How to use Matlab built-in functions “tfdata” and “residue” and partial fraction decomposition

I have a doubt , for a give plant P = (-4220.5s^2+3070.5s+1150)/(16.6s+1)(5s+1)(291.4s^2+27.4s+1)

How can i get them into three partial fraction decomposition in matlab

Answer 1

[Star Strider]

There appear to be several multiplication operators missing, and there could be missing parentheses.

Try this:

syms s 
P1 = (-4220.5*s^2+3070.5*s+1150)/(16.6*s+1)*(5*s+1)*(291.4*s^2+27.4*s+1);
P1pf = partfrac(P1)
P2 = (-4220.5*s^2+3070.5*s+1150)/((16.6*s+1)*(5*s+1)*(291.4*s^2+27.4*s+1));
P2pf = partfrac(P2)

producing:

P1pf =
(2441538294769*s)/94916642 + 5318267004600/(3939040643*(83*s + 5)) + (92013299175*s^2)/571787 + (1259979077*s^3)/6889 - (61492685*s^4)/166 + 3466243338530/3939040643
 
P2pf =
1357585350/(81287*(83*s + 5)) - 16625/(754*(5*s + 1)) - ((20897787425*s)/72878 + 790062525/72878)/(1457*s^2 + 137*s + 5)

alternatively:

syms s 
P1 = (-4220.5*s^2+3070.5*s+1150)/(16.6*s+1)*(5*s+1)*(291.4*s^2+27.4*s+1);
P1pf = vpa(partfrac(P1),5)
P2 = (-4220.5*s^2+3070.5*s+1150)/((16.6*s+1)*(5*s+1)*(291.4*s^2+27.4*s+1));
P2pf = vpa(partfrac(P2),5)

producing:

P1pf =
25723.0*s + 1350.1/(83.0*s + 5.0) + 160920.0*s^2 + 182900.0*s^3 - 370440.0*s^4 + 879.97
 
P2pf =
16701.0/(83.0*s + 5.0) - 22.049/(5.0*s + 1.0) - (1.0*(286750.0*s + 10841.0))/(1457.0*s^2 + 137.0*s + 5.0)

The partfrac function was introduced in R2015a.

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