evelopment in partial fractions
Let the sequence:
\(x\left(n\right)\rightarrow X\left(Z\right)=S\left(Z\right)+\frac{P_0(Z)}{Q_0(Z)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (17)\)
With S\left(Z\right) being a polynomial in Z and \(\frac{P_0(Z)}{Q_0(Z)}\) is a ratio having simple poles at Zi and multiple poles of order 0 at Zn
We can develop \frac{P_0(Z)}{Q_0(Z)} as follows:
\(\frac{P_0\left(Z\right)}{Q_0(Z)}=\sum_{i}{\frac{\alpha_i}{Z-Z_i}+\sum_{j}{\frac{\beta_j}{{(Z-Z_j)}^j}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (18)}}\)
\(\alpha_i=\lim\limits_{Z \to Z_i}(Z- Z_i) \frac{P_0\left(Z\right)}{Q_0(Z)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (19)\)
\(\beta_j=\frac{1}{(Q-j)!} {\lim\limits_{Z \to Z_n}}{\frac{d^{(Q-j)}}{dZ^{(Q-j)}}}\left({(Z-Z_n)}^Q\frac{P_0\left(Z\right)}{Q_0(Z)}\right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ (20)\)
Example : Example :
Determine x(n) given that :
\(X\left(Z\right)=\frac{1}{1-3Z^{-1}+2Z^{-2}}\ \ \ \ \ \left|Z\right|>2\)
\(X\left(Z\right)=\frac{1}{1-3Z^{-1}+2Z^{-2}}=\ \ \ \frac{\frac{1}{2}}{Z^{-2}-\frac{3}{2}Z^{-1}+\frac{1}{2}}=\frac{\frac{1}{2}}{\left(Z^{-1}-1\right)\left(Z^{-1}-\frac{1}{2}\right)}\)
1 and \(\frac{1}{2}\) are the roots of the polynomial.
\(X\left(Z\right)=\frac{\alpha_1}{\left(Z^{-1}-1\right)}+\frac{\alpha_2}{\left(Z^{-1}-\frac{1}{2}\right)}\)
\(\alpha_1=\lim\ limits_{Z^{-1} \to {-1} }{\left(Z^{-1}-1\right)\frac{\frac{1}{2}}{\left(Z^{-1}-1\right)\left(Z^{-1}-\frac{1}{2}\right)}=}\)
\(\alpha_2=\lim\ limits_{Z^{-1} \to \frac{1}{2}}{\left(Z^{-1}-\frac{1}{2}\right)\frac{\frac{1}{2}}{\left(Z^{-1}-1\right)\left(Z^{-1}-\frac{1}{2}\right)}=}-1\)
So
\(X\left(Z\right)=\frac{1}{Z^{-1}-1}+\frac{-1}{Z^{-1}-\frac{1}{2}}=-\frac{1}{{1-Z}^{-1}}+\frac{2}{{1-2Z}^{-1}}\)
When subjected to the inverse Z transform using the Z transform table,
\(x\left(n\right)=\ -u\left(n\right)+2{.2}^nu(n)\)