0
$\begingroup$

so i thought i knew about partial fractions, but apparently i don't. i have the answer to a partial fraction but i can't figure out how you get to that answer.

the value is $$ X(z) = Z*(Z+2)/(Z-1)^2 = 1 + 3z/(z-1)^2+1/(z-1) $$

Where does the "1" comes from? in the 1+ .......

Thanks

$\endgroup$
0
$\begingroup$

The numerator and denominator have the same degree, so the answer will have a constant term. More generally, for any polynomial $f(z)$, you can rewrite $$\frac{f(z)}{(z-1)^2} = g(z) + \frac{Az+B}{(z-1)^2}$$ where the degree of $g(z)$ will be $2$ less than the degree of $f$. This decomposition follows from polynomial long division.

$\endgroup$
  • $\begingroup$ in this case would Az + B be Z+2? now i dont get where the z in "3z" comes from $\endgroup$ – pato.llaguno May 8 '15 at 4:49
  • $\begingroup$ In this case $Az+B$ in this case would be $4z-1 = 3z + (z-1)$. $\endgroup$ – Rolf Hoyer May 8 '15 at 4:52
  • $\begingroup$ that would make the result you said, thats where the 3z comes from, but how did you got that? long division from $z*(z+2)$ divided by $z^2$? $\endgroup$ – pato.llaguno May 8 '15 at 4:58
  • $\begingroup$ @pato.llaguno Division by $(z-1)^2$ yields $z^2+2z = (z-1)^2 + 4z-1$. $\endgroup$ – Rolf Hoyer May 8 '15 at 5:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.