First order low pass filter I am trying to better understand the first-order low pass filter:

Summary:

Per wikipedia,  a first order low pass filter yields the following in discrete time:
$$
\frac{Y(s)}{U(s)}= \frac{\omega_{c}}{s+\omega_{c}}
$$
yields
$$
y[k] = \left(\frac{\omega_c T_{s}}{1+\omega_{c} T_{s}}
\right) u[k]+\left(\frac{1}{1+\omega_c T_{s}}\right) y[k-1]
$$
or
$$
y[k] =\alpha \, u[k] \ + \ (1 - \alpha)y[k-1]
$$

where

$
\begin{array}{cclc}
\omega_{c} 
&amp;:&amp; 
\text{Cutoff angular frequency of filter} 
&amp; 
[\frac{rad}{s}] 
\\
T_{s}
&amp;:&amp;
\text{Sampling period} 
&amp;
[s]
\\
\end{array}
$

Question 1:

Even though this filter is in discrete time, does it still model an analog ($s$-plane) filter?


If I wanted to use a discrete computing system to filter in real-time,
would I need to use the digital ($z$-plane) equivalent?  
If so, what is the general process for performing this?
My best guess is:  


Determine digital cutoff frequency $\omega_d$.
Convert to analog cutoff frequency $\omega_a = 
      \frac{2}{T} \cdot \mathrm{tan}(\omega_d \cdot \frac{T}{2})$.
Determine transfer function for (first-order low-pass) analog filter
using analog cutoff frequency $\omega_a$.
Transform to transfer function for digital filter
using bilinear transform $z = \frac{2}{T} \cdot \frac{z-1}{z+1}$



Relation to exponential smoothing:

On the same page, exponential smoothing is referenced.
The exponential smoothing page describes an exponential weighted average as:
$$
y[k] =\alpha u[k]+\left(1 - \alpha\right)y[k-1]\quad\text{where}\quad\alpha = 1 - e^{-\omega_{c} \cdot T_{s}}$$

Question 2:

How is it possible to relate the first-order low-pass filter alpha
with the exponential smoothing alpha?
To answer your first question, yes, you will need to convert the signal into Z-plane. Bi-linear transformation is one way to achieve the desired result. You can even try Impulse Invariant Transform to convert the analog filter to a digital one.