# Z Transform problem

I have a class exercise of an inverse Z transform and I have some trouble. I will render an example to make my point. Let's asume the Z transform pairs: $$a^n \cdot u[n] \Leftrightarrow \frac{1}{1-az^{-1}} \qquad\qquad x[n-k] \Leftrightarrow z^{-k}X(z) \qquad\qquad \delta[n] \Leftrightarrow 1$$ Given the following function, we want its inverse Z transform: $$H(z) = \frac{p_0p_1z^{-2}}{(1-p_0z^{-1})(1-p_1z^{-1}) }$$

## First Method

Simple fraction expansion:

$$H(z) = z^{-2}\left( \frac{\dfrac{p_0^2p_1}{p_0-p_1}}{1-p_0z^{-1}} + \frac{\dfrac{p_0p_1^2}{p_1-p_0}}{1-p_1z^{-1}} \right)$$

Applying the properties:

\begin{aligned} h[n] &= \frac{p_0^2p_1}{p_0-p_1} p_0^{n-2} \cdot u[n-2] + \frac{p_0p_1^2}{p_1-p_0} p_1^{n-2} \cdot u[n-2] \\ \\ h[n] &= \frac{p_1}{p_0-p_1} p_0^{n} \cdot u[n-2] + \frac{p_0}{p_1-p_0} p_1^{n} \cdot u[n-2] \end{aligned}

## Second Method

\begin{aligned} H(z) &= 1 + \frac{p_0p_1z^{-2}}{(1-p_0z^{-1})(1-p_1z^{-1})} -1 \\ &= 1 + \frac{(p_0+p_1)z^{-1}-1}{(1-p_0z^{-1})(1-p_1z^{-1})} \end{aligned}

Simple fraction expansion:

\begin{aligned} H(z) &= 1 + \frac{\dfrac{p_1}{p_0-p_1}}{1-p_0z^{-1}} + \frac{\dfrac{p_0}{p_1-p_0}}{1-p_1z^{-1}} \\ \\ h[n] &= \delta[n] + \frac{p_1}{p_0-p_1} p_0^{n} \cdot u[n] + \frac{p_0}{p_1-p_0} p_1^{n} \cdot u[n] \end{aligned}

# Conclusion

Both solutions are quite similar, they are the same in the interval [2,$$\infty$$) but different in the interval [0,2]. So, where is the error?

Thanks

Both solutions are correct, and, consequently, identical. Your second solution evaluated at $n=0$ is

$$1+\frac{p_1}{p_0-p_1}+\frac{p_0}{p_1-p_0}=1+\frac{p_1-p_0}{p_0-p_1}=1-1=0$$

And at $n=1$ you have

$$0+\frac{p_0p_1}{p_0-p_1}+\frac{p_0p_1}{p_1-p_0}=\frac{p_0p_1-p_0p_1}{p_0-p_1}=0$$

So the inverse $\mathcal{Z}$-transform can be written as

$$h[n]=\frac{p_0^np_1-p_0p_1^n}{p_0-p_1}u[n-2]$$

• Thank you so much. Sometimes I forget these are discrete functions, nicely demonstrated – Jose Javier Gonzalez Ortiz Mar 24 '15 at 23:37