Obtaining the Impulse response from the frequency of a discrete system I have the following discrete system's transfer function (Z domain):

$$h[z] = {z \over z - \frac{1}{2} }$$

I need to obtain the following:


The frequency response.
The impulse response.
The fourier spectre.




The first one is pretty easy we just substitute $z$ by $ e^{j \omega}$ giving us

$$h[\Omega] = { e^{j \omega} \over  e^{j \omega} - \frac{1}{2} }$$

In the second one is where I have some questions, as a rule the IR is just the anti-transform of the transfer function $h[\omega]$, the original transfer function $h[z]$ maps quite easily to $h[n] = ({1 \over 2})^n u[n]$, however I read or heard, not sure where though that the impulse response obtained from the frequecy response must be continuous, under that assumption we would replace $e^{j \Omega}$ with $\$$ giving us $h(\$) = {\$ \over $ - {1 \over 2}}$, the we would Laplace anti-trasnform to obtain the impulse response, so which approach would be the correct one? maybe I'm confusing concepts.

Now for the sake of completenes for the fourier espectre we just obtain the magnitude ($|h[\omega]|$) and phase of $\theta_{h[\omega]}$ and evaluate them over a set of $\omega$ values, plotting the results.

Thanks in advance.
As the frequency response of the system is asked, the RoC woulbe be $|z|&gt;\frac{1}{2}$. Thus, the impulse response is right sided and is given by $$h[n] = \left(\frac{1}{2}\right)^n u[n]$$ We obtain the frequency response by replacing $z = e^{j\omega}$ in $H(z)$. Thus, $$H(e^{j\omega}) = \dfrac{1}{1-\frac{1}{2}e^{-j\omega}}$$