To find the Z-transform of $f(n) = \frac{2n+3}{(n+1)(n+2)}$, let us proceed step by step:

Step 1: General Definition of Z-transform

The Z-transform of a sequence $f(n)$ is defined as: $F(z) = \sum_{n=0}^\infty f(n) z^{-n}.$

Step 2: Expand $f(n)$ Using Partial Fractions

We decompose $\frac{2n+3}{(n+1)(n+2)}$ into partial fractions: $\frac{2n+3}{(n+1)(n+2)} = \frac{A}{n+1} + \frac{B}{n+2}.$ Multiply through by the denominator $(n+1)(n+2)$: $2n+3 = A(n+2) + B(n+1).$ Expanding and equating coefficients: $2n + 3 = A n + 2A + B n + B.$ $2n + 3 = (A + B)n + (2A + B).$ Equating terms:

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

Solve these equations:

• From $A + B = 2$, $B = 2 - A$.
• Substituting into $2A + B = 3$: $2A + (2 - A) = 3 \implies A = 1.$ $B = 2 - A = 1.$

Thus: $\frac{2n+3}{(n+1)(n+2)} = \frac{1}{n+1} + \frac{1}{n+2}.$

Step 3: Z-transform of Each Term

Using the Z-transform of $\frac{1}{n+k}$ (for $n \geq 0$): $Z\left(\frac{1}{n+k}\right) = \frac{z^{k-1}}{(z-1)^k},$ we compute the Z-transform of both terms.

For $\frac{1}{n+1}$:

$Z\left(\frac{1}{n+1}\right) = \frac{z^0}{(z-1)^1} = \frac{1}{z-1}.$

For $\frac{1}{n+2}$:

$Z\left(\frac{1}{n+2}\right) = \frac{z^1}{(z-1)^2} = \frac{z}{(z-1)^2}.$

Step 4: Combine Results

The Z-transform of $f(n)$ is: $F(z) = \frac{1}{z-1} + \frac{z}{(z-1)^2}.$

Final Answer:

$F(z) = \frac{1}{z-1} + \frac{z}{(z-1)^2}.$

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Related Questions

1. How is the Z-transform of a shifted sequence derived?
2. What is the relationship between the Z-transform and the Laplace Transform?
3. Can you apply the inverse Z-transform to find $f(n)$ back from $F(z)$?
4. How does the region of convergence (ROC) affect the Z-transform of a sequence?
5. How to compute the Z-transform for sequences involving factorial terms?

Tip:

When using the Z-transform, always check the region of convergence (ROC) for stability and causality of the system!