Inverse $\mathcal{Z}$ transform of $\frac{1}{z^2(z+1)}$ without delta function

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While finding inverse $\mathcal{Z}$ transform of $\frac{1}{z^2(z+1)}$ i split it into three parts using partial fractions:

$$\frac{a}{z} + \frac{b}{z^2}+\frac{c}{z+1}$$

and so answer comes as

$$a \delta(n-1) + b\delta(n-2) + c (-1)^{n-1} H(n+1) $$

where $\delta$ is delta function and $H$ is step function

Now I checked with online calculator tool link

It gives answer as $(-1)^{n+1} H(n-3)$. I cannot understand where the $\delta$ went away? And how can my solution arrive at the one given in online calculator, that is, how can I get rid of delta function in my solution

I have verified that online calculator gives correct result also as its $\mathcal{Z}$ transform comes as stated above.

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1 Answer

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We have $$X(z) = \frac{1}{z^2(z+1)}= \frac{1}{z^2}+\frac{1}{z+1} - \frac{1}{z}$$Let $$X(z) = \sum_{n = 0}^{\infty}x[n]z^{-n}$$Also $\mathcal{Z}\{\delta[n - 1]\} = \frac{1}{z}$ and $\mathcal{Z}\{\delta[n - 2]\} = \frac{1}{z^2}$. Note that $$\mathcal{Z}\{a^{n-1}u[n-1]\} = \sum_{n = 1}^{\infty}a^{n-1}z^{-n} = \sum_{n = 1}^{\infty}\frac{1}{a}(\frac{z}{a})^{-n} = \sum_{n = 0}^{\infty}\frac{1}{a}(\frac{z}{a})^{-n} - \frac{1}{a} = \frac{1}{a}\times\frac{1}{1 - \frac{a}{z}} - \frac{1}{a} = $$ $$\frac{1}{a}\times \frac{a}{z-a} = \frac{1}{z-a}$$Here $a = -1$ and we have $$\mathcal{Z}^{-1}\{X(z)\} = -\delta[n-1] + \delta[n-2] + (-1)^{n-1}u[n-1]$$

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