# Transfer Function Help

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• Jul 22nd 2011, 09:37 AM
spazzed
Transfer Function Help
Hi all.

I have to solve for the output, c(t), when I have the following, unfactorable, transfer function.

$\displaystyle C(s) = \frac{100}{s^3+2s^2+10s+100} * \frac{1}{s}$

where $\displaystyle R(s) = \frac{1}{s}$

this is unfactorable, so I tried using matlab to solve for the inverse laplace transform. It doesn't give me favorable results:

1 - sum((10*exp(r3*t) + 2*r3*exp(r3*t) + r3^2*exp(r3*t))/(3*r3^2 + 4*r3 + 10), r3 in RootOf(s3^3 + 2*s3^2 + 10*s3 + 100, s3))

not too helpful.

I used matlab to solve for the roots (poles, actually), and I want to know if you can get the partial fraction expansion from just the poles? Matlab's residue function gives you the roots and poles but it does not match the solution I have.

Roots
-0.915 - j1.1967
-0.915 + j1.1967
1.83

Poles
1.2907 + j4.49
1.2907 - j4.49
-4.5815

The poles are correct, but the roots don't match:

$\displaystyle C(s) = \frac{1}{s} - \frac{0.399436}{s+4.5815} - \frac{1.3994s - 0.2798}{(s-1.29)^2+(4.49)^2}$

$\displaystyle C(s) = \frac{1}{s} - \frac{0.399436}{s+4.5815} - \frac{1.3994s}{(s-1.29)^2+(4.49)^2}- \frac{0.2798}{(s-1.29)^2+(4.49)^2}$

I need to know how to go from the roots and poles calculated by matlab's residue(), to the partial fraction expansion above. Or how I can get a useful inverse laplace transform from matlab.

Thanks guys.