How to Express a Z-Transform as a Generating Function

[hellotheworld]

Homework Statement

For example : How to inverse z-domain function (z²+3z+7)/(z²+4z+3)

The Attempt at a Solution

Whatever I use partial fraction to simply the z-domain function, I cannot continue the next step, such as
1/(z+3)

 

[I like Serena]

Hello,

Let's first get rid of the highest power in the numerator before applying partial fractions:
$$\frac{z^2+3z+7}{z^2+4z+3}
=\frac{(z^2+4z+3)+(-z+4)}{z^2+4z+3}
=1+\frac{-z+4}{z^2+4z+3}
$$
How about applying partial fraction decomposition now?

 

[Ray Vickson]

Homework Statement

For example : How to inverse z-domain function (z²+3z+7)/(z²+4z+3)

The Attempt at a Solution

Whatever I use partial fraction to simply the z-domain function, I cannot continue the next step, such as
1/(z+3)

I am more accustomed to using the generating function
$$G_{a}(x) = \sum_{n=0}^{\infty} a_n x^n, $$
rather than the z-transform
$$T_a(z) = \sum_{n=0}^{\infty} \frac{a_n}{z^n}. $$
So, would substitute ##z = 1/x## into your transform to get the generating function
$$g(x) = \frac{7 x^2 + 3x + 1}{3x^2+4x+1}$$
and then express it in partial fractions. That leaves only the functions
##g_1(x) =1/(3x+1)## and ##g_2(x) = 1/(x+1)## to deal with. All you need to do is expand those as power series in ##x##, and that is just elementary algebra.