Inverse Z-Transform of \frac {Z^2-Zr Cos W0} {Z^2-r^2 Sin^2 W0}

[electronic engineer]

I need to find the inverse z-transform for this function:

\frac {Z^2-Zr Cos W0} {Z^2-r^2 Sin^2 W0}

 

[SGT]

Since \omega_0 is a constant, what you really want is the inverse transform of \frac{z^2-a}{z^2-b^2} = \frac{Az}{z-b} + \frac{Bz}{z+b}.
Find A and B by expanding the fuction in partial fractions. The inverse transform is easy to calculate.

 

[electronic engineer]

thanks but I see that you didn't pay attention to the first order term (zr cos w0), I wonder whether the partial funtions expanded should be changed or it might be the same?!

 

[SGT]

thanks but I see that you didn't pay attention to the first order term (zr cos w0), I wonder whether the partial funtions expanded should be changed or it might be the same?!

You are right. It should be:
\frac{z^2-a}{z^2-b^2} = \frac{Az+B}{z-b} + \frac{Cz+D}{z+b}.

 

[electronic engineer]

I tried to find constants:A,B,C,D but useless, i can't so i invented fiction ways ...but there's one thing which we didn't pay attention on at all, we need to divide P(z) by Q(z) manually till order of P(z) becomes less than Q(z) then find IZT of the function.

\frac {z^2-az} {z^2-b^2}=1 + \frac {-az+b^2}{z^2-b^2}

so now we can expand the resulted function in partial fractionds and find the constansts

=\frac {-az+b^2}{z^2-b^2}=\frac{A}{z-b}+\frac {B}{z+b}

A=(b-a)/2 & B=(-b-a)/2

so IZT function is:
x(n)= \delta(n)+A b^{n-1} u(n-1)+B (-b)^{n-1} u(n-1)

 

[electronic engineer]

and i think your previous fractions is wrong , check this website:

http://mathworld.wolfram.com/PartialFractionDecomposition.html

if you want more details about my solution or more discussing about , do post here please!

thanks.

 

[SGT]

and i think your previous fractions is wrong , check this website:

http://mathworld.wolfram.com/PartialFractionDecomposition.html

if you want more details about my solution or more discussing about , do post here please!

thanks.

You are right.