very large abnormal value of partial fraction expansion using answers of residuez command?

I am making a program on matlab to determine partial fraction expansion of z transform given by G(z)=(18z^3) / (18z^3+3z^2-4z-1)

code m file attached

but whenever i run this program i get abnormally large/garbage values in H that i think are not realizeable

Answer 1

[Adam Danz]

To understand why these large numbers appear, let's look at a small part of the last line of your code.

From your full line

H=r(1)/(1-p(1)*z^(-1) ) +r(2)/(1-p(2)*z^(-1))+r(3)/(1-p(3)*z^(-1))

let's look at

1-p(2)*z

When you evaluate 1-p(2)*z you get 1.333*z. That is converted to a fraction (375299970653903*z)/1125899906842624 + 1. When you evaluate 1-p(2)*z, the output will be in fraction format and if you evaluate that fraction, you'll find that it's the same as 1.333*z. Since your code converts to fraction format, those large numbers stay in the equation.

As a test, reduce the equation manually, assign z a value and compare the result of the simplified and non-simplified equations.

For example, given your other inputs, your last line reduces to ( done quickly, double check...)

H=r(1)/(1-p(1)*z^(-1) ) +r(2)/(1-p(2)*z^(-1))+r(3)/(1-p(3)*z^(-1))
H = 0.36/(0.5*z^(-1)) + 0.24 / (1.333*z^(-1)) + 0.4 / (1.333*z^(-1))
H = (39994*z)/33325

[ABTJ]

Thanks dear sir for your kind reply but i am still very much confused. I have to clear this confusion to my students. I understood your reply to some extent but still i am confused. As far as your word" *When you evaluate 1-p(2)*z you get 1.333*z. That is converted to a fraction (375299970653903*z)/1125899906842624 + 1. When you evaluate 1-p(2)*z, the output will be in fraction format and if you evaluate that fraction, you'll find that it's the same as 1.333*z." that is okay but i am confused regarding the number 8646911323865919 appearing in numerator and how it shapes the overall answer

[ABTJ]

i mean you told about denominator "1-p(2)*z" but numerator is not clear. I am confused regarding numerator & its effect

[ABTJ]

is there any method in matlab,that we can skip this big number form and get simple answer just like

H=r(1)/(1-p(1)*z^(-1) ) +r(2)/(1-p(2)*z^(-1))+r(3)/(1-p(3)*z^(-1)) H = 0.36/(0.5*z^(-1)) + 0.24 / (1.333*z^(-1)) + 0.4 / (1.333*z^(-1))

[Adam Danz]

To answer your first question, regarding the number, 8646911323865919, there are three terms in your equation:

H = r(1)/(1-p(1)*z^(-1) )  
 +  r(2)/(1-p(2)*z^(-1))  
 +  r(3)/(1-p(3)*z^(-1))

Those three terms evaluated individually are

   -9/(25*(1/(2*z) - 1))
   8646911323865919/(36028797018963968*(375299970653903/(1125899906842624*z) + 1))
   3602879692067755/(9007199254740992*(3002399737929439/(9007199254740992*z) + 1))

Notice the number 8646911323865919 is coming from the 2nd term. When the full equation is evaluated, for some reason I haven't looked into the order of the terms are rearranged but that wouldn't affect the result anyway.

I'd have to think about your 2nd question later.