To find the inverse Z-transform of $X(z) = \frac{z^2 - 3z}{(z + 2)(z - 5)}$ using the residue method, we proceed as follows:

Step 1: Partial Fraction Decomposition

The given expression is: $X(z) = \frac{z^2 - 3z}{(z + 2)(z - 5)}$ We want to express this in the form: $X(z) = \frac{A}{z + 2} + \frac{B}{z - 5}$

To find $A$ and $B$, multiply both sides by $(z + 2)(z - 5)$: $z^2 - 3z = A(z - 5) + B(z + 2)$ Expanding this: $z^2 - 3z = (A + B)z + (-5A + 2B)$

Equating coefficients, we get:

1. $A + B = 1$
2. $-5A + 2B = -3$

Solving these equations:

1. $A + B = 1$
2. $-5A + 2B = -3$

From the first equation, $B = 1 - A$. Substituting into the second equation: $-5A + 2(1 - A) = -3$ $-5A + 2 - 2A = -3$ $-7A + 2 = -3$ $-7A = -5 \implies A = \frac{5}{7}$

Substituting $A$ back into $B = 1 - A$: $B = 1 - \frac{5}{7} = \frac{2}{7}$

Thus, the partial fraction decomposition is: $X(z) = \frac{5/7}{z + 2} + \frac{2/7}{z - 5}$

Step 2: Finding the Inverse Z-transform

The inverse Z-transform of $X(z)$ is determined by examining the individual terms.

Term 1: $\frac{5/7}{z + 2}$

We rewrite this as: $\frac{5/7}{1 - (-2/z)}$ The inverse Z-transform of $\frac{1}{1 - a/z}$ is $a^n u[n]$. Thus, the inverse Z-transform of $\frac{5/7}{z + 2}$ corresponds to $\left(\frac{-2}{1}\right)^n u[n]$, giving: $\frac{5}{7}(-2)^n u[n]$

Term 2: $\frac{2/7}{z - 5}$

We rewrite this as: $\frac{2/7}{1 - (5/z)}$ The inverse Z-transform of this is $\left(\frac{5}{1}\right)^n u[n]$, giving: $\frac{2}{7}(5)^n u[n]$

Combining both results

The overall inverse Z-transform is: [ x[n] = \frac{5}{7}(-2)^n u[n] + \frac{2}{7}(5)^n u[n] ]

Final Answer

[ x[n] = \frac{5}{7}(-2)^n u[n] + \frac{2}{7}(5)^n u[n] ]

Would you like me to elaborate on any of these steps or verify the calculations?

Relative Questions

1. What is the significance of using the residue method in inverse Z-transforms?
2. How do we handle partial fraction decomposition for more complex expressions?
3. What are common properties of Z-transforms that help simplify inverse calculations?
4. How does the region of convergence affect the inverse Z-transform?
5. Can the residue method be applied to Z-transforms with higher-order poles?

Tip

Always check the region of convergence when applying the inverse Z-transform, as it ensures the solution corresponds to the correct sequence behavior (causal vs. non-causal).