# How to do this inverse Z-transform? Is this correct?

It was explained to me here that I need to do an inverse z-transform on a z-based equation in order to get something I can use. That example was quite simple. But I'm not sure how to do the same process on a more complex equation.

I am starting with:

$$H(s) = \frac{-ms}{ms+2R}$$

Where m=mass and R=impedance. I can do a backward Euler substitution $$s=\frac{1−z^{−1}}{T}$$:

$$H(z) = \frac{m (z^{-1} - 1)}{-m z^{-1} + m + 2 R T}$$

How do I then get this in terms of $$[n]$$ and $$[n-1]$$? I understand $$H(z) = H[n]$$ and $$H(z)*z^{-1} = H[n-1]$$, but how do I mathematically accomplish this and get rid of the $$z^{-1}$$ terms here?

Here is my best try:

$$-H(z) m z^{-1} + H(z) m + 2 R T H(z) = m z^{-1} - m$$

$$-H[n-1] m - H[n] m + 2 R T H[n] = m z^{-1} - m$$

$$H[n] (-m + 2 R T) = m z^{-1} - m + H[n-1] m$$

$$H[n] = \frac{m z^{-1} - m + H[n-1] m}{2 R T - m}$$

And I'm stuck there with one remaining $$z^{-1}$$ term. Multiplying by $$z^{-1}$$ is supposed to be the same as delaying by a sample right? So if mass is constant, then can I just say for the remaining term $$m z^{-1} = m$$ and get:

$$H[n] = \frac{H[n-1] m}{2 R T - m}$$

Is this the right idea or if I screwed it up what was I supposed to do differently?

Thanks.

A rational transfer function

$$H(z)=\frac{Y(z)}{X(z)}=\frac{\sum_{n=0}^{N}b[n]z^{-n}}{1+\sum_{n=1}^{N}a[n]z^{-n}}\tag{1}$$

corresponds to the following difference equation:

$$y[n]=bx[n]+bx[n-1]+\ldots +b[N]x[n-N]-\\-ay[n-1]-\ldots -a[N]y[n-N]\tag{2}$$

So the current output sample $$y[n]$$ can be computed from the current input sample $$x[n]$$ and $$N$$ past input and output samples. $$N$$ is the order of the difference equation $$(2)$$.

$$H(z)=\frac{Y(z)}{X(z)}=-c\frac{1-z^{-1}}{1-cz^{-1}},\qquad c=\frac{1}{1+\frac{2RT}{m}}\tag{3}$$
$$y[n]=c\big(-x[n]+x[n-1]+y[n-1]\big)\tag{4}$$