# Is the following system stable and causal?

Suppose the following function describes the unit step response of a system, where $u[n]$ is the unit step function. $$y[n]=\left(\frac{1}{2}\right)^{n-1}u[n+1]$$ I want to find out the system function and if the system is stable and causal. My first step is to calculate $Y(z)$. \begin{align} Y(z)&=\sum_{n=-\infty}^{\infty}y[n]z^{-n}\\ &=\sum_{n=-\infty}^{\infty} \left(\frac{1}{2}\right)^{n-1}u[n+1] z^{-n}\\ &=2\sum_{n=-1}^{\infty}\left(\frac{0.5}{z}\right)^n\\ &=2\left( \frac{1}{1-0.5z^{-1}}+\frac{z}{0.5} \right)\\ &=\frac{4z^2}{z-0.5} \end{align}

Above I exploited the fact that I can form a geometric series which converges for $|z|> \frac{1}{2}$. Now calculating $H(z)$ is straight forward, since we know the input is $u[n]$. $$H(z)=\frac{Y(z)}{X(z)}=\frac{\displaystyle \frac{4z^2}{z-0.5}}{\displaystyle \frac{z}{z-1}}=\frac{4z(z-1)}{z-0.5}$$

We see that the system has two zeroes, $z_{0,1}=0$ and $z_{0,2}=1$, as well as one pole at $z_{\infty,1}=0.5$.

My last step was calculating the impulse response for this system. \begin{align} H(z)&=\frac{4z^2}{z-0.5}-4\frac{z}{z-0.5} \\ h[n]&=\left(\frac{1}{2}\right)^{n-1}u[n+1]-4\left(\frac{1}{2}\right)^nu[n] \end{align}

From the above calculations I want to find out if the system is stable and causal. To see if it is causal I need to check if the impulse response disappears for negative time indices. Well, that is not the case, since $h[-1]=0.5^{-2}=4$. That means we have an acausal system, where the region of convergence (ROC) lies in between $|z|=0$ and $|z|=\frac{1}{2}$. Therefore, the system is also not stable, since the unit circle is not in the ROC.

Here I am getting confused: The system converges for $|z|<\frac{1}{2}$, but for calculating $Y(z)$ $|z|$ must be greater than $\frac{1}{2}$. How can that be? Are my calculations and thought process correct?

You are on the right track, but you are confused. First, regarding the calculations you could simply do $$y[n]=2\left(\frac{1}{2}\right)^n(u[n]+\delta[n+1])=2\left(\frac{1}{2}\right)^nu[n]+4\delta[n+1]$$ which gives you
$$Y(z)=\frac{2z}{z-\frac{1}{2}}+4z,\,|z|>\frac{1}{2}$$ The impuse resposne is the first difference of the step response:
\begin{align} h[n]&=y[n]-y[n-1]\\ &=2\left(\frac{1}{2}\right)^nu[n]+4\delta[n+1]-4\left(\frac{1}{2}\right)^nu[n]\\ &=4\delta[n+1]-2\left(\frac{1}{2}\right)^nu[n] \end{align} Obviously, $h[-1]=4$ and the system is not causal. But here you are mixing up finding the inverse $z$-transform of anti-casual signals with system being non-causal. ROC is $|z|>\frac{1}{2}$ and the system is non-causal full stop.