I am following PFE for control system from a book. The author says that first it is best to divide by z for both sides, then multiply back through later. the thing is, that i get a different answer depending on whether or not i follow this.
for example, find inverse z transform for: $F(z) = \frac{z + 1}{z^{2}+0.3z+0.02}$
method 1 $\frac{F(z)}{z} = \frac{z + 1}{z(z^{2}+0.3z+0.02)}$
$z+1=\frac{A}{z}+\frac{B}{z+0.1}+\frac{C}{z+0.2}$
Coefficients, A = 50, B = -90, C = 40.
PFE: $F(z) = \frac{50z}{z} - \frac{90z}{z+0.1}+\frac{40z}{z+0.2}$
Look up table ANSWER (method 1): $f(n)=50\delta(n)-90(-0.1)^{n}+40(-0.2)^{n}$ for $n\ge0$
method 2 no division by z at the start $F(z) = \frac{z + 1}{z^{2}+0.3z+0.02}=\frac{A}{z+0.1}+\frac{B}{z+0.2}$
PF coefficients A = 9, B =-8
PFE is $F(z)=\frac{9}{z+0.1}-\frac{8}{z+0.2}$
This can be written as $F(Z)=\frac{9z}{z+0.1}z^{-1}-\frac{8z}{z+0.2}z^{-1} $
ANSWER from look up table (method 2): $f(n)=9(-0.1)^{n-1}-8(-0.2)^{n-1}$ for $n\ge0$
Any help? Am I missing something? Why don't I get the same answer? The constant $50\delta(n)$ appears to be the main difference
thank u
EDIT: A better approach for Method #2
You correctly have $$F(z)= \frac{9}{z+0.1} - \frac{8}{z+0.2}$$ But to use the look up tables (with positive powers of z), you want the fractions to be in the form $\frac{z}{z+a}$, where $a$ is a constant. The essence of the following step is to express $\frac{a}{z+a}$ as $\left(1-\frac{z}{z+a}\right)$.
$$\begin{align} F(z) &= \frac{9}{z+0.1} - \frac{8}{z+0.2} \\ &= 90 \frac{0.1}{z + 0.1} - 40 \frac{0.2}{z + 0.2} \\ &=90 \left(1 - \frac{z}{z + 0.1}\right) - 40 \left(1 - \frac{z}{z + 0.2}\right) \\ &=90-90\frac{z}{z + 0.1} -40 + 40\frac{z}{z + 0.2} \\ &=50 - 90\frac{z}{z + 0.1}+ 40\frac{z}{z + 0.2} \end{align}$$
Performing the inverse Z-transform: $$\mathcal{Z}^{-1} \{F(z)\} = \mathcal{Z}^{-1}\{50\} - \mathcal{Z}^{-1}\{ 90\frac{z}{z + 0.1} \} + \mathcal{Z}^{-1}\{ 40\frac{z}{z + 0.1} \}$$
Finally: $$ f(n)=50\delta(n)-\left[90(-0.1)^n + 40(-0.2)^n\right]u(n) $$
The previous and longer approach (edited)
You have $$F(z)=\frac{9}{z+0.1} - \frac{8}{z+0.2} = \frac{9z}{z+0.1}z^{-1} - \frac{8z}{z+0.2}z^{-1}$$ This implies that $$zF(z)=\frac{9z}{z+0.1} - \frac{8z}{z+0.2}$$ Taking the inverse $Z$ transform gives: $$\mathcal{Z}^{-1}\left\{zF(z)\right\}=f(n+1)=\left\{ \begin{array}{rl} 9(-0.1)^n - 8(-0.2)^n & \text{if } n \ge 0 \\ 0 & \text{if } n \lt 0 \end{array} \right. $$
Using the unit step function, $$f(n+1)=[9(-0.1)^n - 8(-0.2)^n]u(n)$$ Using a shift of indices and rewriting, $$\begin{align} f(n)&=\left[9(-0.1)^{(n-1)} - 8(-0.2)^{(n-1)}\right]u(n-1) \\ &= \left[\frac{9}{-0.1}(-0.1)^n - \frac{8}{-0.2}(-0.2)^n\right]u(n-1) \end{align}$$
Now, $u(n-1) = u(n) - u(0)=u(n)-\delta(n)$, since $\delta(n)$ is only non-zero when $n = 0$.
This gives: $$f(n)=\left[-90(-0.1)^n + 40(-0.2)^n\right] {\left[u(n) - \delta(n)\right]}$$
Now, $\left[-90(-0.1)^n + 40(-0.2)^n\right]\delta(n) = -50\delta(n)$, again since $\delta(n)$ is only nonzero when $n = 0$. Hence, we can rewrite as follows: $$f(n)=\left[-90(-0.1)^n + 40(-0.2)^n\right] {u(n) + 50\delta(n)}$$
Therefore: $$f(n)=\left\{ \begin{array}{rl} 50\delta(n)-90(-0.1)^n + 40(-0.2)^n & \text{if } n \ge 0 \\ 0 & \text{if } n \lt 0 \end{array} \right. $$
where $\delta(n) = \left\{ \begin{array}{rl} 1 & \text{if } n = 0 \\ 0 & \text{if } n \neq 0 \end{array} \right. $