/(x-2) + (-4)/(x-2)^2 + (1/6)/(x-1/2))

I have checked this with matlab and it is correct. The second two terms are in agreement with what you expected but not the first.

For fliplr(a) (not flipslr) you should have obtained [-2 9 -12 4]. For

   residuez([3 1 -2],[-2 9 -12 4])

I would have thought you would get R = [-5/3 -4 1/6], P = [2 2 1/2]. I can't account for the result you obtained. I do not have the Signal Processing Toolbox to check this. My advice would be to convert R and P back into the form of the ratio of two polynomials using the reverse form of 'residuez' and see if this checks with your original expression or if not, what the difference is. It might give you an idea of the source of the trouble.